6 ! ! ! ! ! ! ! ! ! ! ! 4.6 ! ! ! ! ! ! caused onlyload bydonly the external loads acting on theexternal member plusofacting (vectorially) displacement at this atthe displacement relationship, such as =and PL>AE. support caused support by redundant the external caused only loadsby acting the on member loads plus on (vectorially) the the member the plus displacement • Express the external displacements inthe terms the loadings by using a(vectorially) load– •support Once established, the equation can then be solved for the magnitude of the redundant force. thisdisplace 4compatibility • Once established, the compatibility equation can then be solved for the magnitude of the redundant force. Compatibility. support caused only by the redundant reaction acting on the member. support caused only support by the redundant caused only reaction by the acting redundant on the reaction member. acting on the member. displacement relationship, such as d = PL>AE. 4 • Once established, the compatibility equation can then be solved for the magnitude of the redundant force. 4 Equilibrium. • •Equilibrium. Express the load and redundant in the of loadings bydo using ausing load– • Express external • Express load and the redundant external load displacements and equation redundant terms displacements in terms byforce. of the loadings a load– by us Choose oneexternal ofthe the supports as redundant and the ofof compatibility. To this, the known • Once established, the compatibility equation candisplacements thenwrite be solved forterms thein magnitude ofthe theloadings redundant Equilibrium. displacement relationship, such assuch dsupport, = as PL>AE. Equilibrium. 4 displacement relationship, displacement relationship, d =which PL>AE. as dzero, = PL>AE. displacement at the redundant is such usually is equated to the displacement at the • Draw a free-body diagram and write theaappropriate equations of equilibrium for theappropriate member using equations the •write Draw free-body diagram and write the of •• Draw free-body diagram and the appropriate equations of plus equilibrium for the member using thethisforce. support caused only by•the the external loads acting on the member (vectorially) the displacement at Once the compatibility equation can then beequilibrium solved for the magnitude the redundant force. • aestablished, Once established, Once compatibility established, equation the compatibility can then beequation solved for canreactions. the then magnitude beof solved of for the the redundant magnitude of the redu • Draw Equilibrium. a free-body diagram and write appropriate equations of for the member using the calculated result for thethe redundant. Solve these equations for any other 4.4 S TATICALLY I NDETERMINATE A XIA calculated result on for redundant. result for redundant. Solvereaction these equations for the any other reactions.Solve these equations for any 4 support only the redundant member. 4calculated 4 theby calculated result for thecaused redundant. thesethe equations for acting any other the reactions. • Draw a free-body diagram Solve and write appropriate equations of equilibrium for the member using the Equilibrium. Equilibrium. Equilibrium. • Express thefor external load and redundant displacements terms of the loadings by using a load– Example 4.5’in süperpozisyon prensibi ile equations çözülmesi: calculated result the redundant. Solve these for any in other reactions. displacement relationship, such as d = PL>AE. • Draw free-body diagram and write the appropriate equations of for the for member using the • aDraw a free-body •diagram Draw aand free-body write the diagram appropriate and write equations theequilibrium appropriate of equilibrium equations the of equilibrium member using for the the memb EXAMPLE 4.5 EXAMPLE 4.9 • Once established, the compatibility equation can then be solved for the magnitude of the redundant force. calculated result for redundant. Solve these equations for any other reactions. calculated result for calculated the redundant. result Solve for the these redundant. equations Solve for these any other equations reactions. for any other reactions. ! EXAMPLE 4.9 EXAMPLE 4.9 4 The steel4.9 rod shown in Fig. 4–12a has a diameter of 10 mm. It is fixed EXAMPLE The A-36 steel rodankastre shown in mesnetlidir. Fig. 4–17a hasYükleme a diameter of 10 B mm. It is Çapı 10 mm duvara öncesi Equilibrium. EXAMPLE 4.9olan çelik çubuk A noktasından toshown the wall at A, and before itdiameter is loaded, is a gap 0.2 mm Thefixed A-36 steel rod shown inhas Fig. hasloaded athere diameter mm. It between is to the wall at4–17a A, and before it is there isisof a10gap between The A-36 steel rod in Fig. a4–17a of 10 mm. Itof A the wall at and the rod. Determine the reactions at A and if the in Fi B¿ B¿ fixed to the wall at A, and before it is loaded there is a gap between • Draw a free-body diagram and write the appropriate equations of equilibrium for the member using the The A-36 steel rod shown the wall at and the rod of 0.2 mm. Determine the reactions at A B¿ fixed to the wall at A, and before it is loaded there is a gap between ucu ile B’ arasında 0.2 mm boşluk mevcuttur. Çubuğa C noktasından 20 kN yüklendiğinde A Thethe A-36 steel rod shown in an Fig. 4–17a hasDetermine diameter ofas 10shown. mm. ItNeglect is A the EXAMPLE EXAMPLE 4.9 EXAMPLE 4.9 4.9 rod isB¿ subjected to axial force ofa C. Preactions = 20 E kN wall at and the mm. rod of 0.2other mm. the at P ! 20 kN result0.2 calculated for the Solve these equations for any reactions. and Neglect the size of the collar at Take B¿. =reactions 200 GPa. the redundant. wall fixed at B¿ to and the rod ofA,0.2 Determine the at A mm st fixed to the wall at A, and bef the wall at and before it is loaded there is a gap between C P !mesnetlerinde 20 kN size ofofthe collar C. Take 200 GPa. EstEat=C. and B¿. Neglect the sizeat ofat the collar Take Est = 200 GPa. 0.2 mmmeydana P ! 20 kNve B’ reaksiyonları hesaplayınız.! andB¿B¿. gelecek Neglect the size the collar C. Take = 200 GPa. mmC st A 0.2 A-36 steel rod shown The A-36 in Determine Fig. steel 4–17a rod shown has areactions diameter inof Fig. 4–17a ofand 10 mm. a diameter Itrod is of of A-36 shown Fig. 4–17a has athe diameter mm. Ithas is theThe wall atThe androd the rod ofin 0.2 mm. the at A B¿steel SOLUTION wall at10B¿ the C 800 mm B¿ A P ! 20400 kNmm B¿ fixed to the wall at fixed A, and to before the wall it is at loaded A, and there before is it a is gap loaded between there is a g A fixed to the wall at A, and before it is loaded there is a gap between andSOLUTION Neglect the size of the collar at C. Take B¿. E = 200 GPa. 0.2 mm st SOLUTION PCompatibility. ! 20 kN Here theB¿. support at the B¿ as Neglect size ofthe the 0.2 mm we will considerand 800 C (a) mm 800mm mm the wall at and the the rod wall of at 0.2 mm. and the Determine rod of 0.2 the mm. reactions Determine at A rea B¿ B¿ the wall at and the rod of 0.2 mm. Determine the reactions at A SOLUTION B¿ 400 C Compatibility. Here we will consider the support at as B¿ 400 mm A EXAMPLE redundant. Using principlethe of superposition, Fig.as 4–17b, we have Here we willthe consider support at B¿ 4.9 B¿ Compatibility. SOLUTION (a)20 kN (a) P ! 20 kN P800 ! and the and sizeB¿ of the Neglect collar the at C. size of the C. Takewe B¿. Neglect B¿. Estcollar = we 200athave GPa. Est = 200 mm P ! 20 0.2 mm and B¿.the Neglect the size ofshown the collar atFig. C.free-body Take EstTake = have 200 GPa. AkN mm0.2 mm 0.2 redundant. Using the principle of superposition, Fig. 4–17b, Equilibrium. As on the diagram, Fig. 4–12b, SOLUTION redundant. Using principle of superposition, 4–17b, we + C C C kN 400 mm P ! 20 Compatibility. Here we will consider the support at as B¿ (1) 1 : 2 0.0002 m = d d 0.2 mm 800 mm A (a) A B¿ SOLUTION B¿ A B¿+ A-36 steel rod shown in 4–17a Phas a Bdiameter of 10 mm.end It isB to will assume that force P Fig. ismlarge to cause the rod’s SOLUTION SOLUTION +The : 2mm 0.0002 = dPenough - dCompatibility. P 0.2 ! 20 kN Using0.0002 the principle of- superposition, we have(1) ! 1400 0.2mm mm800 mm Here we w (1) 1: 2 redundant. m = d d P ! 20 kN B Fig. 4–17b, 800 mm mm P B 800 The deflections and are determined from Eq. 4–2. d d fixed to the wall at A, and before it is loaded there is a gap between contact the wall at The problem is statically indeterminate since B¿. (a) P B 400 mm 400 mm Compatibility. Here Compatibility. we will consider Here we the will support consider at the as support B¿ 400 mm Compatibility. Here we will consider the support at as B¿ +The (a) (a)deflections redundant. Using the principle (a) (1) at 1: 2the 0.0002 m =0.2donly -of dDetermine and are determined from Eq. 4–2. dtwo dB the P ! 20 kN at the rod of the reactions A have 0.2 mm InitialThe deflections and are determined from Eq. 4–2. dPwall dUsing there are unknowns and one of equilibrium. Pmm. B equation Pand Çözüm: B B¿ redundant. Using redundant. principle Using superposition, the principle Fig. of 4–17b, superposition, we Fig. 4–1 3 superposition, redundant. the principle of Fig. 4–17b, we have d F P 2 N]10.4 m2Take E = 200 GPa. A position PLAC the size[20110 PInitial ! 20 kN anddB¿.=Neglect of the collar at C. 0.2 mm Initial st 0.5093(10 - 3) m + = = 3 The deflections and are determined from Eq. 4–2. d d + + 3 3 P C + d P Bmm +! dP kN 2P kN P [20110 22AN]10.4 (1) 0.000 1:: 2[20110 :m2 0.0002 m9220110 =N>m P ! PL 20 Uygunluk kN position PL 2 N]10.4 0.2 ©F -F - =FBdm2 2] 2 NdB0.0002 = 0 m- 3 = dP(1)- dB(1) P !position 20 kNP ! 20 0.2 mm 1AC : 2 20 0.0002 d1BdP: AE P ! 20 kN AC p10.005 [200110 x = 0; 0.2 mm 0.2 mm P+denklemi: ! 1m2 ! 2m AP ! 20 kN Initial B¿ SOLUTION dP= = )m = 2 =- 30.5093(10 dP = )m = 90.5093(10 4.6 THERMAL S 2 2 9 2 3 AE 800 mm dP p10.005 m2 2m2 N>m ] Eq. p10.005 [200110 2 N>m ] m2 FinalAE deflections [20110 N]10.4 The The deflections determined and from are Eq. determined 4–2. from Eq. 4–2. dPB and d[200110 dPThe ddeflections position PL F 11.20 P ! 20400 kNmm are determined from dAB BBare B4–2. FBLm2 ACdeflections and d P and - 3 at d dB positionThe Compatibility. Here we will consider the support as B¿ 9 P B are d B =to 76.3944110 move) m to B¿, 2F with d = dCompatibility. = 0.5093(10 = = = The2 force 2P9 causes92 point (a)Final BInitial B no Initial 2 Final F LPredundant. AE Initial F 11.20 m2 F 11.20 m2 p10.005 m2 [200110 23superposition, N>m FBLAB AE Using the principle 4–17b, have the p10.005 m2 of [200110 2]compatibility N>m B 3 ddB position dP B AB further dPB Initial 3[20110 -we 9 m2 -Fig. 9 Therefore the 2 AC N]10.4 m2=][20110 N]10.4 PL PL position P ! 20dBkN Pd P = 2 N]10.4 m2 dB ==position 2FB-for = 2AC [20110 76.3944110 2FB 2condition = 76.3944110 position position PLACdisplacement. P ! 20 kN 3 3 B B ! 20FkN 9 2 9 d2dP= 2 -3 d = ) m2 [20110 = = = 0.5093(10 = 0.5093( AE AE Final + position P P p10.005 m2 [200110 2 N>m ] p10.005 m2 [200110 2 N>m ] PL d! =2BL ) m9 (1) =AE = 0.5093(10 Thermal Stress Fp10.005 m2 P1Substituting kN rod is 2= 2 2 F P 20 AC B11.20 : 0.0002 m P ! 20 kN AB into 2 9 dP - p10.005 29d Eq. 1, we get dB position F 0.2F mm AE B 9 m2 [200110 2 N>m ] m2 [200110 2 N>m ] AE F [200110 2 N>m ]= 76.3944110 B B dB = 2FB= = p10.005 m2 A dP = 2 9 2 (b) AE p10.005 m2 [200110 -23 N>m ]from Eq. 4–2. - AE 9 Substituting into Eq. 1, we get Substituting into Eq. 1, we get p10.005 m22 Final Final The deflections and are determined d d 0.0002 m = 0.5093(10 ) m 76.3944110 2F F 11.20 m2 F 11.20 m2 d = 0.0002 m Final P B F L F L A change in temperature can cause a body to change its dimensions. B ! BB AB B F F 11.20 m2 B>A B AB d d F L B 9 B 20 kN B B F B AB dB position -9 3.39 kN position dB = =position dB body =-33 2will-expand, 2FB= 76.394 = the (b)A (b) 9= 9 = whereas dBtemperature =m into 76.3944110 9 =2 76.3944110 InitialGenerally, 9B if0.0002 the increases, if 2F Substituting Eq. 1,m we =AE 0.5093(10 )=mp10.005 - 276.3944110 2F 9=- 4.05 2B2 N>m-2]m2 AE AE 0.0002 =B- 3get 0.5093(10 )N m 76.3944110 2F m2 [200110 p10.005 [200110 2Ans. N>m2] 3 Final 4.05110 2 kN F B d FB 4.6 T HERMAL S TRESS 1 5 1 p10.005 m2 [200110 2 N>m ] F L P 20 kNFA [20110 2 N]10.4 m2 FA 20 kN position PL P ! 20 kN 3.39 kN Borthe AB This displacement candBOrdinarily be expressed terms of FB decreases, ACit willFcontract. 3.39temperature kNFB the this in expansion - 3 unknown B (b) -33 position dFP 0.0002 == 4.05110 )m ==320.5093(10 =- 92F 0.5093(10 d = = B m ) m 76.3944110 Ans. N = 4.05 kN 2 9 2 Ans. = 4.05110 2 N = 4.05 kN F B From the free-body Fig. reactions load–displacement relationship, Eq. 4–2, applied (c) Substituting into Eq. Substituting 1,m2 we[200110 get diagram, into Eq. 1,] we getB AE AE using B contraction! Substituting is Equilibrium. linearly related top10.005 the temperature increase or4–17c, decrease 2 N>m 20 kN FA into Eq. 1,the we get p10.005 m2 3.39 kN 3 material (b) (b) to segments AC and CB, Fig. 4–12c. Working in units of that occurs. If this is the case, and the is homogeneous and F 3 --Ans. 39newtons and -9 + B (b) Fig. 4–17 Equilibrium. From the free-body diagram, Fig. 4–17c, = 4.05110 2 N = 4.05 kN F (c) Final Equilibrium. From the free-body diagram, Fig. 4–17c, 3 9 (c) : ©F Ans. = 0; F + 20 kN 4.05 kN = 0 F = 16.0 kN B 0.0002 m = 0.5093(10 0.0002 ) m m = 76.3944110 0.5093(10 ) 2F m 76.3944110 2F F 11.20 m2 x F A B 0.0002 m A=experiment 0.5093(10 ) m the - 76.3944110 2Fof BLAB dB position Thermal Stress -B9 B a 20 kN 20 kN F F isotropic, it has been found from that displacement meters, we have A A 3.39 kN 3.39 kN 20 kN d = 2F = = 76.3944110 FA Substituting into Eq. 1, we get B B 3.39 kN + 2 9 2 + ©F 3 = Fig. : ©Fx Equilibrium. =having 0;: -aFxlength 20 - +4.05 =m2 04.05 FkN kN From the free-body AE [200110 2 0kN N>m 3diagram, Fig. 4–17 (c) Fig. 4–17 A =+ 0; -F 20 kN =16.0 F=A] 4.05110 = Ans. 16.0 3kN member LkN can calculated using formula Ans. 4.05110 2the N = 4.05 kN 2 N =Ans. 4.05 kN FkN F4–17c, B =Ans. =p10.005 4.05110 2N =A 4.05 FABbe FB! FABLAC FBLCB Yatay kuvvet dengesi: (b) in temperature can cause a body to change its dimensions. + d = 0.0002 m = 0.0002 m = 4–17c, 0.5093 : ©F 0; -Finto 20From kN -the 4.05 kN = 0From FA151 = 16.0 B>A 4.6 Sfree-body TRESS Fig.the 4–17 A + Substituting Eq. we get Equilibrium. diagram, free-body Fig. kN 4–17c,Ans. diagram, Fig. (c) (c)x20=Equilibrium. AEthe AE Equilibrium. From theT1,HERMAL free-body diagram, Fig. 4–17c, (c) FAwhereas if the temperature increases, body will expand, ifkN 3.39 kN 3 + ©F 0.0002 ++ ©F rature decreases, it will contract.(b) Ordinarily or : dT m = F a A¢TL : 204.05 kN =- kN -0; 4.05 +=20 0 kN F-A(4–4) -92F =4.05 16.0 kN kN = B0 Ans. = 0.5093(10 )m 76.3944110 + Fig. F 10.4 =F4.0511 F Fig. 4–17this expansion x = 0; x m2 A F A = 16.0 B A: ©F4–17 Ans. + -20 kN =--F 0kN 20 kN Fig. 4–17 FA x = 0; m -F A = 16.0 kN 3.39or kNdecrease0.0002 = A n is linearly related to the temperature increase 2 9 2 3 p10.005 m2 [200110 Ans. 2 N 2=N>m 4.05 ]kN FB = 4.05110 s. IfThermal this is the case, and the material is homogeneous and ! Stress Equilibrium. From the free-b (c) Most traffic bridge t has been found from experiment of a From the free-body diagram, Fig. 4–17c, (c) that the displacementEquilibrium. expansion joints to F 10.8 m2 + ©F B = 0; -thermal nge ina temperature can cause a body change its dimensions. 4.6 be Isıcalculated Gerilmeleri aving length L can usingtothe formula - : FA2 +movement 20 kN o Fig. 4–17 x + 2 9 where : ©Fxif = 0; - FA + 20 kN - 4.05 kNp10.005 Ans. = 0 Fm2 16.0 kN2 avoid N>m ] thermal str any A =[200110 ally, if the temperature increases,Fig. the4–17 body will expand, whereas mperature decreases, it will contract. Ordinarily this expansion or or ction is linearly related the temperature increase or (4–4) decrease dT = to a ¢TL a homogeneous = a property of the material,Freferred to as the linear coefficient # ccurs. If this is the case, and the material is and (2) A10.4 m2 - FB10.8 m2 = 3141.59 N m of thermal expansion. The units measure strain per degree of 4 pic, it has been found from experiment that the displacement of a Most traffic bridges are designed with Solving 1 and 2 yields temperature. They Eqs. are 1>°F (Fahrenheit) in the FPS system, er having a length L can be calculated using the formula expansion joints to accommodate the and 1>°C (Celsius) or 1>K (Kelvin) in the system. Ans. F = 16.0 kN FB =Typical 4.05 kN A the deckSI α: Doğrusal ısı genleşme katsayısı! thermal movement of and thus values are given on any thethermal inside stress. back cover avoid Since the answer for FB is positive, indeed end B contacts the wall at δT: Eleman değişim ¢T =cebrik the algebraic change in temperature of the member dT = auzunluğundaki ¢TL (4–4) B¿ as originally assumed. L = the original length of the member ΔT: Elemandaki ısı değişimi 4 operty of the material, referred to ascebrik the linear coefficient dT = the algebraic change in the length of the Most traffic bridges are member designed with NOTE: If were a negative quantity, the problem would be F hermal expansion. The units measure strain per degree of B expansion joints to accommodate the statically determinate, so that and F F = 0 = 20 kN. perature. They are 1>°F (Fahrenheit) in the FPS system, thermal movement of the Bdeck and thusA avoid any thermal stress. 1>°C (Celsius) orStatikçe 1>K (Kelvin) in the SI system. Typical belirli (izostatik) sistemlerde ısı değişimi elemanlarda gerilme oluşturmaz. Aksine ues are given on the inside back cover change in length of a statically determinate member can easily be sistemlerde ısı The değişimi ısı gerilmelerine sebep olur. algebraic change hiperstatik in temperature of the member calculated using Eq. 4–4, since the member is free to expand or contract a property of the material, referred to as the linear when it coefficient undergoes a temperature change. However, in a statically original length of the member of thermal expansion. The units measure strain per degree of indeterminate member, these thermal displacements will be constrained Long extensions of algebraic change in are the 1>°F length of the member temperature. They (Fahrenheit) inby thethe FPSsupports, system, thereby producing thermal stresses that must be carry fluids are subj climate that will ca and 1>°C (Celsius) or 1>K (Kelvin) in the SI system. Typical considered in design. Determining these thermal stresses is possible and contract. Expans values are given on the inside back cover using the methods outlined in the previous sections. The following one shown, are used stress in the material. the algebraic change in temperature of theexamples member illustrate some applications. nge in length of a statically determinate member can easily be the original length the memberis free to expand or contract using Eq. 4–4, sinceof the member ndergoes a temperature in a statically the algebraic change in thechange. length ofHowever, the member PAGE !57 ate member, these thermal displacements will be constrained Long extensions of ducts and pipes that pports, thereby producing thermal stresses that must be carry fluids are subjected to variations in climate that will cause them to expand in design. Determining these thermal stresses is possible and contract. Expansion joints, such as the ! The A betwee R 4 AXIAL LOAD 0.5 in. raised Applying the thermal and load–displacement we the thermal andrelationships, load–displacement relationshi develo 152 C H A P T E R 4 A X I A L Applying LOAD have have 152 CHAPTER 4 AXIAL LOAD A 10 The A-36 steel bar shown in Fig. 4–18a is constrained to just fit FL FL 2 between C H A Ptwo T E R fixed 4 A Xsupports I A L L O A Dwhen T = 60°F. If the temperature is 0 = a¢TL SOLUT 1 0 = a¢TL EXAMPLE 4.10 AE AE raised to T2 = The determine theshown average normal thermal stress 120°F, A-36 steel bar in Fig. 4–18a is constrained to just fit F F Equilib EXAMPLE 4.10 A36 çeliğinden yapılmış bir çubuğun hareketi sıcaklık T =60 Fo iken, 2 ft 152 C H A P TIf E Rthe 4 temperature A Xthe I A L L O A Dis 1 developed in the bar. in two between fixed whenThus, Tto 60°F. from the data on inside back cover, 0.5 in. he A-36 steel bar shown Fig.(b) 4–18asupports is constrained fit 1 =just Thus, from the data on the inside back cover, The A-36 steel bar shown in Fig. 4–18a is const Fig. 4– (b) XAMPLE 4.10 raised determine theThe average normal thermal stress =O 120°F, TL 2L0.5 etween when If the temperature is TA1in. 60°F. A-36 steel bar shown in Fig. 4–18a is constrained to just fit o between two fixed supports when If th T = 60°F. opposi 1 50.5 2 in. two fixed C H A Piki Tsupports E Rankastre 4 Ato X I Amesnet D= arasında 1 engellenmiştir. Sıcaklık T2=120 ye çıkartıldığında çubukta F = a¢TAE F =F a¢TAE C H Ato P T ET R 4= 120°F, A X I A L developed Ldetermine OAD inthe theaverage bar. between two fixed supports when If the temperature is T = 60°F. ised normal thermal stress 0.5 in. raised to determine the average norm = 120°F, T d d 1 2 T 2 T 0.5 in. SOLUTION The A-36 steel bar shown in Fig. -6 4–18a is constrained-6to just 2fitB 3 thermal EXAMPLE 4.10 0.5 in. raised= to determine the average normal 120°F, Thesaplayınız. eveloped in the bar. 2>°F]1120°F - 60°F210.5 in.2 [29110 2 kip>in2] stress [6.60110 2>°F]1120°F in.22[29110+32ckip =Ifdeveloped [6.60110 in the bar. 2 = T ©F gelecek olan ortalama ısı bar gerilmesini E=! 4 between two fixed supports when the temperature is - 60°F210.5 1 = 60°F.α=! Equilibrium.meydana The free-body diagram of the is shown in developed in the bar. EXAMPLE 4.10 d d 0.5 in. F=equal determine the average = normal thermal F load, Ais 2.871The kip A-36stress Fig. 4–18b.C152 Since there isHXno the at A C A P Texternal E R raised 4 A X Ito AL T L 2O A=force D 120°F, 2.871 kip SOLUTION 0.5 in. but steel (a) bar shown in Fig. 4–18a is HAPTER 4 A IAL LOAD LE152 4.10 The A is,developed in the bar. opposite to the force at B; that OLUTION Equilibrium. The free-body diagram of the bar is shown in between two fixed supports when T = 60° SOLUTION 0.5 in. 1 determ Since also represents internal force within the b The A-36 steel Since bar shown Fig. 4–18a isinternal constrained tothe just fit theaxial F also in represents the F axial force within bar, the 2 ft SOLUTION 0.5 in. A The free-body Fig. 4–18b. Since there isthe no load, the force Aaverage is equal but 0.5 The A-36 steel barexternal shown in supports Fig. 4–18a isat in. constrained toIfisraised just fit to TThe determine the averag 120°F, quilibrium. diagram of bar isfixed shown in normal compressive stress is thus 2 = between two when the temperature is T = 60°F. Equilibrium. free-body diagram of the b average normal compressive stress thus F 1 + c ©Fy = 0; FAforce =two Fat =B;Fthat ft EXAMPLE 4.10 Bfixed opposite to the is,A= is120°F, Equilibrium. The diagram of inthe bar shown load, in theComp EXAMPLE 4.10 between when determine If free-body the temperature isSince T1 = 260°F. g. 4–18b. Since there is0.5noin.external load, the force at equal but developed the bar. raised tosupports the average normal thermal stress T Fig. 4–18b. there is noisexternal force SOLUTION 2 2 ft 4–18b. Since there is no external load, force is equal but that oc 0.5 in. pposite to the force at B; that raised is, to T2developed determine the average normal thermal stress = 120°F, in the Fig. bar. opposite thethe force at at B; A that is, kip F The free-body diagram ofF the bar kip is to shown in 2.871 The problem since this force cannot be A steel in. Equilibrium. The bar shown inis2.871 Fig. 4–18a is constrained to =just fitksi 0.5 in. The A-36 steel bar in Fig. 4–18a constrained to just fit +2isftcstatically ©Fy developed = 0.5 0;indeterminate FA = FBopposite =A-36 Fshown to the force at B; that is, to pus s = = 11.5 in the bar. s = the =force at A2is=equal Ans. 11.5 ksi Since there isB notwo external load, A from equilibrium. Fig. 4–18b. 10.5 in.22 between fixed supports when If the temperature is Tthe = temperature 60°F.Abut A 10.5 in.2 1 between two fixed supports when If is T = 60°F. c ©Fdetermined SOLUTION = 0; F = F = F 1 y A B B opposite to the force at B; that is, + c ©Fy = 0; FA = FB = F condit 4 A 0.5 in. raised tosince determine the average normal thermal stress 120°F, T=2 = c 0.5 in. + ©F 0; F = F = F raised to determine the average normal thermal stress = 120°F, T The problem is statically indeterminate this force cannot be SOLUTION y A B 2 4 Equilibrium. The free-body diagram of Compatibility. Since dA>B = 0, the thermal displacement dT at A 2 ft developed in the bar. determined from developed in(c) the bar. B is Fig. (a) 1that + cthi 2c The that problem statically indeterminate since thisforce force be NOTE: ofof From the magnitude of F, it should apparent (c) equilibrium. Equilibrium. The free-body the bar isbeshown in NOTE: theFdiagram magnitude F,The itFig. should apparent that problem is statically indeterminate since occurs, 4–18c, isSOLUTION counteracted by the Fcannot that isFrom required 4–18b. Since there ischanges no be external load, th FA = = F 2 ft (a) + c ©Fy = 0; B statically The problem is indeterminate since this force cannot be etermined from equilibrium. in temperature can cause large reaction forces in st Fig.The 4–18b. Since there is no external the force at in A from isto equal but in atstatically A to its original determined equilibrium. to push theAbar dF back position. The compatibility in temperature cause large reaction forces Equilibrium. free-body diagram of can the load, bar shown opposite the force B; that is, Fig.the 4–18 Compatibility. thermal displacement at A dT is 2 ft Fig. 4–18Since dA>B = 0, determined from equilibrium. App indeterminate members. opposite toisthe force at B; that is, force at A this condition(a) at A becomes Fig. 4–18b.The indeterminate members. Fload, Since there external the is equal isno statically forcebut cannot be SOLUTION occurs, isproblem counteracted by required SOLUTION ompatibility. Sincethat the F4–18c, thermal displacement atBindeterminate Aforce F that is since dA>B = 0, Fig. dT the Compatibility. Since dA>B = 0, the thermal disp have opposite to the force at B; that is, c determined from equilibrium. + ©F 0; displacement = FB = F Compatibility. Since the thermal A d = 0, dTFat tois push the bar by back to its original position. The compatibility dF the y =shown A B A>B at occurs, Fig. 4–18c, counteracted force F that is required 4 Equilibrium. The free-body diagram of the bar is shown in Equilibrium. The free-body diagram of the bar is in that occurs, Fig. 4–18c, is counteracted by the force c c dA>B = 02 ft 1+ 2 =+ dT©F - yd= FA 4–18c, = FB =is F ! occurs, Fig. F 0; F that by the forcebut is required condition A becomes push to2 ftits atoriginal position. The compatibility Fig. 4–18b. Since there isload, nocounteracted external load, force atFAthat is but position. T B the bar dF back Fig. 4–18b. Since there is(a) no0,external the atthe Athe equal toforce push bar to equal its original dF back Compatibility. Since the thermal displacement d = dis c A>B T at A + ©F = 0; F = F = F to push the bar back to its original position. The compatibility d y A B The problem is statically indeterminate sin F ondition at A becomes opposite to the force at B; that is, opposite to the force at B; that is, (a) condition at A becomes Çözüm: Applying the thermal and load–displacement relationships, we by the force that occurs, Fig. 4–18c, is counteracted thatforce is required The problem is statically indeterminate sinceFthis cannot beF determined from equilibrium. dA>B = 0 = condition 1+ c2 dT - dF at A becomes have Thus, f to push the bar back to its original position. The compatibility d determined from equilibrium. (a) F B B (b) statically+indeterminate since this force cannot=beF dA>B = The + c2 0 = problem dT - d+Fcis©F 0;! c ©Fy = 0; F ! FA = FB = FF!A1 +=cF dA>B = 0 = dT - dF 2B yA=becomes 4 condition at F Düşey kuvvet dengesi: c dA>Bwe= Compatibility. 1+ 2 0 = dT - dF Since d = 0, the therma determined from FL equilibrium. Applying the thermal and load–displacement F Compatibility. Since dA>B = relationships, displacement 0, the thermal dT at A d A>B 0 = a¢TL T that occurs, Fig. 4–18c, is counteracted by the F the thermal Applying relationships, we (a) (a)haveand load–displacement AE The problem indeterminate this be forceand cannot be occurs, Fig. 4–18c, counteracted by the force F force that iscannot required problem statically since Applying the thermal load–displacement dis 1 + cthat 2 The = indeterminate 0is=statically ddisplacement A>Bthermal T - dF and load–displacement Compatibility. Since A since =is Applying 0, dthis the thermal we T at Uygunluk denklemi: ! dA>B ! the to push the bar drelationships, ave F back to its original pos to push the bar back to its original position. The compatibility d determined from equilibrium. determined from equilibrium. have F FL by the force F that is required Thus, from the data on the back thatinside occurs, Fig.cover, 4–18c, is counteracted d have condition at A becomes F 0 =ata¢TL condition becomes F F Applying theAthermal and load–displacement relationships, we FL AE push the bar to its original compatibility dF back FL F = a¢TAE 0 = to Compatibility. Since thermal displacement dA>B position. = Since 0, thedThe dT at A 0 =da¢TL Compatibility. the thermal = 0, a¢TL FL displacement A>B T at A have Sinc AE condition at A becomes 0 = a¢TL ! c AE that occurs, Fig. 4–18c, is counteracted by the force F that is required that occurs, Fig. 4–18c, is counteracted by the force F that is required d 1+ 2 = 0 = daverag -6 from the data on 2 3 2 = 0 = d - d Thus, cover, c 2 inside A>B T - d 1 + the AE ) T F 2>°F]1120°FF - 60°F210.5 in.2back [29110 2 kip>in ] = [6.60110 F dA>B FLtoposition. to itsdForiginal The position. compatibility dF back to push the bar back itsThus, original The compatibility hus, from the data on the inside back cover,to push the bar fromcover, the data on the inside back cover, 0= =d a¢TL - on the inside F =1+a¢TAE c 2 (b) (b) Thus, from the data back d = 0 d = 2.871 kip A>B T F condition at A becomes condition A becomes AE Applying thewe thermal and load–displace dT Applying the thermalat and load–displacement relationships, F = a¢TAEF -6 2 3 F = a¢TAE 2>°F]1120°F - within 60°F210.5 in.2 [29110 2 kip>in2] = [6.60110 have have F = a¢TAE Since F also represents the internal axial force the bar, the dT back cover, Thus, from the dataand on the -6 2 inside d the (b) Applying we c thermal 1+in.2 22[2911031+ = 0-6= drelationships, - d= c 2load–displacement 2>°F]1120°F - stress 60°F210.5 2 kip>in ] dA>B dA>B -6 2 2 T F 0 = dT - dF dF= [6.60110 average normal compressive isT thus 2 - 60°F210.5 in.2 = [6.60110 = 2.871 kip FL 2>°F]1120°F - 60°F210.5 in.22>°F]1120°F [2911032 kip>in ] = [6.60110 FL[2 ! have F = a¢TAE 0 = a¢TL 0 = a¢TL = 2.871 kip dT dF load–displacement AE the dF Applying = 2.871 kip we AE theApplying thermal and relationships, also represents the internal axial force within (c) the thermal and = 2.871 kip the bar, NOTE F Since F F F FL 2.871 kip = [6.60110-62>°F]1120°F - 60°F210.5 in.22load–displacement [2911032 kip>in2] relationships, we have Since F also represents the internal axial force within the bar, the average normal compressive stress is thus 0 = a¢TL s = Ans. back cover, Since F also represents the internal axial force = = 11.5the ksihave in tem data on the inside w Thus, axial from force the data on the inside also within bar, theback cover, A AErepresents the internal (b) dF 10.5 Thus, in.22 =from Fig. 4–18 verage normal compressive 2.871 kip Since F (b) F ! stress is thus indete average normal compressive stress is thus FL average normal compressive stress is thus FL F = 0 = a¢TL - 0 = a¢TL - F = a¢TAE Thus, fromSince the data ona¢TAE the inside cover, 2.871 kip back dT (b) FFalso represents internal within theAE bar, the AE s = Ans. = apparent = the 11.5 ksi dT axial force NOTE: From the magnitude of F, it should be that changes -6 2 3 2 F F 2 2.871Fkip 2>°F]1120°F 60°F210.5 in.2 [29110 2=kip>in ] -62>°F]1120°F = [6.60110 A average normal compressive stress is thus 10.5 in.2 2.871 kip- 60°F210.5 [6.60110 F =2 a¢TAE 2.871 kip s = can =cause F large Ans. = 11.5 ksi F in temperature reaction forces in statically Thus, from the data on the inside back cover, s = dT Thus, from the data on s the= inside Ans. = 11.5 ksi = back cover, = 11.5 ksi = (b) AdF 10.5(b) in.2 2 A indeterminate members. -6= 2.871 kip 2 A 32 kip>in 10.5 in.22 dF 2[29110 10.5 in.2 = 2.871 kip 2>°F]1120°F 60°F210.5 in.2 ] = [6.60110F 2.871 kipthat changes NOTE: From the magnitude=ofa¢TAE F, it should apparent FF =be a¢TAE dT s = =the internal = axial 11.5 ksi Since F also represents force within theAns. bar, the d d T 2in statically 2 Since3 F also 2represents the internal axial f F OTE: From the magnitude of F, it (c) should because apparent thatreaction changes -6 A in temperature forces = can 2.871 kip 10.5 in.2 -6-is 2magnitude 3 that (c) 2>°F]1120°F 60°F210.5 in.2 [29110 2 kip>in ] = large [6.60110 NOTE: From the ofchanges F,2]it should be appa NOTE: From the magnitude of F, it should be average normal compressive stress thus 4–18 [29110 2 kip>in = [6.60110 2>°F]1120°F - 60°F210.5 in.2apparent temperature can indeterminate cause large members. reaction forces in statically average normal compressive stress is thus for in temperature can cause large reaction forces in statically in temperature can cause large reaction dF Since F also thekip internal axial force within the bar, the dF represents = 2.871 determinate members. Fig. 4–18 Fig. 4–18 = 2.871 kip (c) indeterminate NOTE: compressive From the magnitude it should that changes indeterminate members. average normal stress is thus 2.871 kip be apparent Fof F, members. s = = = 11.5 ksi Since F alsocan represents the internal axial force within the bar,Ans. the F the2.871 kip in temperature cause large reaction forces in statically 2 the internal axial force within the bar, Since F also A represents Fig. 4–18 10.5 in.2 s = = = 1 average normal compressive stress is thus indeterminate members. normal A 10.5 in.22 2.871 kip compressive stress is thus Faverage s = Ans. = = 11.5 ksi 2 (c) 10.5 in.2F NOTE: FromAthe magnitude of2.871 F, it should apparent that changes kipF be 2.871 kip in From = (c) = large = 11.5 ksi NOTE: the magnitude of F, it should b in temperature cans cause forces statically 2 = =reaction Ans. = 11.5 ksiAns. A 10.5sin.2 Fig. 4–18 2 A 10.5in in.2temperature can cause large reactio indeterminate members. (c) NOTE: From the magnitude of F, itFig. should 4–18be apparent that changes indeterminate members. in temperature can cause large reaction forces in statically (c) NOTE: From the magnitude of F, it should be apparent that changes Fig. 4–18 (c) indeterminate members. NOTE: From the magnitude of F, it should be apparent that changes in temperature can cause large reaction forces in statically in temperature can cause large reaction forces in statically Fig. 4–18 Fig. 4–18 indeterminate members. indeterminate members. 0.5 in. 1+ c 2 1+ c d2 A>B = 0 = dT - dF dA>B = 0 = dT - dF PAGE !58 4.6 THERMAL STRESS 4.6 153 THERMAL STRESS 153 AMPLE 4.11 EXAMPLE 4.11 ! 300 mm 300 mm 150 kN/m e rigid beam shown in Fig. 4–19a is fixed to the top of the three mm The rigid beam in Fig.ve4–19a is fixed to the top three sts made of A-36 steel and 2014-T6 aluminum. The posts each have 4.6of Tthe HERMAL STRESS 153 300 mm 150 kN/m Şekildeki rijit kirişshown A36 çeliği aluminyumdan yapılmış direklerin üstüne300 mesnetlidir. 4.6 THERMAL STRESS 4.6 153 THERMAL S made A-36 and 2014-T6 ength of 250 mmposts when no of load is steel applied to the aluminum. beam, andThe theposts each have 4.6 T HERMAL S TRESS 153 o length of 250Cmm when no load is applied the beam, thedir. Rijit kirişe 150 kN/m T1Determine =20 başlangıç uzunlukları 250and mm mperature is T1Sıcaklık the force bytoeach =a 20°C. 4.6 THERMAL STRESS de dikmelerin 153 supported 60 mm 4.6 THERMAL temperature is T1 = 20°C. Determine the force supported by each AMPLE 4.11 mm STRESS 4.6 T250 HERMAL 153 60 mm st if the bar is subjected to a uniform distributed load of4.11 150 kN>m o ye çıkarılması halinde EXAMPLE 4.11 250 mm 4.6 T HERMAL S TRESS 1 5 3 üniform yayılı yük uygulanması ve sıcalığın T =80 C dikme EXAMPLE 40 mm 40 mm 2 post if the bar is subjected to a uniform distributed load of 150 kN>m 40 mm 40 mm d the temperatureand is raised to T2 = 80°C. the temperature is raised to T = 80°C. 300 mm 300 mm 150 kN/m o e rigid beam shown in Fig. 4–19a is fixed to the top 2of the three kuvvetlerini Est=200 GPa, Ealis=73.1 ve αal=23µ/C 300 mm st=12µ/C The rigidbulunuz. beam shown in Fig. 4–19a fixed GPa, to theαtop of the othree EXAMPLE 4.11 Theposts rigid each beam shown4.6in Fig. 4–19a is fixed to the1 5top of the three 150 kN/m300 mm 300 Steel Steel Aluminum sts made of A-36 steel and 2014-T6 aluminum. The have THERMAL STRESS OLUTION EXAMPLE 4.11 Steel 3Aluminum Steel SOLUTION posts made of A-36 steel and 2014-T6 aluminum. The posts each have EXAMPLE 4.11 posts made of A-36 steel and 2014-T6 aluminum. The posts each have ength of 250ismm when no isofapplied to the beam, 300 andmm the 300 mm 150 kN/m Fig. 4–19a fixed tomm theload top the three 300 300 mm 300 mm 3 quilibrium. diagram of the beam is shown in Fig.4–19b. the three The free-body The rigid beam shown in Fig. 4–19a is fixed to the top of the three Equilibrium. Thethe free-body diagram ofis the beam is shown inload Fig.4–19b. a length of 250 mm when noFig. load applied to the the 150shown kN/m (a)to the beam, asupported length of 250 when nobeam, isthe applied and the (a)mm 300 300 mm 150 kN/m mperature is Taluminum. Determine force by each rigid beam in 4–19a ismm fixed toand the top ofand three and 2014-T6 The posts each have 1 = 20°C. The 60 mm posts made of A-36 steel 2014-T6 aluminum. The posts each have 300 mm 300 mm each have TheMoment rigid beam shown in 4–19a is fixed to the top ofDetermine the three oment equilibrium about the beam’s requires the forces the equilibrium about the beam’s center requires the forces in the temperature isof Determine the supported by each TA-36 = Fig. 20°C. 150 kN/m temperature is thehave force250 supported T = in 20°C. mm 1center 1 force PLE 4.11 60 mm by each st if the bar is is subjected to athe uniform distributed load 150 kN>m posts made steel and of 2014-T6 aluminum. The posts each n no load applied to beam, and the apost length ofbar 250is mm no load is applied to the beam, and the 250 mm60 mm 40of mm 40 mm posts made of A-36 steel andon 2014-T6 aluminum. Thewhen posts each have m, and the steel posts to be equal. Summing forces on the free-body diagram, el posts to be equal. Summing forces the free-body diagram, post if the bar is subjected to a uniform distributed load 150 kN>m if the subjected to a uniform distributed load of 150 kN>m d the temperature is raised to = 80°C. T a length of 250 mm when no load is applied to the beam, and the 90 kN 40 mm 40 mm kN 2 40 mm 40 Determine force supported by iseach temperature Determine the force90supported by each Tto 20°C. ainlength of 250 mm when no load isTtemperature applied beam, the80°C. 1 =the we have d have by eachshownthe 60 mm mm the temperature raised tothe =60is 80°C. 300 mm eC. gid beam Fig. 4–19a is fixed top ofthe and raised toand T300 2 three 60and mm 2 =mm temperature istoTthe Determine theisforce supported by each 20°C. 150 kN/m load 60 4 1 =mm 250 mm mm 4 Steel Steel post if the bar is subjected to a uniform distributed of 150 kN>m Aluminum d to a uniform distributed load of 150 kN>m 250 is0;T1 = 20°C. Determine the force by each (1) 3 OLUTION kN>m c2014-T6 40 mm 3 st made of = A-36 steeltemperature and The posts each have mm 250 mm + 2F ©F + = Fal0to - 90110 2 Ndistributed =supported 0 (1) 40 mm mm load of 150 y+=if post the is 2F subjected uniform kN>m 60Steel c150 Steel Aluminum 250 mm 40 mm ©Fto Faluminum. -bar 90110 2N Aluminum mm Steel the atemperature is 40 raised to kN>m T2 = 80°C. y T 0;= 80°C. st SOLUTION 40 mm 40 mm post40 ifSOLUTION the baris isal subjected to aand uniform distributed load of 150 2mm hsed of 250 when no load applied to the beam, and the uilibrium. The free-body diagram of the beam is shown in Fig.4–19b. 40 mm 40 mm (a) the and the temperature is raised to T2 =and 80°C. Compatibility. Due to load, geometry, material symmetry, and the temperature is raised to = 80°C. T Equilibrium. The free-body diagram of in the beam is shown in Fig.4–19b. Steel 2theby Alumin Equilibrium. The free-body diagram of the beam is shown in Fig.4–19b. ature is Determine thecenter force supported each T1 = 20°C. (a) SOLUTION oment equilibrium about the beam’s requires forces the ompatibility. Due toAluminum load, geometry, and material symmetry, the (a) Steel Aluminum 60 mmSteel Steel top of eachequilibrium post isSteel displaced by an equal amount. Hence, Steel Steel Aluminum ! SOLUTION Moment about the beam’s center requires the forces in the 250 mm Aluminum Moment equilibrium about the beam’s center requires the forces in the the bar is subjected to a uniform distributed load of 150 kN>m Steel Steel el posts to be equal. Summing forces on the free-body diagram, p of each post isSOLUTION displaced by an equal amount. Hence, The free-body Equilibrium. diagram of the beam is shown in Fig.4–19b. 40 mm 40 mm90 kN (a) toinbe equal. Summing forces the free-body diagram, ody diagram ofisthe beam is shown Fig.4–19b. (2) on requires 1+steel T2Equilibrium. =equilibrium dalto be steeldposts equal. Summing forces the free-body diagram, eFig.4–19b. temperature raised to = 80°C. T2posts stdiagram The free-body ofon the beam isthe shown in Fig.4–19b. have (a) (a) Moment about beam’s center the forces inkN the Fst 90 Fst Fal4 (a) 90 Çözüm: Equilibrium. The free-body diagram of the beam is shown in Fig.4–19b. (2) T 2 d = d (a) stforces al we have the ut theinbeam’s center have 3 of Moment equilibrium about the beam’s requires the forces inon thethe The final position the top of each post iscenter equal toSumming its displacement F Fal free-body Fstdiagram, steel posts to be equal. forces crces st Steel Steel Aluminum (1) ©F 2Frequires Fal -the 90110 2 Nin =the 0we 4 Moment about the beam’s center requires the forces in the TION y = 0; st +equilibrium 3 The final position of the top post isKuvvet equal todengesi: displacement diagram, (b) to diagram, beSumming equal. Summing on the csteel Summing forcessteel on the free-body caused by the temperature increase, plus its causeddiagram, by32 N = 0 + ©F =of 0;each 2F +chave Fforces 90110 2displacement Nfree-body =2F 0 +free-body we (1) ©F =its 0; Fdiagram, y posts st+ aly al - 90110(1) posts to be equal. onforces the 90 kN 90 kN 90st kN rium. The free-body diagram of the beam is shown in Fig.4–19b. ompatibility. Due to load, geometry, and material symmetry, the 90 kN (b) (a) we have used by the temperature increase, plus its displacement caused by 3 the internal axial compressive force, Fig. 4–19c. Thus, for the steel and c (1) + ©F = 0; 2F + F 90110 2 N = 0 we have 4 y st al 4 Compatibility. Due to load, geometry, and material symmetry, the nt equilibrium about the beam’s center requires the forces in the 4 ! Hence, Compatibility. Due to load, geometry, and material symmetry, the each post is displaced by anpost, equal amount. 4 3 we have ep of internal 4–19c. for steel 3 aluminum (dal)T (1) +=0c0; ©Fforce, =on 0;Fig. 2FThus, +90110 Fal 90110 2 Nand = 0 3- the y st (1) + F - axial 90110compressive 2c N = ( d ) (1) +Summing ©F 2F + F 2 N = 0 osts toal(1) be equal. forces the free-body diagram, top equal amount. st T top of an each post isDue displaced by geometry, an equal Hence,symmetry, y of each post is displaced st al by Compatibility. toHence, load, and material the Uygunluk denklemi: T2 dst = dal 90 kN amount. uminum post, we have 1+ T2 dst = -1dst2T + 1dst(2) 2F (d ) Fstby an Fal amount. F e Initial Position Compatibility. Due to load, geometry, and material symmetry, the stHence, al T ( d ) top of each post is displaced equal metry, the st T (2) 1+ T2 d = d (2) (dal)F dst = Due tois load, geometry, thedal st and al material symmetry, 4 load, geometry, and symmetry, the The final positionCompatibility. of thematerial top of= each equal toT2 its ! by displacement ! 21+ Fst Fal Fst Fst 3 post Fal T 2 d 1d 2 + 1d top of each post is displaced an equal amount. Hence, st st T st F 1+ T2 d = -1d 2 + 1d 2 (1) = 0; 2F + F 90110 2 N = 0 al al T al F 1 + T 2 d = d y by an equal top st of each al post displaced by an equal amount. Final Position st Position al is equal (b) (dst)(Fdal)F (2) The final isposition the top offinal each post isHence, equal itsInitial displacement ed amount. Hence, The position of thetotop of each post used by the temperature increase, plus its of displacement caused by dst !to dalits displacement Fst F Sıcaklık artışı dikmelerde 1d+ force, T 2Eq. 1d dthe = temperature dand 2Fig. gives (b) by st al position the top of each post is(2) equal to its displacement (b) increase, plus its of displacement caused by Tinternal 2 d(2)axial 24–19c. + 1d 2symmetry, (2) TApplying 2caused dThe =final dby eatibility. compressive Thus, for the Due to1 +load, geometry, and the increase, plus its displacement caused al =by-the altemperature Tmaterial alcaused F st al steel F F F Final Position st al st (2) = d (dstst)F Fal Fst uzama deplasmanı Fal st al Fst d st st !plus (b by temperature increase, its Fdisplacement caused Fby The final position the top each is equal toforce, itsFddisplacement FThus, the internal axial compressive Fig. for the 2top +caused 1deach 2force, =ofFthe -1d 2post + to 1d each postpost, is displaced an equal amount. uminum we have the axial Fig.aland 4–19c. st al ststeel stHence, T of stinternal F post al4–19c. Tcompressive al2F (Thus, dal)T for the steel and Theby final position of-1d the of is equal its displacement placement pplying Eq. 2 gives ( d ) st T oluşturacaktır. top of each post is equal to(b) itspost, displacement (b) the internal axial compressive force, Fig. 4–19c. Thus, for the steel and caused by the temperature increase, plus its displacement caused by aluminum we have aluminum we haveon the (b) (dal)T caused the temperature pluspost, its displacement caused Using 4–4 the material properties insideby back (2) d=stEqs. = d4–2 caused by T2 d2stby 2Tand + 1d 2and st F increase, (dst)T (dst)T (b) (dal)F -its 1dstdisplacement + -1d 1dstalst2aşağı = 1d + 1dalkabul 2force, Initial Position Fst Thus, Fst aluminum post,Fig. we 4–19c. have Deplasmanlar doğru pozitif edildiğinden re increase, plusthe caused Tthe F al2by T F4–19c. internal axial compressive forFalthe steel and internal axial compressive force, Fig. Thus, for the steel and cover, we get (c) (dst)T 1+ T2 d = -1d 2 + 1d 2 steelposition and of the top of each post is equal to its 1+ T2 dst = -1dst2T + 1dst2FInitial Position stdisplacement st T st F final (dPosition Initial T2 Eqs. dand =post, -1d 2the + steel 1dwe 2Fhave al)F aluminum sive force, Fig. 4–19c. Thus, for and al al Tpost, alproperties sıcaklık deformasyonunun negatif işaret 1önünde + T 2 on ddst !=d - 1d 2 + 1d 2 ( d ) sing 4–2 and 4–4 the material the inside back Final Position st T st F aluminum we have al T ( d ) (dst()dTal)T Initial Position by the temperature increase, plus its displacement caused2 by+ 1d 2 al (b) (dal)T Fst 10.250 m2 st F 1+ Dikmelerdeki T2 d1+ = T2 -1d 2F (c) (dst)T Fig. al d al1d T 2 +ve al F st2)T pplying Eq. 2 gives FinaldPosition deplasmanlar sıcaklık yayılı 1 +(dT-6 = +2 dalst = -1d ver, we getcompressive (dal)alT22T ++ 1d (dst)4–19 2>°C]180°C m2 stst2+ TF (1d (dst)F dst ! dPosition F 1vardır. + T-[12110 2 force, dst- =20°C210.250 1d ernal axial Fig. 4–19c. Thus, for the and !ddalal)F 1+ 1d T 2stst2steel 1dal2alPosition 2F Initial al dstst )T Fdal = T 2 - 1d al 9T Initial (dal)F stdst(! (dst p10.020 m2 [200110 2 N>m ] dal Applying Eq. 2 gives (+ dal)1d Initialyük Position Eq. 2 gives FApplying etkisinden kaynaklanan iki farklı deplasmanın toplamı olarak yazılabilir. +1 1d 2 = -1d 2 2 dum - 1d 1d post, we stst22T st F al T al F st = st2have T +-1d F ( d ) + T2 dInitial = + -Eq. 1d 22Tgives + m2 1dal2F Applying 2 F 10.250 al T al al ( d ) Position st ( d ) 1 + T 2 d = 1d 2 1d Final Position al m2 F (dstFinal )F Position al al T al F -6 F (dstd)alF -1d 1d = -1d 1deşit 2 1d al10.250 !Fig. dal al4–19 2stFT = -1d 1d 2Fdst ! Bu deplasman birbirine olduğundan, [12110 2>°C]180°C - Applying 20°C210.250 m2 +2iki st T + st2Finside al2-1d T +st2 al+ ing Eqs. 4–2 and d4–4 the properties on the back -6 1d Position al22Tdst++ dst dand = 1dst(2material d )+ 2 m2 2 st st2 F 2TTN>m +F1d Eq. 2Final gives =alF -[23110 2>°C]180°C -p10.020 20°C210.250 +1d9st st ! st22F = - 1d al T (dal2)F1dal2F dal = - 1dal2T + Applying 1d Initial Position Eq.Tst2 Fgives m2 [200110 2 ] al2 9 Position (p10.030 d ) on m2 ver, we get (c) [73.1110 2 N>m ] Using Eqs. 4–2 and 4–4 and the material properties theFinal inside back 300 mm dst !4–2 dal and 4–4st and F UsingEqs. Eqs. material properties on 4–2 and the properties onthe theinside insideback back - 1d 1d = - 1d 2 Fand + 1dthe dal = - 1dal2T + 1d stst22TF + F al al2Fmaterial al2stF2T + - 1d 1dUsing = we - st1d2get 1d4–4 Final Position al2F T al+ al3d2T (dst)F ! 10.250 m2 cover, we get (c) ! dal cover, st F 10.250 m2 -6 cover, we get = 1.216F 165.9110 2 F (3) st st al -6 ng-Eq. 22>°C]180°C gives2>°C]180°C Fig. 4–19 Using Eqs. 4–2+and andm2 4–4+ and the material properties on the inside back =[12110 [23110 -4–2 20°C210.250 Using Eqs. and m2 4–4 the material properties on the9 inside back - 20°C210.250 2 m2 + 1d 2 9 2[73.1110 2 st2F = - 1dal2T + 1dal2F FFstst10.250 m2 p10.030 m2 2FN>m ] nside back st10.250 p10.020 m2 [200110 2 N>m ] 10.250 m24–19 cover, we get (c) -6 -6 we get (c) Fig. - 1dst2cover, + 1d 2 = 1d 2 + 1d 2 -6 2>°C]180°C T st be F consistent, al T all numerical al -F 20°C210.250 -[12110 2>°C]180°C m2 been + - 20°C210.250 20°C210.250 m2 To data has expressed --[12110 [12110 2>°C]180°C m2 + 2 in terms 9 + 2 2 99 2 d the material properties on(c) the inside back 2 3 F 10.250 m2 p10.020 m2F[200110 2 N>m ] p10.020 m2 [200110 2 N>m p10.020 m2 [200110 2 N>m]2] m2 = 1.216F - 165.9110 2 alCelsius. Fnewtons, (3) Eqs. 4–2 and the properties on the inside back Solving meters, and degrees Eqs. and 3 st110.250 F 10.250 m2 stmaterial al -6 4–4 andof st -6 Fig. 4–19 (c) 2 =50-[23110 2>°C]180°C - -6 m2 + Fig. 4–19 -20°C210.250 [12110 2>°C]180°C - 20°C210.250 - [12110 2>°C]180°C - 20°C210.250 m22[73.1110 + m292+N>m 2(c) F m2 we m2 get FFalal10.250 simultaneously al910.250 2] 2 92 N>m2] 10.250m2 m2 p10.030 m2 p10.020 m2 [200110 Fig. 4–19 -6yields -6p10.020 m2 [200110 2 N>m ] -6 = -[23110 2>°C]180°C 20°C210.250 m2 + = -[23110 2>°C]180°C 20°C210.250 m2 + = [23110 2>°C]180°C 20°C210.250 m2 + 9 consistent, 2 To be all numerical data has been expressed in terms F 10.250 m2 2 9 2 2 99 22 2 st 00110 2 N>m ] F 10.250 m2 p10.030 m2 [73.1110 2 N>m ] 3 F = -16.4 kN F = 123 kN Ans. 2 2N>m ]] Fig. F 4–19 p10.030m2 m2[73.1110 [73.1110 N>m st st al Fm2 10.250 m2 p10.030 -6 = 1.216F 165.9110 2 Fstand (3) al4–19 10.250 Fig. 20°C210.250 m2- + al2 al -6 10 2>°C]180°C 20°C210.250 m2 + newtons, meters, degrees Celsius. Solving Eqs. 1 and 3 -6 9 = -m2 [23110 2>°C]180°C 20°C210.250 2 29 m2 + 2 m2 3+ = The [23110 2>°C]180°C p10.020 [200110 2F20°C210.250 N>m ] 2 9 2 3 3 negative value for-F indicates that this force acts opposite to 10.250 m2 p10.020 m2 [200110 2 N>m ] 2 9 2 st = 1.216Fal - 165.9110 (3) ] 2 N>m 1.216F -[73.1110 165.9110 Fst2 = (3) p10.030 m2 1.216F 22 ] (3) st p10.030 m2 [73.1110 2 165.9110 N>m al al multaneously yields To numerical data has been in the terms shown in Fig.10.250 4–19b. In expressed other words, steel posts are in tension 2 be consistent, 9 2allthat F m2 F 10.250 m2 [73.1110 2 N>m ] 3 al al -6 =kN 1.216F - 3165.9110 2 (3) newtons, meters,m2 and degrees Celsius. Eqs. 1 aland 32 all =Solving - consistent, 165.9110 (3)terms st the aluminum post is=1.216F inF compression. -be 16.4 kN F 123 Ans. 23110 2>°C]180°C -F 20°C210.250 m22 F + To be numerical data has been been expressed expressedininterms terms 20°C210.250 += To consistent, all numerical data expressed in st al al To all data has st 2be2 consistent, 9has been 2 numerical 9 m2 ! p10.030 [73.1110 2 N>m ] m2 [73.1110 2 N>m ] multaneously yields of p10.030 of newtons, meters, and degrees Celsius. Solving Eqs. 1 and (3) newtons, meters, and degrees Celsius. Solving Eqs. 1 and 3 ofnumerical newtons, meters, andto degrees Celsius. Solving Eqs. 1 and 3 3 e negative valueTo forbeFconsistent, indicates this force opposite stTo be consistent, all has been expressed allthat numerical data acts hasdata been expressed in terms in terms 3 simultaneously yields 3 (3)Ans. simultaneously yields F=st 1.216F =simultaneously -16.4 kN165.9110 Fyields = 2123(3) kN st! 4–19b. almeters, = 1.216Fin - F165.9110 2 alother In words, steel posts are in tension of newtons, and degrees Celsius. Solving Eqs. 31 and 3 newtons, meters, and the degrees Celsius. Solving Eqs. 1 and datinshown terms al Fig.of F = 16.4 kN kN F FAns. Ans. st kN al = 123 kN negative valuesimultaneously for Fstissimultaneously indicates that this force acts kN opposite to123 F = -16.4 F = yields F = -16.4 Ans. dee1the aluminum post in compression. st al yields st al = 123 kN consistent, and 3 all numerical data has been expressed in terms umerical has been expressed in terms The negative value for F indicates that this this force acts opposite to at shown indata Fig.and 4–19b. In other words, the steel posts are in tension The negative value indicates force acts opposite negative value for FststkN indicates wtons, meters, degrees Celsius. Eqs. 116.4 3123 FFstThe Fthis 123 Ans.force acts opposite to st = FstSolving = for - 16.4 kN Fand =that kN= Ans.that to al al kN that shown in Fig. 4–19b. In other words, the steel posts are in tension the aluminum post is in compression. ddneously degrees Celsius. Solving Eqs. 1 and 3 thatThe shown in Fig. 4–19b. In other words, the4–19b. steel posts are inopposite tension to yields The negative that shown in Fig. Inopposite other posts are in tension negative for that this actswords, value forvalue Fst indicates that this force acts to the steel st indicates andFthe aluminum post isforce in compression. Ans. and the aluminum post is in compression. and the aluminum post is in compression. Fig.kN otherthe words, steel are in tension Fst that = -shown 16.4that kNinshown Fal 4–19b. =in123 Ans. Fig. In4–19b. other In words, steel the posts areposts in tension pposite to 16.4 kN F = 123 kN Ans. and the aluminum post is in compression. and the aluminum post is in compression. al gative value for F indicates that this force acts opposite to in tension st own in Fig. 4–19b. other words, steel posts that In this force actsthe opposite to are in tension st indicates eInaluminum post is in steel compression. other words, the posts are in tension in compression. PAGE !59 (c) FigF A D 154 4 AXIAL R CHAPTER 4 L O A D 154 AXIAL LOAD CHAPTER 4 154 154 AXIAL LOAD CHAPTER 4 AXIAL LOAD P LT ELRO4 4C154 APXTI EARL 4 L O ACADH154 HA XAI A AD AXIAL LOAD EXAMPLE 4.12 C H A P T E R EXAMPLE 4 A X I A L L O A D 4.12 54 C APTER 4 CH HAPTER 4 A AD A XX II AA LL LL O OAD EXAMPLE EXAMPLE 4.12 4.12 2 2 A 2014-T6 aluminum tube having area of 600 is a cross-sectional 2 EXAMPLE 4.12a cross-sectional A 2014-T6 aluminum tubemm having area of 600 mm2 is EXAMPLE 4.12 2having a 12aluminum PLE 4.12tube! having T6 a cross-sectional area of 600amm is A tu 600 mm2aluminum A 2014-T6 aluminum tube cross-sectional area of is 600 mm A 2014-T6 aluminum tube having cross-sectional area of is A 2014-T6 2014-T6 aluminum tu used as a sleeve for an A-36 steel boltused having a sleeve cross-sectional area of bolt having a cross-sectional area of as a for an A-36 steel EXAMPLE 4.12 sleeve for an2A-36used steelas bolt having a cross-sectional area of used as a sleeve for an A used as a sleeve for an A-36 steel bolt having a cross-sectional area of a sleeve for an A-36 steel bolt having a cross-sectional area of used as a sleeve for an A 2 2 2 2 2 400 mm , Fig. 4–20a.2When the Talanı = 2014-T6 15°C, temperature isFig. thehaving nut 2 is tüp, 2 of 600 1nut 22 is 600 mm enkesit bulunan bir aluminyum enkesit alanı 400 mm 400 mm , tube T = 15°C, 4–20a. When the temperature the nut 600 mm A tube having a cross-sectional area of mm A 2014-T6 aluminum tube a cross-sectional area is 2 aluminum 1 , Fig. 4–20a. When400 T = 15°C, the temperature is the 600 mm A 2014-T6 aluminum tube having a cross-sectional area of is 600 mm A 2014-T6 aluminum having a cross-sectional area of is 400 mm , Fig. 4–20a. W 1 400 mm , T = 15°C, Fig. 4–20a. When the temperature is the nut mm , T = 15°C, Fig. 4–20a. When the temperature is the nut 400 mm , Fig. 4–20a. W 1 1 2 holds the assembly in a such snug that position suchasthe that the axial force insteel the o de, holds assembly in athe snug position such that the axial inT the used a sleeve for an having aposition cross-sectional area ofaxial used as sleeve for an steel bolt a cross-sectional area of 600 mm 2014-T6 aluminum tube having abulonun cross-sectional area ofhaving isthe e assembly in a snug position axial force in the olan çeliğinden yapılmış biraxial üzerine geçirilmiştir. C used as a assembly sleeve for an A-36 steel bolt having aaA-36 cross-sectional of used as aA sleeve for an A-36 steel bolt having abolt cross-sectional area offorce 1=15 holds the in holds assembly inA-36 aarea snug such that force in assembly the holds the inthe aA36 snug position such that the force in the holds the assembly in aa 2 2 = T 80°C, bolt is negligible. If the temperature increases to 2 2 2 400 mm , T = 15°C, Fig. 4–20a. When the temperature is the nut 400 mm , T = 15°C, Fig. 4–20a. When the temperature is the nut T = 80°C, bolt is negligible. If the temperature increases to used as a sleeve for an A-36 steel bolt having a cross-sectional area of 150 mm T = 80°C, negligible. If the temperature increases to 1 1 400 mm , T = 15°C, Fig. 4–20a. When the temperature is the nut 2 400 mm , T = 15°C, Fig. 4–20a. When the temperature is the nut 2 bolt is negligible. 150 mm somun, aksamda bir iç kuvvet oluşturmayacak derecede sıkıştırılmıştır. 1 T2 =is 80°C, bolt is negligible. If 1 the temperature 150 increases tobolt mm bolt isin negligible. Ifsleeve. the150 150 mm negligible. If If mm 2 temperature increases to T2 = 80°C, force the bolt and holds the assembly in a snug position such that the axial force in the 400 mm , T = 15°C, Fig. 4–20a. When the temperature is the nut holds the assembly in a snug position such that the axial force in the determine the force in the bolt and sleeve. ne thedetermine force in thethe bolt and sleeve. o 1 holds the assembly in a snug position such that the axial force in the holds the assembly in a snug position such that the axial force in the determine determine the bulonda force in the andmeydana sleeve. determine the force in the bolt and C sleeve. Sıcaklık T2=80 ye çıkarılırsa vebolt tüpte gelecek kuvveti determine the the force force in in tt 80°C,too T2 = 80°C, is negligible. If negligible. the temperature to T 150 mm is holds themm assembly intemperature a snug position that the is temperature increases 150 2in=the TIf2 =the 80°C, bolt is negligible. If bolt the temperature increases to such T2axial = 80°C, bolt negligible. If thebolt increases toincreases 150 mm 150 mm oforce hesaplayınız. E =200 GPa, E =73.1 GPa, α ve α stthe albolt st=12µ/C al=23µ/C determine force in the and sleeve. T = 80°C, bolt is negligible.determine If the temperature increases 150 mm the force in the bolt and to sleeve. 2 the force inthe the bolt in and force thesleeve. bolt and sleeve. ON SOLUTION determinedetermine SOLUTION SOLUTION SOLUTION SOLUTION SOLUTION determine the force in the bolt and sleeve. free-body of a top segment of the um. Equilibrium. The free-bodyThe diagram of adiagram top Equilibrium. segment of the The free-body diagram of a top segment of the Equilibrium. Equilibrium. The The free-body diagram of a top segment of the diagram of a top segment Equilibrium. The fre fre The free-body of the The FEquilibrium. assembly is shown in Fig. 4–20b. forces b andSOLUTION s are produced Fb andSOLUTION Fs are Fproduced y is shown in Fig. 4–20b. The forces SOLUTION SOLUTION Fb and Fsforces assembly is shown inFFig. 4–20b. Thein forces are produced assembly is shown in Fi F assembly is shown in Fig. 4–20b. The forces and are produced assembly is shown in F F F assembly is shown Fig. 4–20b. The and are produced (a) b s (a) b s since sleeve has a higher coefficient of thermal than thediagram of a top (a)segment of (a) SOLUTION sleeve has the a higher coefficient of thermal expansion thanexpansion the Equilibrium. The free-body the Equilibrium. The free-body diagram ofthermal athan topthe segment of the since the sleeve has a higher coefficient of thermal expansion Equilibrium. The free-body diagram of a top segment of the Equilibrium. The free-body diagram of a top segment of the since the sleeve has a hig since the sleeve has a higher coefficient of thermal expansion than the hi since the sleeve has a higher coefficient of expansion than the 4therefore 4 bolt, and therefore sleeve willwhen expand more when the temperature 4 4 the sleeve will the expand more the temperature Fbwhen F assembly is shown in Fig. 4–20b. The forces and Equilibrium. The free-body diagram of as are top segment ofproduced s are Fbthe assembly shown in Fig. 4–20b. The forces and produced F Fs F assembly isassembly shown inis sleeve Fig. 4–20b. The forces and are produced F shown in therefore Fig. 4–20b. The forces and produced bolt, and the sleeve will expand more the temperature bolt, and therefore the s s are and therefore will expand more when the temperature b is bthe bolt, and therefore sleeve will expand more whenFthe temperature a) (a) Itbolt, is(a) required that (a)the Çözüm: sed. Itisisincreased. required that since the sleeve has a higher coefficient of thermal expansion than the F F assembly is shown in Fig. 4–20b. The forces and are produced since the sleeve has a higher coefficient of thermal expansion than the It is require b s since the sleeve has a higher coefficient of thermal expansion than the since the sleeve has a higher coefficient of thermal expansion than the is increased. is increased. It is required that (a) is increased. It is required that require is increased. It is required that 4 bolt, and therefore the sleeve will expand more when the temperature since the sleeve has a higher coefficient of thermal expansion than the bolt, and therefore the sleeve will expand more when the temperature bolt, and therefore sleeve will expandwill more whenmore the temperature bolt, andthe therefore the sleeve expand when the temperature Fbolt, F increased. It is is required that = 0; + c ©Fy = 0; is increased. F = Fb increased. (1) s = is b required and therefore the sleeve will expand more when the temperature It (1) is required that It + c ©Fyy = (1) 0; + c ©Fs y = is 0; It is required = that Fb+increased. + cisthat ©Fy F=s 0; (1) c ©F = 0; Fs = Fb (1) Fs = Fb y is increased. It is required that Düşey kuvvet dengesi: ! (1) y = + cThe ©Fytemperature = +0;c ©Fy =increase Fs0;=the F 0;+ c ©Fcauses Fsy ==and F0;b boltFs = Fb (1) Fs = Fb(1) + bcsleeve ©F (1) ibility.Compatibility. The temperature increase causes the sleeve and Bulonun bolt Uygunluk denklemi: deplasmanı = Tüpün deplasmanı (1) c + ©F = 0; F = F y s b Compatibility. The te Compatibility. Thethe temperature causes the sleeve and bolt Compatibility. The temperature increase causes the sleeve causes and bolt Compatibility. The temperature increase the sleeve and bolt 2T and 1db2However, expand However, increase the redundant forces 1db2T1d, sFig. d 1ds2to 4–20c. redundant forces T , Fig. 4–20c. T and F 1d 2 1dbb to expand and s 1dsthe 2Tand 2to to expand and Fig. 4–20c. However, redundant forces Fshorten F1d 1ds2Consequently, 1db2Tthe , uzamasına and Fig. 4–20c. However, the redundant forces T ,expand s theBulondaki çekme içsleeve. kuvveti bulonun olacağından (+) 1d 1d 2T , Fig. to expand and 4–20c. However, the redundant forcesss TT s bConsequently, Fb and Fsbolt elongate bolt shorten the the T Fss elongate the and sleeve. the s2T bsebep Compatibility. The temperature increase causes the sleeve and bolt F F and elongate the b Compatibility. The temperature increase causes the sleeve and bolt Compatibility. The temperature increase causes the sleeve and bolt Fab final Fs elongate and reaches theposition, bolt and shorten theand sleeve. Consequently, Compatibility. Thethe temperature increase causes the sleeve bb and bolt ss the Fbnot F and elongate bolt andasshorten thethe sleeve. Consequently, the F Fsame bolt and the shorten the sleeve. Consequently, end of reaches the assembly a final which is not the he assembly which is the as s same b the s, elongate Fs position, 1d 2 1d 2 to expand and Fig. 4–20c. However, redundant forces Compatibility. The temperature increase causes the sleeve and bolt s T b T Fs Fs F end of the assembly rea 1d 2 1d 2 , to expand and Fig. 4–20c. However, the redundant forces 1d 2 1d 2 , to expand and Fig. 4–20c. However, the redundant forces deplasmana,Tüpteki basınç iç kuvveti tüpün kısalmasına sebep olacağından (-) end of the assembly acondition final position, which as 1dassembly 2aTnot 1dreaches 2T , Fig. expand andthe However, theasredundant forcesas s reaches s Tcondition bend s T T of T toend the bassembly reaches final position, which not the same sis bsame position. Hence, the compatibility becomes of the a 4–20c. finalisConsequently, position, which is not the same its initial position. Hence, the compatibility becomes Fits Fbolt Fss2shorten and elongate the bolt and shorten theand sleeve. 1d 1dthe ,FFig. to expand and 4–20c. However, theshorten redundant forces the sF initial bthe T b2 Tsleeve. its initial position. Henc F and elongate the and Consequently, the F F and elongate the bolt and shorten the sleeve. Consequently, the position. Hence, compatibility condition becomes F and elongate the bolt the sleeve. Consequently, the b s b s its initial position. Hence, the compatibility condition becomes b s its initial position. Hence, the compatibility condition becomes Fb Fb Fdeplasmana end of the assembly reaches a final position, which is not the same as F sebep olur. Sıcaklık artışı heriki malzemede de uzamaya sebep olur ve F F and elongate the bolt and shorten the sleeve. Consequently, the b end b b s end of the assembly reaches a final position, which is not the same as of the assembly reaches a final position, which is not the same as which is not the same as end of the assembly reaches a final position, = b1d -end 2of 1 + Td2 = 1db2T + 1d db2=F 1d 2Ts2+T 1d 2Fsits =F initial 1ds2assembly 1ds2Freaches position. Hence, compatibility condition b1d T the athe final position, which is not becomes the same as becomes its initial position. Hence, the compatibility condition becomes its initial position. Hence, the compatibility condition becomes F (b) (b) F its initial position. Hence, the compatibility condition b 1+ d = 1 1 + T 2 d = 1d 2 + 1d 2 = 1d 2 1d 2 1 + T2 T2 Fb b T b F T 1d 2 s + F 1d 2d == 1d (b) (+) deplasmandır. (b) b F1+ = 1dbs22TT +- 1d 1dbecomes 1+dsT2 b T2position. Hence, bcompatibility T b F bs2F = 1ds2T - 1ds2F its initial the condition Fb g Eqs.Applying 4–2 and 4–4, the mechanical properties from Eqs.and 4–2using and and using1+ the mechanical properties (b) 4–4, d1d = 21d 2Tfrom 1db2F = 21ds2T - 1ds2F b1d (b) (b) 1+ T2 we Eqs. d (b) = 4–4, 1dbT2 2and += 1d1d = +1d 2+Fproperties 1 + T4–2 2 and T d b2F s2T s1d b2the T b F = s2Td-=1d 1+ T2 1ds bFfrom 2T + 1db2F = 1ds2T - 1ds2F Applying Eqs. 4–2 and 44 Applying using mechanical on the cover, have theinside tableback on the inside back we have ! cover, Applying Eqs. 4–2 4–4, and usingproperties the mechanical Applying Eqs. 4–2 and 4–4, and using mechanical from properties from (b) 1+ T2 = 1db2T + 1db2F =and 1dthe s2T - 1ds2F the table on the inside back cover, we dhave the table on the inside b the table on the inside back cover, we have the table on the inside back cover, we have Applying Eqs. 4–2 and 4–4, and using the properties mechanicalfrom properties from -6 15°C210.150 [12110-62>°C]180°C m2 + Applying Eqs. 4–2 and 4–4, and using the mechanical properties from Applying Eqs. 4–2 and 4–4, and using the mechanical [12110 2>°C]180°C -6- 15°C210.150 m2 + Applying Eqs. 4–2 and 4–4, and using the mechanical properties from -6 the inside back cover, we mechanical have Applying Eqs.on 4–2 and 4–4, and using properties from [12110 -the15°C210.150 m2 + the [12110-6 2 -6 theFtable on the inside back cover, we have the table on 2>°C]180°C the table inside back cover, have -6we the table on the inside back cover, we [12110 2>°C]180°C -have 15°C210.150 m2 + 10.150 m2 [12110 2>°C]180°C 15°C210.150 m2 + b Fbthe 10.150 m2 table onFthe inside back cover, we have 10.150 m22 -62>°C]180°C - 15°C210.150 m2 + 2 9 -6 2 2 -6 9 [12110 1400 mm22110-6 m2>mm 2[200110 2 N>m ] b-62>°C]180°C m2 Fb+10.150 m2 + Fb10.150 -6 m2 1400 mm22110 m2>mm 2[200110 2 N>m ] - 915°C210.150 [12110 2>°C]180°C 15°C210.150 m2[12110 [12110 2>°C]180°C - 15°C210.150 m2 + 2 -6 2 2 -6 2 1400 mm 2110 m [12110 >mm 2[200110 2 N>m ] mm 1400 mm -6 2>°C]180°C 15°C210.150 m2 + 2 -6 2 2 2 -6 1400 2 mm 9 >mm2 2[20011092 N>m2] -6 15°C210.150 m2 = [23110 2>°C]180°C 2110 m b10.150 2110 >mm2m2 2[200110 2 N>mm2 ] Fb10.150 m2 = [231102>°C]180°C -6- 15°C210.150 m21400 Fbmm 10.150 m2Fm F 10.150 Initial -6 (db)T b = m2 [23110 2>°C]180°C - 15°C210.150 2 m2 -6 2 m2 2 9 2 = [23110 -62 Initial -6 b10.150 Initial 1400 mm 2110 mposition 2[200110 2 -N>m ] 2 -6 22 2-6 -6 9F2= 2 >mm 2 9 2>°C]180°C 2 Fs10.150 (db)T (dbb)TT 2 [23110 15°C210.150 m2 2 -6 2 2 9 = [23110 2>°C]180°C 15°C210.150 m2 1400 mm 2110 m >mm 2[200110 2 N>m ] F 10.150 m2 1400 mm 2110 m >mm 2[200110 2 N>m ] (db)F position position 1400 mm 2110 ] (db)T (ds)9T m >mm - Final (dbs)T 9 -6 2 2[200110 d d 2 N>m m22-6 2 F 10.150 Initial - -6 m2>mm ((ddb))F22110 mm m2>mm-22[200110 2 N>m ] (dbb)FF (ds2)2T2[73.1110 1600 mm -6 -6 2 2 N>m 2 d] -6 s 1400 9[23110 22110 = 2>°C]180°C 15°C210.150 m2 s T Initial Final F 10.150 m2 Final position d 1600 mm 2110 m >mm 2[73.1110 2 N>m ] ( d ) F 10.150 m2 s -6 = [23110 2>°C]180°C position [23110 2>°C]180°C m2 ((ddbb))TF = s 2 2>°C]180°C 2 2-b F15°C210.150 2Final - 15°C210.150 9 [23110 =m2 - 15°C210.150 m2 Initial -6 (db)T position position 1600 mmFinal 2110-6 m >mm 2[73.1110 2 N>m ] Tposition 1600 m m (d(sd)bT! )gives Eq. 1 and solving = [23110 2>°C]180°C 15°C210.150 m2 ( d ) 2 -6 2 2 9 2 d b T 2 -6F1600 2 mm 2 m2 9 >mm 22[73.1110 2 N>m ] (d ) Using Eq. 1 and position position s10.150 2110 m (db)T (db)F position d d solving 1600 mm 2110 m >mm 2[73.1110 2 N>m ] F 10.150 m2 (ds)Tgives )F s T F 10.150 m2 ( d ) ( d ) ( d ) Final s s s F d b F b F Fs10.150 m2 Using Eq. and solvi (ds)T Using Eq. 1 and solvin (db)FAns. (ds)F position 2 F 10.150 -6 2 m2 2 9 2 d 1Final - solving -gives Fs =Final (dF Final s)Fb = 20.3 (kN 1600 2110 2 N>m ] gives 2 kN-6 2Using 91 2>mm 2and -6 mm 2 Eq. 2-sandmsolving 9 2[73.1110 2 Using Eq. 12>mm solving gives Fb Final =mm 20.3 Fs db=)F1600 Ans. ! position position 2110 m 2[73.1110 2 N>m ] 1600 mm 2110 m >mm 2[73.1110 2 N>m ] 2 -6 2 2 9 2 position (c) 2110 92mN>m >mm 2 N>m ] (ds)F -6 2 2 (c) 20.3 kN 22110gives Fs = F Ans. b =11600 position m21600 >mmmm 2[73.1110 ] 2[73.1110 Using Eq. and mm solving (ds)F elastic material s)Since F linear behavior was1(c) assumed in gives this = F = 20.3 kN F Ans. Using Eq. 1 and gives Using Eq. and solving (ds)solving (c) s b F = F = 20.3 kN F Ans. s b NOTE: Since in1this Usinggives Eq. and solving gives (ds)F linear elastic material behavior was assumed Using Eq. 1 and solving Fig. the4–20 average normalNOTE: stresses(c) should be checked to make sure Fig. 4–20 Since linear elastic material behavior was assumed in this =kN Fb = 20.3 kN Fssure Ans. NOTE: Since linear e (c) (c) average normal stresses analysis, the should kN Fs =beFchecked Ans. = to Fbmake = Since 20.3 Ans. Fig.the 4–20 b =F20.3 sNOTE: linear elastic behavior was assumed inAns. thisaverage no (c) Fig. 4–20 do not exceed the analysis, proportional limits for material. =material Fwas = assumed 20.3 kN Ans. F NOTE: Since linear elastic material behavior in this the average normal stresses should be checked to make sure s b analysis, the (c) = F = 20.3 kN F that they do not exceed the proportional limits for theanalysis, material. s b Fig. 4–20 the average normal stresses should be checked to make sure NOTE: Since linear elastic material behavior was assumed in this analysis, the average normal stresses should be checked to make sure that they do not exceed the proportional limits for the material. that they do not exceed Fig. 4–20 Fig. 4–20 NOTE: Since linear elastic material assumed this NOTE: Since linear elasticbehavior materialwas behavior wasinassumed in this Fig. 4–20 NOTE: Since linear elastic material behavior was assumed in this that they do not exceed the proportional limits for the material. analysis, the average normal stresses should be checked to make sure Fig. 4–20 NOTE: Since elastic material was assumed in this that they dolinear not exceed proportional for thesure material. analysis, the average normal stresses should bethe checked tobehavior makelimits sure analysis, the average normal stresses should be checked to make analysis, the average normal stresses should besure checked to make sure that they do not exceed proportional limits for the analysis, the average normal stresses should be checked to material. make that they do notthey exceed theexceed proportional limits the for the material. that do not the proportional limits for the material. that they do not exceed the proportional limits for the material. that they do not exceed the proportional limits for the material. PAGE !60 ing. The chapter begins with a discussion of r and moment diagrams for a beam or shaft. d torque diagrams, the shear and moment means for determining the largest shear and d they specify where these maximums occur. nt at a section is determined, the bending Eğilme ted. First we will 6. consider members that are cross section, and are made of homogeneous 6.1 Kayma ve Moment Diyagramları erward we will discuss special cases involving Boyuna eksenlerine dik doğrultuda yük taşıyan ince uzun elemanlara kiriş denmektedir. Basit d members made of composite materials. be given to curved members, stress mesnetli kiriş, konsol kiriş, çıkmalı kiriş gibi mesnetlenme türüne göre isimler verilmiştir. ending, and residual stresses. 256 C H A P T E R 6 B E N D I N G256 CHAPTER 6 BENDING Moment Diagrams er and support loadings that are applied ngitudinal axis are called beams. In general, bars having a constant cross-sectional area. s to how they are supported. For example, a pinned at one end and roller supported at the red beam is fixed at one end and free at the ng beam has one or both of its ends freely rts. Beams are considered among the most elements. They are used to support the floor a bridge, or the wing of an aircraft. Also, the boom of a crane, even many of the bones of P w0 A A D B x1 6 C x2 x3 6 Fig. 6–2 w(x) Positive external distributed load V V Positive internal shear M M Positive internal moment Beam sign convention Fig. 6–3 Because of the applied loadings,Because beams develop an internal shearbeams dev of the applied loadings, Simplyand supported beammoment that, force force bending in general, vary from point that, to point and bending moment in general, va along the axis of the beam. In along order the to properly design a beam it to pro axis of the beam. In order therefore becomes necessary to therefore determinebecomes the maximum shear necessary to and determine th moment in the beam. One way moment to do thisinisthe to beam. expressOne V and wayMtoasdo this is functions of their arbitrary position x along the beam’s axis. These x along functions of their arbitrary position shearCantilevered and moment functions can then be plotted and represented by be plot shear and moment functions can then beam graphs called shear and moment graphs diagrams. Theshear maximum values of V called and moment diagrams. Th and M can then be obtained from these graphs. since the shear and M can thenAlso, be obtained from these graph and moment diagrams provide detailed information about the variation and moment diagrams provide detailed informa of the shear and moment along the axis, they are often used of beam’s the shear and moment along theby beam’s axis engineers to decide reinforcement materials within engineers to decide where to placethe reinforcem Overhanging beam where to place beam or how to proportion thebeam size of at variousthe points or the howbeam to proportion size of the b Fig.length. 6–1 along its along its length. In order to formulate V and M in terms of xtowe must choose theMorigin In order formulate V and in terms of x w Kirişlerin tasarımı için for iç kuvvetlerin (moment veis kesme and the positive direction x. Although the choice arbitrary, most and the positive direction for x. Although the often the origin is located at the left end of the beam and the positive often the origin is located at the left end of the kuvveti) maksimum değerlerine gerek vardır. İç kuvvetler 255 P w0 direction is to the right. direction is to the right. kiriş enkesit değişimine, mesnetlenme ve yüklenme In general, the internal shear and moment functions of shear x will and be moment general, the internal durumlarına göre kiriş boyuncaIndeğişiklik discontinuous, or their slope willdiscontinuous, be discontinuous, at points or their slopewhere will bea discontin B D iç kuvvetlerin kiriş boyunca göstermektedir. Bu nedenle distributed load changes or where concentrated forces or distributed load changes or couple where concentr C x1değişimini gösteren diyagramların çizilmesi moments are applied. Because of moments this, the shear and moment functions are applied. Because of this, the shea x2 gerekmektedir. must be determined for each region of the beam between any two of the b must be determined for each region x3 x1, x2, and discontinuities of loading. For example, coordinates discontinuities of loading. For x example, coordi 3 will Fig. 6–2 to describe the variation have to be used of used V and throughout the have to be to M describe the variation of V length of the beam in Fig. 6–2. These be valid only coordin length coordinates of the beamwill in Fig. 6–2. These x1, from within the regions from A to B for B to Cfrom for xA within the regions to Bfrom for x1, from B 2, and C toİşaret D for xKuralı: C to D for x3 . 3. w(x) Beam Sign Convention. Beam Before Sign presenting a method for Convention. Before pre determining the shear and moment as functions of x and later plotting determining the shear and moment as function Positive external distributed load these functions diagrams), is firstand necessary these functionsit(shear momenttodiagrams) V (shear and moment V establish a sign convention so asestablish to definea “positive” and “negative” sign convention so as to define “p values for V and M. Although values the choice of a sign conventionthe is choice o for V and M. Although Positive here internal shear arbitrary, we will use the one often used in engineering practice arbitrary, here we will use the one often used M Fig. 6–3. The positive directions are as follows: the M in and shown and shown in Fig. 6–3. The positive directio distributed load acts upward ondistributed the beam; load the internal shearonforce acts upward the beam; t causes a clockwise rotation of the beam segment on which acts; causes a clockwise rotationit of theand beam segme Positive internal moment the Beam internal moment causes compression in themoment top fibers of thecompression segment in the t the internal causes sign convention such that it bends the segment so such that that it holds water.the Loadings are it holds w it bends segmentthat so that opposite Fig. to these 6–3 are considered negative. opposite to these are considered negative. PAGE !61 Shear and Moment Fun segment of the beam is sh this segment, wx, is repre 258 C H A P T E R 6 B E N D I2N5G8 (a) CHAPTER 6 BENDING segment is isolated as a fre centroid of the area comp wx x>2 from the right end. A x $ EXAMPLE 6.1 ŞekildekiEXAMPLE üniform yayılı6.1 yüklü basit mesnetli kirişin moment ve kesme yields 2 L M Draw the and shown moment Draw the shear and moment for shear the beam in diagrams Fig. 6–4a. for the beam A diagrams + c ©Fy = 0; kuvveti diyagramlarını çiziniz. x w 258 CHAPTER 6 BENDING SOLUTION Support Reactions. w wL 2 V SOLUTION (b) Support Reactions. The reactions are The support reactions are shown in support Fig. 6–4c. wL Shear Moment Functions. A free-body Shear and Moment Functions. A and free-body diagram of the left d+ ©M = 0; -a L 2 segment of the beam is shown in Fig. 6–4b. The di segment of the 6 beam is shown in Fig. 6–4b. The distributed loading on wx, this segment, is represented by its resultant this segment, (a) wx, is represented by its resultant force only after the (a) segment is isolated as acts a free-body This fo is isolated as afor free-body diagram. This throughdiagram. the AMPLE 6.1 Çözüm: Draw the shearsegment and moment diagrams the beam shown in Fig.force 6–4a. centroid of the area comprising the distributed lo EXAMPLE 6.1 EXAMPLE 6.1 centroidwxof the area comprising the distributed loading, a distance of wx x>2 from the right end. Applying the two equa Draw the shear and moment diagrams for the beam shown in Fig. 6–4a. PLE 6.1 x Öncelikle mesnet xreaksiyonlarını hesaplayalım. x>2 from the right end. Applying the two wequations of equilibrium w Shear and Moment Dia SOLUTION yields 2and Draw thethe shear moment diagrams forfor the beam shown inin Fig. 6–4a. Draw shear and moment diagrams the beam shown Fig. 6–4a. yields 2 shown in Fig. 6–4c are obt M Support Reactions. The reactions Draw the shear diagrams forsupport the beam shown inare Fig.shown 6–4a. in Fig. 6–4c. M and moment wL w A wL + Lc ©Fy = 0; wL SOLUTION zero shear can-beVfound - wx = 0 fr A wL c ©Fy x= 0; + wx V = 0 w w 2 SOLUTION xSupportShear SOLUTION 2 2 2 and Moment Functions. A free-body diagram of the left V Reactions. The support reactions are shown in Fig. 6–4c. V V L w wLReactions. SOLUTION V segment ofSupport the beam isReactions. shown in Fig. 6–4b. The reactions distributed loading onininFig. The support are 6–4c. Support The support reactions areshown shown Fig. 6–4c. L wL L the left 2 V = wa - xb (b) Shear and Moment Functions. A free-body diagram of 2Support wx, this segment, is represented by its resultant force only after the wL Reactions. The support reactions are shown in Fig. 6–4c. V = wa - xb (1) (b) 2 L (a) 2 acts through Shear and Moment Functions. A ofofthe segmentsegment of the beam is shown in Fig. 6–4b. distributed loading ondiagram Shear and Moment Functions. Afree-body free-body diagram theleft left 2The is isolated as a free-body diagram. This force the Kesme kuvvet fonksiyonlarıni elde edebilmek için kirişi mesnetler arası Lve L moment x birloading wL x wx,ofFunctions. this segment, is represented by its resultant force after the segment the beam isthe shown in Fig. 6–4b. The distributed onon Shear and Moment A free-body diagram ofonly the left segment of the beam is shown 6–4b. The distributed loading centroid the areaofcomprising distributed loading, wL Lin Fig. (a) wx d+ ©M x= a0;distance -ofa b x + 1wx2 a b + M L wL segment is isolated as a free-body diagram. This force acts through the wx, this segment, is represented by its resultant force only after the segment of the beam is shown in Fig. 6–4b. The distributed loading on d+ ©M = 0; a bx + 1wx2a b + M = 0 wx, this segment, is represented by its resultant force only after the 2 2 2 " x>2 from the right end. Applying the two equations of equilibrium noktadan keselim. (x=0) olarak sol mesneti alalım. x (a)(a) Referans 2 2loading, 2 force 6thenoktası NOTE: From the mome centroid of area comprising the distributed a distance of segment is isolated as a free-body diagram. This acts through the wx, this segment, is represented by its resultant force only after the segment is isolated as a free-body diagram. This force acts through the 6 yields 2 wx (a) 2 M w wL point on the beam wher x>2 is from the right end. Applying the two equations of equilibrium ofdiagram. thethe area comprising the distributed a adistance of segment isolated as a centroid free-body This force acts through the loading, centroid of area comprising the distributed loading, distance M x M max ! wx wx w M = of1Lx - x22 wL 2 A2 8 Eq. 6–2 (see - equations x2 from thethe right end. Applying the two equations ofofequilibrium centroid of the the distributed of 2 (2)Sec. 6.2) th from right the1Lx two equilibrium + carea ©Fy comprising = x>2 0;x>2 - end. wxloading, -Applying VM = a0=distance x yields x wx xM 2 2 equations of equilibrium have yields x>2 the right end. Applying the two 2 from yields V 2 wL x A Shear and Moment Diagrams. The shear an - wx - V =wL 0 MM x y = 0; w yields+ c ©F 2 x wL wL wLL 2 Shear and Moment Diagrams. Theinshear and are moment diagrams A A 2 MV (b) V = wa xb (1) shown Fig. 6–4c obtained by plotting Eqs. c ©F + c+©F = 0; wx V = 0 = 0; wx V = 0 y y Mmax1 2 wL x x 2 obtained shown in Fig. 6–4c are by plotting Eqs. 1 and 2. The point of 2 2 wL A (c) zero shear can be found from Eq. 1: L V yV= 0; + c ©F wx V = 0 L wL x2 =wL wcan a be - found x b wL (1) shear from Eq. 1: 2zero V (b) wL2 L wL wLwL x 2 V LL 2 L d+ ©M = 0; a bx + 1wx2a b + M = 0 2 2 Fig. 6–4 V = wa xb V 2 2 V = wa xb (b)(b) wL V = (1) w(1) a - xb = 0 L 2 2 2L 2 - (1) V 2 2 wL x V = wa xb V = wa xb = 0 (b) wL- a d + ©M = 0; b x + 21wx2 a b + M = 0 2 xx 2 2 wwL wL wL 2 2 1wx2ab b+ + d+ ©M = =0; 0; M = - a- a1Lx bxbx MM= =0 0 (2) x = L -x+x+1wx2a 2L 2 wLd+ ©M x 2 22 2 2 L 2 d+ ©M = 0; a bx + 1wx2a b + M = 0 x = x 6 6 w L 2M = 2 21Lx - x22 " wL 2 (2) wL 2 w wNOTE: From the moment diagram, this value 2 2 w Shear "and Moment Diagrams. The shear and moment diagrams MM= = diagram, 1Lx (2) 1Lx- -x2this x2 22 value of x represents (2)the 2 Diyagramları çizelim: shown in 2Fig. NOTE: From the moment M 6–4c w wL2 by plotting 2 point on the beam where the maximum mom are obtained Eqs. 1 and 2. The point of 2 M ! w max - x 2 M = 1Lx (2) 2 ShearwLand Moment Diagrams. and moment diagrams M 8 shear point on from theThe beam where theEq. maximum moment occurs, sinceVby= dM>dx = 6–2 (see Sec. 6.2) the slope zero shear can be found Eq. 1: 2 M ! max wLmoment L wL Kesme kuvveti ve x’e sınır değerleri Fonksiyonların shown in wFig.fonksiyonlarını 6–4c Shear are obtained by plotting Eqs. 1 vererek and 2.The The point of 8w Vçizelim. =shear dM>dx =moment 0. From Eq.and 6–2 (see Sec. 6.2) the slope Eq. 2, we Moment Diagrams. The shear and moment diagrams Shear and Moment Diagrams. and diagrams have 2 2 w L zero shear can be Diagrams. found from Eq. 1:6–4c V shown in Fig. 6–4c are obtained by plotting Eqs. 1 and 2. The point ofof have Shear and Moment The shear and moment diagrams shown in Fig. are obtained by plotting Eqs. 1 and 2. The point wL L x V wa - xb = 0 maksimum ve sıfır noktalarını grafik üzerinde not= edelim. L w L L 2 2 2 shown in Fig. 6–4c arex obtained zero shear can be found from Eq. 1: by plotting Eqs. 1 and 2. The point of zero shear can be found from Eq. 1: M = L a b a b 2 B wL L L wL wL L wL max 2 wL L w L L V 2 2 2 V = a - x sıfır b =Molduğu 0 from Eq. 1:w(c) 2 2 wL 2 = La b a b Kesme kuvvetinin x mesafesi: B R 2 L max 2zero shear can be2 found L 2 LL 2 = =0 0 2 V V (c) x = x V V= =wawa2 - -xbxb 2 L L 2 Fig. 6–4 2 2 wL2 wL V = x wa= L- xb = 0 2 wLwL " = 2 x Fig. 6–4 2 2 wL of x represents the 8 L 2 2 NOTE: From the moment diagram, this value 2 LL wL x x= = =moment x xon the beam where the maximum 2M "2 8 wL point occurs, since by L Mmax !L L 2 NOTE: From moment this value 2of2x represents the wL wLthe xSec. = 6.2)diagram, x 2 82 V = dM>dx = 0. From Eq. 2, we Eq. 6–2 (see the slope " " L 2 2 M point on the2 beam where the maximum moment occurs, since byofof 2 NOTE: From the moment diagram, this value x xrepresents Mmax ! wL " wL NOTE: From the moment diagram, this value representsthe the have 2 8 V = dM>dx = 0. Eq. (see Sec. 6.2) the slope From Eq. 2, we 2 6–2 2 MM 2 wL point on the beam where the maximum moment occurs, since by wL point on the beam where the maximum moment occurs, since by NOTE: From the moment diagram, this value of x represents the x Momentin maksimum değeri: L 2 Mmax ! ! Mmax L2 have wmoment Lslope 8 8 the beam where V = dM>dx = 0. Eq. 6–2 (see Sec. 6.2) the From Eq. 2, we V = dM>dx = 0. Eq. 6–2 (see Sec. 6.2) the slope From Eq. 2, we wL point on the maximum occurs, since by Mmax = B La b - a b R Mmax ! 2 x 2 8 2Eq. 2, we (c) have = 2L 0. From Eq. 6–2 (see Sec. 6.2) have the slope V L w = dM>dx L2 M = L a b a b B R max 2 x x have 2 2 2 2 ww LL L L (c) L L2 2 Fig. 6–4 wL x2 2 M = La b a M = La b a b bR R B B 2 L w L = Lmaxmax 2 2 22 22 (c) (c) 2b - a 8 M = La b B R Fig. 6–4 max 2 2 = wL2 2 (c) 2 2 8 Fig. 6–4 Fig. 6–4 wL wL = = 2 Fig. 6–4 wL 88 = 8 CHAPTER 6 BENDING 258 258 6.1 C HCAHP AT EP RT E6R 6LB EBNEDNI NDGI N G EXAMPLE CHAPTER 6 BENDING PAGE !62 6.1 6.1 SHEAR AND MOMENT DIAGRA 259 SHEAR AND MOMENT DIAGRAMS EXAMPLE 6.2 6.1 Draw the shear and moment diagrams for the beam shown in Fig. 6–5a. ! EXAMPLE 6.2 Şekildeki üçgen yayılı yüklü konsol kirişin kesme kuvveti ve moment w0 L 2 Draw the shearçiziniz. and moment diagrams for the beam shown w0 in Fig. 6–5a. diyagramlarını EXAMPLE 6.2 w0 L EXAMPLE 6.2 w0 L w0 6.1 6.1 SHEAR AND MOMENT DIAGRAMS Çözüm: w0 L2 3 L (a) 2 L 3 Mesnet r and moment diagrams forreaksiyonları: the beam shown in Fig. 6–5a. SOLUTION wn in Fig. 6–5a. L Support Reactions. w0 L w0 2 force and w0 L Fig. 6–5a. ownw0in 6.1 SHEAR AND MOMENT D 2 w0 Draw the shear and moment diagrams for the beam shown in Fig. 6–5a. DrawSthe and moment diagrams for the wshear 6.1 HEAR AND M OMENT DIAGRAMS 259beam shown in Fig. 6–5a. 0L EXAMPLE 6.2 2 2 2 w0 L 259 w0 L L L w0 L w0 3 w0 w 20 Draw the shear and moment for the beam shown2in Fig. 6–5a. w0diagrams 3 259 (a) (b) w0 L SHEAR AND MOMENT DIAGRAMS 6.2 2 w0 6.1 SHEAR AND M SHEAR AND MOMENT DIA w0 L (b) w0 L 2 w0 L 2 2 w0 L 2 2 w0 L22 2 w0 L L 3 0 L is replaced The distributed wload 3 3 by its resultant (b) 3 (a) the reactions have been determined as shown in Fig. 6–5b. (b) (a) 2 1 w0 x x 2 L SOLUTION w0 w0 L2 w0 L L L w0 L 2 3 Support Reactions. TheShear distributed load is replaced by its resultant x 2 1 w 3 w0 L and Moment A free-body diagram(b)of a 0 beam x 2 (a) asFunctions. w0 w0 x L force and the reactions have been determined shown in Fig. 6–5b. 2 2 segment of length x is shown in Fig. 6–5c. Note wthat the intensity wof" L SOLUTION 0L SOLUTION the triangular load at the section is found by proportion, that is, 2 1x Shear and Moment Functions. A Support free-bodyReactions. diagram of The a beam 2 distributed load is replaced by itswresultant 2 0L w0 L2 L L Support Reactions. The distributed is replaced by its resultant or With the load intensity known, the M w>x = w >L w = w x>L. 1 3w 0 6–5c. Note that 0 3 segment of length x is shown SOLUTION in Fig. the intensity of 3 6–5b. force and the reactions have been determined as shown in Fig. x 2 L 3 force and thedistributed reactions have been determinedfrom as shown in Fig. 6–5b. ofis the loading isthat determined the area under w0 L a) the triangular load(b)at theresultant section found by proportion, is,load is2 replaced w L Support Reactions. The distributed by its1resultant 0 V x w L 2 1(c) the diagram. Thus, 0 Shear and Moment Functions. A free-body diagram of a beam b) With and the load intensity known, thefree-body w>x = w0>L or w = w0x>L.Shear 3a beam 2 force the reactions have been determined asdiagram shown in 6–5b. 2 6 and Moment Functions. A of Fig. 3 Kesme kuvveti moment loading fonksiyonlarını elde edelim. Referans noktası olarak ankastre x that the intensity of w0 L segment ofw length xarea is shown in Fig. 6–5c. Note resultant of thevedistributed is determined from the under segment of length x is shown in Fig. 6–5c. Note that the intensity of L w x 1 0 0 b) Shear and Moment Functions. A free-body diagram of(c)a beam that is,2 the triangular load by proportion, + c ©F = 0; - section -found V = is 0byfound ¢at the ≤ xissection the diagram. Thus, ytriangular loadyayılı at the proportion, that is, mesneti seçelim, eğer boş ucuthe seçersek trapez yük ile uğraşmak zorunda kalırız. 2 2 L w0 L2 2 segment of length x is shown in Fig. 6–5c. Note that the intensity of w0 Lthe orx>L. With the load intensity known, w = With w0x>L. 0>L or ww the load intensity known, the w>xw= xww>x >L = = 1w 3 tions. The distributed load is replaced bywitsLresultant 0 0 0x 1the triangular 3 load atwdistributed the sectionloading is found by proportion, that is, under 0 resultant of0 the is determined from the area xw x 2 c ©Fy = 0;as shown in0 Fig. resultant of the distributed loading determined from the area under x V = 0 w02is ¢ ≤ w L x L reactions have been +determined 6–5b. 2 2 by its resultant 0 1 w0 x or w =Thus, w>xL= w w0x>L. (1) the w0 L3 V = w "With 1L - the x 2 load intensity known, 0>L 2 2 the diagram. x w L 0 the diagram. 2L L is determined from the area under 2 w0 x w0 x wn in Fig. 6–5b. 2 L resultant of2Thus, the distributed loading w " of a beam doment by its resultant Functions. wA L 0 L free-body 1 wdiagram 0x w0diagram. Thus, wL wx the x w2 L w0L 1 0wM0x-(1)11 ¢ w00xw≤0 x L -1 V = 0 w0 xV = of 1L2+-c ©F L the intensity wn in Fig. 6–5b.in Fig.26–5c. Note ngth xofis x22y =w0; 2 that 0L 0 ram ashown beam c2L + ©F = 0; x V = 0 xb + M = 0 ¢ ≤ w " M d+ ©M = 0; 1x2 + ¢ ≤2xa y 2 2 L w L L 0 found by proportion, w0 L2 2 2 L load at the section is that is, w L w x he intensity of 3 0 1 x 2 V1 L 3 0 2 2 2 c + ©F = 0; x V = 0 w L ¢ ≤ V 3 gram a 0beam y0 or w of = that w x>L. 3 ortion, is, With2 the load intensity 2 L w0 2 6 2 M2 1 x V wknown, wthe 3 1 w0x x 12 0L 0L w w0 L w0 L w the intensity of 0 V w L = 1L x 2 2 2 0 0 d+ ©M = 0; 1x2 + xa xb + M = 0 ¢ ≤ ety distributed loading is determined from the area under 2 3 2 3 known, the 3 0L (1) 2 w(1) V1-2L = 1L -2L 6 wx0 L2 x 2 (2) M = + 3L x 3 2 2 L 3 2 3 w0 2 portion, x 2w L 6L(c)V =2L 1x V hus, the area that underis, w L2 0 (1) 1L - 3x22 V 0 ity known, the 3 2L 2 6 (c) w0L L2 Eqs. w0L6–1 w0 results 1 1wof 1 2 applying 3 0xSec. 6.2, 0 Land 3 can be 2 checked 3w0Lby x w w 1w w0under L 1 w0x 0 0x+ 6–2 d + ©M = 0; 1x2 x a x b + M = 0 ¢ ≤ (2) M =These 1-2L + 3L x x 2 m the area 2 w0 L2 d+ xb + M - ¢ ≤ x - V = 0 (c) 30L1x2 + 221¢ L 3 = 0 w0 L2 6L that is,©M = 0; w30L2- w w0x≤ x2 a 3 1L 2 2 2 L V 3 M d + ©M = 0; 1x2 + ¢w x ≤ x a x b + M = 0 w3 L2 V dV 36–2ww0 2 0 0 Sec. 0L 2 6.2, 3 x These results can be checked by applying Eqs.w6–1 and of w w L OK = = 10 2x2 = 0 3 2 3 w0 M = 3 0 1 - 2L w30(2) L V w0 2 + x3L w0 dx 3 x - x 2 2 that w0 L (2) M = 2L 1 - 2L 6L+ 3L2x L -M 2 (1) V = Diyagramları 1L is,- x22 çizelim: w 2 w L 0 0 2L w L 2 dM w0w (2) 2 2 x =6L 12- 2L3 2+ 3L2wx0 - x2 32 2 w xM dV 0 0 (1) w0 0 10 w = = 2x2 = V = = 10 +6L 3L -OK 3x = 1LEqs. - x 2 andOK 0L These results can be checked by2applying 6–1 6–2 ofwSec. 6.2, 2 ! 2 dx 2L L These results can be checked by applying Eqs. 6–1 and 6–2 of Sec. 6.2, dx 6L 2L Bu sonuçları bölüm 6.2’deki bilgilerle kontrol edelim: 3 w0L w0L w L 1 w0x 1 that is, These can be checked by applying Eqs. 6–1 and 6–2 of Sec. 6.2, (d) (1) 1x20 + ¢ + Mthat = 0is, wresults ≤xa xb w 2 M wDiagrams. 0 0 1 M 2 2 2 The and Moment graphs 1 and 3 2 2 V L = dM 3 = Shear 2 w0 w0 Lof2 Eqs. w0x2 are dVOK is,2 0-L 10 that + 3L 3x 2 = 1L x 2 w w x dV x b + M = 0 w L2 V 0 0 OKM = ! 10=3 -- 2x2 dx 6L 2L shown in Fig.3 6–5d. 0 OK w = dVw= = wdx w0x= - L 3 010 - 2x2 Fig 2L w0 3 3V (d) dx 2L L w L OK w = = 10 2x2 = 1 0 2 3 dx 2L L (2) - 2L + 3L x x 2 xb +M M== 0 1 w 2 Shear and Moment Diagrams. The graphs of Eqs. 1 and 2 are w w 2 0 0 2 2 6L 0wL0 L 3 dMV =wdM w0 L2 0 0 22 2= = - w3x 1L -OK x22 w OK 2 210 + 3L 2 (2) w 3shown V = dM = wdx x2Fig. 22 6–5 0L 2 ! 010 + 3L 01L 2-2L 2 V in Fig. 6–5d. 2- 3x 22 = 6L 3 V = dx =6L 10 + 3L - 3xx 2 =2L 1L - x 2 OK !! w3 0 L an be checked by applying Eqs. 6–1 and 6–2 of Sec. 6.2, w0 L dx 6L 2L 3 (2) Shear and Moment Diagrams. The graphs 1ofand Eqs. 1 and 2 are x 2 6–2 of Sec. 6.2, Shear Moment Diagrams. M Shear and and Moment Diagrams. The Thegraphs graphsofofEqs. Eqs. 1 and2 2are are shown in Fig. 6–5d. w0 w0x dV x shown ininFig. 6–5d. shown Fig. 6–5d. M OK w = = 10 2x2 = x d 6–2 of Sec. dx6.2, 2L L x OK w0 w M dM 0 = OKx ! w0 L2 = 10 + 3L2 - 3x22 = 1L2 - x22 dx2 6LOK w L2 2L 2 3 OK ! 0 - x2 (d) 3 oment Diagrams. The graphs of(d) Eqs. 1 and 2 are 2 OK ! w0 L2 L x22 2 are –5d. qs. 1 and qs. 1 and 2 are 3 Fig. 6–5 (d)6–5 Fig. Fig. 6–5 PAGE !63 BENDING EXAMPLE EXAMPLE 6.3 6.3 EXAMPLE 6.3 EXAMPLE 6.3 2 kip/ft 6 kip/ft SOLUTION EXAMPLE 6.3 Draw the shear and moment diagrams the beam shown in Fig. 6–6a. Draw the shear and momentDraw diagrams forfor the beam shown indistributed Fig. 6–6a. the Draw shear and moment diagrams for theload bea Support Reactions. The the shear and moment diagrams f 260 CHAPTER 6 BENDING 260 CHAPTER 6 BENDING 6 kip/ ft 6 kip/ft 6 kip/ ft and rectangular component loadings and 6 kip/ ft 260 CHAPTER 6 BENDING 18 ft 2 kip/ ft replaced by their resultant forces. Th 2 kip/ft 2 kip/ ft SOLUTION 2 SOLUTION kip/ ft SOLUTION SOLUTION 6 ki (a) Draw and moment for the beam shown Fig. 6–6a. determined as shown on the beam’s free-bo B E N Dthe I N Gshear EXAMPLE 6.3 Draw the sheardiagrams and moment diagrams for theinbeam shown inReactions. Fig. 6–6a. The Support The distributed load is divided into triangular Support Reactions. The into distributed load is d Support Reactions. distributed load is Reactions. divided triangular EXAMPLE 6.3 Support The distributed p/ ft 2loadings kip/ ft 36 kip 36 kip 6 kip/ ft EXAMPLE 6.3 Şekildeki trapez yayılı and rectangular component and these loadings then and rectangular component loadings and thes and rectangular component loadings and these loadings areare then and rectangular component loadings ! yüklü basit mesnetli kirişin kesme kuvveti vemoment Shear and Moment Functions. AforfreeDraw the shear and diagrams the 18 ft 4resultant kip/ft 18 ft 18 ft 18 ft replaced by their forces. The reactions have been SOLUTION SOLUTION replaced by their resultant forces. The ret Draw the shear and moment diagrams for the beam shown in Fig. 6–6a. replaced by their resultant forces. The reactions have been replaced by their resultant forces segment is in Fig. 6–6c. As above, moment diyagramlarını çiziniz. (a) 6the kip/beam’s ft for thefree-body Draw the shear moment diagrams beam shown in Fig. 6–6a. (a) and as he shear and moment diagrams for the beam(a) shown in Fig. 6–6a. 2on kip/ft (a) determined shown diagram, Fig. 6–6b. determined as shown on the beam’s free-body d determined as shown on the beam’s free-body diagram, Fig. 6–6b. determined as shown on the beam’s fre 6 kip/ft Support Reactions. The distributed load is divided into triangular by rectangular and triangular dis Support Reactions. The distributed load divided into triangular C H A P T E R 6 B E N D Ireplaced NG 6 kip/ ft is 2 kip/ ft 2 6 0 SOLUTION 36 kip 36 kip 36 kip 36 kip and rectangular component loadings and these loadings are then intensity of the triangular at the sectio 18 ft load 36 kip kip and rectangular component loadings and these loadings then 36 kip 36 kip 2 kip/ft Draw the shear and moment diagrams for the beam shown inare Fig.36 6–6a. and Moment Functions. A free-body diagram of left SOLUTION Shear and Moment Functions. Athe free-body 2 kip/ ft Shear and Moment Functions. A 9Shear ft and Moment Functions. free-body diagram of the left 4 SOLUTION kip/ ft Shear 4 kip/ ft ION 4 kip/ ft A replaced by their resultant The forces. reactions have been 4 kip/ft Support Reactions. The distributed load The resultant force and the location of eac replaced by theirforces. resultant The reactions have been (a) 6 kip/ft segment is shown in Fig. 6–6c. As above, the trapezoidal loading is segment is shown in Fig. 6–6c. As above, the t 12 ft segment is shown in Fig. 6–6c. As abo Support Reactions. The distributed load is divided into triangular segment is shown in Fig. As above, the trapezoidal loading is equat 2 Support kip/ ft6–6b. 2 kip/ ft 6–6c. shown on the beam’s free-body diagram, Fig. and rectangular component loadings and as shown on the beam’s free-body diagram, Fig. 6–6b. Reactions. The distributed isrectangular divided intoand triangular also shown. Applying the equilibrium 2 kip/ ftload rtdetermined Reactions.asdetermined The distributed load is divided into triangular 2 kip/ft EXAMPLE 6.3 replaced by rectangular and triangular distributions. Note that the replaced by triangular distribu 18 ft replaced by rectangular and triangula and rectangular component loadings and these loadings are then SOLUTION ft replaced by18component rectangular and triangular distributions. Note that the The 36 kip resultant 36 kipthen replaced bytriangular their forces. and these loadings are loadings are30and then kip rectangular 42 kip loadings pctangular component loadings and18 these intensity of the triangular load at the section is found by ft intensity of the load atproportion. the section intensity offound the triangular load at1theiss replaced byleft their resultant forces. The reactions have been (a) 18 ftThe Shear and Moment Functions. A free-body diagram of the intensity of the triangular load at the section is by proportion. Shear and Moment Functions. A free-body diagram of the left Support Reactions. distributed load is divided into triangular 9 ft 9 ft kip/ ftk determined as shown on the beam’s free-bod replaced by their resultant forces. The reactions have been c ©Fy of d by 4their (b) kip/ ft resultant forces.9 ftThe =resultant 0; 30 14di + kip - 12 kip>ft2x -4each ft Draw the shear and moment diagra (a) reactions have been The 9resultant force and the location each distributed loading are The resultant force and the location of The force and the location o Çözüm: determined as shown on the beam’s free-body diagram, Fig. 6–6b. (a) segment is shown in Fig. 6–6c. As above, the trapezoidal loading is segment is shown in Fig. 6–6c. As above, the trapezoidal loading is 2 The resultant force and the location of each distributed loading are 12 ft 12 ft and rectangular component loadings and these loadings are then 12 ft determined as shown on the beam’s free-body diagram, Fig. 6–6b. ined as 2shown 36 kip 36 kip kip/ ft on the beam’s free-body 12 ft diagram, Fig. 6–6b. 6also kip/ ftshown. Applying also shown. Applying the equilibrium equations, we have 2 kip/ ft also shown. Applying the equilibrium equations, the equilibrium e t replaced by rectangular 18 ft 1shown. x Applying replaced rectangular and triangular thethe 18 ft Note distributions. Note that the Shear equations, and Moment Functions. A free-b equilibrium we have 36 triangular kip 36 kip 18that ft replacedbyand by their forces. distributions. The also reactions been 4 x have 4 kip/ ft 18resultant ft kip x2 30 kip 36 kip 36 42kip kip Shear and Moment Functions. A free-body diagram of the left 30 42 kip 2x 2 kip/ ft 2 18 30 kip 42 kip intensity of the triangular load at the section is found by proportion. Shear and Moment Functions. A free-body diagram ofV segment is in Fig. 6–6c. SOLUTION intensity of the triangular at theon section by proportion. 4iskip/ft and Moment Functions. A free-body diagram of left x the 1 shown = ¢left 30As-1above, 2x - th 30 kipload 42 kip determined as shown the beam’s free-body diagram, 6–6b. 1 4found kip/ ftthe 2-kip/ ftabove, c ©Fy4Fig. (b) 1-0; cabove, =xin(b) 0; 14yrectangular kip>ft2 ax kip +(b) 30 kip 12 kip>ft2x -is V 14 = kip>ft 0 dist =+the + ©F 30 kip 12 kip>ft2x is Fig. 6–6c. As loading ctrapezoidal = 0;- 30 ©F -b-xis 12 kip>ft2x -9 y the kip/ft The resultant force and the location ofsegment each distributed loading are segment isshown shown in Fig. 6–6c. As trapezoidal loading replaced by and triangular The resultant force and the location of each distributed loading are nt is shown in Fig. 6–6c. As above, the trapezoidal loading is c (b) 9 ft 2 kip/ft = 0; 14 kip>ft2a + ©F 30 kip 12 kip>ft2x bx V = 0 2 18 ft 2 Support Reactions. The distribu 18 y 2 kip 2 kip/ ft replaced by rectangular 6 A and distributions. Note thethe 2 of 12 18 that ft that also shown. Applying equations, replaced by rectangular and triangular distributions. Note Shear andequilibrium Moment Functions. free-body of the lefttriangular intensity triangular load at the section d also by rectangular and triangular distributions. Note that the1 wediagram shown. Applying the equations, we have 1 the xequilibrium xhave d+ ©M =and 0;the2ft rectangular component 4 kip/ft 12 kip/ft x 2loadi 4 x 4 x 9 ft 4 loading x load intensity of18 the triangular thethe section is is found byby proportion. x force x each x found 2 x1 in is intensity of load section proportion. 2 xthe segment isthe shown Fig. 6–6c.byAs above, isft atVat 18 kip triangular ty of42the load at section proportion. and location of 2 trapezoidal 2 the x 2 triangular 18 ft by fo = ¢The 30 -resultant 2x replaced -2found kip ≤ 18 V = their 30are ki x 9 ft9 ft 2x 2 24 18118x x ¢the 1xThe resultant =2x¢1 30(1) - ≤2x Mand 2 kip/ft 12 ftx force xV- resultant x force the location of each distributed loading x (1) 9 9kip>ft x c V = 30 2x kip ¢ ≤ The resultant and the location of each distributed loading are (a) sultant force and the location of each distributed loading are = 0; 14 kip>ft2 a + ©F 30 kip 12 kip>ft2x b x V = 0 replaced by rectangular and triangular distributions. Note that the also shown. Applying the equilibrium equati 30 kip 42 kip 4x kip>ft2a kip/ ft 4 018Vftkip/ ft4 -30 kip1x2 + 12 kip>ft2xa b + 14 14 + c ©Fy = 0; 30 kip y- 1212kip>ft2x b x V = ft kip/ ft determined as shown on the beam’ 12 ft 18kip/ft 18 9 have 2 also 18 ft 18 equilibrium equations, we 2 2 shown. Applying the 24 we 18 ft6 also 3 Applying 6 6 at the the42equilibrium equations, we have own. Applying theintensity equilibrium equations, have of the triangular load section is dfound 30 kip xshown. kip 1818 ft ft (b) 218kip/ ft kip/by ft= proportion. x2©M + 0; d + ©M = 0; 2 kip/ ft 6 36 kip 36 kip 1 d + ©M = 0; 2 3030 kipkip 42242 kip 3 2 kip/ft kip The resultant force and the location of xeach distributed Functions. ©M = 0; loading are 0; x ©Ffty =Shear 30xand kip -Moment 12 kip>ft2x - 2 14xki 2d+ 2 (b) x¢M 1+41ckip/ x 1 (1) M x 1 x 1 x x V = 30 2x kip ≤ 2 M 12 30 kip x 1 (1) c= (b) M = 30x x ¢ V = 30 2x kip = 0; 14 kip>ft2a + ©F 30 kip 12 kip>ft2x bx V = 0 ¢ ≤ c (b) x y = 0; 14 kip>ft2 a + ©F 30 kip kip>ft2x b x V = 0 x also shown. Applying the equilibrium equations, we have 1 x = 0; 14 kip>ft2 a 30 kip 12 kip>ft2x b x V 0 is 12 shown As x y- 30 kip1x2 1218kip>ft2x abkip>ft2x 14M kip>ft2 a 9 a302bkip1x2 + 14+segment kip>ft2 xxba +in b Fig. y ft M18 ft a+ b 6–6c. += 014 p/ 271k 4ftax+ V x (c) V+ 12 kip>ft2xx221 kip/2ft 30 kip1x2 2 x 3 V 9 2 b 3+ 2M2 =and 18 ftxrectangular 3 1 kip1x2 2 2replaced x 2 18 ft -30 kip>ft2a bxa 0 2trian 4 x3 + 12 kip>ft2xa b2 + 14 18 by 42 kip x V 1 x 2 x x 2 x x18 x 1 x 2 Equation 2x2 2 2 may 18be ft checked 3= ¢by 0; 14 4x xx23xx x V 30 2x ≤ 6 kip/ft 3 noting that d + ©M = 0; d ++©M 3 c ©F=y2x= x x - V = 0 intensity of xx - 12 kip>ft2x - 14 kip bx 2 kip>ft2a 2 4 the triangular kip/ ft 2 load x at29 td 21818 x 2 x2 0; 2x 30 (1) 4 29 ft kip/ 2≤ kip VftV= =¢ 30 kip 2(1) 2x-- ≤xw # ¢ 30-= -2x V = ¢ 30 - 2x ≤2x kip x 18 2 (1) 30xkip 2 18 ft 3 dV>dx 30 kip (2) 30 2x kip M = 30x x kip ¢ ≤ M 30x kip ft ¢ ≤ 1 x 18 M = 30x x ¢ Also, This equa = = 2 x. 2 x 27resultant force 9and the27locatio 9 29 The 2x kip/ft 4 491 kip/ft # ft2 kip/ 30kip>ft2xa kip1x2 +x12 kip>ft2x a(c)ftb + x614bkip>ft2 a b(c) x a b 12+ft(c) M =M 0 =6¢ 30xdKontroller: 30 kip ft when x(2)= 18 ft, - ©M kip 1818 - 30 kip1x26 6+ -12 b + 14 kip/ kip>ft2a =also 0; and x+ =x 0,-w =≤ -shown. 2 kip>ft, 2 ft x a 3xb2 +18M ft = 0 23 kip/ ft (c) 2 Applying the equilibriu 27 2 2 18 18 ft (1) 2 kip/ft 2 kip/ ftV = ¢ 30 -6 2x d+ ©M = 0; d + ©M = 0; = 0; 6 kip/be ft checked Equation 2 may that dM>d kip/ft3 ≤ kip Equation by noting thatbedM>dx that is,1by Eq. 1. t = by V,xnoting 6 kip/ ft kip Equation 2checked may be checked noting M2 may 30 kip 42 M 9 x x 3 2that kip/ft 2noting -Shear 30 kip1x2 + 12 kip>ft2x ax. -bsince +-Eq. 14 kip>ft2 2 equation x 2 6 kip/ft and Moment Diagrams. Equati Equation 2 may be checked by that is, 1. dM>dx = V, M x x 1 x x # M x 1 x x V Also, This w = dV>dx = 2 x 1 x x Also, This equation checks, when w = dV>dx = 2 x. (2) Also, This w = dV>dx = 2 x. 2 kip/ ft - x - -30 kip ≤ kip1x2 (b) 2 kip/ ft kip/ftft(2) 9 kip 30 12 2 = 29 kip>ft2x 9kip>ft2a 3+ + Vb 0; x -M x2a =- ¢ 30x 1212 kip>ft2xa b b+ + 14 bxa M = 0 -0of ¢14x30x kip>ft2x 14 kip>ft2 a+ c ©F b yx a= b+the + M p1x2 + 12 kip>ft2x bM+= kip>ft2 b≤xkip a b# ft+ M 0x2kip1x2 27-=30 t Fig. 6–6d. point maximu d+a ©M Also, This equation since =2a42-2 -2x292x. = in 0, w -18 2ftkip>ft, x Fig. =when 18 ft,xw= =1 when 6–6a. x x= w 0, = w dV>dx = - 2 kip>ft, xx== ft, w =and 6when kip>ft, and when = 0, w = 3checks, 18 2kipand 18 ftx3Since 32-kip>ft, 2 2=kip/ft 20;3 3 V V 18 27 ft 3 30 kip x x x x x 2 x = 20, w = 1-2 kip>ft, x3 1 x (Eq. 6–2), then, from Eq. dM>dx = V = 0 and when Fig. 6–6a. x = 18 ft, w = -6 kip>ft, 2 4 x 6 kip/ ft V(kip) Equation22 may be noting that that is, Eq. 1. dM>dx = V, x 1 x x 3 3 30 kip 3checked by M = 30x x ¢ ≤ 2 2 2and x b2 +Moment 2 noting Equation 2 may-30 be checked that Eq.bxa 1. (c) = kip>ft2a V, that is,Shear 2 2 by xShear x and x 18 = 0 Diagrams. Shear Moment Diagrams. Equations 1 and 2(2) are plotted and Moment Diagrams. kip1x2 b + 14 M 2 2 30 2dM>dx = ¢ 3027E # # # kip (2)VEquations kip (2) 18 kip M = 30x x kip ft MAlso, =30¢30 30x x2+2 -12(c)kip>ft2xa kip ft ¢ ≤ M = 30x x kip ft ≤ ¢ ≤ 30 x This equation checks, since when w =- dV>dx = 2 x. 2 2 ft 3 9 (c) (c) 4 kip/ Fig.27 Since point of0plotted maximum Also, w = dV>dx = - 2 - 9 x. 27 This equation checks, since 27Equations inwhen Fig. 6–6d. Since theftin point of6–6d. moment occurs when Shear and Moment Diagrams. 1the and 2Vare inmaximum Fig. 6–6d. Since the point of mam = = 30 - 2x 18 ft kip kip kip 42 kip Equation 2 may be checked 6–6a. x = 0, w =30 2 kip>ft, and when x =3042 18 ft, 36 w 30= kip - 6 kip>ft, Fig. 42 6kip kip/ by noting that d (Eq. 6–2), then, from Eq. 1, dM>dx = V = 0 when x = 18 ft, w = - 6V(kip) Fig. 6–6a. x = 0, w = - 2 kip>ft, andV(kip) kip>ft, (Eq. 6–2), then, from Eq. 1, dM>dx = V = 0 (Eq. 6–2), then, from dM>dx = V = 0 in Fig. 6–6d. Since the point of maximum moment occurs when 2 kip/ ft x(ft) x 1. V(kip) d + ©M =that 0; 21. 1. 630x kip/ Equation 2 may be checked by noting that that is, Eq. dM>dx = V, on 2 may be checked by noting is,Eq. = V, ftthat 30 kip that dM>dx # Equation 2 may be checked by noting that is, Eq. dM>dx = V, (2) M = 6 ¢kip/ft - 42 x2kip kip ft ≤ Also, This equat w = dV>dx = 2 x. 9.735 ft = V = 0 (Eq. 6–2), then, 2 kip/ ft dM>dx Choosing root, 9 6 kip/ ft from Eq.the 1, positive 2 and Moment Diagrams. Equations 1w and 2 are =plotted 30 x12 27are 30 30 V(kip) x2since x 18- ft, Thisx equation checks, since when dV>dx -29 x.29 x.This This equation checks, 1since when w Shear = dV>dx -Shear 22 kip/ft equation when wplotted = =dV>dx = x- 2- 2- M 2 kip/ and= Moment Equations andAlso, 2Also, and when == 0, w--=checks, 2-kip>ft, x = 9 x.ftDiagrams. V = 0 = 30 2x V = 0 = 30 V = 0 30 2x 30 kip1x2 + 12 kip>ft2x a b + in Fig. 6–6d. the of maximum moment when 2 18 kip/ x = 9.735 Vand 30 xpoint x62kip>ft, when Fig.that win=642 -kip 2 kip>ft, =Since 18be ft, checked w =point - 6 by kip>ft, when Fig. 6–6a. x= = =-=2-V, 2occurs kip>ft, =18 #42 9 Fig. and when 6–6a. x 6–6a. 0, 0, w w=when kip>ft, x x= x(ft) ft,ft, wftw = =-6-kip>ft, 2 ft 92 Fig. 6–6d. and Since the of maximum moment occurs 3 is, kip/ft x(ft) Equation 2 may noting that Eq. 1. dM>dx x(ft) V = 0 Shear =Thus, 30 -from 2x from Eq.M(kip"ft) 1, dM>dx = V = 0 (Eq. 6–2), then, x ft Mxmax ! 163 kip"ft and Moment Diagrams. Equatio 9.735 ftChoosing Eq. 2, root, 9.735 9.735 Choosing the positive 9the then, from 1,2 x. This equation dM>dx = V = Also, 0 (Eq.w6–2), Choosing positive root, the positive checks, since when root, = dV>dx =ft -Eq. 21 -and x(ft) 9 2 are plotted 2 2 Diagrams. 2and and Moment Diagrams. Equations Shear and Moment Diagrams. Equations 1and and 2 the are plotted Shear Moment Equations 1 2 are plotted in Fig. 6–6d. Since point of maximum x 30 kip M = 30x ¢ 9.735 ft V Diyagram x when = 9.735xft= and Fig. 6–6a. = 0, wof=maximum -çizimi: 2 kip>ft, = 2x 18 ft, wChoosing = 6–6d. - 6 kip>ft, 30 42 kip of maximum the positive root, 9.735 x = 9.735 ft = when 0 =x2occurs 30x - Fig. (c) #42 6–6d. Since thexpoint moment when inkip Since the point moment occurs #42 #42 in Fig. 6–6d. Since the point of maximum moment occurs when (Eq. 6–2), then, from Eq. 1 dM>dx = V = 0 30 kip 42 kip 9 V = 0 = 30 2x V(kip) M(kip"ft) M(kip"ft) M 30 kip 42 kip! 163 M(kip"ft) 163M kip"ft !then, 163from kip"ft kip"ft M Thus, from 2, M (Eq. 6–2), then, from Eq. 1, = V = 0x(ft) max2, = 3019.7352 - 19.735 maxthen, Thus, Eq. (Eq. 6–2), from Eq. 1,Eq. dM>dx=Thus, =V V=from =0max 0! x =Eq. 9.735 ft from max Eq. 2, 9 (Eq. 6–2), 1, dM>dx V(kip) #42 x(ft) Shear and Moment Diagrams. Equations 1 and 2 are plotted V(kip) 30 6 kip/30 ft kip Equation 2 may be checked by not Choosing M(kip"ft) the positive root, 42 kip 163 maximum kip"ft Mmax ! of =3 kip 0 =# ft 30 -2 2x 2 x2 Thus, from(d) Eq. 2, when Choosing the positive = V163 x2 point in 6–6d. Since the moment occurs 3030Fig.root, x Also, w = dV>dx = 2 V(kip) 19.7352 2 kip/ ft 42#42 kip x = 9.735 ft V = 0 = 30 - 2x 9 x.2 Th V = 0=x(ft) =3030- -2x2x - -- 19.7352 2 = 3019.7352 M0, -and 19.7352 M = 3019.7352 -19 Eq. 1, 9.735 ft Fig. 6–6 VM=max0 = dM>dx = V =x 0= (Eq. 3019.7352 96–2), maxw9 9.735 ft then, from max when x = = 2 kip>ft, x 9 3 x(ft) x(ft) x(ft) ! 163 kip"ft 30 Choosing the positive root, x(ft) 27 19.7352 Thus, from Eq. 2, x(ft) 2 2 9.735 # (d) # ngThus, the positive root, Choosing the positive root, (d) # from Eq. 2, = 163 kip ft 9.735 ft ft M = 3019.7352 19.7352 (d) x = x163 kip = 163 kip ft Choosing the positive root, max =Diagrams. 9.735ftft Shear and27Moment V = 0 = 30 - x(ft) 2x 3 #42 19.7352 x(ft) x = 9.735 ft x = 9.735 ft Fig. 6–6 9 Fig. 6–6 # M(kip"ft) Fig. 6–6 (d) = 9.735 ftft from #42 3 2 =x 163 kipThus, Mmax ! 163 kip"ft x(ft) inEq. Fig. #42 Mmax = 3019.7352 19.7352 - 30 kip 19.7352 9.735 ft2, 6–6d. Since the point of M(kip"ft) 42 kip 2 kip"ft ! 163 M 27 M(kip"ft) rom Eq. 2,x(ft) max Thus, from Eq. 2, Choosing the positive root, Mmax = 3019.7352 19.7352 ! 6–6 163 kip"ft M-max dM>dx = V = 0 (Eq. 6–2), then, fr Thus, from Eq. 2, Fig. V(kip) = 163 kip # ft 27 x = 9.735 ft 30 = 163 kip # ft #42 19.73523 19.7352 M3 3 = 3019.7352 - #42 19.7352 2 2 19.7352 max M(kip"ft) = 0 = 3 x(ft) - 19.7352 Mmax = Thus, 3019.7352 19.7352 M = 3019.7352 Mmax ! 163Vkip"ft 2 ax ! 163 kip"ft from-Eq. 2, max M = 3019.7352 19.7352 max 27 x(ft) x(ft) 27 (d) = 163 kip # ft 27 # ft x(ft) Choosing (d) = 163 kip # ft 9.735 ft = 163 kip 3 the positive root, # (d) = 163 kip ft 19.7352 Fig. 6–6 M = 3019.7352 - 19.73522 Fig. 6–6 max x = 9 Fig. 6–6 #42 x(ft) 27 x(ft) M(kip"ft) Mmax ! 163 kip"ft Thus, from Eq. 2, = 163 kip # ft (d) x(ft) (d) Fig. 6–6 PAGE !64 Fig. 6–6M max = 3019.7352 = 163 kip # ft 6.1 261 6.1 SHEAR AND MOMENT DIAGRAMS 261 SHEAR AND MOMENT DIAGRAMS EXAMPLE 6.4 EXAMPLE 6.4 6.1 SHEAR AND MOMENT DIAGRAMS 261 Draw the shear and moment diagrams for the beam shown in Fig. 6–7a. nt diagrams for theDraw beamthe shown in and Fig. 6–7a. shear moment diagrams for the beam shown in Fig. 6–7a. 15 kN 6.1 SHEAR AND MOMENT DIAGRAMS 261 EXAMPLE 6.4 5 kN/ m ! Ölçü ve yükleme durumu şekilde 15 kN 6.1 SHEAR AND MOMENT DIAGRAMS 261 15AND kN MOMENT DIAGRAMS 80 kN!m 6.1 SHEAR 261 5 kN/m 6.1 SHEAR AND M OMENT DIAGRAMSkirişin 2 6 1kuvveti ve5 kN/ m 80 kN!m verilen basit mesnetli kesme moment 80 kN!m Draw the shear80and moment for the beamDshown 6.1diagrams SHEAR AND MOMENT IAGRAMSin Fig. 6–7a. 261 6.1 SHEAR AND OMENT DIAGRAMS 2 16 1 kN!m 6.1 SHEAR AND MM OMENT DIAGRAMS 26 C M diyagramlarını çiziniz. A C C EXAMPLE 6.4 A 15 kN V 5 kN/ m ms for the in Fig. 6–7a. B x1 EXAMPLE 6.4 .4 B beam shownEXAMPLE 6.1 80 kN!m 6.4 Draw the shear and moment diagrams for the beam shown in Fig. 6–7a. m m Çözüm: 5.75 kN 5m 5m ment diagrams5for the beam shown in Fig. 6.4 6–7a. EXAMPLE M 6.1 261 SHEAR B AND MOMENT DIAGRAMS x1 V5 m 580 m kN!m SHEAR AND MOMENT DIAGRAMS M (a) 5.75 kN 261 5 Nwn in Fig. 6–7a. C (a) (b) Draw the shear and moment for the beam shown Fig. 6–7a. (a)diagrams (b) nd moment diagrams the beam shown inmoment Fig. 6–7a. 5 kN/ m for Draw the shear for the beam shown inin Fig. 6–7a. A80and kN!m 15 kN diagrams x1 V AB ve BC bölgelerinden kesimler yapalım: B SOLUTION EXAMPLE 15 kN 5 kN/ m Draw the6.4 shear and moment diagrams for the beam shown in Fig. 6–7a. M 80 kN!m 5 kN/m 15 kN 80 kN!m 15 kN 5(x2 " 5) 15 kN 515 5(x2supports " 5) 5 m 80 kN!m m kN Support m 5.75 kN Reactions. The reactions have been 80 kN!m M 15atkNthe C SOLUTION 5 kN/ 5 kN/m 5 kN/ mm kN!m Draw the shear and moment diagrams the beam in Fig. 6–7a. 80 for kN!m 8080 kN!m M shown 15 kN (a) 80 kN!m (b) x M determined and are shown on the free-body diagram of the beam, V 80 kN!m 80 kN!m e reactions at the supports have beenThe1 80 C supports5 kN/ Support Reactions. reactions at the kN!m m been 80 kN!m 80 kN!m M have MM CA x V 80 kN!m Fig. 6–7d. 1 n on5 the diagram ofand the beam, B on the 15 kN determined are shown free-body diagram C the m free-body M beam, M kN M Cof CV5.75 A x1 V 5 kN/ m B x1 15 kNV 5(x2 " 5) SOLUTION A 80 kN!m x Fig. 6–7d. 6.1 SHEAR AND MOMENT Dx IAGRAMS 261 V x V 80 kN!m 5 m 5 m B C 1 1 5.75 kN 1 (b) B B Shear and Moment Functions. Since there is a discontinuity of M 5m 5 5.75 m Support kN5 m B at the supports x1 V Reactions.A The reactions V have been 5m kN V (a)55.75 (b) 80 kN!m 5 m 5.75 kN 52m x " 5 x " 5 5.75 kN x " 5 x " 5 m 5 m C 5.75 kN 2 5 m 5 m distributed load and also a concentrated load at the beam’s center, 2 2 ctions. (a) Since there is aand discontinuity Shear Moment Functions. Since there is a discontinuity of A are of (b) determined shown on515 the free-body of the beam, (b) and 5(x2 " 5) 25diagram x1 kN(b)(b) V 2 2 m regions 2 5.75 Bm kN (a)(a) (a) ofcenter, x must be considered in order to describe theMshear and (b)attwo a concentrated load at the beam’s center, distributed load and also a concentrated load the beam’s x x Fig. 6–7d.6.4 2 2 " 5) 15 kN 5(x EXAMPLE SOLUTION (a) 2 (b) ons at the supports have been 5 m 5 m moment functions for the entire beam. 5.75kN kN nsidered in order two to describe 80and kN!m regionsthe of xshear must be describe 5.75 kN in order to 15 kN 5(x2the " 5)shear and 5.75 15 kN 5(xconsidered 5m V5) 2 " 5) 15 kNkN5(x5(x 5) SOLUTION 25" Support Reactions. Thethe reactions at the supports have been x15 (a) 2" free-body ofSOLUTION the beam, (b) 15 kN 5(x 2 " 5 x2 " 80of kN!m 2 " 5) entire beam.diagram Shear and Moment Functions. Since there is a discontinuity moment functions for entire beam. M (c) (c) Draw the shear and moment diagrams for the beam shown in Fig. 6–7a. Fig. 6–7b: 0 … x 6 5 m, 15 kN 5(x SOLUTION 2 " 5) 1 Thehave reactions at 80 the supports have on been 2 2 rts been determined and are shown free-body diagram the beam, 80reactions kN!m kN!m Reactions. The reactions atthe the supports have been 8080 Support Reactions. The at have been M distributed load and have alsothe abeen concentrated load atofsupports the beam’s center, kN!m kN!m ions. The reactions atSupport the supports x2 80 kN!m 15 kN 5(x " 5) Fig.the 6–7b: 0 SOLUTION … determined xdiagram 6 5 m, of own free-body beam, Support Reactions. reactions at the supports have been 15free-body kN m of on thethe beam, 2 15 kN Fig. 6–7d. 1determined 80 k 80 kN!m and are shown onthe the free-body diagram ofthe the beam, + cdiagram ©F =the 0; 5.75 515 mkNThe are shown on ofshear M V 6beam, M 6 two regions of xand must be considered order and are shown free-body diagram of the beam, MM 5.75 kN kN - V = 0 5ykN/m x2in "the 5 x2 free-body " 5to describe 5 kN/ m Mbeam, Since thereon is the a discontinuity of determined and 80 are shown on diagram of the 5 m Support Reactions. The reactions at the supports have been 5 kN/ m V Fig. 6–7d. M Fig. 6–7d. !m kNbeam. 80 kN!m kN!m 8080kN!m cmoment (1) 5.75 kN - VShear = 0+ and for thekN entire ©FyMoment = functions 0;80 kN!m 5.75 - V 2 = 02 V = 5.75 x2 "kN 5 (c) x2 " 5 Since of Fig. trated load at the beam’s center, determined andFunctions. are shown on free-body diagram of the x2 there 5 m6–7d. 5 m is a discontinuity V the V beam, 52Mm 5m V M V 2 x " 5 x " 5 x " 5 x " 5 A (1) 5 m (1) 2 a concentrated V "2 " 5= 5x0 " 5 5 V = 5.75 kN V = 5.75 kN dbeam’s +2 ©M kN 2" 2x 2x discontinuity of distributed load also load at the center, unctions. Since there a and discontinuity C0; C 1 + xM 5xm in order to describe shear Shear Moment Functions. Since there a" discontinuity of m - 5.75 Shear Since there is isxa22= discontinuity x2 kN Fig. 6–7b:2ofFunctions. 0the … xis 5and V Fig. 6–7d. 5 x2 " 5 - 80 of 5.75 15 kN 1 6 and ment Functions. Since there ism, aMoment discontinuity of x2 " 25 2x2 " 5 C 2kN 2 6 2 2 2 2 Shear and Moment Functions. Since there is a discontinuity of 5 kN/ m (2) A also AkN A beam’s two of= x 0; must be in order toBxdescribe shear and beam’s center, 2 beam’s 2 5.75 so at distributed and a aconcentrated load at the center, and also concentrated load the beam’s center, ma -concentrated 5.75 x1regions + =distributed 0the m. 51V mB xload d load +M ©M -considered 80center, kN m 5.75 kN + xM =the 0at V 2 M = 15.75x + 802 kN (c) xx2 x" 2 load 2 m 1 and also a kN concentrated load the beam’s center, 5 x "2 5 B5.75 80 kNx!m c ©F xa2 discontinuity +functions = at 0;Moment - V 1=there 0 25 mis distributed load and also5kN a concentrated load at the beam’s center, m 5m 52 m 5.75 Shear and Functions. Since of y regions x2 2 moment for the entire beam. e the shear and considered in order to describe the shear and two regions of x must be considered in order to describe the shear and two of x must be considered in order to describe the shear and (c) 5.75 kN 5.75 kN+ 802 kN m5 m 6 x … (2) 5.75 kNkN 5.75 2 2 (2) = 15.75x m M = a15.75x Fig. 6–7c: 10(1) m, 5.75 must be considered ordertwo to describe the shear and 1 + 802 kNin 1 55.75 kN 2 regions of xalso must be considered in order to describe the shear and distributed load and concentrated load at the beam’s center, 5 m m 15 kN 5.75 kN V kN x V = 5.75 kN (c) 5.75 kN 34.25 6kN 2 5.75 kN moment functions for the entire beam. moment functions for the entire beam. he entire beam. (c) (c)(c) 34.25 kNC m (c) the Fig. of 6–7b: 0 …beam. x51 two m, for the entire beam.5tokN/ 15 kN xfunctions must be considered in order describe shear and c:s for the entire Fig. 6–7c: m6 65regions xmoment m, (c) 5.75 kN (a) (b) 2 … 10 + c ©F - 15 kN - 5 kN>m1x - 5m m2 -6V = 0 V (kN) VA(kN) kN!-m80 kN B - V = 0 d +0©M 0;65805m, m - beam. 5.75 kN x1 + M = 0y = 0; 5.75 kN 52 kN/ moment for the entire Fig. 6–7b: Fig. 6–7b: 0… …x=1functions x16 15(c) kNkN5 m 15 kN m, :g. 6–7b: 15 5m 15 kN 15 kN 80 kN!m 6 65.75 6 6 + c ©F = 0; 5.75 kN V = 0 Fig. 6–7b: 0 … x 6 5 m, c 15 kN y 15 kN 5 kN>m1x 5 m2 V = 0 1 + ©F = 0; 5.75 kN 15 kN 5 kN>m1x 5 m2 V = 0 5 kN/ m(3) 5 kN/ m 5 kN/ m 2 (2) y(1) 2 5 kN/m .75 kN 6 115.7515 6 M = 15.75x m = 5x5(x C 1 + 802 kN 222kN 5 kN/ m 5 kN/m5.75V kNkN " 5) Fig. 6–7b: 5y=m, 80 kN ! m c1©F 80 = kN00!m…+ x 80 kN ! m c6 kN 34.25 kN 0; 5.75 kN V = 0 15 + ©F = 0; 5.75 kN V = 0 80 kN ! m (1) y 5.75 kN5.75 -SOLUTION V V = 5.75 kN 6 5.75 5.75 80 kN ! m c C 80 kN ! m A + ©F = 0; 5.75 kN V = 0 kN V = 0 5 kN/ m B y (3) (3)- 80 kN m - 5.75 kN x + 15 kN1x - x(m) 5=kN x1 + -M5x=220kN5 m 6 x2 … 10 115.75 6–7c: = 115.75 5x22 supports kN d + ©M m, Fig.V = x(m) 0;been 5 m2 VkN (kN) Reactions. reactions at=--=5+ the have 2 2 80 !m B (1)80 VV 5.75 kN (1) m m A (1) +=c 0; ©F 5.75 kN V 5.75 kN!m CC y = 0; (1)m5 dVSupport + ©M - 80 kNThe 5.75 x M= =0kN 0kN CkN V = 5.75 kN (1) V = 5.75 1 (1) C = 5.75 kN C (2) 802 kN C 5 m 5 m determined and are shown on the free-body diagram of the beam, c A -+ 5.75 kN xm + 15 kN1x 5 m2 d + ©M = 0; m 5.75 kN x + 15 kN1x 5 m2 80 kN + ©F = 0; 5.75 kN 15 kN 5 kN>m1x 5 m2 V = 0 "9.25 "9.25 A B (1) 2 A 2©M 25.75 2 y©M=B= 2 M V = 5.75 kN d + 0; 80 kN m 5.75 kN x + M = 0 B 5.75 kN 34.25 kN x 5 m d + 0; 80 kN m kN x + M = 0 A C 1 1x + M A 2 (2) M = 15.75x + 802 kN m B B A kN-80 m kN - 5.75 kN x + M = 0 d + ©M = 0; 80 kN m 5.75 kN = 0 1 B1 1 kN 5xm 5m m Fig. - 5.75 5¢m5 m m = 0 + 5 kN>m1x ≤ 5+m5 M 6–7d. 1 + M = 0 5m kN 5.75 2 - 5 m2 A 5xmkN 5m B 2 34.25 kN 5 m d + ©M = 0; V (kN) - 80 kNV m -15.75x 5.75 + M =m 0 5 m 5.75 m MM kN (3)(2) (2)(2) 5 6mkN 55+x802 12 ==M115.75 kN (2)x2 5+-m802 = 15.75x 802 kNkN m 2 1--1+5x 2m Fig. 6–7c: x … 10 m, = 15.75x + 802 m x(m) (2) 5 m 5 m M = 15.75x m 2 1 5.75 kN 34.25 kNkN (2) 5 m M V M ¢= 115.75x + kN 5.755.75 kNkN 34.25 V (kN) 5 m2 +V M + 5 kN>m1x - 15.75x 5kN m2 ¢ + 802 kN ≤ 802 ≤ +m M = 034.25 kN 5.751kN kN 25 234.25 34.25 kN x2 " 5 x2 " 5 (2) 5.75 kN>m1x =m 0=6x0m M = 6–7c: 5.751kN 34.25 2 - Shear 2 Fig. 6–7c: 5©M m 6m 10 m, Moment Functions. Since there discontinuity ofM kN 25 m2 d-and 2 + is15akN1x Fig. 5 … 10 m, 2x2… (4) Fig. 6–7c: 5 6 x … 10 m, = 1 2.5x + 15.75x + 92.52 kN m V (kN) 5.75 kN 34.25 kN + = 0; m 5.75 kN x 5 m2 80 kN "9.25 2 V2 (kN) c ©F 2 = 0 2 V (kN) = 0; x5.75 kN2m, - Fig. 15 kN - 5 kN>m1x22 - 5 m2 - V M (kN!m) M (kN!m) V (kN) Fig. 6–7c: +distributed m,–7c: y "34.25 "34.25center, 5.75 6–7c:2V (kN) V (kN) 5 m 6load 10 also a concentrated loadkN>m1x at the beam’s 2 …and x2 -2 5x215.75x 2 kN 2 + 92.52 V (kN) c (3) +kN ©F =M 0;(4) 5.75 kN - -15 kN -x-25 5 m2 V = 0 (4) c+©F = 1 2.5x + 15.75x + 92.52 kN m 5x m x(m) + 0; 5.75 kN 15 kN 5 kN>m1x 5 m2 V = 0 y©F 2 c 2 5.75 2- 2-15+ VkN = 0- 5 two = 0; 5.75 kN 15 kN 5 kN>m1x 5 m2 V = 0 y = 2 108.75 108.75 These results can be checked in part by noting that w = dV>dx 5 m y 2 (3) 2 regions of x must be considered in order to describe the shear and kN>m1x 5 m2 V = 0 V = 115.75 5x 2 kN 80 75 kN - 15 kN - 5+kN>m1x - 5+5.75 m2 -kN V -= 15 0 5 m2 c2©Fy =2 0; x(m) -2 ¢5 kN>m1x2 ≤- +5 M m2 =- 0V = 0 5.75 kN 5 kN>m1x 5.75 2V-kN 5.75 N x2in+ part 15 kN1x 5 m2 5.75 (3) and Also, when Eqs. 1 and 2x(m) give V = dM>dx. x = 0, "9.25 5.75 = 115.75 5x 2 kN 2 (3) 1 These results can be checked in part by noting that w = dV>dx moment functions for the entire beam. 2 ked by noting that w = dV>dx V = 115.75 5x 2 kN 2 (3) (3) d + ©M = 0; - 80 kN m(3) V = 115.75 5x 2 kN (c) 2 5.75 2 5.75 x(m) 80 80 x(m) 5.75 x(m) - 5.75 x2 + 15 kN1x m25.75 kN and (3)kN V = 115.75 - 5x22-kN "9.25m; when x = 10 m, Eqs. 3 and 4 (3) Mx(m) V == 115.75 2=kN 2 -V5 = V = 115.75 -0, 5xEqs. 2and = M 80(kN!m) kN 22 kN x(m) "34.25 x(m) 2 and d5x when 1kN1x and -2 5give dM>dx. x1 =kN when 1V give 0, Eqs. 2 5.75 1 x2 - 5xm +d5©M =2=0; mm x +x92.52 15+ m2 -Also, 80 kN "9.25 (4) 2 2 + ©M 0; 5.75 kN x + 15 kN1x 5 m2 80 kN M = 1 2.5x + 15.75x + kN m Fig. 6–7b: 0 … x 6 m, d + ©M = 0; m 5.75 kN 15 kN1x 5 m2 80 kN "9.25 "9.25 15 kN 2 2 2 2 2 2 1 -80 5 m2 "9.25 108.75 give V = 34.25 kN and M = 0. These values check with + M = 0 ¢m ≤ 6 the x 5 m and when x Eqs. 3 and 4 V = 5.75 kN M = 80 kN m; = 10 m, when x Eqs. 3 and 4 kNkN m; = 10 m, d + ©M = 0; m 5.75 kN x + 15 kN1x 5 m2 80 kN "9.25 - 25.75 x22 kN + 15 - 5 m2 m -kN5.75 x2kN1x + 152 kN1x 2 "9.25 2 "9.25 2 2 5 kN/ m 2 - 5 m2 + 5 kN>m1x - 5 m2 ¢ +-xM =m 0m reactions ≤xvalues 5 m 2 and support shown on the free-body diagram, Fig. 6–7d. 2 5 x 5 80 kN ! m c ©F M (kN!m) give V = 34.25 kN M = 0. These check with the These results can be checked in part by noting that w = dV>dx = 0; 5.75 kN V = 0 M = 0. These+values check with the 2 2 y x(m) 80 + +5 + kN>m1x 525 m2 =M x m ≤ ≤+ ≤+M+ 5 kN>m1x 5¢2m2 0 (kN!m) 2 2- 5 kN>m1x m2 Mx(m) =0 = 0M ¢- ¢5"34.25 xm 5(4) m xkN -support 5and 2 2 - V 2 diagram, =kN>m1x 02 free-body 2Eqs. +5 92.52 m + 5shown kN>m1x -5.75 5free-body m2 Mand =(1) 0 2 give ¢1 = 0, ≤2 +1Fig. reactions on the diagram, 6–7d. Also, when = dM>dx. x n the Fig. 6–7d. "34.25 2= 2 108.75 5.75x 5 m2 + M = 0 ¢ ≤ 1x m2 + M = 0 ¢ ≤ V kN 2 2 2 MM (kN!m) C "34.25 and Equations 1 through 4 "34.25 are plotted (4) Moment Diagrams. M (kN!m) = and 1 - 2.5x 15.75x + 92.52 m m, Eqs. (kN!m) (d) (d)2 kN Shear 2 (kN!m) 108.75 M "34.25 x2 = 4 (4)M (kN!m) V =25.75MkN M 2= =+ kN m; 2 22 when 2(kN!m) "34.25 "34.25 A(4) M2+ M 180 2.5x + 15.75x + +210 92.52 kN m3mand rt by noting that w = dV>dx M (kN!m) M = 1 2.5x + 15.75x + 92.52 kN m B (4) 2 2 d + ©M = 0; 80 kN m 5.75 kN x + M = 0 2 in Fig. 6–7d. M = 1 2.5x 15.75x 92.52 kN 108.75 "34.25 108.75 1 80 2 2 and Equations 1w through 4"34.25 are (4) rams. Equations 1give through 4-be are plotted M =Diagrams. 1 -in2.5x + 0. 15.75x +values 92.52 kN m m= 1-2.5x 2TheseShear 2 that V = Moment kN M2 =by These withplotted the 2 (4) 5m m 108.75 108.75 can checked part noting =check dV>dx (4) Fig.5108.75 6–7 x(m) M12.5x 92.52 kN m Fig. 6–7 (4) and 15.75x +results 92.52 kN m34.25 80 Eqs. and 2These give = 0, 2 1+ 215.75x 2 + 108.75 2 + 108.75 (2) in Fig. 6–7d. M = 15.75x + 802 kN m results can be checked in part by noting that w = dV>dx These results can be checked in part by noting that w = dV>dx 1 support reactions shown on the free-body diagram, Fig. 6–7d. These results can be checked in part by noting that w = dV>dx 8080 80 and Also, when Eqs. 1 and 2 give V = dM>dx. x = 0, These results can be checked in part by noting that w = dV>dx 1 5.75 kN 34.25 kN w = dV>dx when x2 = 10 in Eqs.by 3 and 4 =that m,80part 80 (d) necked be checked noting w =kN dV>dx Also, when Eqs. 2give V dM>dx. x=1x10 =xm, and Also, when 14and 2give give V=Fig. dM>dx. = 0,Eqs. in partV5by that wdM>dx. dV>dx 10, and Also, when 1 and 2 give V 0,Eqs. 6–7c: m noting 6and x2and … 10 m, 80x and when 3 1and = 5.75 kN M ==dM>dx. 80 m; Also, Eqs. 1Eqs. 2and Vthe = x21 = 0, 1 = 80when nd 2 give V (kN) These values check with x(m) and when x Eqs. 3 and 4 V = 5.75 kN M = 80 kN m; = 10 m, and when x Eqs. 3 and 4 V = 5.75 kN M = 80 kN m; = 10 m, Also, when 1and give x. x1V 0,V Shear and Moment Equations through 2with o, when Eqs. 1Eqs. 2and give = 0,V and 3plotted and 4 = and 5.75 kN M 80 kN m; when m, x2check 3are and 4 5.75 kN M m; when =2x1 10 m,10Eqs. == =34.25 kN and M2=Diagrams. =80 0.=kN These values the4Eqs. 2 = Fig. 6–7 x(m) qs. 3=and 4x1give e-body diagram, 6–7d. cFig. +m; ©F =xV 0; 5.75 15 kN -and 5 kN>m1x - 5values m2 - check V =check 0with give V =334.25 -kN 34.25 kN = 0. values with the VV =shown -kN kN and M =M=These 0. These the y give 2 These d when Eqs. 3 and 4 MkN 80 kN = 10 m, in Fig. 6–7d. x(m) x(m) 2 give = 34.25 and M = 0. values check with the give = 34.25 kN and M 0. These values check with the when x Eqs. and 4 80 m; = 10 m, (d) support reactions on the free-body diagram, Fig. 6–7d. x(m) 2 x(m) kkN with the x(m) support reactions shown on the free-body diagram, Fig. 6–7d. support reactions shown on the free-body diagram, Fig. 6–7d. 5.75 (d) and M = 0. These values check with the support reactions shown on the free-body diagram, Fig. 6–7d. support reactions shown on the free-body diagram, Fig. 6–7d. (3) x(m) nd M = 0. These values check with the V = 115.75 5x 2 kN Equations through 4 are plotted 2 x(m) x(m) g. 6–7d. on1 the (d) (d)(d)(d) Fig. 6–7 1 through 4 are plotted free-body diagram, Fig. 6–7d. Equations Shear and Moment Diagrams. (d) ns shown on the free-body diagram, Fig. 6–7d. Shear and Moment Diagrams. Equations 1 through 4 are plotted Fig. 6–7 Shear and Moment Diagrams. Equations 1 through 4 are plotted (d) Shear Moment Equations through 4 are4 plotted ©M = Shear 0; and m -Diagrams. 5.75 kN x2 + 15Equations kN1x 51 m2 - 80 kNMoment and Diagrams. through are plotted "9.25 (d) 21 Fig. 6–7 ind + Fig. 6–7d. 6–7 Fig.Fig. 6–7 gh 4 are plotted Fig. 6–7 in Fig. 6–7d. # # # # # # # # # # # # # # # ## # # # # # ## # # # # # # # # # # # ## # # # # # # # # # # # # # # # # # # # # inin Fig. 6–7d. in Fig. 6–7d. ment Diagrams. Equations 1 6–7d. through 4 are plotted Fig. 6–7plotted Fig. agrams. Equations 1 through 4 are + 5 kN>m1x2 - 5 m2 ¢ Fig. 6–7 x2 - 5 m Fig. 6–7 ≤ + M = 0 2 M = 1 - 2.5x22 + 15.75x2 + 92.52 kN # m M (kN!m) (4) These results can be checked in part by noting that w = dV>dx and V = dM>dx. Also, when x1 = 0, Eqs. 1 and 2 give V = 5.75 kN and M = 80 kN # m; when x2 = 10 m, Eqs. 3 and 4 give V = - 34.25 kN and M = 0. These values check with the support reactions shown on the free-body diagram, Fig. 6–7d. "34.25 108.75 80 Shear and Moment Diagrams. Equations 1 through 4 are plotted in Fig. 6–7d. PAGE !65 x(m) (d) Fig. 6–7 right side, where 0 6 and k 6moment, 1 [for example, if w(x) is uniform, both thethe internal resultant shear acting on the right face of k = 2 ]. Applying of equilibrium to the have the segment, mustthe beequations changed by a small amount in segment, order to we keep the segment in equilibrium. The distributed load has been replaced by a resultant force w1x2 ¢x that acts at a fractional distance k1¢x2 from the right side, where 0 6 k 6 1 [for example, if w(x) is uniform, k = 12 ]. Applying the equations of equilibrium to the segment, wex have w(x)" w(x) yöntemle elde edilmesi 6.2. Kesme kuvveti ve moment diyagramlarını grafik Yayılı yük bölgeleri: w(x)"x F w(x) k(" x) w(x) 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS V + c ©Fy = 0; F 6.2 M w(x) V + w1x2 ¢x - 1V + ¢V2 = 0 ¢V = w1x2 ¢x M0 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS " x V x M x2 ¢x - 1V + ¢V2 = 0 ¢V = w1x2 ¢x x M0 d+ ©MO = 0; "x 263 k("x) O M ! "M V ! "V "x M ! "M -V ¢x - M - w1x2 ¢x[k1¢x2] + 1M + ¢M2 = 0 Free-body diagram ¢M = V ¢x + w1x2 k1¢x22 O of segment " x V !M"V (b) GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND OMENT DIAGRAMS 263 6.2 GRAPHICAL M ETHOD FOR CONSTRUCTING S HEAR AND MOMENT DIAGRAMS 263 "x Fig. 6–8 6.2G GRAPHICAL M ETHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS M - w1x2 ¢x[k1¢x2] + 1M + ¢M2 = 0by ¢x and taking6.2 RAPHICAL M: ETHOD FORabove Cdiagram ONSTRUCTING SHEAR AND M OMENT DIAGRAMS ¢x 0, Dividing the limit as the two equations Free-body c (a) 6.2 263 2 + ©Fy = 0; V + w1x2 ¢x - 1V + ¢V2 = 0 of segment"x become 2 V ¢x + w1x2 k1¢x2 + c ©Fy = 0;(a) V + =w1x2 ¢V w1x2¢x ¢x- 1V + ¢V2 = 0 (b) c ©F=y 0; = 0; V w1x2 + w1x2 ¢x - 1V + ¢V2 + c+©F V¢V + ¢x¢x - 1V + ¢V2 = 0= 0 y = w1x2 Fig. 6–8¢V = w1x2 ¢x ¢V = w1x2 ¢x he limit as ¢x : 0, the above two equations dV d + ©MO = 0; - V ¢x - M - w1x2 ¢x[k1¢x2] + 1M += ¢M2 w1x2 = 0 dx 2 d+ ©MO = 0;¢M-V - M - w1x2 ¢x[k1¢x2] + 1M + ¢M2 = 0 = ¢x V ¢x + w1x2 k1¢x2 + ©M = 0;- V-¢x V ¢x - M - w1x2 ¢x[k1¢x2] + 1M + ¢M2 d +d©M - M - w1x2 ¢x[k1¢x2] + 1M + ¢M2 = 0 = 0 (6–1) O 0; O = slope of 2 distributed ¢M = V ¢x + w1x2 k1¢x2 2 2 = ¢x V ¢x + w1x2 k1¢x2 ¢M¢M = V + w1x2 shear diagram load intensity = k1¢x2 dV by ¢x and taking the limit as ¢x : 0, the above two equations = Dividing w1x2 at each point at each point dx become Dividing by ¢x and taking the limit as ¢x : 0, the above two equations ¢x :giderken 0, the Dividing taking the limit as ¢x above equations ¢x : 0, Dividing by by andand taking theböler limit as ¢x the above twotwo equations Denge değerine ve limit alırsak; (6–1) becomedenklemlerini ope of distributed become become agram = load intensity dV dM point at each point = w1x2 = V dx dx dV dVdV = w1x2 = w1x2 = w1x2 dx (6–1) (6–2) slope of distributed dx dx slope of shear w ! w(x 6 shear diagram slope intensity = load (6–1) dM diagram = at each of moment distributed (6–1) (a) (6–1) Bu iki denklem bize kesme kuvveti ve diyagramlarını hızlı bir biçimde elde etme slope of distributed distributed slope ofmoment = V at each point at each point dx at each point point shear diagram load intensity =diagram shear intensity shear diagram intensity = load = load imkanı sağlar. at each at point at each at point A at each point at each point each point each point (6–2) C D slope of shear w ! w(x) wB 6 nt diagram = at each w = negative incr These dM two(a)equations provide a convenient means for quickly V slope = negative in V and moment diagrams for a beam. Equation 6–1 0 6 obtainingdx the =shear each point point "wC dMthe slope states that at a point of the shear diagram equals the intensity dMdM "wD = V A VA =example, V= V consider theBbeam in Fig. 6–9a. dx of the distributed loading. For (6–2) w B slope of sheardx Cdx D (b) w ! w(x) The distributed loading is negative and increases from zero to wB . w = negative increasing provide a convenient meansmoment for quickly diagram = V at each (a) (6–2) (6–2) Therefore, the shear willslope curve has a negative slope, wB (6–2) =anegative increasing slope0diagram ofslope shear w ! w(x)w w of !Bw(x) slope ofbeshear shear that w ! w(x) oment diagrams for a beam. Equation V = positive decreasing at each6–1 point point "w C -w . w = 0, -w , -w , increasing from zero to Specific slopes and B at each at each C D momentmoment diagram slope = positive decreasing =diagram ope of the shear diagram equals the intensity diagram = = at each A moment (a) (a) (a) "w D -wB are shown inVAFig. 6–9b. A M B " For example, consider the beam in Fig. 6–9a.at each at point point each point point atEq. each point point D x the slope Cof the In a similar(b)manner, 6–2 states that at a point negative and increases from zero to wB . VD 0 A negative increasing Aw = A moment diagram is equal to the shear. that the shear diagram in These two equations provide a convenient means forNotice quickly B B V "VB V 0 m will be a curve that has a negative slope, C negative D C increasing DC D slope = C obtaining the shear and diagrams for a= positive beam. decreasing Equation 6–1then becomes negative , Vdecreases Fig.moment 6–9b starts at +VA to zero, and "wC wB . Specific slopes wA = 0, - wC, - wD, and VA w = negative increasing slope =moment positive decreasing increasing These two equations a convenient means for have quickly two equations provide convenient means for quickly w = negative increa states that These at a point the slope of the shear diagram equals the intensity and decreases to -Vprovide The diagram will then anVinitial These two equations a convenient means for quickly V V w = negative Ba. provide "w D0 VA M slope slope = negative increasing (c) slope = negative increasing 0 = negative incr 0 "w obtaining the shear and moment diagrams for a beam. Equation 6–1 obtaining the slope shear and moment diagrams for a beam. Equation 6–1 B of the distributed loading. For example, consider the beam in Fig. 6–9a. obtaining and moment diagrams for a beam. Equation 6–1 +Vshear ofthe which decreases to zero, then the slope becomes negative "wC "wCx "wC 6–2 states that at a point the slope of theA (b) states that atnegative aatpoint the slope of the shear diagram equals intensity states that at aand point the ofthe the equals the The distributed loading isdecreases and increases from zero states that aslope point slope ofdiagram the shear equals -V Vto V VDthe , 0,the to Specific andintensity -V are VDslopes 0 diagram BC., intensity B .shear A, w "wD "wD "wD o the shear. Notice that the diagram inloading. For VA B VA VA Vhas C ofshear thediagram distributed example, the beam in Fig. 6–9a. "VFig. Therefore, a example, curve For that a consider negative slope, of the shear distributed loading. For consider the beam inthe Fig. 6–9a. of the distributed loading. example, consider beam in Fig. 6–9a. 6–9 shown inwill Fig.be 6–9c. B x (b) x creases to zero, and thenThe becomes negative (b)V = positive(b) wtoB . wB . decreasing distributed loading isVAnegative increases from -loading wB . Specific 0,and -w , wD, and increasing from zero to distributed slopes wBzero . toslope The distributed is loading negative and from zero tozero The isw negative and increases from A =increases C = positive decreasing "V e moment diagram will then have anthe initial B "VB Therefore, shear be be athat curve has has a negative - wB are Therefore, shown in Fig. Therefore, the shear diagram will a curve that ax negative slope, "VB the6–9b. shear diagram will be will a curve hasthat a negative slope, M slope, (c)diagram V =decreasing positive B decreasing ses to zero, then the slopeincreasing becomes negative V"w = positive decreasing V = positive w . w = 0, w , w , from zero to Specific slopes and In a similar manner, that at a point the slope of the B A C D w . w = 0, w , w , increasing from to Specific slopes and -wzero . w = 0, -w , -w , increasing fromEq. zero6–2 to states Specific slopes and B A C D B A C D slopeslope =decreasing positive decreasing = positive decreasing slope ecific slopes VA, VC, VD-, w 0, and - VB are VD M= positive 0 in Fig. 6–9b. moment -w diagram is-Bequal toFig. the6–9b. shear. Notice ware shown in Fig. 6–9b.that the shear diagram in inshown M MC V B are "wB "w B are shown "w Fig.that 6–9 at a point the slope of the B aAsimilar manner, Eq. 6–26–2 states +V ,manner, Fig. 6–9b starts at In to6–2 zero, and then becomes In adecreases similar manner, Eq. states thatnegative at pointof the slope of the In a similar Eq. states that at a point thea slope the V 0 V D VD 0 VD V 0 "V diagram istoequal to the shear. Notice that thediagram shear diagram in in -moment VB . The and decreases tomoment moment diagram then have initial isthe equal towill the shear. Notice that the shear moment diagram isdiagram equal shear. Notice that thean shear inAdiagram C VC B VC (c) x + V , Fig. 6–9b starts at decreases to zero, and then becomes negative PAGE ! 6 6 + VA6–9b slope of Fig. which decreases to zero, slopeand becomes negative VA, the Fig. 6–9b starts at +Athen decreases to then zero,becomes and thennegative becomes negative V starts at +V to zero, A, decreases A VA "VB VA - V-moment . The then have an initial Vdecreases V. AThe , moment Vdiagram , VD, diagram and decreases toand slopes 0,will andthen -will Vhave arean BV and to diagram then have an initial "VB B .decreases Cmoment Bwill -VBto . The and decreases toSpecific initial B (c) (c) x (c) + V slope of which decreases to zero, then the slope becomes negative x Fig. 6–9 shown inslope Fig. 6–9c. slope of +AVA which to decreases to the zero, thenbecomes the slopenegative becomes negative of +V zero, then slope A which decreases - V-BV . Specific VA,VV,C,VVD and decreases to to slopes - VB- V are are . Specific , ,V0, , and and decreases slopes 0, and 264 V CHAPTER 6 = w1x2 dx C H A P T E R 6andBdM E N D I= N GV dx. Noting that w(x) dx and V dx¢V represent L differential 264 CHAPTER 6 BEND ING Equationsloading 6–1 andand 6–2 shear may also be rewritten in the form areas diagram, respectively, we dV = w1x2 dx (6–3) C E6R B 6BEunder BD"V EI N N DGIthe N G distributed 2266442 6 4 CCHHAA PP THT EAE RPRT6 NN G E D IN change in dx area under C(e) HAPTER 6 B E N D I can N G integrate these andareas = V dx. any Noting that w(x) and Vthe dxbeam, represent differential x dM between two points C and D on = Equations 6–1 and 6–2 may shear also be rewritten in the form dV = w1x2 dx C H A P T ED loading 264 C R 6 and B E Nwrite D Iareas NG under the distributed shear respectively, we = w1x2 dx Fig. 6–9d, Equations 6–1loading and 6–2 and maydistributed also bediagram, rewritten in the form dV 264 264 BENDING Equations 6–1 and 6–2 may also be rewritten in the form dV = w1x2 dx andEquations dM = V6–1 dx. Noting that w(x) dx and Vinin dx represent differential Equations 6–1 and 6–2 may also rewritten inthe the form dV w1x2 dx and 6–2 may also be rewritten the form dV === w1x2 dx 6–1 and 6–2 may also bebe rewritten form dV w1x2 dx can Equations integrate these areas any two points and D on the beam, and dM = Vbetween dx. Noting that w(x) dxC and V dx represent differential Equations 6–1 and 6–2 may also be rewritten in the form dV = w1x2 dx areas under the distributed loading and shear diagram, respectively, we and dM = V dx. Noting that w(x) dx and V dx represent differential and dM ==Vwrite Noting that w(x) and dx differential and dM Vdx. dx. Noting that w(x)dx dx andVVand dxrepresent represent differential 6–9d, and areas under the distributed loading shear diagram, respectively, we C D and dMFig. = V dx. Noting that w(x) dx and V dx represent differential can integrate these areas between any two points C and D on thedV beam, Equations 6–1 and 6–2 may also be rewritten in the form =we w1x2 areas under the distributed loading and shear diagram, respectively, we dx areas under the distributed loading and shear diagram, respectively, areas under the distributed loading and shear diagram, respectively, we ¢V = w1x2 dx (d) ¢M = V1x2 dx can integrate these areas between any two points C and D on the beam, areas under the distributed loading and shear diagram, respectively, we M (d) C Fig. 6–9d, and write and dM = V dx. Noting that w(x) dxpoints and V dxDD represent differential integrate these areas between any two Cand and D on the beam, can integrate these areas between any two points CCand on the beam, L D cancan integrate these areas between any two points on the beam, V L (d) Fig. 6–9d, and write can D integrate these areas between any two points C and D on the beam,respectively, (d) (d) "V (6–4)we (6–3) C areas under the distributed loading and shear diagram, Fig. 6–9d, and write Fig. 6–9d, and write d) = in w1x2area dx under Fig. 6–9d,inand write C"M DD D change area under¢V C Cx change VC Fig. 6–9d, and write can integrate these areas between D = L = any two points C and D on the beam, V C D (d) shear distributed loading ¢V = w1x2 dx diagram (6–3) "V VV V moment shear Fig. 6–9d, and write C D change in area under ¢V = dxw1x2 dx L V (e) x (f) ¢V = w1x2 ¢V = w1x2 dx x = = ¢V w1x2 dx (6–3) L "VD C CD ¢V =shearin w1x2 dx distributed loading L change area under V LL Equation 6–3 states that the change in shear between C and D is equal "V x (e) L (6–3) to(6–3) (6–3) = "V change in area under (6–3) "V "V D x change in area under ¢V = w1x2 dx (e) C"V change in area under (6–3) shear distributed loading the area the distributed-loading ¢M under = V1x2 dx = curve between these two points, under (e) (e) Fig. 6–9 (cont.) == =area change in change areainunder C D xx x (e) L shear distributed Lthis e) C C D Dx shear distributed loading Fig. 6–9d. In case the change is negative sinceloading the distributed load shear distributed loading = (6–4) C D (6–3) shear distributed loading "V C D shear distributed loading change area under ¢M =frominV1x2 dx the "M in areaSimilarly, under acts downward. Eq. 6–4, change in moment between M (e) xchange = L = area 6 C D Cmoment and D, Fig.shear 6–9f,diagram is ¢M equalshear shear diagram distributed loading (6–4)within =to the V1x2 dxunder the M "M ¢M =dxdx V1x2 dx change in area under L the region from C to D. Here the change is positive. M x ¢M = V1x2 ¢M = V1x2 (6–4) M M ==dx V1x2 dx L ¢M D ¢M moment = V1x2 M L L M C Since the above equations not apply at points shear "M change in do area under Equation 6–3 states that the change in shear between C anddiagram D is equal to where a concentrated (6–4) (6–4) (6–4) L L F "M = (6–4) change in area under (6–4) (f) "M "M force or couple moment acts, we will now consider x the distributed-loading change area under change in in¢M area the area under curve between these two points, =under dx each of these cases. moment diagram =V1x2 Fig. 6–9 (cont.) M "MC "M D change in under change in the area under =area = inshear shearload diagram L states change shear between C and D is equal to (6–4) Fig. 6–9d. In this case the 6–3 change is that negative since moment the distributed = (f) = moment shear diagram moment shear diagram x Equation moment shear diagram moment shear diagram (f) C D x the area under the distributed-loading curve between two points, (f) (f) "M acts Cdownward. Similarly, from Eq. 6–4, the change in moment between Regions of Concentrated Force and Moment. Atofreex x change in area under Equation 6–3 states that the change in shear between Cthese and D is equal Fig. 6–9 (cont.) D (f) (f) C C D Dx xFig. 6–9d. InEquation =shear 6–3 states that the change in shear between Cequal and D this case the change is negative since the distributed load C and D, Fig. 6–9f, is equal to the area under the shear diagram within Equation 6–3 states that the change in shear between C and D is tois Equation 6–3 states that the change in between C and D is equal toequal to body diagram of a small segment of the beam in Fig. 6–10a taken from C D VC D the area under the distributed-loading curve between these two points, moment shear diagram 6–3 states that the change in shear between C and D is equal to MFig. 6–9 (cont.) MEquation ! "M acts Equation 6–3 states that the change in shear between C and D is equal to the area under the distributed-loading curve between these two points, Similarly, from Eq. 6–4, the change initthe moment between the area under the distributed-loading curve between these points, from CFig. to downward. D. Here the change isshown positive. the area under the distributed-loading curve between these two points, Fig.region 6–9 (cont.) under the force isthe in Fig. 6–10a. Here can be two seen that (f) In this case change is negative since distributed loadforce x6–9d. Fig. 6–9 Fig. (cont.) 6–9the (cont.) the area under the distributed-loading curve between these two points, the area under the distributed-loading curve between these two points, 6 Fig. 6–9d. In this case the change is negative since the distributed load C D Fig. 6–9 (cont.) Fig. 6–9 (cont.) CFig. and D,Equation Fig. is equal to the area the shear diagram within 6–9d. In this case the change ischange negative since thethe distributed Fig. 6–9d. In this case change isunder since distributed Since the above equations do6–9f, not apply atthe points where anegative concentrated equilibrium requires downward. Similarly, from Eq. 6–4, the change in moment between 6–3 states that the in shear between C and Dload isload equal to Fig. 6–9d.acts In this case the change is negative since the distributed load Fig. 6–9d. In this case the change is negative since the distributed load F acts downward. Similarly, from Eq. 6–4, the change in moment between the region from C tonow D.equal Here the change is positive. 6 acts downward. Similarly, from Eq. 6–4, the change inbetween moment between acts downward. Similarly, from Eq. 6–4, the change in moment between force or couple moment acts, we6–9f, will consider each of these cases. C and D, Fig. is to the area under the shear diagram within the area under the distributed-loading curve these two points, acts downward. from Eq. 6–4, the change in¢V2 moment between 6 cSimilarly, Fig. 6–9 (cont.) +downward. ©F =6–9f, 0; V +the Fthe -6–4, 1V +points =under 0in acts Similarly, from Eq. the change moment between Cy and D, Fig. 6–9f, is equal tounder area the shear diagram within 6 6 the above equations do not apply at where a concentrated CSince and D, Fig. is equal to area the shear diagram within C and D, Fig. 6–9f, is equal to area under the shear diagram within load the region from C to D. Here the change is positive. Fig. 6–9d. In this case the change is negative since the distributed 6 Fig. 6–9f, is equal to the area under the shear diagram within Tekil yük bölgeleri:C and D, F C and D, Fig. 6–9f, is equal to the area under the shear diagram within the region from C to D. Here the change is positive. force or couple moment we will now consider each of athese cases. between theSince region from CForce to Here the change is positive. the region from C D. toacts, D.and Here the change is positive. ¢VEq. = F (6–5) the above equations do not apply at6–4, points where concentrated Regions Concentrated Moment. Athe freeacts Similarly, from change inwhere moment theof region from C to downward. D. Here change is positive. the region from Cequations tothe D. Here the change is points positive. Since the above equations doat not apply atwhere points a concentrated F Since the above do not apply where a concentrated Since the above equations do not apply at points a concentrated V ! "V 6 V force or couple moment acts, we will now consider each of these cases. "x of the a small segment the beam in Fig. taken C and D,of is equal to6–10a the area under the shear diagram Since above equations do6–9f, not apply at the points where afrom F F bodyFdiagram Thus, when FFig. acts upward beam, ¢V isconcentrated positive so the shear will Since the above equations doon not apply at points where athese concentrated force or couple moment acts, weconsider will now consider each of thesewithin cases. force orthe couple moment acts, we will now each of of force or couple moment acts, we will now consider each these cases. FM ! "M F of Concentrated Force and Moment. Acases. freeunder the force is shown in Fig. 6–10a. Here it can be seen that force region from C to D. Here the change is positive. force orRegions couple moment acts, we will now consider each of these cases. “jump” upward. Likewise, if F acts downward, the jump 1¢V2 will be force or couple moment acts, we will now consider each of these cases. (a) V body diagram a small ofForce the beam in Fig. 6–10awhere takenA equilibrium requires Regions ofofthe Concentrated and freeSince abovesegment equations do not apply atMoment. points a from concentrated M downward. Regions of Concentrated Force and Moment. A freeFM ! "M Regions of Concentrated Force and Moment. A freeRegions of Concentrated Force and Moment. A freeunder the force is shown in Fig. 6–10a. Here it can be seen that force V body diagram of a small segment of the beam in Fig. 6–10a taken from force or couple moment acts, we will now consider each of these cases. V Regions of Concentrated Force and Moment. A freeWhen the beam segment includes the couple moment M , Fig. 6–10b, c V + ©F = 0; V + F 1V + ¢V2 = 0 M Regions of Concentrated Force and Moment. A free0 M ! "M body diagram of a small segment of the beam in Fig. 6–10a taken M ! "M M V V y body diagram ofsegment small segment of the beam Fig. 6–10a taken from body a small ofinthe beam in Fig. 6–10a taken from from equilibrium requires under the force isaofequilibrium shown insegment Fig. 6–10a. Here itincan be seen that force V M ! "M then body diagram of adiagram small of the beam Fig. 6–10a taken from M MV M !M "M ! "M M moment requires the change in moment to be body diagram of a small segment of the beam in Fig. 6–10a taken from under the force is shown in Fig. 6–10a. Here it can be seen that force M M ! "M under the force isin=shown in Fig. 6–10a. Here can be seen that force under the force is shown in Here Fig. 6–10a. Here it can be seen that force ¢V FV (6–5) requires M M ! "M under the force is 0; shown Fig. 6–10a. it¢V2 can be0it seen that force Regions of Concentrated Force and Moment. A freec +equilibrium ©F = + F 1V + = y equilibrium requires under the force is shown in Fig. 6–10a. Here it can be seen that force equilibrium equilibrium V ! "V d+ ©MOrequires =requires 0; Ma +small ¢Msegment - M0 -ofVthe ¢x - M in = 0 O equilibrium "x V diagram taken from crequires Thus, when F acts+upward on0;the beam,ofV ¢V+isFpositive so¢V2 the shear will Fig. 6–10a ©Fybody = - 1V + = 0beam equilibrium requires M M ! "M ¢V = F (6–5) c M0 + ©F = 0; V + F 1V + ¢V2 = 0 c c + ©F = 0; V + F 1V + ¢V2 = 0 + ©F = 0; V + F 1V + ¢V2 = 0 y under the force is shown in Fig. 6–10a. Here it can be seen that force y ify F¢x 0, get + the upward. acts + c ©Fy Likewise, = 0; Letting V: + downward, F we - 1V ¢V2jump = 0 1¢V2 will be (a) V ! “jump” "V "x c ¢V = F (6–5) + ©F = 0; V + F 1V + ¢V2 = 0 equilibrium requires y Thus, when F acts upward on the¢V beam, is positive so the shear (6–5) will downward. = F =¢V (6–5) ¢M =F¢V M0 = F (6–6)(6–5) V ! "V ¢V = F ¢V (6–5) V ! "VV ! "V “jump” upward. Likewise, "x "x V if F acts downward, the jump 1¢V2 will be V ! "V V ! "V Thus, when F acts upward on the beam, ¢V is positive so the shear will When the beam segment includes the couple moment M , Fig. 6–10b, ¢V = F (6–5) (a) "x c "x "x 0 + ¢V2 = 0 + when ©F 0; V the +beam, Fbeam, - the 1V ! "V y F=when Thus, F acts on beam, is positive so the shear will "xM !V"M Thus, when Fcase, acts upward onupward the ¢V is¢M positive sokuvveti the shear acts upward ¢V isthe positive so the shear will In this if M applied clockwise, is¢V positive so the moment Vthen ! "Vmoment Thus,equilibrium when FThus, acts upward on the ¢V is positive so the shear will F downward. yukarı doğru uygulandığında pozitiftir ve kesme “jump” upward. Likewise, ifinFon acts downward, jump 1¢V2 willwill be requires the change moment to be 0 isbeam, "x (b) (a) “jump” upward. Likewise, if F acts downward, the jump 1¢V2 will be “jump” upward. Likewise, if F acts downward, the jump 1¢V2 will be Thus, when F acts upward on the beam, ¢V is positive so the shear will “jump” upward. Likewise, if F acts downward, the jump 1¢V2 will be V (a) “jump” upward. ¢V = F (a) (a) diagram will segment “jump” upward. Likewise, when M0 acts When the beam includes the couple moment M Fig. Likewise, if çıkış F acts downward, the jump 1¢V2 will be 6–10b, (6–5) downward. 0,counterclockwise, (a) M ! "M M diyagramında yukarı olacaktır. downward. + ©M = 0; M + ¢M M V ¢x M = 0 downward. “jump” upward. Likewise, if F acts downward, the jump 1¢V2 will be V ! "V downward. O O 0 V (a) Fig.d "x downward. theThus, jump 1¢M2 will be downward. thenWhen moment requires the change in moment to the equilibrium beam segment includes couple moment Mbe 0, Fig. M ! "M V6–10 M when Fthe acts upward onthe the beam, ¢Vmoment is positive so 6–10b, the shear will V V When beam segment includes the couple moment M When the beam segment includes the couple moment , Fig. 6–10b, When the beam segment includes the couple 6–10b, downward. M0 V 0, Fig. 6–10b, 0M M ! the "M M"M M !M 0, Fig. "M M M ! "M When beam segment includes the couple moment M , Fig. M 6–10b, then moment equilibrium requires the change in moment to be 0 M ! Letting ¢x : 0, we get M “jump” upward. Likewise, if F acts downward, the jump 1¢V2 V d + ©M = 0; M + ¢M M V ¢x M = 0 (a) then moment equilibrium requires the change in moment to bewill be then moment equilibrium requires the change in moment to be O then equilibrium requires themoment change be 6–10b, O moment 0 the beam segment couple in Mto 0, Fig. M ! "M then momentWhen equilibrium requires theincludes change in tomoment be M downward. ¢M = M (6–6) d + ©M = 0; M + ¢M M V ¢x M = 0 M O moment requires the -change inVmoment 0 0 O : 0, equilibrium V ! "V "x ¢x we get =M M + 00M ¢M M¢x Mto=be 0 M , Fig. 6–10b, dthen + ©M = 0;+ M +0; =¢x0= -0moment V O d + ©M =©M 0; +0 segment ¢M - ¢x V= -couple M O d + ©M Letting O O 0- -M O M Od+ When the beam includes the ¢M M¢M VM ¢x -0 V M 0 O 0 O = 0; M ! "M M M0 Letting ¢x : 0, we get In this case, if M is applied clockwise, ¢M is positive so the moment M M M d + ©M = 0; M + ¢M M V ¢x M = 0 0 0 0 O 0 then moment equilibrium requires be O Letting 0 M0 the change in moment to M0 ¢M = (6–6) Letting ¢x : 0, we get ¢x : 0, we get (b) Letting ¢x : 0, we get V ! "V ¢x : 0, we get "x diagram Letting will “jump” upward. Likewise, when M0¢M acts=counterclockwise, M0 M (6–6) 0¢M 0, we get M clockwise, V ! "V d +¢x ©M ¢M M V=positive ¢x 0 moment "x =- M (6–6) M0 - Mso=(6–6) In Letting this case, if:M ¢M the ¢M = (6–6) (6–6) O 0= is0;applied 0 0M 1¢M2 be downward. Fig. 6–10 V the ! "V 0 is ! "V will ¢M = +M¢M VOjump !"x"V V "x "x 0 V !(b) "V "x M0 diagram will “jump” upward. Likewise, when M acts counterclockwise, In case, if M is applied clockwise, ¢M is positive so the moment ¢M = Mclockwise, 0 ¢M V ! "V Letting ¢x : 0, In this this case, ifthis M is0clockwise, applied clockwise, ¢M is0 so positive sopositive the moment Inapplied Mget applied ¢M isolur so(6–6) the moment In this if00case, M isifwe applied clockwise, is positive so the moment (b) Burada yönünde uygulandığında pozitif ve "x o0 saat 0 is if M M iscase, ¢M is positive thecounterclockwise, moment (b)6–10 (b) In this case, (b) the jump 1¢M2 will be downward. diagram will “jump” upward. Likewise, when M acts Fig. (b) 0 diagram will “jump” upward. Likewise, when M acts counterclockwise, diagram will “jump” upward. Likewise, when M acts counterclockwise, diagram will “jump” upward. Likewise, when M acts counterclockwise, 0 0 0 diagram will “jump” upward. Likewise, when M acts counterclockwise, In this case, if M is applied clockwise, ¢M is positive so the moment 0 ¢M = M0 (6–6) 0 be downward. the jump 1¢M2 will Fig. V ! "V (b)6–10 diyagramda artış, aksi taktirde azalış olacaktır. the jump 1¢M2 will be downward. the jump 1¢M2 will Likewise, be downward. the jump will be downward. Fig. "x 6–10Fig. 6–10 the jump 1¢M2 will be1¢M2 downward. Fig. 6–10 Fig. 6–10 diagram will “jump” upward. when M0 acts counterclockwise, In this case, Mdownward. 0 is applied clockwise, ¢M is positive so the moment the jump 1¢M2 willifbe Fig. 6–10 (b) diagram will “jump” upward. Likewise, when M0 acts counterclockwise, the jump 1¢M2 will be downward. Fig. 6–10 C D (d) PAGE !67 266 CHAPTER 6 BENDING 266 C Hdiagrams A P T E R 6 for B Ethe N D Ibeam NG Draw the shear and moment shown in Fig. 6–11a. SOLUTION ! EXAMPLE 6.5 Ölçü ve yükleme durumu şekilde verilen konsol kirişin kesme kuvveti ve EXAMPLE 6.5 at the fixed support is shown on Support Reactions. The reaction Draw the shear and moment diagrams for the beam shown in Fig. 6–11a. the free-body diagram, 6–11b. momentFig. diyagramlarını çiziniz. Draw the shear and moment diagrams for the beam shown in Fig. 6–11a. Shear Diagram. The shear at each end of the beam is plotted first, SOLUTION Fig. 6–11c. Since there is no distributed loading Pon theSOLUTION beam, the slope P P how theSupport Reactions. The reaction at the fixed support is shown on of the shear diagram is zero as Pindicated. Note force P at the Support Reactions. The the free-body Fig.reaction 6–11b. at the fixed support is shown on center of the beam causes the shear diagram to jump downward andiagram, the free-body diagram, Fig. 6–11b. amount P, since this force acts downward. Shear Diagram. The shear at each end of the beam is plotted first, Diagram. The shear at each endloading of the beam plotted L the ends ofShear L moments at Fig. Since is no distributed on theisbeam, thefirst, slope Moment Diagram. L The the 6–11c. beam are there L Fig. 6–11c. Since there is no distributed loading on the beam, the diagram is zero as indicated. Note how the force Pslope at the plotted, Fig. 6–11d. Here the moment diagram consists of of the twoshear sloping ofcenter the diagram zero as indicated. Note how the force P at thean (a) of the beamiscauses the shear diagram to jump downward lines, one with a slope of !2P and of shear !P. (a) the other with a slope center of the beam causes the shear diagram to jump downward an P, since The value of the moment in the center of the amount beam can be this force acts downward. amount P, since this force acts downward. determined byÇözüm: the method of sections, or from the area underDiagram. the Moment The moments at the ends of the beam are Diagram. The the moments the ends of the are shear diagram. If we choose the left half of the shear Moment diagram, plotted, Fig. 6–11d. Here momentatdiagram consists of beam two sloping plotted, Fig. 6–11d. Here the moment diagram consists of two sloping lines, one with a slope of !2P and the other with a slope of !P. M ƒ x = L = M ƒ x = 0 + ¢M lines,The onevalue with a of slope !2P andinthethe other with of a slope !P. can be theofmoment center the of beam The value of the moment in the center of the beam can be determined by the method of sections, or from the area under the determined by the method of sections, or from the area under the shear diagram. If we choose the left half of the shear diagram, M ƒ x = L = -3PL + (2P)(L) = -PL shear diagram. If we choose the left half of the shear diagram, Mƒ = Mƒ + ¢M M ƒ x =xL= L= M ƒ x =x0= 0+ ¢M 6 P 6 P Mƒ = -3PL + (2P)(L) = -PL M ƒ x =xL= L= -3PL + (2P)(L) = -PL 2P 3PL V 2P w#0 slope # 0 (b) P downward force P downward jump P x P 3PL 3PL (b) w#0 wslope # 0 # 0 (b) V slope # 0 downward force P V downward force PP downward jump downward jump P 2P 2P P P x x (c) (c) (c) V # positive constant slope # positive constant x "PL "3PL P 2P 2P P M P M (d) M V # positive constant # positive constant Vslope # positive constant slope # positive constant x Fig. 6–11 "PL "PL "3PL "3PL (d) (d) Fig. 6–11 Fig. 6–11 PAGE !68 x M0 LETHOD 6.2 L GRAPHICAL RAPHICAL M ETHOD FOR FOR C CONSTRUCTING ONSTRUCTING SSHEAR HEAR AND ANDM MOMENT OMENTD DIAGRAMS IAGRAMS 267 267 (a) EXAMPLE SOLUTION EXAMPLE 6.6 $ Ölçü ve yükleme durumu şekilde verilen basit mesnetli kirişin kesme Support Reactions. The reactions are shown on the free-body Drawdiagram the shear and moment diagrams for the Draw the beam beam shown shown in in in Fig. 6–12b. kuvveti vethe moment diyagramlarını çiziniz. Fig. 6–12a. Fig. 6–12a. Shear Diagram. The shear at each end is plotted first, Fig. 6–12c. M00 Since there is no distributed load on the beam, the shear diagram has zero slope and is therefore a horizontal line. L moment L L L Moment Diagram. The is zero at each end, Fig. 6–12d. The moment diagram has a constant negative slope of !M0!2L since this (a) (a) is the shear in the beam at each point. Note that the couple moment Çözüm: SOLUTION SOLUTION M0 causes a jump in the moment diagram at the beam’s center, but it does not affect the The shearreactions diagram at this point.on Support Reactions. are shown Support shown on the the free-body free-body diagram in Fig. 6–12b. diagram M0 6 Shear Diagram. Diagram. The The shear shear at Shear at each each end end is is plotted plotted first, first, Fig. Fig. 6–12c. 6–12c. Since there there is is no no distributed distributedLload Since load on on the the Lbeam, beam, the the shear shear diagram diagram has has zero slope and is therefore a horizontal line. M /2L M /2L 0 zero slope and is therefore a horizontal line. 0 (b) Moment Diagram. Diagram. The The moment moment is Moment is zero zero at at each each end, end, Fig. Fig.6–12d. 6–12d.The The V moment diagram has a constant negative slope of !M !2L w "negative 0 moment diagram has a constant slope of !M00!2L since since this this slopepoint. "0 is the the shear shear in in the the beam beam at at each each is point. Note Note that that the the couple couple moment moment M00 causes causes aa jump jump in in the the moment moment diagram M diagram at at the the beam’s beam’s center, center, but but itit does not not affect affect the the shear shear diagram diagram at does at this this point. point. x M M00 !M0 /2L (c) L clockwise moment L L L M0 positive jump M0M0 /2L M M /2L (b) M00/2L M /2L (b) V 0" negative constant V V w" " 00 w slope " slope " 00 M0 /2 slope " negative constant x – M0 /2 !M M0/2L /2L ! 0 M M (c) (c) xx (d) Fig. 6–12 clockwise moment clockwise moment M M00 positive jump positive jump M M00 V V" " negative negative constant constant slope " negative slope " negative constant constant M /2 0 M /2 0 xx –– M /2 M00 /2 (d) (d) Fig. Fig. 6–12 6–12 PAGE !69 66 Shear Diagram. The shear at each end point is plotted first, Figs. 6–13c SOLUTION and 6–14c. The distributed loading on each beam indicates the slope and 6–14c. The distributed loading on each beam indicates the slope Support Reactions. The reactions atthe the fixed support are shown of the shear diagram thus produces shapes Support Reactions. Theand reactions at the fixed supportshown. are shown of the shear diagram and thus produces the shapes shown. on each free-body diagram, Figs. 6–13b and Fig. 6–14b. on each free-body diagram, Figs. andatFig. 6–14b. Moment Diagram. The 6–13b moment each end point is plotted first, MomentShear Diagram. The moment atateach end point pointisisplotted plottedfirst, first, Diagram. The shear each Figs.on 6–13c Figs. 6–13d and 6–14d. Various values of the shear at each the Shear Diagram. The shear at each end point is plotted Figs. point 6–13c 268 C H A P T E R 6 B E N D I N G Figs. 6–13d and 6–14d. Various values of the shear at eachfirst, point on thethe slope and 6–14c. The distributed loading on each beam indicates beamThe indicate the slope of the diagram at the and 6–14c. distributed loading onmoment each beam indicates thepoint. slopeNotice beam indicate the slope of theand moment diagramthe at the point. Notice of the shear diagram thus shapes this variation theproduces curves shown. of the how shear diagram andproduces thus produces the shapes shown.shown. how this variation produces the curves shown. Moment Diagram. The moment atend each end ispoint iswplotted NOTE: Observe how the degree of the curves from to V tofirst, M Moment Diagram. The moment each point plotted first, $ EXAMPLE 6.7 Ölçü ve yükleme durumu şekilde verilen konsol kirişlerin kesme NOTE: Observe how the degree of at the curves from w kuvveti to V topoint M on the Figs. 6–13d and 6–14d. Various values of the shear at each increases exponentially due toofthe integration ofpoint dV on = w dx and Figs. 6–13d and 6–14d. Various values the shear at each the increases exponentially dueslope to the integration of ofdV dx point. and Notice beam indicate of the moment diagram atwthe Draw the= shear moment diagrams for each the=beams shown inload dM Vdx .and Forthe example, in Fig. 6–14, the linear distributed beam indicate the slope of the moment diagram atdistributed the point. Notice ve moment diyagramlarını çiziniz. dM = Vdx . For example, in Fig. 6–14, the linear load how this variation produces the curves shown. Figs. 6–13a and 6–14a. produces a parabolic shear diagram and cubic moment diagram. how this variation produces the curves shown. produces a parabolic shear diagram and cubic moment diagram. NOTE: Observe how the the curves w0 degree NOTE: Observe how wthe degree of the of curves from wfrom to VwtotoMV to M SOLUTION 0 increases exponentially due to the integration of dV = w dx and increases w0 exponentially due to the integration of dV = w dx and w0 6 Support Reactions. The reactions at6–14, the fixed support are shown dM = Vdx . For example, in Fig. the linear distributed dM = Vdx. For example, in Fig. 6–14, the linear distributed load load on each free-body diagram, Figs.diagram 6–13b and Fig. 6–14b. produces a parabolic shear and cubic moment diagram. produces a parabolic shear diagram and cubic moment diagram. 6 6 L (a) L (a) 6 w0 L is plotted first, Figs. 6–13c Shear Diagram. The shear atweach end point 0 w0 L (a) and w6–14c. The distributed loading on each beam indicates the slope 0 (a) of the shear diagram and thus w0 produces the shapes shown. w0 w0 w0Lmoment at each end point is plotted first, Moment Diagram. The w0L 2 w0 L 2 Figs. 6–13d and 6–14d. Various valuesLof theLshear at each point on the w0 L Çözüm: L L (a) (a) diagram at the point. Notice beam indicate the slope of the moment (a) (a) w0 curves shown. 2 how this variation produces the ww0L (b) 0 w0 L2 w0L2 (b) w0 w0L (b) w0 L2 w0 NOTE: Observe whow (b) 0L 6the V 2 w L 6 2 degree of the curves from w to V to M 0 2 2V w0 L w ! negative constant ("w wdue w ! negative of decreasing 0) 0L increases exponentially to the integration w ! negative constant ("w0 ) w L w! negative decreasing dV = w dx and 0 V slope = negative constant ("w0 ) slope ! negative decreasing 2 V slope = negative constant ("w0 ) slope ! negative decreasing 2 dM = Vdx. For example, in Fig. 6–14, the linear distributed load w L w0L2 diagram and cubic (b) moment diagram. w0 L w L20 produces a parabolic w0L2 shear (b) 0 2 w0 (b) (b) w ! negative w ! negative constantconstant ("w0x) ("w0 ) x V w0 0 ) slope = negative constant V slope = negative constant ("w0 ) ("w 6 (c) (c) w L 0 w0 L V! positive decreasing VM ! positive decreasing M slope = positive decreasing slope = positive decreasing L xx x x (a) (c) (c) V ! positive decreasing w0 V decreasing M! positive ML slope = positive decreasing 2 w w L slope = positive decreasing w0 L2 "0 0 " x 2 2 (d) x (d) w0 L 2 2 6–13 w0 L2 Fig. 6–13 Fig. (b) 2 2 w0 L2 " w0 L w ! negative constant ("w0 ) " 2 V slope = negative(d) constant ("w0 ) 2 (d) w0 L 6 6 0 V wwL0 V 0 decreasing x w ! negative 0 w0L x w ! negative decreasing slope ! negative decreasing 2 (c) slope ! negative decreasing 2 (c) V! positive decreasing V! positive decreasing M slopeM! positive slope ! positive decreasing decreasing 0 L 0 xx x x (a) (c) (c) w0 V ! positive decreasing VM! positive decreasing w0L M slope ! slope ! positive decreasing positive decreasing 2 x 2 x w0 L2 " w0 L " (d) 6 (d) 6 w0L2 (b) Fig. 6–14 Fig. 6–14 6 V w0L w ! negative decreasing 2 slope ! negative decreasing 2 w0 L w0 L2 " (d) " 6 (d) 6 Fig. 6–14 Fig. 6–14 Fig. 6–13 Fig. 6–13 0 x (c) M V ! positive decreasing slope = positive decreasing M V ! positive decreasing slope ! positive decreasing x x " w0 L2 2 x (c) (d) " Fig. 6–13 PAGE !70 w0 L2 6 (d) Fig. 6–14 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT EXAMPLE 6.8 6.2 269 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS Draw the shear and moment diagrams for the cantilever beam in Fig. 6–15a. 2 kN $ EXAMPLE 6.8 Ölçü ve yükleme durumu şekilde verilen konsol kirişin kesme kuvveti ve moment diyagramlarını çiziniz. MB " 11 kN#m Draw the shear and moment diagrams for the cantilever beam 2 kN 2 kN in Fig. 6–15a. 1.5 kN/m 1.5 kN/m 2m OD FOR 269 CONSTRUCTING SHEAR AND MOMENT2 D IAGRAMS kN Çözüm: 2 kN B By " 5 kN 2m w"0 slope " 0 (b) 2V6(kN) 9 w " negative constant slope " negative constant B CONSTRUCTING SHEAR 6.2 (a) AND MOMENT DIAGRAMS 11RAPHICAL kN#m METHOD FOR MB "G 1.5 kN/m 2m w"0 slope " 0 2m 2 EXAMPLE 6.8 B 2m 2m 2m A ntilever beam (b) 1.5 kN/m A V (kN) (a) !2 SOLUTION MB " 11 kN#m Draw the shear and moment diagrams for the cantilever beam 2 kN 2 4 By " 5 kN x (m) Support Reactions. The support reactions at the1.5 fixed kN/msupport in Fig. 6–15a. 2m 2m (c) B are shown in Fig. 6–15b. !2 SOLUTION (b) Shear Diagram. The shear at end is M !2 kN. D This value is G RAPHICAL METHOD FOR CONSTRUCTING SHEARAAND OMENT IAGRAMS 269 Support Reactions. The 6.2 support reactions at the fixed support w "2 0kN w " negative constant plotted at x " 0, Fig. 6–15c. Notice how the shear diagram is (c) 1.5 kN/m B are shown in0Fig. 6–15b. slope " By " !5 5VkN slope " negative constant " negative constant constructed by following the slopes defined by the 2loading w. 2m m slope " negative constant Shear Diagram. The shear at end A is !2 kN. This value is The shear at x = 4 m is !5 kN, the reaction on the beam. This EXAMPLE 6.8 6 V (kN) plotted at A x " 0, Fig. 6–15c. Notice theverified shear diagram is the area under (b) the distributed value how can be by finding M (k N#m ) V " negative constant constructed following slopes defined by the loading MB " 11 kN#m B Draw the shearbyand momentthe diagrams for the beam w. 2slope kN " negative loading, Eq.cantilever 6–3. V " negative increasing w " 0 constant w " negative constant 2 4 shear at x = 4 m is !5 kN, the reaction on the beam. This 1.5 kN/m in The Fig. 6–15a. x (m) slope " negative increasing 2 slope " 0 slope " 2m 2m negative constant 0 value can be verified by finding the area under the distributed M (kN#m) V ƒ x = 4 m = V ƒ x = 2 m + ¢V = -2 VkN !2 loading, Eq. 6–3. (kN)- (1.5 kN>m)(2 m) = -5 kN he fixed support 2 kN (c) (a) 1.5 kN/m 2 0 4 x (m) Moment Diagram. of zero at x2 = 0 is plotted4 in By " 5 kN V ƒ x = 4 m = V ƒ x = 2 m + ¢V = - 2!5 kN - (1.5 kN>m)(2 m) The = - 5moment kN 2m 2m x (m) Fig. 6–15d. Notice how the moment diagram is constructed based !4 N. This value is 6 its slope, which is equal to the shear on knowing (b) at each point. A SOLUTION 2 kN Moment Diagram. The moment of zero at x = 0 is plotted !2 in hear diagram is The change of moment from to m is determined !11 x = 0 x = 2 B VFig. " negative constant 6–15d. Notice The how support the moment diagram is constructed based w " 0 (d) w " negative constant the loading w. Support Reactions. reactions at the fixed support slope " negative constant V " negative increasing from themarea under the shear diagram. Hence, moment slope(c) " 0 the 2m slope " negativeat constant knowing its6–15b. slope, which is equal to2the shear at each point. n the beam. This B areon shown in Fig. slope " negative increasing !5 2 kN m is x = 2 change of moment from x = 0 to x = 2 m is determined the distributed MThe (kN#m ) Shear Diagram. shear end A is !2Hence, kN. This is at V (kN) (a) from the area The under the at shear diagram. thevalue moment 6 V " 2 kN plotted 6–15c. Notice how is = 0 + [-2 kN(2 m)] = -4 kN # m 2m M ƒ xthe = M ƒ xdiagram is 0, Fig. x =at2 xm " = 2 m shear = 0 + ¢M 2 4 M " 4 kN#m 0 x (m) constructed by following the slopes defined by the loading w. V " negative constant 2 4 (e) (2 m) = - 5 kNThe shear at x = 4 m is !5 kN, the reaction on the beam. This# slope " negative constant V " negative increasing x (m) slope "of negative increasing value determined from the2 mmethod sections, M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 This + [ -same 2 kN(2 m)] can = -be 4 kN m !4 by finding the area under the distributed value can be verified M !2 (kN#m) Fig. 6–15e. SOLUTION (e) loading, Eq. 6–3. = 0 is plotted in This same value can The be determined from the sections, Support Reactions. support reactions at method the fixedofsupport 2 (c) 4 !11 onstructed based B are (d) 6–15b. Fig.shown 6–15e. in Fig. 0 x (m) Fig. 6–15 !5 ar at each point.V ƒ x = 4 m = V ƒ x = 2 m + ¢V = - 2 kN - (1.5 kN>m)(2 m) = - 5 kN 2 kN m is determined Shear Diagram. The shear at end A is !2 kN. This value is 6 !4 Fig. 6–15c. Notice how the shear diagram is the moment at plotted at x " 0,V " 2 kN V " negative constant Moment Diagram. The moment of zero at x =by0 the is plotted constructed by following the slopes defined loadingin w. slope " negative constant V " negative increasing !11 M "!5 4 kN#m Fig. 6–15d. Notice the is constructed based (d) The shear at x how ismoment kN, diagram the reaction on the beam. This = 4m slope " negative increasing on knowing its slope, which is equal to the shear at each point. value can be verified by finding the area under the distributed 2M kN(kN#m) 2m = -4 kN # m The change of6–3. moment from x = 0 to x = 2 m is determined loading, Eq. from the area (e) under the shear diagram. Hence, the moment at 4 V "22 kN 0 x (m) thod of sections, x = 2 m is V ƒ x = 4 m = V ƒ x =Fig. + ¢V = - 2 kN - (1.5 kN>m)(2 m) = - 5 kN M " 4 kN#m 2 m 6–15 M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(2 m)] = - 4 kN # m Moment Diagram. The moment of zero at x = 0 is plotted in Fig. 6–15d. Notice how the moment diagram is constructed based This same value can be determined from the method of sections, on knowing its slope, which is equal to the shear at each point. Fig. 6–15e. The change of moment from x = 0 to x = 2 m is determined from the area under the shear diagram. Hence, the moment at x = 2 m is M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(2 m)] = - 4 kN # m PAGE !71 This same value can be determined from the method of sections, Fig. 6–15e. 2m (e) !4 !11 (d) 2 kN Fig. 6–15 V " 2 kN M " 4 kN#m 2m (e) Fig. 6–15 !4 (d) V" Fig. 270 DING $ CHAPTER 6 EXAMPLE 6.9 BENDING Ölçü ve yükleme durumu şekilde verilen çıkmalı kirişin kesme kuvveti ve Draw the shear and moment diagrams for the overhang beam in moment çiziniz.for the Draw the shear diyagramlarını and moment diagrams Fig.overhang 6–16a. beam in Fig. 6–16a. 4 kN/m 4 kN/m A 270 A 270 CHAPTER 6 CHAPTER 6 " 10 kN B B BENDING 2m (a) (a) slope " 0 w " negative constant slope " negative constant gative constant NDING egative constant 2m 4m 2m 4m A BENDING 4m EXAMPLE (b)6.9 Ay " 2 kN B " 10 kN EXAMPLE w " 0 6.9 y 2m 4 kN/m 4 kN/m Draw the shear and moment diagrams for the overhang beam in Fig. 6–16a. Draw the shear and moment diagrams for the overhang beam in SOLUTION Fig. 6–16a. Support Reactions. The support reactions are shown in V (kN) SOLUTION 4 kN/m 8 Support Reactions. The support4 kN/m reactions are Fig. shown 6–16b. in 4 kN/m 4 kN/m Fig. 6–16b. Shear Diagram. The shear of !2 kN at end A of the beam A Shear Diagram. The shear of !2 kN at end Aisofplotted the beam at x " 0, Fig. 6–16c. The slopes are determined A B A x (m) Draw the moment diagrams for the overhang beamloading in 0shear and is plotted at x " 0, Fig. 6–16c. The slopes are determined from the and from this the shear diagram is 4 6 A B x (m) Fig. 6–16a. from the loading4 m and !2 from this2 m the shearconstructed, diagram isas indicated in the figure. In particular, notice 6 2 m 4m 6 the positive jump of 10 kN at m due to the force By, as x = 4 constructed, as indicated in the figure. In particular, notice 2m 4 kN/m 4m 2m 4m (b) Ay "jump 2 kN of(c) indicated 4 kN/m the positive 10 kN at x =By4"m10due force Byin , asthe figure. kN to the (a) w " 0 (b) indicatedAin the figure. 2 kN y" By " 10 kN (a) slope "0" 0 negative V decreasing w" V " negative constant w " negative constant Moment Diagram. The moment of zero at x " 0 is A slope " negative decreasing plotted, Fig. 6–16d. Then following the behavior of the slope slope " 0 slope " negative constant SOLUTION Moment Diagram. The moment of zero at x " 0 is slope negative constant negative decreasing B w ""negative constant V (kN) " negative decreasing plotted, Fig. found from the shear diagram, the moment diagram is 6–16d. Then following the behavior of the slope SOLUTION slope " negative constant Support Reactions. The support reactions are shown in M (kN#mV) (kN) constructed. The moment at x " 4 m is found from the area found from the shear diagram, the moment diagram is Fig. 6–16b. Reactions. The support reactions are shown in Support 4 m 8 slope " 0 2 m 2m under thearea shear diagram. constructed. The moment at x " the 8 4 m is found from Fig. 6–16b. 4 slope " 0 Shear Diagram. The shear of !2 kN at end A# of the beam under the shear diagram. 0 By " 10 kN (a) x (m) M ƒ x = 4 is = M ƒ x = 0at+ x¢M = 0Fig. + [6–16c. - 2 kN(4 m)]slopes = - 8 kN m plotted " 0, areA m determined 6 Shear Diagram. The shear of The !2 kN at end of the beam x (m) # x (m) M = kN(4 m)] = 8 kN m 0 M ƒ x = 0 + ¢M = 0 + [-2 ƒ x = 4 m from the loading and from this the shear diagram is We canisalso obtain by using method sections, plotted atthis x "value 0, Fig. 6–16c.the The slopesofare determined 4 6 6 constant negative x (m) SOLUTION 0 " negative constant constructed, as indicated in the figure. In particular, notice !24 as of shown in Fig. from the6–16e. loading and from this the shear diagram is We the6 method sections, !8 by using 6 can also obtain this value the positive of 10 kN in at x the force B 4 m due constructed, as indicated the= figure. Intoparticular, notice The support reactions are shown in jump !2 y, as as6Support shown in Reactions. Fig. 6–16e. (d) indicated in the figure. the positive jump of 10 kN at x = 4 m due to the force By, as Fig. 6–16b. (c) indicated in the figure. V " 2 kN (c) Moment Shear Diagram. The shearVVof !2 kN decreasing at end A of the beam Diagram. The moment of zero at x " 0 is " negative V" negative constant " 2 kN M " 8 the kN#m slope "The negative decreasing plotted, Fig.Diagram. 6–16d. ThenThe following behavior of xthe"slope A Moment moment of zero at 0 is slope " negative constant is plotted at x " 0, Fig. 6–16c. slopes are determined V " negative decreasing V " negative constant x (m) slope " negative decreasing found from the shear diagram, the moment diagram is M " 8 k N#m plotted, Fig. 6–16d. Then following the behavior of the slope from the loading and from this the shear diagram is slope " negative constant 4 6 A 4m M (kN#m) constructed. at x " 4 m from the areais found themoment shear diagram, theis found moment diagram constructed, as indicated in the figure. In particular, noticefromThe 2 M (kN#m) 4 m under constructed. Thediagram. moment at x " 4 m is found from the area the positive jump of 10 kN at x = 4 mslope due "to0 the force By,the as shear slope "0 4 2 kN the shear diagram. under indicated in the figure. x (m) 0 M ƒ x = 4 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m 4 6 (e) 20kN x (m) Moment Diagram. The moment of zero at x M "ƒ x =04 mis = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m " negative decreasing 6 We can also obtain this value by using the method of sections, (e) " negative decreasing plotted, Fig. 6–16d. Then Fig. 6–16 following the behavior of the slope as shown in Fig. 6–16e. We can also obtain this value by using the method of sections, !8 found from the shear the moment diagram is Fig. diagram, 6–16 as shown in Fig. 6–16e. (d) at x!8 constructed. The moment " 4 m is found from the area slope " 0 under the shear diagram.(d) V " 2 kN 8 6 x (m) M ƒ x = 4 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m A We can also obtain this value by using the method of sections,A as shown in Fig. 6–16e. 2 kN 2 kN V " 2 kN M " 8 kN#m A 4m 2 kN (e) Fig. 6–16 PAGE !72 V " 2 kN M " 8 kN#m M " 8 kN#m 4m 4m (e) (e) Fig. 6–16 Fig. 6–16 6.2 EXAMPLE 6.10 $ EXAMPLE 6.10 Ölçü ve yükleme durumu şekilde verilen şaft için kesme kuvveti ve The shaft in çiziniz. Fig. 6–17a supported byyatay a thrust bearing at A and a moment diyagramlarını A ismesnedinde hareket The shaft injournal Fig. 6–17a is supported by athe thrust at A and aengellenmiş, B mesnedi ise bearing at B. Draw shearbearing and moment diagrams. kayıcı mesnettir. journal bearing at B. Draw the shear and moment diagrams. 120 lb/ft A A D FOR CAL CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 120 lb/ft A B 271 METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 12 ft Çözüm: 2 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 271 B 12 ft 271 (a) (a) 120 lb/ft A B 120 lb/ft B 12 ft 12 ft (b) Ay = 240 lb (b)w # negative increasing By # 480 lb Ay = 240 lb # negative increasing w # negativeslope increasing By # 480 lb V (lb) slope # negative increasing V (lb) 240 SOLUTION 12 6.93 240 aring at A and a 120 lb/ft x (f SOLUTIONSupport Reactions. 0 12 The support reactions are shown in 6.93 diagrams. thrust bearing at A and a x IAGRAMS (ft) 120 lb/ft 6.2 G RAPHICAL M ETHOD FOR C ONSTRUCTING S HEAR AND M OMENT D 0 Support Reactions. Fig. 6–17b. The support reactions are shown in (c) moment diagrams. /ft Fig. 6–17b. Shear Diagram. (c) As shown in Fig. 6–17c, the shear at is x = 0 6.2 B GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT IAGRAMS 271 V #D positive decreasing A 120 lb/ft " 480 Shear Diagram. As shown the in Fig. 6–17c, the shear at xloading, = 0 is the Following slope defined by the shear slope # positive decreasing V # positive decreasing A !240. EXAMPLE 6.10 B V # negative " 480 increasing !240. Following the slope defined by the loading, the shear slope # positive decreasing diagram is constructed, where at B its value is "480 lb. Since the 12 ft # negative increasing B V # negativeslope increasing diagram is constructed, where at Bpoint its value is "480 shear changes sign, the6–17a bethe located. doand M V = by 0lb.must EXAMPLE 6.10 (b)The 12 ft shaft in Fig. iswhere supported aSince thrust bearing To at A a (lb$ft) 120 lb slope # negative increasing V#0 B Ashear lb y = 240 changes thissign, we will use the method free-body of the point where must beGThe located. To do diagram VDraw =of0 sections. M (lb$ft) 6.2shear RAPHICAL METHOD FOR CONSTRUCTING SHEARslope AND M DIAGRAMS # OMENT 0 (b) journal bearing at lb B. the and moment diagrams. w # negative increasing B # 480 y V#0 AyFig. = the 2406–17a lb left of the shaft, sectioned at at andiagram arbitrary this we will use the segment method sections. The free-body The shaft in is supported by a thrust bearing A and a ofposition x, is slope 120 lb/ft slope # negative #0 w # increasing negative increasing By # 480 lb V (lb) 120 lb/ft 1109 6 shown in Fig. 6–17e. Notice that the intensity of the distributed the left segment of the shaft, sectioned at an arbitrary position x, is journal bearing at B. # Draw the shear and moment diagrams. slope negative increasing A 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 271 (lb) load at x is w # 10x, which has been found by proportional 1109 shown inV Fig. 6–17e. Notice that the intensity of the distributed EXAMPLE 6.10 120 lb/ft by proportional i.e.,which 120!12has # w!x. load at x istriangles, w # 10x, been found 240 A B 12 ft x( 0 A 12 6–17a Thus, V #in0,Fig. B 6.93 Thefor shaft is supported by a thrust bearing at A and a triangles, i.e., 120!12 # w!x. 240 6.93 12 120 lb x (ft) (b) EXAMPLE 6.10 0 are shown in x (ft) 0 12 6.93 at B. Draw journal bearing the A = 240 lb Thus, for V # 0, 1 shear and moment diagrams. 12 ft y 6.93w # negative (ft)2 (10x)x = 0 240 lbx = 0; (d) 12 increasing B # 0 + c ©F 12 ft A y eactions are shown in y B (c) The+shaft Fig. 6–17a is supported thrust bearing at A and a 120 lb/ft 120 lb/ft slope # negative increasing (b)(d) c ©Fin 240 lb by - 12a(10x)x = 0 y = 0; V (lb) x = 6.93 ft 1 A = 240 lb (c) A hear at x = 0 is journal y [10 x ] x V # positive decreasing bearing at B. Draw the shear and moment diagrams. w # negative increasing By # 480 lb2 12 ft x " 480 7c, the shear at xslope = 0#ispositive x = 6.93 ft (a) ding, the shear 1 [10 x ] x V #decreasing positive decreasing slope # negative increasing Moment Diagram. diagram starts at 0 V since 3 10 x 2 (lb) there V # negative increasingThe moment "120 480lb/ft decreasing 12 ft 80the lb. loading, Since thethe shear slope # positive 240 Bx A it is constructed based on Athe is slope no SOLUTION moment atincreasing A; then (a) # negative Moment Diagram. The diagram starts at 0 since there B slope as 3 10 (b) V #moment negative increasing 12 alue is "480 the e located. To lb. doSince 6.93 V x M (lb$ft) determined from the shear diagram. The maximum moment Ay0= 240 lb slope # negative increasing is no at A; then it is constructed based on the slope as V # 0 Support Reactions. The support reactions are shown in A be located. To do moment 0 must ody diagram of V M (lb$ft) w # negative increasing By # 12 ft is equal to 240 slope #0 M occurs at the shear zero, since 12 ft = 6.93diagram. ft, whereThe A Fig. V6–17b. #x0shear determined from the slope #12negative The free-body of 6 Bmaximum moment ary position x, diagram is SOLUTION (c) increasing A(b) 6.93 slope # 0 V (lb) Fig. 6–17d, dM>dx = V = 0, x (ft) occurs where the shear is equal zero, since x = 6.93 ft,The 0 lbat x = 0 is = 240 x M 6to(a) an position x, is atReactions. 1109 thearbitrary distributed Support support reactions shown in Shear Diagram. As shown inare Fig. 6–17c, theAyshear V # positive decreasing w # negative increasing By # 480 lb 12 ft Fig. 6–17d, dM>dx = V = 0, 1109 ensity of the distributed d + ©M = 0; by proportional Aypositive # 240 lbdecreasing Fig. 6–17b. " !240. Following the slope defined by the loading, shear(c) slope x# slopethe # negative increasing 1 1 V (lb) 240 n found by proportional M + [(10)(6.93)] 6.93 (6.93) 240(6.93) = 0 V # negative increa A B diagram is constructed, where at B its value is "480 lb. Since the (e) max SOLUTION 2 3 d+ ©M = 0; A # 240 lb Shear Diagram. As shown in Fig. (a) 6–17c, the shear at x = 0 is y 12 V # positive decreasing slope #6.93 negative incre x (ft) 0 1 changes sign, 1 point where V = 0 must be located. To do shear the M0 (lb$ft) # Mmax +6.93 [(10)(6.93)] 6.93 (6.93) 240(6.93) = 0 slope A B "6–17 480 !240. Following the slope defined by the loading, the shear # positive decreasing (e) Support Reactions. The support reactions are shown in 12 M = 1109 lb ft 2 3 Fig. max x (ft) 0 V#0 240 this we will at use the method of sections. Thethe free-body diagram ofV # negative increasing 6.93 12 diagram is constructed, where B its value is "480 lb. Since slope #(c) 0 SOLUTION # (d)Fig. 6–17b. M = 1109 lb ft 12 6.93 Fig. 6–17 max Finally, notice how integration, first of the loading w which is linear, slope # negative increasing the left segment of the shaft, sectioned atTo an do arbitrary position x, is x (ft) x = 0 shear changes sign, the point where must be located. V = 0 0 (d) M (lb$ft) Shear Diagram. As shown in Fig. 6–17c, the shear at is x = 0 Support Reactions. The support reactions shown and in then a moment V # positive decreasing 1109 produces shear diagram which isare parabolic, ft 1a[10 shown inxof 6–17e. Notice that intensity of the distributed V#0 notice how integration, first of the loading w the which linear, ] Fig. x sections. thisFinally, we will use the method free-body diagram of loading, " !240. theThe slope defined by isthe the slope shear slope # positive decreasing Fig. 6–17b. 2 Following # 0 (c) x = 6.93 ft diagram which is cubic. 1 load at xxsectioned is w[10is #x]parabolic, which has been found by proportional x10x, produces a shear diagram which and then avalue moment 6 the left segment of the shaft, at an arbitrary position x, is V # negative increas 2 diagram is constructed, where at B its is "480 lb. Since the at 0 since there Shear Diagram. As shown 3in shear at x = 0 is V # positive decreasing x # the 10 x 6–17c, triangles, i.e.,Fig. 120!12 w!x. diagram which is shear cubic. slope # negative incre NOTE: Having studied these examples, test= yourself bylocated. coveringTo do 1109 shown in Fig. 6–17e. that the of thethe distributed am the starts at 0 as since there changes sign, the point where be V 0 must 3intensity M0(lb$ft) " 480 on slope !240. Following the Notice slope defined by the loading, shear slope # positive decreasing 10 x V# 0, Thus, for V 6.93 the shear and moment diagrams in Examples 6–1 through 6–4 V#0 12 load at isx constructed, isHaving wover # this 10x, which has been found by ed based on the slope as we will the method of sections. # negative increasing imum moment NOTE: studied these yourself byThe covering diagram atuse Bexamples, its value "480 lb.proportional Since the free-body diagramVof V istest slope # 0 A and see ifwhere 1 concepts discussed here. you can construct them using the c slope # negative increasing triangles, i.e., 120!12 # w!x. 240 lb + ©F = 0; (10x)x = 0 The maximum moment the left segment of the shaft, sectioned at an arbitrary position x, is (d) M= 0 to zero, since shear over the shear diagrams inmust Examples 6–1 2 through changes sign,and theAmoment pointy where be located. To do 6–4 V M (lb$ft) x (ft)1109 0 distributed M V#0 forifuse Vyou #the 0, method is equal to zero, since shown in Fig. 6–17e. Notice that diagram the intensity of the and can construct them using the concepts discussed here. thisThus, we see will of sections. The free-body of 6.93 12 x = 6.93 ft x slope # 0 1 [10 x ] x load at x is w # 10x, which has been found by proportional 6 2 the of the shaft, sectioned at an arbitrary position x, is x - 1 (10x)x 240 lb + cleft ©Fsegment = 0 x (d) y =A0; # 240 lb y i.e., 120!12 # w!x. 1109 shown in Fig. 6–17e.triangles, Notice that the2 intensity of the distributed Moment 3 10 A # 240 lb Diagram. The moment diagram starts at 0 since there 240(6.93) = 0 y(e) 0 forhas V #at 0,A; then x = 6.93 1 [10 x ] x load= at been found proportionalbased on the slope as is Thus, nowhich moment it by is ft constructed 6.93 12 6.93) B - 240(6.93) 0 x is w # 10x, V (e) 2 1 triangles, i.e., 120!12 # w!x. from the shear diagram. The maximum moment x Fig.determined 6–17 c 240 lb - 2(10x)x = 0 + ©Fy = 0; (d) A Moment TheFig. moment diagram starts the at 0 shear since there 3 10x x(ft) 0 to zero, since t Thus, forDiagram. V # 0, occurs at 6–17 is equal x = 6.93 ft, where 6.93 12 w which is linear, is no moment at A; then it is constructed based on the slope as x = 6.93 ft 1 V dM>dx = V = 1 0, Fig. 6–17d, loading w which is linear, c ©Fy = 0; from the x 2 [10 x ] x 240 lb +determined (10x)x = 0 then a moment (d) 2 shear diagram. The maximum moment x bolic, and then a moment Theequal moment diagram starts at 0 sinceAthere dft, + ©M = Diagram. 0;the shear 3 10 Ay # 240 M lb occurs at x = 6.93Moment where x 1 =is 6.93 ft to zero,1 since 1 [10 x ] x is no moment at A; it is constructed based the slope Mmax + then 6.93 A 3(6.93) 240(6.93) = 0as 2 B - on (e) V 2 [(10)(6.93)] self by covering dM>dx = V = 0, Fig. 6–17d, x x determined from the shear diagram. The maximum moment Moment Diagram. The moment diagram starts at 0 since there 3 test yourself by covering # A 10 x 6–1 through 6–4 d+ ©M = 0; Mmax = 1109 lb ft Fig. 6–17 Ay # 240 lb whereonthe since x = 6.93 ft, based is no6–4 moment at A; occurs then it at is constructed the shear slope is as equal to zero, Examples through V s discussed6–1 here. 1 1 Mmax the +dM>dx 6.93 240(6.93) = 0 AThe B - first 6–17d, = diagram. V =how 0, Fig. Finally, integration, ofmoment the loading w which is linear, (e) determined from shearnotice maximum 2 [(10)(6.93)] 3 (6.93) e concepts discussed here. x A produces a shear diagram which is parabolic, and then a moment M occurs at x = 6.93 ft, where the shear is equal to zero, since # Mmax= =0;1109 lb ft d + ©M Ay # 240 lb Fig. 6–17 diagram is cubic. 1 6–17d, which dM>dx = V = 0, Fig. Mmax + 12[(10)(6.93)] 6.93 (6.93) B - 240(6.93) = 0 x A (e) PAGE ! 7 3 3 Finally, notice how integration, first of the loading w which is linear, NOTE: Having studied these examples, test yourself by covering d+ ©M = 0; A # 240 lb # ft y Mmax = then 1109 albmoment produces a shear diagram which is parabolic, and Fig. 6–17 1 the shear6.93 andA 13moment 6–1 through 6–4 (e) M + over (6.93) B -diagrams 240(6.93)in=Examples 0 2 [(10)(6.93)] diagram whichmax is cubic. Finally, howconstruct first of the w which is linear, and seenotice if you can them using theloading concepts discussed here. # integration,

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