5 in. of the cross-sectional area of the Determine the moment of inertia T-beam shown Fig. A–7a x¿ axis. area of the Determine theinmoment of about inertiathe of centroidal the cross-sectional T-beam shown in Fig. A–7a about the centroidal x¿ axis. 8 in. 2 in. 8 in. (a) 1.5 in. 1.5 1.5 in. in. 4.45 in. 4.45 in. 1.5 in. C A–7 Fig. A.2 MOMENT OF INERTIA AN AREA 789 A.2 FOR MOMENT OF INERTIA FOR AN AREA 789 x¿ x¿ EXAMPLE A.2 C in. ! EXAMPLE 10A.2 Şekildeki T kiriş enkesit alanının x’ ağırlık merkezinden geçen eksenine 8.55 in. 10 in. 5 in. SOLUTION I momentini 8.55 in. area of the göre atalet Determine the moment of inertia of the hesaplayınız. cross-sectional Determine the moment of inertia of the cross-sectional area of the The area is segmented into as shown in Fig. A–7a, and 5 in.two rectangles T-beam shown in Fig. A–7a shown about the centroidal x¿ axis. T-beam A–7a the centroidal the distance from theinx¿Fig. axis andabout each centroidal axisx¿is axis. determined. Using the table on8 in. the inside2 front cover, the moment of inertia of a in. 8 in. 1 3 1.5 in. about its centroidal2axis rectangle is Applying the parallelI = bh . in. 12 1.5 in. 4.45(a) in. in. axis1.5 theorem, Eq. A–5,1.5 toin. each rectangle and4.45 adding the results, we in. (a) x¿ have x¿ Fig. A–7 C C A–7 I = ©(I x¿ + Ady2) 8.55Fig. 10 in. 10 in. in. 8.55 in. SOLUTION I 15 in. 5 in. 3 2 The area is =segmented into two as shown in in. Fig.-A–7a, SOLUTION I c 12 in.2110 in.2 rectangles + 12 in.2110 in.218.55 5 in.2and d 12 the from the x¿into axistwo and each centroidal axis determined. Thedistance area is segmented rectangles as shown in is Fig. A–7a, and Using the table on1the the inside cover, the moment of determined. inertia of a the distance from axis front and each centroidal axis is x¿ 2 in. 3 2 2 1in.the + its c centroidal 18 in.213 in.2 in. - parallel1.5 in.2 rectangle axis is+cover, Applying the I18=in.213 bh3in.214.45 .moment Using theabout table on the inside front of inertia of ad 12 12 (a) 1 axis theorem, Eq. to each rectangle and3 adding the results, we rectangle about itsA–5, centroidal axis is I = (a) 12 bh . Applying the parallel4 Ans. I = 646 in have ÇözümEq. 1. A–5, to each rectangle and adding the results, we axis theorem, Fig. A–7 have Fig. A–7 I = ©(I x¿ + Ady2) SOLUTION II 2 I = ©(I x¿ + Ady ) SOLUTION I 4.45 in. 1can be Iconsidered The area as in.2110 one large rectangle two 2 SOLUTION 12two in.2110 in.23 + as 12 in.218.55 in. and - less 5 in.2 d small = c into The area is segmented rectangles shown in Fig. A–7a, x¿ 1shown 12 rectangles, shaded in Fig. A–7b. We have as shown in Fig. The isand segmented twoaxis rectangles C 12 in.2110 in.23centroidal +into 12 in.2110 in.218.55 in. - 5 in.22 d A–7a,13and = cx¿area the distance from the axis each is determined. in. 12 2the x¿ axis and each centroidal axis is determined. 1 x¿ front Using the table on the the inside I =distance ©(I +from Adcover, 10 in. y ) 3 the moment of inertia of a 2 8.55 in. + c 18 in.213 in.2 + 18 in.213 in.214.45 in. 1.5 in.2 d 1 Using 12 the tableison inside front cover, the moment of inertia of a 6.5 in. rectangle about its centroidal the1parallelI =the12 bh3. Applying 1 axis 3 2 5 in. 3 - 1.5 in.2 d 1 + c 18 in.213 in.2 + 18 in.213 in.214.45 in. 3and adding rectangle about its centroidal axis isthe Applying the2 dparallelI =results, axis theorem, Eq. A–5, rectangle we- 6.5 in.2 12 bh .in. = to c 12each 418 in.2113 in.2 + 18 in.2113 in.218.55 Ans. I = 646 in 12 axis theorem, Eq. A–5, to each rectangle and adding the results, we have Ans. I have = 646 in4 3 in. 2 in. 3 in. A I = SOLUTION ©(I x¿ + AdyII2) 2 c 1 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d 2 I =12©(I x¿ + Ady ) 4.45 in. The can as one large rectangle less two small SOLUTION II be 1 area Çözüm 2.3considered 4+ 12 in.2110 in.218.55 in. - 5 in.22 d in.2110 in.2 = rectangles, c 12area x¿ 4.45 in. (b) I = 646 in Ans.2 shown shaded in Fig. We have 1 The can be considered asA–7b. one large rectangle less two small 12 C d = c 12 in.2110 in.23 + 12 in.2110 in.218.55 in. - 5 in.2 13 in. x¿ rectangles, shown shaded 12 2 in Fig. A–7b. We have C 10 in. 1 I = ©(I x¿ + 3Ady ) 13 in. 8.55 in. 2 2 in.213 in.214.45 in. - 1.5 in.2 d + c I18=in.213 in.2 + 18 6.5 in. ©(I 1 x¿ + +Adc y1) 18 3in.213 in.23 + 18 in.213 in.214.45 in. -2 1.5 in.22 d10 in. 5 in. 12 8.55 in. 6.5 in. = c 18 in.2113 in.2 + 18 in.2113 in.218.55 in. - 6.5 in.2 d 5 in. 12 1 I = 646 in4 = c 12 18 in.2113 in.23 + 18 in.2113 in.218.55 in.Ans. - 6.5 in.22 d 12 I 1= 646 in4 Ans. 3 in. 2 in. 3 in. A - 2 c 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d 1 SOLUTION II 3 in. 2 in. 3 in. 12 A 3 2 - 2 c 13 in.2110 in.2 + 13 in.2110 in.218.55 in. - 5 in.2 d 4.45 in. The area can be considered SOLUTION 12 4 asII one large rectangle less two small (b) I = 646 in Ans. x¿ 4.45 in. rectangles, shown shaded in Fig. We have as one large rectangle less two small The area canA–7b. be considered C (b) I = 646 in4 Ans. 13 in. x¿ shown shaded in Fig. A–7b. We have I = ©(I x¿ + rectangles, Ady2) C 10 in. 8.55 in. 13 in. 6.5 in. I = ©(I x¿ + Ady2) 10 in. 5 in. 1 8.55 in. = c 18 in.2113 in.23 + 18 in.2113 in.218.55 in. - 6.5 in.22 d 6.5 in. 5 in. 12 1 = c 18 in.2113 in.23 + 18 in.2113 in.218.55 in. - 6.5 in.22 d 12 1 3 in. 2 in. 3 in. A - 2 c 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d 12 1 3 in. 2 in. 3 in. - 2c 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d 4 12 (b) I = 646 in Ans. I = 646 in4 Ans. PAGE !90 (b) A 400 mm 790 APPENDIX 100 mm A P P E N D790 IX A 790 SOLUTION The cross section can be considered as Athree composite rectangular 790 PPENDIX A GEOMETRIC PROPERTIES OF A areas A, B, and D shown in Fig. A–8b. For the calculation, the centroid x of each of these rectangles is located in the figure. From the table on A G E O M E T R I C the P R Oinside P E R T I Efront S OF A N A R Ethe A moment of inertia of a rectangle about its cover, EXAMPLE A.3 400 mm 1 centroidal axis is I = 12 bh3. Hence, using the parallel-axis theorem for rectangles A and D, the calculations are as y follows: Determine the 100 mm shown in Fig. A ! EXAMPLE A.3 Şekildeki kiriş enkesitinin ağırlık 100 mm Rectangle A : 600 mm merkezinden geçen eksenlerine E O MxE Tve R I Cy P R O P E R T I E SDetermine Ogöre F A N atalet A R Ethe A moments of inertia of the beam’s cross-sectional area G E O MAEPTPREI CN DPI RX OAP E RyG TIES OF AN AREA A P P E N D I X A hesaplayınız. G E O M E T R I C P R O P E R T I E S O F A N A R1E A momentlerini shown in Fig. A–8a about the x and y centroidal axes. 2 3 2SOLUTION 100 mm (a) 1100 mm21300 mm2 400 mm+ 1100 mm21300 mm21200 mm2 The cross secti 12 EXAMPLE A.3 areas A, B, and 9 E A.3 Çözüm: = 1.425110 2 mm4 SOLUTION EXAMPLE A.3 x of each of thes 400 mm 1section The can beof considered ascross-sectional three composite rectangular 2moments 3 y thecross ofmm21100 inertia the beam’s area IDetermine + Ad mm2 +area 1100 mm21300 mm21250 mm22the inside fron y Determine the moments beam’s cross-sectional y = Iof y¿ inertia x of=the 1300 y 400 area mmthe centroid y Determine the moments of inertia of the beam’s cross-sectional areas A, B, and D shown in Fig. A–8b. For the calculation, 12 shownthe in Fig. A–8a about theaxes. x and y centroidal axes. 100 mm centroidal axis 100 mm shown in Fig. A–8a about xofand y centroidal xin 9each 4about Fig. A–8a xarea and y centroidal axes. thesethe rectangles is located in the figure. From the table on mmmoments of inertiashown Determine the of =the beam’s 1.90110 2 cross-sectional mmof 100 mm 100 for rectangles A theaxes. inside front cover, the moment of inertia of a rectangle about its shown in Fig. A–8a about the x400 and y centroidal mm 100 mm 1 SOLUTION 100 mm 200 mm centroidal axis is I = 12 theorem bh3. Hence, using the parallel-axis A SOLUTION SOLUTION 400 mm Rectangle B : Rectangle A: The cross section can be considered as three composite rectangular 250 mm 600 mm for rectangles A and D, the calculations are as follows: 300 mm The cross section can be considered as three composite rectangular 400 mm The cross section can beinconsidered three composite rectangular 1 Fig. A–8b. as SOLUTION areas A, B, and D shown For the calculation, the centroid 3 9 4 100 mm in Fig. A–8b. For Ix the = incalculation, 1600 mm21100 mm2 = (a) 0.05110 2centroid mm areas A, B, and D shown the the centroid 1 B,these and Drectangles shown A–8b. For calculation, x areas The cross section can be considered as A, three composite rectangular of each of is located in the figure. From the the table on 12Fig. Ix = I x¿ + Ady2 = 1 x B of each x Rectangle A : of thesexrectangles is located in the figure. From in thethe table on From the table on 12 600 mm of of these rectangles is located figure. areas A, B, and D shown in Fig. A–8b. For the calculation, the theeach inside front cover, the centroid moment of inertia of a rectangle about its 400 mmfront cover, the inside the moment of inertia aboutofits 1 1of 3 a rectangle the inside front cover, the moment ofusing inertia 3a rectangle 9about 4 its = 1.42511092 mm4 x100 mm 400each mm of these rectangles of is located in figure. From on centroidal axis is I I=y1the the parallel-axis bhtable . Hence, = mm21600 mm2 = 1.80110 2theorem mm 1 the 400 3 (a) mmaxis 11100 3 1100 300ismm centroidal Icentroidal =Ix12 the parallel-axis theorem =bhI3x¿. Hence, +axis Adisy2 using = = 12 mm21300 mm2 + parallel-axis 1100 mm21300 mm21200 mm22 100 mm 250 mm 12 I Hence, using the theorem bh . the inside front cover, the momentfor of rectangles inertia of aArectangle about its 1 and D, calculations are as follows: 1212the mm 200 mm D,using the calculations astheorem follows: 1 D3 A and Iy = I y¿ + Adx2 = 1 for rectangles A9 and are D, the calculations are as follows: y centroidal axis for is Irectangles = 12 the parallel-axis bhmm . Hence, 4 100 12 = 1.425110 2 mm 100 mm 100 mm 100 mm for rectangles A and D, the calculations are Rectangle as A follows: 9 100 mm Rectangle : 2 D1: = 1.90110 2 mm4 600 mm Rectangle A: 3 2 Rectangle A : I = I + Ad = 1300 mm21100 mm2 + 1100 mm21300 mm21250 mm2 600 mm mm 100 600 mm y y¿ x 1 (b)y 1 y2 = 121100 mm21300 Ix = I x¿2 + Ad mm23 + 1100 mm21300 mm21200 mm22 3 2 (a) 200 mm Rectangle A : A 1 I = I + Ad = 1100 mm21300 mm2 + 1100 mm21300 mm21200 mm2 12 9 4 1 x x¿ y 3 2 mm (a) 3 Rectangle B: (a)100 mmAd 2 = = 21.90110 1100 + 1100 mm21300 mm22 12 1100 250 mm Ix = I x¿ mm21300 + Ad mm21300 mm2mm21200 + 1100 mm21300 mm21200 mm22 x = I x¿ + y AIdikdörtgeni: Fig. A–8 y =mm2 9 4 300 mm 12 = 1.425110 2 mm 12 1 9 4 3 = 1.425110 2 mm Ix = I x¿ + Ady2 = 1100 mm21300 mm2 + 1100 mm21300 mm21200 mm22 Ix = 9 4 A 200 mm 1 B: = 12 1.425110 2 mm4 = 1.42511092 mm 2 3 Rectangle 1 x 2 B I = I + Ad = 1300 mm21100 mm2 + 1100 mm21300 mm21250 mm2 250 mm y y¿ x 2 3 2 300 mm I1y = I y¿ + Adx2 = 1 1300 mm21100 + 1100 mm21300 mm21250 mm22 12 y4 1 mm2mm21250 = 1.42511092I mm 2 3 23 3 1291300 + 1100 mm2 = I y¿ mm21100 + Adx =mm2 mm21100 mm2 + 1100 mm21300 mm21250 y =yI y¿ + Adx = Iy 1300 = 1600 mm21100 mm2 = 0.0511092mm2 mm4 y 4 Ixmm21300 Iy = 12 =9 1.90110 1 12 412 2 mm 300 mm 100 mm 2 3 2 250 mm = 1.901109 2 xmm B mm21100 Iy = I y¿ +100 Ad 1300 mm21300 mm21250 mm2 9 4 mm2 + 1100 4 x = mm = 12 1.90110 2 mm = 1.90110 2 mm 200 mm D 1 The moments of inertia for the entire cross thus4 3 section are 9 200 mm = 1100 mm21600 mm2 = 1.8011092 mm I 100 mm y = 1.90110A 2 mm4 300 mm Rectangle D: 250 mm 129 Rectangle B: I = 1.425110 9 9 mm250 mm 00 mm B200 dikdörtgeni: 2 + 0.05110 2 + 1.425110 2 300 mm A x Rectangle B : Rectangle B : 200 mm D 250 mm 1 250 mm 1 300 mm 3 9(b) 4 2 9 4mm2 = 0.05110 2 mm Ix = 1= 1600 mm21100 1100 mm Ix = IAns. 1 x¿ + Ady = 9 4 Rectangle B: 2 mm 32.90110 9 43 12 Rectangle D : I = 1600 mm21100 mm2 = 0.05110 2 mm Ix = x 1600 mm21100 12 x mm2 = 0.05110 2 mm B A 12 12 1 Fig. A–8 9 9 9 3 x B Iy9221=mm 1.90110 2 + 1.8011092 +3 1.90110 2 = 1.425110 2 mm4 14 Ixx = 1600 (b) mm21100 mm2 = 0.05110 2 3 9 mm21300 4 I = I + Ad = 1100 mm21300 mm2 + 1100 mm21200 mm2 12 x x¿ y = 1100 mm21600 mm2 = 1.80110 2 mm I 1 3 12 1 y x 3 9 4 300 mm 9 1 9 4mm2 250 mm 12 2 mm21600 Iy =mm2 2 mm = 1100 5.60110 1100 mm21600 = 1.80110 2 mm4 = 1.80110 2 mm Iy = IAns. 1 300 mmIy = 300 mm Fig. A–8 y¿ + Adx = 250 mm200 mm 1 9 4 12 mm 12 D = 12 3 1.425110 2 9 mm 4 = 1100 mm21600 mm2 = 1.80110 2 mm I 200 mm D y 200 mm D mm 9 12 100 mm 1 = 1.90110 2 mm4 Rectangle D: 2 3 2 Dmm dikdörtgeni: 100 mm 100 Iy = I y¿ +DAd Rectangle : x = 12 1300 mm21100 mm2 + 1100 mm21300 mm21250 mm2 Rectangle D: 1 (b) 2 100 mm The moments o I1x = I x¿ + Ad = 1 1100 mm21300 mm233 + 1100 mm21300 mm21200 mm222 9 4 Rectangle(b)D: = y21.90110 (b) 2 3 2 mm 2 12 I = I + Ad = 1100 mm21300 mm2 + 1100 mm21300 mm21200 mm2 Ix = I x¿ + Ady = x 1100x¿ mm21300 mm2 + 1100 mm21300 mm21200 mm2 y 12 1 Ix = 1.42 12 = 1.425110 9 4 2Fig. A–8 = A–81100 mm21300 mm23 + 1100 mm21300 mm21200 mm22 for the entire cross section are thus 9 2 mm 4 x = I x¿ + Ady Fig. The moments of inertia Fig. IA–8 9 4 = 1.425110 2 mm = 12 1.425110 2 mm = 2.90 1 3 9 9 I1y = I y¿ + Adx22 = 1 1300 mm2 1100 9mm21300 mm21250 mm222 Ix mm21100 = 1.425110 2 A+3 + 0.05110 2 + 1.425110 2 = 1.42511092 mm4 2 3 2 12 1300 = I y¿ mm21100 + Adx =mm2 mm21100 mm2mm21250 + 1100 mm21300 mm21250 mm2 Iy = 1.90 Iy = I y¿ + Adx = Iy 1300 + 1100 mm21300 mm2 9 1 9 412 12 = 1.90110 3 2 4 = 2.90110mm2 2 mm Ans. 2 mm Iy = I y¿ + Adx2 = 1300 mm21100 mm2 + 1100 mm21300 mm21250 9 4 A = 5.60 = 12 1.9011092 mm4 = 1.90110 2 mm 9 9 9 Iinertia 2 entire + 1.80110 2 section + 1.90110 2 y = 1.90110 9 4 The moments of for the cross are thus = 1.90110 2 mm The moments of inertia for the Sonuç olarak: 9 entire 4 cross section are thus The moments of inertia for the entire cross section are 2 mm =9 5.60110 Ans. 9 thus 9 I = 1.425110 2 + 0.05110 2 x 9 + 1.4251109 2 The moments of inertia for the entire section are 9thus 9 cross 9 9 I = 1.425110 2 + 0.05110 2 + 1.425110 2 x Ix = 1.425110 2 + 0.05110 2 + 1.425110 9 4 2 = 2.90110 Ans. 9 9 2 mm4 Ix = 1.42511092 + 0.05110929 + 1.425110 2 4 = 2.90110 2 mm Ans. A = 2.90110 2 mm Ans. 9 9 A Iy = 1.90110992 + 1.80110 9 2 + 1.901109 2 = 2.9011092 mm4 Ans. I = 1.90110 2 + 1.80110 2 + 1.90110 2 9 9 9 y Iy = 1.90110 2 + 1.80110 2 + 1.90110 2 =92 5.60110992 mm44 Ans. Iy = 1.9011092 + 1.8011092 9+ 1.90110 Ans. = 5.60110 2 mm4 = 5.60110 2 mm Ans. = 5.6011092 mm4 Ans. GEOMETRIC PROPERTIES OF AN AREA Ix = I x¿ + Ady = PAGE !91 4 A.3 PRODUCT OFA.3 INERTIA FOR AN OF AREA 9REA 3 PRODUCT INERTIA FOR AN7A A.3 PRODUCT OF INERTIA FOR AN AREA 793 793 EXAMPLE A.4 EXAMPLE A.4 EXAMPLE EXAMPLE A.4 A.4 A.3 Determine PRODUCT OF Ithe NERTIA FOR AN AREA 793 product of inertia of the beam’s cro Determine the product of inertia of the beam’s cross-sectional area, EXAMPLE A.4 Determine the product of inertia of the beam’s cross-sectional area, shownarea, in Fig. A–12a, about the x and y centroidal ax Determine the product of inertia of the beam’s cross-sectional shown in Fig. A–12a, about the x and y centroidal axes. shown in Fig. A–12a, about the x and about y centroidal axes. shown in Fig. the x and y centroidal axes. y Determine theA–12a, product of inertia of the beam’s cross-sectional area, y EXAMPLE A.4 y y ! Şekilde verilen kiriş enkesit alanının PRODUCT NERTIA FOR AN AREAaxes. 793 100 mm yA.3 about y x shown in Fig. A–12a, theOFx Iand y centroidal 100 mm 100 mm 100Determine ve ymm eksenlerine göre atalet hesaplayınız. 100 mm mm çarpım the100 product ofmomentini the beam’s 100 cross-sectional area, y of inertia mm y y y 100 mm shown in Fig.100 A–12a, about the x and y centroidal axes. A 200 mm mm A 200 mm 100 mm 400 mm A 200 mm 250 mm 200 mm 300 mm A 300 mm mm y 250 300 mm 250 mm 300 mm area, A 200 mm 100 mm x B x x 250 mm B 300 mm x B x x B 400 mm 400 mm A 200 mm 100 mm 300 mm 250 mm x 250 mm 250 mm x B 400 300 mmmm 300 mm 20 250 mm 250 mm200 mm D 300 mm 200 mm D 400 mm 100 mm 200 mm D 300 mm 100 mm 100 mm 100 mm x 250 100 mmmm x B 100 mm A 200 mm 100 mm 100 mm 600 mm 200 mm D 600 mm 300 mm 600 mm 250 mm 400 mm 100 mm 100 mm 600 mm 100 mm 300 mm (a) (b 250 mm (a) (b) (a) 200 mm (b) D 600(a) mm x x B (b) Fig. A–12 Fig. A–12 100 mm 100 mm Fig. A–12 (a) 400 mm (b) Fig. A–12 SOLUTION 300 mm SOLUTION 250 mm600 mm SOLUTION Asasinthree Example A.3, the cross section can be co 200 mm Fig. A–12 As in Example A.3, the cross section can be considered D SOLUTION As in Example composite A.3, the sectionareas can A, beB,considered as threeThe coordinates (a) cross (b) composite rectangular and D, Fig. A–12b. 100 mm As in Example 100 A.3, mmthe cross section can be considered as three rectangular areas A, B, and D, Fig. A–12 SOLUTION 400 mm y 400 mm 400 mm Çözüm: oduct of inertia of the beam’s 100 mmcross-sectional 400 mm 2a, about the x and y centroidal axes. x y 100 mm 400 mm 100 mm 400 mm 100 mm100 mm composite rectangular areas A, B, and D, A–12b. The are coordinates for theDue centroid of each of these rectangles are shown for in theExample centroid of each of Fig. these rectangles shown inThe thecoordinates figure. composite rectangular areas A, B, and D,can Fig.be A–12b. A.3, the cross section considered as three Fig. A–12 for the centroidAs of each of these rectangles are shown in the figure. Due is zero to symmetry, to symmetry, the product of inertia of each rectangle about for the centroid of eachareas of these rectangles are A–12b. shown in thecoordinates figure. Duea the product of inertia of each rectang composite rectangular A, B, and D, Fig. to symmetry, the product of inertia of each rectangle is zero about aThe SOLUTION set about of x¿,ay¿ axes that pass through the rectangle’s (b) set of axes that pass through the rectangle’s centroid. Hence, x¿, y¿ to the symmetry,kendi the product of inertia of each rectangle is zero bir dikdörtgenin ağırlık merkezinden geçen eksenlere göre çarpım atalet momenti for of each of these rectangles are shown in the figure. Due set ofHer thatcentroid pass through the rectangle’s centroid. Hence, x¿, Asy¿inaxes Example A.3, the cross section can be considered as three application of the parallel-axis theorem to each of th application of the parallel-axis theorem to each of the rectangles yields set of axes that pass through the rectangle’s centroid. Hence, x¿, y¿ to symmetry, the product of inertia of each rectangle is zero about a simetriden sıfırdır. Fig. A–12 application of the dolayı parallel-axis theorem to each of the rectangles yields composite rectangular areas A, B, and D, Fig. A–12b. The coordinates application the parallel-axis theorem each of thecentroid. rectangles yields set of x¿, y¿of that pass through thetorectangle’s Hence, Rectangle A: Rectangle Aaxes : of these for the centroid of each rectangles are shown in the figure. Due application of the parallel-axis theorem to each of the rectangles yields Rectangle A : to symmetry, theIproduct of+inertia A dikdörgeni: A.3, the cross section can be considered asx¿y¿three Ixy = I x¿y¿ + Adxdy Rectangle Adxdyof each rectangle is zero about a xyA=: I ID, = Ix¿, Ad that pass through the rectangle’s centroid. Hence, xy of x¿y¿y¿+ axes xd ycoordinates ular areas A, B, andset Fig. A–12b. The Rectangle = 0 + 1300 mm21100 mm21- 250 mm21 IxyA:= I0x¿y¿ + Admm21100 + 1300 mm21 - 250 mm21200 mm2 xdy application the parallel-axis theorem the rectangles yields each of these rectangles are in mm21100 the figure.mm21 Due = 0shown + of 1300 9- 250 mm21200 4to each ofmm2 Ixy == I40Admm21100 dy = - 1.5011092 mm4 1.50110 + +1300 mm21-250 mm21200 mm2 product of inertia of each zero x¿y¿ about a x2 mm = rectangle - 1.501109is2 mm 9 = 0 + 1300 mm21100 Rectangle A: centroid. 2 mm4 mm21-250 mm21200 mm2 that pass through the rectangle’s Hence, Rectangle B: Rectangle B=: -1.50110 9 4 : Iof parallel-axis Rectangle theorem to Beach the yields = Ad -1.50110 2Imm = Irectangles + dy Ixy = I x¿y¿ + Adxdy = I + Ad d xy x¿y¿ x xy x¿y¿ x y Rectangle B=: I B dikdörgeni: I+xy1300 + Adxmm21-250 dy x¿y¿ = 0 mm21100 mm21200 mm2 = 0 + 0 + 0+ Adxdy Ixy = I0x¿y¿ Rectangle B: = 0 + 0 = I00 + +0 Ad d = 0 = -1.5011092 mm4 Ixy == x¿y¿ x y = 0 ¿ + Adxdy Rectangle D: Rectangle D: == 00 + 0 + 1300 mm21100 mm21-250 mm2 Rectangle B: Rectangle D: mm21200 = 0 IxyD=: I x¿y¿ Ixy = I x¿y¿ + Adxdy Rectangle Ixy += Ad I x¿y¿ xdy+ Adxdy .5011092 mm4 Ixy = I x¿y¿ + Adxdy = 0 + 1300 mm21100 mm21250 mm21 200 mm2 = 0 + 1300 mm21100 mm21250 mm21Rectangle : I x¿y¿ += Ad D dikdörgeni: IxyD= 0 + xdy0 = 0 + 1300 mm21100 mm21250 mm21 - 200 mm2 9 4 = 0 2 mm =I01.50110 = - 1.5011092 mm4 I9xy == + Ad d mm21250 mm21-200 mm2 x y Ixy = I x¿y¿ + Adx= dy - 1.50110 2 mm4 x¿y¿+ 1300 mm21100 9 for the == 0-1.50110 1300 mm21100 mm21250 mm2 The product of inertia for the entireAcross section of+ inertia crossmm21-200 section is thus 2 mm4 entire RectangleThe D: product = 0The + 0product of inertia for the entire cross is thus 9 9 section 4 4 9 2for mm = Ad =-1.50110 2 the mmentire ] + 0 cross + [ -section 1.50110is2 thus mm4] Ixy = [-1.5011092 mm4] + 0 + [- 1.50110 IxyThe = Iproduct + d1.50110 = 0 of[x-inertia x¿y¿Ixy y 9 4 Ixy = [ - 1.5011092 mm4] + 0 + 9[9- 1.50110 2 mm ] 44 =of[-1.50110 -inertia 3.00110 Ans. = -3.0011092 mm4 22the mm The formm21250 section is92thus = product 0 +I 1300 mm21100 mm2 = mmentire ] +mm21-200 0cross + [-1.50110 mm4] = - 3.0011092xymm49 Ans. Sonuç olarak: Ixy == [-1.50110 2 mm4 9922 mm44] + 0 + [-1.5011092 mm4] = -1.50110 -3.00110 Ans. ¿ + Adxdy A 9 4 = -3.00110 Ans. 2 mmcross section is thus The mm21-200 product of inertia for the entire + 1300 mm21100 mm21250 mm2 Ixy = [-1.5011092 mm4] + 0 + [-1.5011092 mm4] .5011092 mm4 m -3.0011092 mm4 inertia for the entire cross section =is thus -1.5011092 mm4] + 0 + [-1.5011092 mm4] 3.0011092 mm4 Ans. Ans. PAGE !92 A C 796 GEOMETRIC PROPERTIES OF AN AREA 7 9 6 principal APPENDIX A GEOMETRIC PROPERTIES OF AN AREA moments of inertia for the beam’s crossP R O P E RDetermine T I E S O F A N the AREA APPENDIX A GEOMETRIC PROPERTIES OF APPENDIX A LE sectional area shown in Fig. A–15 with respect to an axis passing A.5 through the centroid C. EXAMPLE A.5 Determine the y A.5A RŞekilde EA x¿ sectional area ! EXAMPLE gösterilen kiriş enkesitinin C 100 mm Determine the principal moments of inertia for the beam’s crossy through the ce geçen x’ veiny’ Fig. asal eksenlerine göre y¿ axis passing x¿ ağırlık merkezinden sectional area shown A–15 withthe respect to an Determine principal moments of inertia for the beam’s crossy SOLUTION mm Determine theasal principal moments of inertia for the beam’s crossmomentlerini x¿ hesaplayınız. area shown in Fig. A–15 with respect to an axis passing through centroid The moments atalet and of the inertia of theC.cross sectional section with respect 5 100product mm E O MG E TEROI C I EPSE ROTFI EASNOA I XGA M EPTRRO I CP EPRRTO F RAENA sectional area shown in Fig. A–15 with respect to an axis passing x 400 mm SOLUTION through the centroid C. to the yy¿axes have up1 ! 57.1# through the x, centroid C. been determined in Examples A.3 and A.4. The The moments the principal moments of inertia for the beam’s cross-crossthe principal moments of inertia for the beam’s resultsDetermine are Determine x SOLUTION up1 ! 57.1# to the x, y axe x¿ sectional area shown in Fig. A–15 with respect to an axis passing x¿ sectional area shown in Fig. A–15 with respect to an axis passing C 400 mm The moments and product ofSOLUTION inertia of the cross section with respect 400 mm 57.1# SOLUTION through results are 100 mm u ! "32.9# the C.up !C. 9 through 4x centroid 9 4 9 4 the centroid p2 the cross section with respect The moments inertia Iofythe = 5.60110 2 have mm I determined = -3.00110 2and mmproduct x, y axes been in Examples A.3ofand A.4. of The m CIx = 2.90110 2 mm The moments and productto inertia of the cross section x xy with respect 400 mm toand the A.4. x, y axes and A.4. The u p2 ! C are in Examples A.3 to"32.9# the x, y axesÇözüm: have beenresults determined The have been determined in Examples A.3 Ix = 2.9011092 m 100 mm 400 mm results are 100 mm the SOLUTION x¿ Using Eq. A–11, angles of inclination of the principal axes up ! "32.9# SOLUTION results are 7.1# up1 ! 57.1# 9 of inertia 9section 4 with respect 9 4 mm 600 moments product the y¿ The and are Ix and =x 2.90110 2 mm4 of göre Iof 5.60110 2 mmsection Ixywith = -3.00110 100 mm Theönce moments product inertia ofcross themomentleri cross respect 2 mm y = x ve and y eksenlerine atalet Using Eq. A 9 4 9 4 xDaha x, y axes have been determined in Examples A.3 and A.4. The I = 2.90110 2 mm I = 5.60110 2 mm Ixy = - 3.0011092 mm4 9 to the 4 to 9 4 9 4 100 determined mm the axes have been in Examples Ix 600 = 2.90110 2 mm Iy x,= y5.60110 2 mm Ixy = x- 3.00110 2 mm A.3 andyA.4. The mm Fig. A–15 y¿ are and hesaplanmıştı: 400 mm400 mm 9 resultsresults are are Using the angles -Ixy 2 of inclination of the principal axes x¿ 600 mm Eq. A–11,3.00110 9# tanA–11, 2up =the angles = = Using Eq.-2.22 A–11, the angles of inclination of the principal axes x¿ Fig. A–15Eq. and y¿ 9 x¿ Using of are inclination principal axes 1I [2.901109of 29 -the 5.60110 y2>2A–15 9 x - I 4 Fig. 4 and 2]>2 9 4 y¿ are 9 4 y = 5.60110 2 mm9 4 xy = -3.00110 2 mm9 I = 2.90110 2 mm I I 100 mmy¿ and are x tan 2up = Ix = 2.90110 2 mm Iy = 5.60110 2 mm Ixy = -3.00110 2 mm4 100 mm 9 3.00110 2 2up1 = 114.2° and -I 2uxyp2 = -65.8° tan 2u = of inclination = of the principal = -2.22 3.0011092 9 - Ixy9x¿ p 3.00110 UsingUsing the angles axes 9 -Eq. Ixy A–11, Eq. A–11, the angles inclination of2uthe axes=x¿ 1Ix - 2of Iy2>2 [2.90110 2 =-principal 5.60110 2]>2 tan = - 2.22 p tan 2upand =Asal = konumu9 belirleyelim: = - 2.22 y¿ are 1Ix - Iy2>2 eksenlerin –15 [2.9011092 - 5.6011092]>2 1Iand 2>2A–15, Thus, as shown iny¿IFig. [2.90110 2 - 5.6011092]>2 x yare 2up1 = 114.2° and 2up2 = -65.8° Thus, as shown 9 2up1 = 114.2° and 2up2 = - 65.8° 2up1 = 114.2° -Iand 2u = 65.8° 3.00110 2 9 xy -I p2 3.00110 2 xy u = 57.1° and u = -32.9° tan 2utan = = -2.22 pp1=2u = 9 9 = p2 Thus, in[2.90110 Fig. A–15, 1Ip x as - 1I Ishown 2 - 5.60110 2]>2 92]>2 = -2.22 9 y2>2 [2.90110 2 5.60110 x - Iy2>2 Thus, as shown in Fig. A–15, Thus, as shown in Fig. A–15, The principal moments inertia with and respect to the x¿-65.8° and y¿ axes 2up1of=2u 114.2° 2u up1 = 57.1° up2 = -32.9° and p2 =and 2up2 = -65.8° p1 = 114.2° are determined by using Eq. A–12. The principa up1 = 57.1° and up2 = - 32.9° up1 = 57.1° and up2 = - 32.9° are determined Thus, Thus, as shown inThe Fig.in A–15, principal moments of inertia with respect to the x¿ and y¿ axes as shown Fig. A–15, 2 x¿The y¿ axes moments of inertia with respect to the x¿ and y¿ axes The moments with respect to Eq. the A–12. andprincipal Iy Ix - Iy of Ix +principal areinertia determined by using 2 max = ¢ using ≤momentleri: ; Asal + Ixy are determined by using Eq. A–12. Ix Ix + Iy by Eq. A–12. atalet u = 57.1° and u = -32.9° min are determined p1 2 C 2 up1 = 57.1° and p2 up2 = -32.9° ¢ ; Imax = min 2 C I9x - Iy 2 9 Ix + Iy 9 22 2axes 292 x¿ y¿ principal moments of inertia with respect to the and 2 I+xThe 5.60110 2.90110 2 5.60110 2 ¢ ≤ ; = + I Iy -Imax I Ix + 2.90110 + I I I I y xy x y x y minThe principal with axes 9 x¿ 2 B moments 2 CEq. A–12. 2 ofIinertia max = R +respect = [-3.00110 2]2 ≤and+y¿Ixy ¢ determined ≤ + ;I2xy ¢to the ; ; 2.9011092 + 5.60 by min 2 2 are determined C using 2 A–12. 2 C are by using Eq. 2 C 2 = 2 9 9 2.9011092 - 5.6011092 2 9 9 2.90110 2 + 5.60110 2 9 9 = 4.25110 2 ; 3.29110 2 2.90110292 - 5.60110;92 2 B2.90110 9 9 9 92]2 9 2 R = + [-3.00110 2.90110 2 + I5.60110 2 2 + 5.60110 2 2.90110 2 - 5.60110 2 9 I - Iy 2 x + I = 4.25110 2 ; 3.291 9 2 C 2 92]2 R= + [ - 3.00110 + I;y ¢ Bx Ix ≤- + Iy I 2 2 Iyx ; B R + [ - 3.00110 ; 2] Imax =2 max 2 xy C 2 2 C 2 min ¢ ≤ ; I = + I 2 C 2 xy or min 2 = 4.25110 C 92 2; 3.2911092 or 4.2511092 ; 3.2911092 = 4.25110992 ;2 3.2911092 9 2.9011092 + 5.60110 2 9 9 4 2.9011092 - 5.60110 92 9 9 94 2 Imax = 27.54110 2; mm Imin = 0.960110 2 Rmm+2 [-3.00110 Ans. 92]2 9 2 2.90110 +or5.60110 2B 2.90110 2 - 5.60110 = Ima B R + [-3.00110 = 2] 2 C; 2 or or 2 C 2 9 4 9 9 moment = 47.54110 , Specifically, 7.54110I9max 29 mm Imin2 =mm 0.960110 2 mm44 Ans. = 4.2511092the ; maximum 3.29110 2 4 9 Imaxof= inertia, 9 4 9 Specifically, th 9 I = 7.54110 2 mm I = 0.960110 2 mm Ans. = 4.25110 2 ; 3.29110 2 max min I = 7.54110 2 mm Imin = 0.96011092 mm4 Ans. occurs with respect to the x¿ axis (major axis), since by inspection maxmost A occurs with res of the cross-sectional area is farthest from this axis. To show this, Specifically, the away maximum moment Imax = 7.5411092 mm4, 9of inertia, 4 or of 9 2 mm Specifically, the maximum moment of inertia, Imax = 7.54110 Specifically, the, maximum moment of inertia, Imax = 7.54110 2 mm4,the cross-sec substitute theordata with u = with 57.1°respect into thetofirst occurs theof x¿ Eq. axisA–10. (major axis), since by inspection most substitute the d occurs withArespect to the x¿ axis (major axis), since by inspection most with x¿ To axisshow (major 9 4 9respect 4 to the cross-sectional area is occurs farthest away from thisthe axis. this,axis), since by inspection most Imaxof 7.54110 2 mm I = 0.960110 2 mm Ans. of the cross-sectional area is = farthest away from this axis. To show this, 9 9 4 of theinto area isAns. farthest away from this axis. To show this, Imax = 7.54110 2 mmu4 min Imin =cross-sectional 0.960110 the data 57.1° the first2ofmm Eq. A–10. substitute the data with u =substitute 57.1° into the firstwith of Eq.=A–10. substitute the data with u = 57.1° into the first of Eq. A–10. Specifically, the maximum moment of inertia, Imax = 7.5411092 mm94, 4 = 7.54110 2 mm , Specifically, the maximum moment of inertia, I occurs with respect to the x¿ axis (major axis), since bymax inspection most occurs with respect to the x¿ axis (major axis), since by inspection most of the cross-sectional area is farthest away from this axis. To show this, of the cross-sectional area is farthest away from this axis. To show this, substitute the data with u = 57.1° into the first of Eq. A–10. substitute the data with u = 57.1° into the first of Eq. A–10. 1 2 PAGE !93 308 CHAPTER 6 BENDING 08 CHAPTER 6 BIENNGD I N G 3EXAMPLE B6E N D 6.16 BENDING G ER A.4 shown, such A.4 or or A.5), A.5), the the principal principal axes axes yy and and zz are are oriented oriented as asthat shown, theysuch represent the m that they represent the minimum and maximum principal principal moments moments of inertia, I = 0.960110-32 m44 and I = 7.54110-3 -32 m44, respectively. of inertia, Iyy = 0.960110-3 respectively.y zz Determine the normal s Determine the normal stress at point P and the orientation orientation of of the the neutral axis. neutral axis. SOLUTION SOLUTION The Z-section shown in Fig. 6–34a is subjected to the bending moment The Z-section shown in Fig.6.16 6–34a is subjected to the bending moment For use of Eq. 6–19, it EXAMPLE # For use of Eq. 6–19, it is important that the z axis represent the ! Şekilde gösterilen Z enkesite M=20 kNm eğilme momenti tesir represent the of Using the methods of Appendix A (see Example M = 20 kN m. of M = 20 kN # m. Using the methods of Appendix A (see Example principal axis for the max principal axisy for maximum moment of inertia. (Note most of that most of A.4 A.5), the principal axes y and zasal are oriented as that shown, such etmektedir. veshown zthe asal atalet momentleri The Z-section shown inprincipal Fig. 6–34aaxes is subjected the bending moment The or Z-section ineksenlerdir Fig. 6–34a isve subjected to the bending moment A.4 or A.5), the yFig. and z to are oriented as shown, such the area is located furthes The Z-section shown in Fig. 6–34a is subjected to the bending moment The Z-section shown in 6–34a is subjected to the bending moment z area is represent located furthest from this axis.) z¿ and that they the Example minimum maximum principal of M that Using the methods ofand Athe (see = 20they kN #represent m. of #(see z Appendix #Using z¿ of Using the methods of Appendix A (seemoments Example M = 20 kN m. maximum principal moments # the methods ofExample Appendix A (see M20=the 20minimum kN m. Using of the methods of Appendix A Example M = kN m. 4 -3 100-3mm TheA.4 Z-section shown in 100 Fig.mm 6–34a subjected the bending moment 4 z aretooriented -3 4 A.5), or theIprincipal axes-3is2ythe and as such of inertia, and respectively. Iy the = 0.960110 2axes m 7.54110 2 m4as ,Internal Iz z= are Comp A.4 or principal y and oriented shown, Moment such ofA.5), inertia, and Iaxes =or0.960110 m 2shown, , respectively. or A.5), principal y and zmare oriented as shown, such y A.4 z =axes Moment Components. From Fig. 6–34a, A.4 A.5), the principal y7.54110 and zInternal are oriented as shown, such # of that Using the methods of Appendix A (see Example M =they 20 kN m. M P represent the minimum and maximum principal moments z Determine the normal stress at point P and the orientation of the 32.9# that they represent the minimum and maximum principal moments Determine the normal stress atMpoint P and the orientation of the moments P that they represent minimum and maximum principal moments zthe that they represent minimum and principal 32.9#the -3 -3 4maximum A.4oforinertia, A.5), the axes zIzare oriented such -3 4 -3 4 and Iy axis. =principal 0.960110 2 myI4 and =-37.54110 mshown, ,I respectively. -34 4 2as 4 neutral axis. -3Iy -3 42 0.960110 of inertia, and =M 2# m m Iz == 7.54110 neutral of inertia, and respectively. = 0.960110 2 m = 7.54110 m , = 20 mnoktasındaki sin 57.1° 16.79 kN2# gerilmeyi m , respectively. My = 20 k of inertia, and respectively. I = 0.960110 2 m = 7.54110 2 m , I y z 400 mm kN P y olarak verilmiştir. normal y z hat they represent the minimum and maximum principal moments u ! 57.1# mm stress at point Determine the400 normal P and theatorientation of the the u ! 57.1# theorientation normalofstress at point P and theM orientation of the M = 20 k ! 20 kN"m Determine the normal stress point P and of the -3 4 Determine normal stress at-3M2point PDetermine and the orientation the kN"m ! # # z of neutral inertia, axis. and Iy = 0.960110 2 mthe m420 , respectively. Iz = 7.54110 ve tarafsız eksenin yönelimini belirleyiniz. y¿ M = 20 kN m cos 57.1° = 10.86 kN m z y¿ neutral axis. neutral neutral axis.axis. SOLUTION SOLUTION Determine the normal stress at point P and the orientation of the 100 mm 100 mm For use of Eq. 6–19, it is important that the zFor axisuse represent the it is important that the z axis represent ofStress. Eq. 6–19, the neutral axis. Bending Stress. The point P must must Bending The y300and z coordinates of point P SOLUTION mmmoment of inertia. (Notebe 6 SOLUTION principal axis for the maximum moment of inertia. (Note that most of principal axis for the maximum that most of first. No 300 mm SOLUTION determined 6 SOLUTION of P P are are determined first. Note that (a)the y¿, z¿Mycoordinates of For use Eq. 6–19, it furthest is important thatM the yz axisbe represent the y axis 1-0.2 For use of Eq. 6–19, it is important that the z represent the yis important (a) the of area isFor located from this axis.) For use of Eq. 6–19, it that the z axis represent the m, 0.35 the area is located furthest from this axis.) use of Eq. 6–19, it isz important 1that the 0.35 z axis represent thecolored triangles from the construction construction m2. Using Using the - 0.2 z¿ SOLUTION principal axis for the maximum moment of inertia. (Note thatm, most ofm2. principal axis for the maximum moment of inertia. (Note that most principal axis for the maximum moment of inertia. (Note that most of shown in Fig.of6–34b, we h principal axis for the maximum of inertia. (Note that shown in the Fig. 6–34b, wemost haveof Forthe use of isEq. 6–19, it 100 is Components. important the z moment axis6–34a, represent mm area located furthest from this that axis.)From Internal Moment Fig. the area is located furthest from this axis.) Internal Moment Components. From Fig. 6–34a, the area is located furthest from this axis.) zfrom this axis.) themaximum area is located z¿furthest zaxis for the z principal of inertia. (Note that most of yP = - 0.35 sin P moment m 32.9# Mz Çözüm: yP = - 0.35 sin 32.9° - 0.2 cos 32.9° = - 0.3580 m 100 mm Internal Moment Components. From Fig. 6–34a, # # 6.5 U NSYMMETRIC BENDING he area is located furthest from this axis.) Internal Moment Components. From Fig. 6–34a, M = 20 kN m sin 57.1° = 16.79 kN m Internal Moment Components. From 6–34a,My = 20 kN # m sin 57.1° = 16.79 kN # m y Moment Internal Components. From Fig. Fig. 6–34a, zP = 0.35 cos Mz P zP = 0.35 cos 32.9° - 0.2 sin 32.9° = 0.1852 m 400 mm 32.9# z kN"m Mz !M 20 M u ! 57.1# 2.9# # # M = 20 kN m cos 57.1° = 10.86 kN m # # M ! 20 kN"m z nternal Components. From Fig. 6–34a, = 20 kN m sin 57.1° = 16.79 kN m y¿ MomentM # # # y # mform #zym== 2020and M kN the mcos sin 57.1° 57.1°algebraic 16.79 kN kN m mApplying Eq. kN m == 10.86 #kN M =x,kN 20 siny¿57.1° = 16.79 y 6–17, y, z#kN a=right-handed system, proper My The = yu20! maxes sin 57.1° 16.79 kN6–17, mM 400 mm Applying Eq. 57.1# 20 u ! 57.1# # # ! kN"m 57.1# M ! 20 kN"m M = 20 kN m cos 57.1° = 10.86 kN m # # M ! 20 kN"m signs must be assigned to the moment components and the coordinates z y¿ # # Mzm= 20 kN m cos 57.1° = 10.86 kN m M !M 20 kN"m #The #kN M 20 m #57.1° cos = 10.86 kN = 20 kN100 m sin 16.79 m y¿57.1° mm Bending Stress. y57.1° coordinates of= M point P #M must M z Mz yP M = z20=zkN m kN cos 10.86 kN m z and zPPthe y¿ y y¿ yis P Stress. y zz yP Bending Thecase, y the andresulting z coordinates P must+ y P when applying this equation. When this stress ofsPpoint = s = + be determined first. Note that the coordinates of P are y¿, z¿ kN"m P #ymmm #m P 300 I are Iy 6 Mz = 20 kN100 coswill 57.1° = 10.86ifkN tensile it isof positive and compressive ifNote it isy negative. be the determined that y¿, z¿ coordinates Iconstruction Ifirst. yBending y The and z be coordinates point P [(sx)max zz must Stress. The and the z coordinates of pointof PPz must Using the triangles from 1 - 0.2Stress. m, 0.35 m2. M Bending Stress. The z Bending coordinates of y point P must y y and (a) colored Bending Stress. The y and z coordinates of point P must y 3 3 3 # m21300 mm P noktasının y-z eksenlerindeki koordinatları: z¿ # # 3 3 Using the colored triangles from the construction 1-0.2 m, 0.35 m2. be determined first. Note that the coordinates of P are y¿, z¿ 110.86(10 ) N z¿ z [(s ) " (s ¿ ) ] 116.79(10 110.86(10 m21 m210.1852 m2 z z¿ be determined first. Note that the coordinates of P are y¿, z¿ shown6in Fig. 6–34b, we have x max x max 116.79(10 110.86(10 ) N m210.3580 m2 ) N m210.1852 m2 z that be determined first. Note the of P0.200 arem + y¿, z¿ coordinates Mthat be determined first. Note the coordinates of P are y¿, z¿ a Orientation of the Neutral Axis. The angle of the = y (a) = M 0.200 m 0.350 m y Using the colored triangles from the construction 1 0.2 m, 0.35 m2. Bending The y 0.200 and mz coordinates point P m, must y 0.350 shown in Fig. 6–34b, we have My Stress. y 0.350 m mthe of -3 4 -3 construction colored triangles from the 10.2 0.35 m2. Using -3 4the y 7.54110-32 m Using colored triangles from construction 1-0.2 m, 0.35 m2. 7.54110 m44 N Using the colored triangles from the the construction 1-0.2 m, 0.35 m2.the 22 m neutral in 32.9° Fig. 6–32d can be by applying Eq. 6–17 0.960110-322 m 32.9# yP = -32.9# 0.35 sin 32.9° - axis 0.2 cos = shown -of0.3580 mdetermined 32.9# in Fig. 6–34b, we have be shown determined first. Note that coordinates P are y¿, z¿ P P in Fig. 6–34b, we have P Fig. shown in 6–34b, we in 6–34b, we N stress s have = 0,have with since the by definition noyPnormal acts on A = -0.35 sin 32.9° - the 0.2 neutral cos 32.9° = -0.3580 m MPa N = 3.76 = 3.76 3.76 MPa MPa Ans. N Using theFig. colored triangles construction -0.2 m, 0.35 m2. shown = Ans. x z-P0.35 = 0.35 cos 32.9° 0.2 sinfrom 32.9° =0.3580 0.1852 m z y = sin 32.9° 0.2 cos 32.9° = m axis. We have P y = 0.35 sin 32.9° 0.2 cos 32.9° = 0.3580 m [(sx)max 32.9# 32.9# hown in Fig. 6–34b, we havey =yP-=0.35 -0.35 sin 32.9° - cos 0.2 32.9° cos 32.9° =P -0.3580 32.9# a sin 32.9° - 0.2 = z-0.3580 mcosm32.9° 0.35 - 0.2The sin 32.9° P P =of Orientation of Neutral Axis. The angle =uu 0.1852 isOrientation shown in of Neut = 57.1° 57.1°mis Orientation Neutral Axis. angle shown in = Applying 6–17, z =Eq. 0.35 cos 32.9° - 0.2 sin 32.9° = 0.1852 mFig. 6–34a.zPThus, [(sx)max ! ¿x)max6–34a. ] 0.35 cos 32.9° - 0.2 sin 32.9° = 0.1852 m(sFig. Thus, a ! 85.3# I= MThus, ! = cos 0.35 cos 32.9° - sin 0.2 32.9° sin 32.9° = 0.1852 yP =P -0.35 sin 32.9° 0.2 =- -0.3580 m Fig. 6–34a. y zm m ! 85.3# 85.3# zP =-zP0.35 cosaa32.9° 32.9° 0.2 = 0.1852 Applying Eq. 6–17, y¿ y¿ z y = M z y¿ M y y P PP noktasındaki normal gerilme: -3 Applying Eq.z 6–17, MzI6–17, 7.54110-3 m44 y 32.9° 0.26–17, sin 32.9° = 0.1852 m Applying Eq. sP = z-P = 0.35 +cos 7.54110 22 m Applying (d) Applying Eq.-Eq. 6–17, M z M y tan a = tan 57.1° 57.1° tan a = B y P Iz M I z P tan a = B y RR tan -3 -3 44 Mz yP y zP 0.960110 2 m M z s = + M y y P P3 0.960110 2 m z P M z M y 3 y P Applying Eq. 6–17, M z z P- 0.3580 + M IzM sin Ithen y P M Since M cos usand z y# Pm21 110.86(10 )N m2z = 116.79(10 )=NM m2 zP = -# m210.1852 + u, y y = P A A sP =sIPy-= - I+ + (b) z Iz 4 3 Iy a A + = Ia = = 85.3° 85.3° Ans. a = I y y -3 4 -3 Ans. z y Iz Iy (b)z¿ My3 zP 7.54110 y Mz yP 110.86(10 ) N # m21-0.3580 m2 (b) 116.79(1033) N # m210.1852 m2 z3 0.960110 2 m 2 m # # 3 116.79(10 0.3580 m2 ) N m210.1852 m2 # # z¿ = -z¿ 110.86(10 + z ) N m213 3 116.79(10 110.86(10 ) N m21 0.3580 m2 ) N m210.1852 m2 z # # + = 0.200 3 m #) N 0.350 3 I ) m The neutral axis is oriented oriented as shown in Fig. 6–34b. -3 The 116.79(10 110.86(10 N m210.1852 -3 m2 4 m2 + m21-0.3580 = - Iz= z3.76 MPa 110.86(10 )m N m21-0.3580 m2 m2116.79(10 ) Nz#axis m210.1852 Iy The is as shown Fig. 6–34 Fig. 6–34+ in Fig. 6–34b. - 4neutral 0.200 0.960110 -32 m44 neutral axis(sis) orie Ans. m 0.960110-3= 4 tan u -3 z -3224m = ¢ 7.54110 ≤ m4 PFig. 6–34-3 0.350 + +2 my = - = -3-232.9# x max -34 43 7.54110 m 0.960110 2 m 0.3500.350 m m 3 7.54110 7.54110 Iy -32 m42 m 32.9# 7.54110 2 m 2 m ) N # m210.1852 0.960110 116.79(10 110.86(10 ) N N# m21-0.3580 m20.960110 P m2 = 3.76 MPa Ans. Orientation Axis. The angle u == 57.1° isAns. shown in + = -= 3.76 MPa N -3 of4 Neutral (6–18) 3.76 MPa Ans. =2 m 3.76 0.960110-32 m4 = 3.76 MPaMPa Ans.Ans. Fig.7.54110 6–34a. Thus, 32.9# 32.9# Orientation of Neutral Axis. The angle is shown u = 57.1° Orientation of Neutral Axis. The angle u = 57.1° isOrientation shown in of Neutral Axis. The angle u = 57.1° is showninin y¿ 32.9# -3 Axis. 4Axis. = 3.7632.9# MPa Ans. Orientation of Neutral The angle is shown = 57.1° Orientation of Neutral The angle is shown in in u = u57.1° 7.54110 2 m Fig. 6–34a. Thus, a ! 85.3# Fig. 6–34a. Thus, Fig. 6–34a. Thus, z a ! 85.3# This equation defines the neutral axis for the cross section. Since the Fig. 6–34a. Thus, tan a = tan 57.1° B R y¿ a ! 85.3# Fig. 6–34a. Thus, y¿ a ! 85.3# -3 y¿a Orientation y¿ ofTarafsız Neutral Axis. The angle is = shown in 57.1° 0.960110 m=4 is -3 44 -3of 4 2u y¿ slope this line tan then y>z, eksenin yönelimi: -3 7.54110 7.54110 2 m 7.54110 22m m -3 -342 m4 (sx)max Fig. 6–34a. Thus, tan a = B tan aa == BB tan7.54110 57.1° R tan 57.1° R7.54110 2m A tan -3 4 -3 44 R tan 57.1° a = 85.3° Ans. tan a = tan 57.1° -3 B R y 0.960110 2=mB 0.960110 2 m tan a tan 57.1° R -3 4 0.960110 2 m -3 0.960110 0.960110 2 m42 m 7.54110-32 m4 A (e) The neutral axis is oriented as shown in Fig. 6–34b. Iz AA tan a = tan 57.1° R a B= 85.3° -3(b) 85.3° Ans. y 85.3° Ans. 4 a = 85.3° yy (6–19) tan aAns. = tan u aa ==Ans. A yA Ans. 0.960110 (b) 2am= (b) 85.3° y Iy The neutral axis is oriented as shown in Fig. 6–34b. ! The neutral axis The neutral axisisisoriented orientedas asshown shownin inFig. Fig.6–34b. 6–34b. Fig.6–34 6–34 Fig. The neutral axis is oriented as shown in Fig. 6–34b. a = 85.3° Ans. The neutral axis is oriented as shown in Fig. 6–34b. 6–34 y 6.16 " The neutral axis is oriented as shown in Fig. 6–34b. Here it can be seen that unless Iz = Iy the angle u, defining the direction of the moment M, Fig. 6–32a, will not equal a, the angle defining the inclination of the neutral axis, Fig. 6–32d. (s¿x)max Important Points (s¿x)max • The flexure formula can be applied only when bending occurs • about axes that represent the principal axes of inertia for the cross section. These axes have their origin at the centroid and are oriented along an axis of symmetry, if there is one, and perpendicular to it. !94 some arbitrary axis, then the If the moment is appliedPAGE about moment must be resolved into components along each of the principal axes, and the stress at a point is determined by superposition of the stress caused by each of the moment (f) Fig. 6–32 (cont.)

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