Scilab Textbook Companion for
Vector Mechanics for Engineers: Stastics And
Dynamics
by F. P. Beer, E. R. Johnston, D. F. Mazurek,
P. J. Cornwell And E. R. Eisenberg1
Created by
Akshatha Nayak
M.Tech
Electrical Engineering
Bits Pilani
College Teacher
None
Cross-Checked by
Lavitha
June 9, 2016
1 Funded
by a grant from the National Mission on Education through ICT,
http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab
codes written in it can be downloaded from the ”Textbook Companion Project”
section at the website http://scilab.in
Book Description
Title: Vector Mechanics for Engineers: Stastics And Dynamics
Author: F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell And E.
R. Eisenberg
Publisher: McGraw-Hill, NY
Edition: 9
Year: 2007
ISBN: 9780073529400
1
Scilab numbering policy used in this document and the relation to the
above book.
Exa Example (Solved example)
Eqn Equation (Particular equation of the above book)
AP Appendix to Example(Scilab Code that is an Appednix to a particular
Example of the above book)
For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means
a scilab code whose theory is explained in Section 2.3 of the book.
2
Contents
List of Scilab Codes
4
2 Statics of particle
5
3 Rigid bodies equivalent systems of forces
16
4 Equilibrium of rigid bodies
22
5 Distrubuted forces centroids and centers of gravity
31
6 Analysis of structures
39
7 Forces in beams and cable
47
8 Friction
60
9 Distributed forces Moment of Inertia
67
10 Method of virtual work
70
3
List of Scilab Codes
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
3.1
3.2
3.3
3.4
3.6
3.7
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
5.1
5.2
5.7
5.9
Determine the resultant . . . . . .
Tension in ropw . . . . . . . . . .
Resultant of forces . . . . . . . . .
Tension of Tab and Tac . . . . . .
Force . . . . . . . . . . . . . . . .
Drag force . . . . . . . . . . . . .
Resultant force on AB and Ac . .
resultant of AB and AC . . . . . .
Vertical force . . . . . . . . . . . .
Moment of force . . . . . . . . . .
Moment of force . . . . . . . . . .
magnitude of force and lambda . .
Couple M equivalent to two couple
Distance from the shaft . . . . . .
Angle and degee . . . . . . . . . .
Angle and degee . . . . . . . . . .
Reaction and direction . . . . . . .
Reaction and direction . . . . . . .
Angle and degee . . . . . . . . . .
Tension and angle . . . . . . . . .
Reaction . . . . . . . . . . . . . .
Reaction and direction . . . . . . .
Tension in vector form . . . . . . .
coordinates . . . . . . . . . . . . .
centroid . . . . . . . . . . . . . . .
Coordinates of centroid . . . . . .
Mass of steel . . . . . . . . . . . .
equivalent concentrated mass . . .
4
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5
6
7
8
9
9
10
12
16
17
18
19
19
20
22
23
24
24
25
26
27
28
29
29
31
32
33
34
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
5.10
5.11
5.12
6.1
6.2
6.3
6.4
6.5
6.6
7.1
7.2
7.3
7.4
7.5
7.8
7.9
7.10
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
9.4
9.5
9.7
10.3
10.4
Reaction and direction . . . . . .
Coordinates of centroid . . . . .
components of centroids . . . . .
force . . . . . . . . . . . . . . . .
force . . . . . . . . . . . . . . . .
Calculation of force . . . . . . .
components of force . . . . . . .
components of force . . . . . . .
Force . . . . . . . . . . . . . . .
free body diagram we . . . . . .
free body diagram . . . . . . . .
free body diagram . . . . . . . .
free body diagram . . . . . . . .
free body diagram . . . . . . . .
free body diagram . . . . . . . .
free body diagram . . . . . . . .
free body diagram . . . . . . . .
value of friction force . . . . . .
Force P to prevent block . . . .
Minimum distance . . . . . . . .
force required . . . . . . . . . .
Couple required to loosen clamp
force required . . . . . . . . . . .
Tension . . . . . . . . . . . . . .
Torque . . . . . . . . . . . . . .
Area of plate . . . . . . . . . . .
Principle moment of inertia . . .
Principle moment of inertia . . .
Force exerted by each cylinder .
Angle . . . . . . . . . . . . . . .
5
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35
36
37
39
40
41
42
43
44
47
48
50
52
54
56
57
58
60
61
62
62
63
64
65
66
67
68
69
70
70
Chapter 2
Statics of particle
Scilab code Exa 2.1 Determine the resultant
1
2
3
4
5
6
7
8
clc ;
// Page 22
// G e t t i n g r e s u l t a n t o f two v e c t o r s
P =40; // N
Magnitude o f v e c t o r P
Q =60 // N Magnitude o f v e c t o r Q
// i m a g i n e t r i a n g l e f o r t r i a n g l e law o f v e c t o r s
B =180 -25; // d e g r e e , A n g l e b e t w e e n v e c t o r P and
vector Q
9
10 //R− R e s u l t a n t v e c t o r
11 B = B * %pi /180; // c o n v e r s i o n i n t o r a d i a n
12 //Rˆ2=Pˆ2+Qˆ2−2∗P∗Q∗ c o s (B) ;
C o s i n e Law
13 R = sqrt ( P ^2+ Q ^2 -2* P * Q * cos ( B ) ) ; // N
14
15 printf ( ” M a g i n i t u d e o f R e s u l t a n t i s R= %. 2 f N\n ” ,R ) ;
16
17
18 //A− A n g l e b e t w e e n R e s u l t a n t and P v e c t o r , Unknown
19
20 // s i n (A) /Q == s i n (B) /R s i n e law
6
21
22 A = asin ( Q * sin ( B ) / R ) ; // r a d i a n
23
24
25 A = A *180/ %pi ; // // C o n v e r s i o n i n t o d e g r e e
26
27 alpha = A +20; // d e g r e e
28 printf ( ” A n g l e o f R e s u l t a n t
v e c t o r R with x a x i s
is
%. 2 f D e g r e e s \n ” , alpha ) ;
Scilab code Exa 2.2 Tension in ropw
1 clc ;
2 // Page 29
3 R =25; // kN
Magnitude o f R e s u l t a n t v e c t o r
4 alpha =45; // d e g r e e
5 // T1 and T2 a r e t e n s i o n s i n r o p e 1 and r o p e 2
respectively
6 A =30; // d e g r e e , A n g l e b e t w e e n v e c t o r T1 and
resultant
7 B = alpha ; // d e g r e e , A n g l e b e t w e e n v e c t o r T2 and
resultant
8 C =180 -( A + B ) ; // d e g r e e , A n g l e b e t w e e n v e c t o r T1 and
T2
9
10
11 // c o n v e r s i o n o f a n g l e s i n t o r a d i a n
12 A = A * %pi /180;
13 B = B * %pi /180;
14 C = C * %pi /180;
15
16
17 // s i n (A) /T2 == s i n (B) /T1 == s i n (C) /R . . . . . . . . . . . . . .
s i n e law
18
7
19 T1 =( R * sin ( B ) ) / sin ( C ) ; //kN
20 T2 =( R * sin ( A ) ) / sin ( C ) ; //kN
21
22
23 printf ( ” T e n s i o n i n r o p e 1 i s T1=%. 2 f kN and i n r o p e
2 i s T2=%. 2 f kN \n ” ,T1 , T2 ) ;
24
25
26
27
28
29
30
31
32
33
34
35
36
// Minimum v a l u e o f T2 o c c c u r s when T1 and T2 a r e
p e r p e n d i c u l a r t o e a c h o t h e r i . e C=90 d e g r e e
C =90; // d e g r e e
A =30; // d e g r e e
B =180 -( A + C ) ; // d e g r e e s
alpha = B ; // d e g r e e s
B = B * %pi /180; // r a d i a n
T2 = R * sin ( B ) ; // kN
T1 = R * cos ( B ) ; //kN
printf ( ”Minimum t e n s i o n i n r o p e 2 i s T2=%. 2 f kN \n ” ,
T2 ) ;
printf ( ” c o r r o s p o n d i n g T1=%. 2 f kN \n ” , T1 ) ;
printf ( ” a l p h a=%. 2 f d e g r e e s ” , alpha ) ;
Scilab code Exa 2.3 Resultant of forces
1
2
3
4
5
6
7
8
9
10
11
clc ;
// p a g e 31
F1 =150; // N
F2 =80; // N
F3 =110; //N
F4 =100 // i n N
F1x =129 // i n N
F2x = -27.4
F3x =0
F4x =96.6
8
12
13
14
15
16
17
F1y =75
F2y =75.2
F3y = -110
F4y = -25.9
Rx = F1x + F2x + F3x + F4x ; //N H o r i z o n t a l component o f R−
resultant
18 Ry = F1y + F2y + F3y + F4y ; //N V e r t i c a l component o f R−
resultant
19
20
21
22
23
24
//R=Rx i +Ry j
printf ( ”R= %. 2 f i + %. 2 f j \n ” , Rx , Ry ) ;
alpha = atan ( Ry / Rx ) ; // Radian , A n g l e made by r e s u l t a n t
w i t h +ve x a x i s
25 alpha = alpha *180/ %pi ; // C o n v e r s i o n i n t o d e g r e e s
26
27 R = sqrt ( Rx ^2+ Ry ^2) ; // N , Magnitude o f r e s u l t a n t
28 printf ( ” a l p h a= %. 2 f d e g r e e s and R= %. 2 f N” , alpha , R ) ;
Scilab code Exa 2.4 Tension of Tab and Tac
1 clc ;
2 // p a g e 38
3 W =3500; // l b
weight of automobile
4 alpha =2; // d e g r e e
5 // TAB and TAC a r e t e n s i o n s i n c a b l e AB and c a b l e AC
respectively
6 A =90+30; // d e g r e e
, A n g l e b e t w e e n v e c t o r T1 and
resultant
7 B = alpha ; // d e g r e e
, A n g l e b e t w e e n v e c t o r T2 and
resultant
8 C =180 -( A + B ) ; // d e g r e e , A n g l e b e t w e e n v e c t o r T1 and
T2
9
9
10
11 // c o n v e r s i o n o f a n g l e s i n t o r a d i a n
12 A = A * %pi /180;
13 B = B * %pi /180;
14 C = C * %pi /180;
15
16
17 // s i n (A) /TAB == s i n (B) /TAC == s i n (C) /W
..............
s i n e law
18
19
20 TAB =( W * sin ( A ) ) / sin ( C ) ; //N
21 TAC =( W * sin ( B ) ) / sin ( C ) ; //N
22
23 printf ( ” T e n s i o n i n c a b l e AB i s TAB=%. 2 f
C a b l e AC
l b and i n
i s TAC=%. 2 f l b \n ” ,TAB , TAC ) ;
Scilab code Exa 2.5 Force
1 clc
2 // p a g e 39
3 mass =30; // kg
4 W = mass *9.81; // N, Weight o f p a c k a g e
5 alpha =15; // d e g r e e
6 alpha = alpha * %pi /180; // C o n v e r s i o n i n t o
7 F = W * sin ( alpha ) ; //N
8 printf ( ”F= %. 2 f N” ,F ) ;
Scilab code Exa 2.6 Drag force
1 clc ;
2 // p a g e 39
10
radian
3
4
5
6
7
8
9
10
11
12
13
14
15
16
alpha = atan (7/4) ; // r a d
beta = atan (1.5/4) ; // r a d
T_AB =200; //N t e n s i o n i n c a b l e AB
T_AE = -300; //N, t e n s i o n i n c a b l e AE
. . . Equillibrium
// R= T AB+T AC+T AE+F D=0
Condition . . . . . . . . . . . 1
T_ABx = - T_AB * sin ( alpha ) ; // Xcomponent o f T AB
T_ABy = T_AB * cos ( alpha ) ; //Y component o f T AB
// T ACx=T AC∗ s i n ( b e t a ) ; Xcomponent o f T AC
// T ACy=T AC∗ c o s ( b e t a ) ; Y component o f T AC
// Sum Fx =0 g i v e s −T AB∗ s i n ( a l p h a ) N + T AC∗ s i n (
b e t a ) +F D = 0 . . . . . . . . . . 2
17 //Sum Fy=0 g i v e s T AB∗ c o s ( a l p h a ) N +T AC∗ c o s ( b e t a ) +
T AE = 0 . . . . . . . . . . . . . . . . 3
18
19 T_AC =( - T_AB * cos ( alpha ) - T_AE ) / cos ( beta ) ; //N, From 3
20
21 F_D = T_AB * sin ( alpha ) - T_AC * sin ( beta ) ; //N, From 2
22
23 printf ( ” V a l u e o f d r a g f o r c e i s F D=%. 2 f N and
t e n s i o n i n c a b l e AC i s T AC= %. 2 f N” ,F_D , T_AC ) ;
Scilab code Exa 2.7 Resultant force on AB and Ac
1
2
3
4
5
6
7
// p a g e 50
clc ;
dx = -40; //m
dy =80; //m
dz =30; //m
f =2500; //N, M a f n i t u d e o f f o r c e F
d = sqrt ( dx ^2+ dy ^2+ dz ^2) ; //m, t o t a l d i s t a n c e o f v e c t o r
11
8
9
10
11
12
13
AB
//F=f ∗ lambda , lambda − u n i t
c a l c u l a t e e a c h component
vector
Fx = f * dx / d ; //N , X component
Fy = f * dy / d ; //N , Y component
Fz = f * dz / d ; //N , Z component
printf ( ” Component o f F
;
14 printf ( ” Component o f F
;
15 printf ( ” Component o f F
;
16 printf ( ”We may w r i t e F
. 2 f k \n ” ,Fx , Fy , Fz ) ;
17
18
19
20
21
22
23
24
25
26
27
v e c t o r= AB/ d . So we can
by m u l t i p l y i n g t h i s u n i t
of F
of F
of F
a l o n g X a x i s i s %. 2 f N\n ” , Fx )
a l o n g Y a x i s i s %. 2 f N\n ” , Fy )
a l o n g Z a x i s i s %. 2 f N\n ” , Fz )
a s \n F = %. 2 f i + %. 2 f j + %
thetax = acos ( Fx / f ) ; // r a d i a n , a n g l e w i t h +ve x a x i s
thetay = acos ( Fy / f ) ; // r a d i a n , a n g l e w i t h +ve y a x i s
thetaz = acos ( Fz / f ) ; // r a d i a n , a n g l e w i t h +ve z a x i s
// C o n v e r s i o n o f a n g l e s i n t o d e g r e e
thetax = thetax *180/ %pi ; // d e g r e e
thetay = thetay *180/ %pi ; // d e g r e e
thetaz = thetaz *180/ %pi ; // d e g r e e
printf ( ” A n g l e made by F w i t h +ve X a x i s %. 2 f d e g r e e \
n ” , thetax ) ;
28
29
printf ( ” A n g l e made by F w i t h +ve Y a x i s %. 2 f d e g r e e \
n ” , thetay ) ;
30 printf ( ” A n g l e made by F w i t h +ve Z a x i s %. 2 f d e g r e e \
n ” , thetaz ) ;
31 printf ( ” \n\n ” )
32 F =800 // N , g i v e n f o r c e
33
34
35
theta =145 // D e g r e e s , a n g l e w i t h p o s i y i v e X a x i s
12
36
37
38 theta = theta * %pi /180; // C o n v e r s i o n i n t o r a d i a n
39
40
41
42 Fx = F * sin ( theta ) ; //N, H o r i z o n t a l component
43
44 Fy = F * cos ( theta ) ; // N, V e r t i c a l Component
45
46
47 printf ( ” \n\n ” )
48 printf ( ” H o r i z o n t a l component o f F i s %. 2 f N\n ” , Fx ) ;
49
50 printf ( ” V e r t i a l component o f F i s %. 2 f N\n ” , Fy ) ;
51
52 printf ( ”We may w r i t e F a s \n F = %. 2 f i + %. 2 f j ” ,Fx
, Fy ) ;
53
54
55
56
57
58
59
60
61
62
63
64
F =300 // N , g i v e n f o r c e
AB = sqrt (8^2+6^2) ; // m Length o f AB
cos_alpha =8/ AB ;
sin_alpha = -6/ AB ;
Fx = F * cos_alpha ; //N, H o r i z o n t a l component
Fy = F * sin_alpha ; // N, V e r t i c a l Component
Scilab code Exa 2.8 resultant of AB and AC
1 clc ;
2 // p a g e 51
13
3
4
5
6
7
8
9
10
11
12
13
14
15
16
T_AB =4200; //N , T e n s i o n i n c a b l e AB
T_AC =6000; //N , T e n s i o n i n c a b l e AC
// V e c t o r AB=−(5m) i +(3m) j +(4m) k
// V e c t o r Ac= −(5m) i +(3m) j +(5m) k
ABx = -5; //m
ABy =3; //m
ABz =4; //m
ACx = -5; //m
ACy =3; //m
ACz = -5; //m
AB = sqrt (( -5) ^2+3^2+4^2) ; //m, Magnitude o f v e c t o r AB
AC = sqrt (( -5) ^2+3^2+5^2) ; //m, Magnitude o f v e c t o r AC
//vT AB=T AB∗lambdaAB , lambdaAB − u n i t v e c t o r= vAB/
AB . So we can c a l c u l a t e e a c h component by
multiplying this unit vector
17 T_ABx = T_AB * ABx / AB ; //N , X component o f T AB
18 T_ABy = T_AB * ABy / AB ; //N , Y component o f T AB
19 T_ABz = T_AB * ABz / AB ; //N , Z component o f T AB
20
21
printf ( ” Component o f T AB a l o n g X a x i s
T_ABx ) ;
22 printf ( ” Component o f T AB a l o n g Y a x i s
T_ABy ) ;
23 printf ( ” Component o f T AB a l o n g Z a x i s
T_ABz ) ;
24 printf ( ”We may w r i t e T AB a s \n T AB =
j + %. 2 f k \n ” , T_ABx , T_ABy , T_ABz ) ;
i s %. 2 f N\n ” ,
i s %. 2 f N\n ” ,
i s %. 2 f N\n ” ,
%. 2 f i + %. 2 f
25
26
27
//vT AC=T AC∗lambdaAC , lambdaAC − u n i t v e c t o r= vAC/
AC . So we can c a l c u l a t e e a c h component by
multiplying this unit vector
28 T_ACx = T_AC * ACx / AC ; //N , X component o f T AC
29 T_ACy = T_AC * ACy / AC ; //N , Y component o f T AC
30 T_ACz = T_AC * ACz / AC ; //N , Z component o f T AC
31
32
printf ( ” Component o f T AC a l o n g X a x i s i s %. 2 f N\n ” ,
14
T_ACx ) ;
33 printf ( ” Component o f T AC a l o n g Y a x i s i s %. 2 f N\n ” ,
T_ACy ) ;
34 printf ( ” Component o f T AC a l o n g Z a x i s i s %. 2 f N\n ” ,
T_ACz ) ;
35 printf ( ”We may w r i t e T AC a s \n T AC = %. 2 f i + %. 2 f
j + %. 2 f k \n ” , T_ACx , T_ACy , T_ACz ) ;
36
37 Rx = T_ABx + T_ACx ; //N ,X component o f R
38 Ry = T_ABy + T_ACy ; //N ,Y component o f R
39 Rz = T_ABz + T_ACz ; //N , Z component o f R
40
41 printf ( ” Component o f R a l o n g X a x i s i s %. 2 f N\n ” , Rx )
;
printf ( ” Component o f R a l o n g Y a x i s i s %. 2 f N\n ” , Ry )
;
43 printf ( ” Component o f R a l o n g Z a x i s i s %. 2 f N\n ” , Rz )
;
44 printf ( ”We may w r i t e R a s \n R = %. 2 f i + %. 2 f j + %
. 2 f k \n ” ,Rx , Ry , Rz ) ;
42
45
46 R = sqrt ( Rx ^2+ Ry ^2+ Rz ^2) ; //N, Magnitude o f r e s u l t a n t
47
48 thetax = acos ( Rx / R ) ; // r a d i a n , a n g l e w i t h +ve x a x i s
49 thetay = acos ( Ry / R ) ; // r a d i a n , a n g l e w i t h +ve y a x i s
50 thetaz = acos ( Rz / R ) ; // r a d i a n , a n g l e w i t h +ve z a x i s
51
52 // C o n v e r s i o n o f a n g l e s i n t o d e g r e e
53 thetax = thetax *180/ %pi ; // d e g r e e
54 thetay = thetay *180/ %pi ; // d e g r e e
55 thetaz = thetaz *180/ %pi ; // d e g r e e
56
57 printf ( ” A n g l e made by R w i t h +ve X a x i s %. 2 f d e g r e e \
n ” , thetax ) ;
58
59
printf ( ” A n g l e made by R w i t h +ve Y a x i s %. 2 f d e g r e e \
n ” , thetay ) ;
60 printf ( ” A n g l e made by F w i t h +ve Z a x i s %. 2 f d e g r e e \
15
n ” , thetaz ) ;
16
Chapter 3
Rigid bodies equivalent systems
of forces
Scilab code Exa 3.1 Vertical force
1 clc ;
2 // Given d a t a
3 // p a g e 85
4 F =100; // l b , V e r t i c a l
f o r c e a p p l i e d t o end o f
lever
5 theta =60; // d e g r e e , a n g l e made by l e v e r w i t h +ve X
axis
6 l =24; //
, length of lever
7
8 // a ) Momemt a b o u t O
9 d = l * cosd ( theta ) ; // mm , p e r p e n d i c u l a r
d i s t a n c e from
o to the l i n e of action
10
11 Mo = F * d ; // N . m, Magnitude o f moment a b o u t O
12 printf ( ” Magnitude o f moment a b o u t O o f t h e 500 N i s
%d l b . i n and i t i s i n c l o c k w i s e d i r e c t i o n a s
f o r c e t e n d s t o r o t a t e l e v e r c l o c k w i s e \n ” , Mo ) ;
13
14
// b ) H o r i z o n t a l f o r c e
17
15
16 d = l * sind ( theta ) ; // i n ,
p e r p e n d i c u l a r d i s t a n c e from
to the l i n e of action
o
17
18 F = Mo / d ; // N,
Ho ri zon ta l Force at A r e q u i r e d to
p r o d u c e same Moment a b o u t O
19 printf ( ” Magnitude o f H o r i z o n t a l F o r c e a t A r e q u i r e d
t o p r o d u c e same Moment a b o u t O i s %f l b \n ” ,F ) ;
20
21
22
23
24
25
26
27
// c ) S m a l l e s t f o r c e
// F i s s m a l l e r when d i s maximum i n e x p r e s s i o n Mo=F
∗d , s o we c h o o s e f o r c e p e r p e n d i c u l a r t o OA
Mo =1200 // i n l b
d =24 // i n , p e r p e n d i c u l a r d i s t a n c e from o t o t h e
line of action
F = Mo / d ; // N, S m a l l e s t F o r c e a t A r e q u i r e d t o p r o d u c e
same Moment a b o u t O
printf ( ” Magnitude o f s m a l l e s t F o r c e a t A r e q u i r e d t o
p r o d u c e same Moment a b o u t O i s %f l b \n ” ,F ) ;
28
29 // d ) 1 2 0 0 N v e r t i c a l f o r c e
30 Mo =1200; // l b −i n ,
31 F =240 // i n l b
32 d = Mo / F ; // m, p e r p e n d i c u l a r
d i s t a n c e from o t o t h e
line of action of force
33 OB = d / cosd ( theta ) ; //m, d i s t a n c e o f p o i n t B from O
34
35
printf ( ” V e r i c a l f o r c e o f 1 2 0 0 N must a c t a t %f i n
f a r from t h e s h a f t t o c r e a t e same moment a b o u t O\
n ” , OB ) ;
Scilab code Exa 3.2 Moment of force
1 clc ;
18
2 // Page 86
3 // Given d a t a
4 F =800; // N , F o r c e a p p l i e d on b r a c k e t
5 theta =60; // d e g r e e , a n g l e made by l e v e r w i t h +ve X
6
7
8
axis
theta = theta * %pi /180; // C o n v e r s i o n o f a n g l e i n t o
radian
r_AB =[ -0.2 , 0.16]; //m v e c t o r drawn from B t o A
r e s o l v e d i n r e c t a n g u l a r component
F =[ F * cos ( theta ) , F * sin ( theta ) ] //N , v e c t o r F
r e s o l v e d i n r e c t a n g u l a r component
k =1; // U n i t v e c t o r a l o n g Z a x i s
9
10
11 // M B=r AB ∗ F r e l a t i o n 3 . 7 from s e c t i o n 3 . 5
12 M_B = det ([ r_AB ; F ]) * k ; // N .m
13 printf ( ” The moment o f f o r c e 800 N a b o u t B i s %. 2 f N .
m . −ve s i g n shows i t s a c t i n g c l o c k w i s e \n ” , M_B ) ;
Scilab code Exa 3.3 Moment of force
1 clc ;
2 // p a g e 86
3 // Given d a t a
4 P =30; // l b , F o r c e a p p l i e d t o s h i f t l e v e r
5 alpha =20; // d e g r e e , a n g l e made by f o r c e P w i t h −ve X
axis
6 Q = P * sind ( alpha ) // i n d e g r e e
7
8 d =3 // i n f t
9 M_o = Q * d //N .m , h e r e n e g a t i v e
s i g n s are taken as each
component c r e a t e s moment c l o c k w i s e
10 printf ( ” The moment o f f o r c e P a b o u t B i s %. 2 f l b − f t
. −ve s i g n \n shows i t s a c t i n g c l o c k w i s e \n ” , M_o ) ;
19
Scilab code Exa 3.4 magnitude of force and lambda
1 clc ;
2 // p a g e 87
3 // Given d a t a
4 // M A=r CA ∗ F r e l a t i o n 3 . 7 from s e c t i o n 3 . 5
5 f =200; // N , Magnitude o f F o r c e d i r e c t e d a l o n g CD
6 r_CA =[0.3 ,0 , 0.08]; //m, v e c t o r AC r e p r e c s e n t e d i n
r e c t a n g u l a r component
7 // lambda=CD/ norm (CD)−m, U n i t v e c t o r a l o n g CD
8 //F=f ∗ lambda ; / /m, F o r c e
9 CD =[ -0.3 , 0.24 , -0.32]; // V e c t o r CD r e s o l v e d i n t o
r e c t a n g u l a r component
10 // norm (CD) ; m, m a g n i t u d e o f v e c t o r CD
11
12 lambda = CD / norm ( CD ) ; //m, U n i t v e c t o r a l o n g CD
13 F = f * lambda ; //m, F o r c e
14 // M A=r CA ∗ F r e l a t i o n 3 . 7 from s e c t i o n 3 . 5
15 // i =1; j =1; k =1; U n i t v e c t o r s a l o n g X, Y and Z
direction respectively
16
17
// Componenets o f moment M A a l o n g X, Y and Z
direction respectively
18 M_Ax = det ([ r_CA (2) , r_CA (3) ; F (2) , F (3) ]) ; //N .m
19 M_Ay = - det ([ r_CA (1) , r_CA (3) ; F (1) ,F (3) ]) ; //N .m
20 M_Az = det ([ r_CA (1) , r_CA (2) ; F (1) , F (2) ]) ; // N .m
21
22
printf ( ” Answer can be w r i t t e n a s M B = %. 2 f N .m i +
%. 2 f N .m j + %. 2 f N .m k \n ” , M_Ax , M_Ay , M_Az ) ;
Scilab code Exa 3.6 Couple M equivalent to two couple
20
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
clc ;
// Given d a t a
// Moment arms
Fx = -30; // i n l b
Fy =20; // i n l b
Fz =20; // i n l b
// c o u p l e F o r c e s
x =18; // iN
y =12; // iN
z =9; // iN
Mx = Fx * x ; //N . m, Component o f Moment a l o n g X a x i s
My = Fy * y ; //N . m, Component o f Moment a l o n g Y a x i s
Mz = Fz * z ; //N . m, Component o f Moment a l o n g Z a x i s
// T h i s t h r e e moments r e p r e s e n t component o f s i n g l e
couple M
17 printf ( ” C o u p l e M e q u i v a l e n t t o two c o u p l e can be
w r i t t e n a s \n M = %. 2 f l b −i n i + %. 2 f l b −i n j + %
. 2 f l b −i n k \n ” ,Mx , My , Mz ) ;
Scilab code Exa 3.7 Distance from the shaft
1
2
3
4
5
6
7
8
9
10
11
clc ;
// p a g e 113
Mo =24; //N .m ∗k , C o u p l e o f moment
f = -400; //N, Magnitude o f f o r c e
OB =300; //mm, D i s t a n c e o f f o r c e from p o i n t O
theta =60; // d e g r e e , a n g l e made by l e v e r w i t h +ve X
axis
x = cosd ( theta )
BC = Mo /( - f * x ) ; //m
BC = BC *1000; //mm, C o n v e r s i o n i n t o m i l l i m e t e r
disp ( BC )
OC = OB + BC ; //mm, D i s t a n c e from t h e s h a f t t o t h e p o i n t
21
of application of this equivalenet force
12
13
printf ( ” D i s t a n c e from t h e s h a f t t o t h e p o i n t o f
application of this equivalenet single force is
%f mm” , OC )
22
Chapter 4
Equilibrium of rigid bodies
Scilab code Exa 4.1 Angle and degee
1 clc ;
2 // p a g e 166
3 // D e t e r m i n a t i o n o f B
4 // At e q u i l l i b r i u m +sum (M A) =0
5 //B ∗ 1 . 5m− ( 9 . 8 1 kN ) ( 2 m) − ( 23 . 5 kN ) ( 6 m) =0 , B assumed
6
7
8
9
10
11
12
13
14
15
16
17
18
19
t o be i n +ve X d i r e c t i o n
B =(9.81*2+23.5*6) /1.5 //kN
printf ( ”B=%. 2 f kN \n +ve s i g n shows r e a c t i o n i s
d i r e c t e d a s assumed ” ,B ) ;
// D e t e r m i n a t i o n o f Ax
//Sum Fx=0
//Ax+B=0
Ax = - B ; //kN
printf ( ”Ax=%. 2 f kN\n ” , Ax ) ;
// D e t e r m i n a t i o n o f Ay
//Sum Fy=0
//Ay−9.81 kN−23.5kN=0
Ay =9.81+23.5; //kN
printf ( ”Ay=%. 2 f kN\n ” , Ay ) ;
A =[ Ax , Ay ]; //kN Adding component
A = norm ( A ) ; // Magnitude o f f o r c e A
23
theta = atan ( Ay / Ax ) ; // r a d i a n s
theta = theta *180/ %pi ; // d e g r e e s , c o n v e r s i o n i n t o
degrees
22 printf ( ” R e a c t i o n a t A i s A=%. 2 f kN making a n g l e %. 2 f
d e g r e e s \n w i t h + ve x a x i s ” ,A , theta ) ;
23 // S l i g h t v a r i a t i o n i n t h e a n s w e r b e c a u s e o f r o u n d o f f
error
20
21
Scilab code Exa 4.2 Angle and degee
1 clc ;
2 // Page 148
3
4 // At e q u i l l i b r i u m
5
6
7
8
9
10
e q u a t i o n s a r e +−> sum Fx=0 , +sum (
M A) =0 , +sum (M B) =0
//Sum Fx=0 g i v e s
Bx =0; //kN
printf ( ”Bx=%. 0 f kN \n ” , Bx ) ;
//+sum (M A) =0 g i v e s −(70kN ) ( 0 . 9m)+By ( 2 . 7m) −(27kN )
( 3 . 3m) −(27kN ) ( 3 . 9m) =0 , B assumed t o be i n +ve Y
direction
By =(70*0.9+27*3.3+27*3.9) /2.7 //kN
printf ( ”By=%. 2 f kN +ve s i g n shows r e a c t i o n i s
d i r e c t e d a s assumed \n ” , By ) ;
11
12
//+sum (M B) =0 g i v e s −A ( 2 . 7m) +(70kN ) ( 1 . 8m) −(27kN ) ( 0 . 6
m) −(27kN ) ( 1 . 2m) =0 , A assumed t o be i n +ve Y
direction
13 A =(70*1.8 -27*0.6 -27*1.2) /2.7 //kN
14 printf ( ”A=%. 2 f kN +ve s i g n shows r e a c t i o n i s
d i r e c t e d a s assumed \n ” ,A ) ;
15 // Answer d i s p l a y e d i n KN
24
Scilab code Exa 4.3 Reaction and direction
1 clc ;
2 // p a g e 168
3 // Take x a x i s
4
5
6
7
8
p a r a l l e l t o t r a c k and Y a x i s
perpendicular to track
W =25; //kN
// R e s o l v i n g w e i g h t
Wx = W * cos (25* %pi /180) ; //kN
Wy = - W * sin (25* %pi /180) ; //kN
// At e q u i l l i b r i u m e q u a t i o n s a r e +−> sum Fx=0 , +sum (
M A) =0 , +sum (M B) =0
9
10
//+sum (M A) =0 g i v e s − ( 10 . 5kN ) ( 6 2 5 mm) − ( 2 2 . 6 5 kN ) ( 1 5 0
mm)+ R2 ( 1 2 5 0 mm) =0 , R2 assumed t o be i n +ve Y
direction
11 R2 =(10.5*625+22.65*150) /1250; //kN
12 printf ( ”R2=%. 0 f kN +ve s i g n shows r e a c t i o n i s
d i r e c t e d a s assumed \n ” , R2 ) ;
13
14
//+sum (M B) =0 g i v e s ( 1 0 . 5 kN ) ( 6 2 5 mm) − ( 2 2 . 6 5 kN ) ( 1 5 0
mm)+ R1 ( 1 2 5 0 mm) =0 , R1 assumed t o be i n +ve Y
direction
15 R1 =(10.5*625 -22.65*150) /1250; //kN
16 printf ( ”R1=%. 1 f kN +ve s i g n shows r e a c t i o n i s
d i r e c t e d a s assumed \n ” , R1 ) ;
17
18 //Sum Fx=0 g i v e s , 2 2 . 6 5 N−T=0
19 T =22.65; //kN
20 printf ( ”T=%. 2 f kN +ve s i g n shows r e a c t i o n
d i r e c t e d a s assumed \n ” ,T ) ;
Scilab code Exa 4.4 Reaction and direction
1 clc ;
25
is
2
3
4
5
6
7
8
9
10
11
12
// // p a g e 168
Ax =4.5 // i n m
Ay =6 // i n m
DF = sqrt (( Ax ^2) +( Ay ^2) )
F =150 // i n KN
Ex = -( Ax / DF ) * F
printf ( ”Ex=%. 2 f kN \n ” , Ex ) ;
Ey =(( Ay / DF ) * F ) +(4*20)
printf ( ”Ey=%. 2 f kN \n ” , Ey ) ;
M_E = -((20*7.2) +(20*5.4) +(20*3.6) +(20*1.8) -(( Ay / DF ) * F
* Ax ) )
13 printf ( ”M E=%. 0 f kN +ve s i g n shows r e a c t i o n i s
d i r e c t e d a s assumed \n ” , M_E ) ;
Scilab code Exa 4.5 Angle and degee
1
2
3
4
5
6
7
8
9
10
11
12
13
14
clc ;
// p a g e 169
// At e q u i l l i b r i u m +sum (Mo) =0 ,
// s=r ∗ t h e t a ;
//F=k ∗ s=k ∗ r ∗ t h e t a ;
k =45; //N/mm
r =75; //mm
W =1800; //N
l =200; //mm
// t r i a l and e r r o r
printf ( ” P r o b a b l e a n s w e r s by t r i a l and e r r o r method
a r e \n ” ) ;
15 for i =0:0.1: %pi /2 // from 0 t o 90 d e g r e e s
16
17
difference =( sin ( i ) -k * r ^2*( i ) /( W * l ) ) ;
26
18 if difference <0.01 then
// A p p r o x i m a t i o n
19
theta = i ;
20
theta = theta *180/ %pi ; // D e g r e e s , c o n v e r s i o n
degrees
21 printf ( ” Theta=%. 2 f
22 end
23 end
d e g r e e s \n ” , theta ) ;
Scilab code Exa 4.6 Tension and angle
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
clc ;
// p a g e 185
m =10; // kg mass o f j o i s t
g =9.81; //m/ s ˆ2 g r a v i t a t i o n a l a c c e l e r a t i o n
W = m * g ; //N
AB =4; //m
// Three f o r c e body
BF = AB * cos (45* %pi /180) ; //m
AF = BF ; //m
AE =1/2* AF ; //m
EF = AE ; //m
CD = AE ; //m
BD = CD / tan ((45+25) * %pi /180) ; //m
DF = BF - BD ; //m
CE = DF ; //m
alpha = atan ( CE / AE ) ; // r a d i a n s
alpha = alpha *180/ %pi ; // d e g r e e s
// From g e o m e t r y
G =90 - alpha ; // d e g r e e s
B = alpha -(90 -(45+25) ) ; // d e g r e e s
C =180 -( G + B ) ; // D e g r e e s
27
into
26
27 // F o r c e t r i a n g l e
28 //T/ s i n (G)=R/ s i n (C)=W/ s i n (B) . . . . . s i n e law
29
30 T = W / sin ( B * %pi /180) * sin ( G * %pi /180) ; //N
31 R = W / sin ( B * %pi /180) * sin ( C * %pi /180) ; //N
32 printf ( ” T e n s i o n i n c a b l e T= %. 1 f N\n R e a c t i o n At A
i s \n R= %. 1 f N w i t h a n g l e a l p h a= %. 1 f d e g r e e s
w i t h +ve X a x i s ” ,T ,R , alpha ) ;
Scilab code Exa 4.7 Reaction
1
2
3
4
5
6
7
8
9
10
11
12
13
clc ;
// p a g e 194
m1 =80; // kg mass o f man
m2 =20; // kg , mass o f l a d d e r
m = m1 + m2 ; // kg
g =9.81; //m/ s ˆ2 g r a v i t a t i o n a l a c c e l e r a t i o n
W = - m * g ; //N, j
C = -0.6* W /3; //N
Bz = -0.6* C /1.2; //N
By = -0.9* W /1.2; //N
printf ( ” R e a c t i o n At B i s B= (%. 0 f ) N j +(%. 1 f N) k \n
” ,By , Bz ) ;
14 printf ( ” R e a c t i o n At C i s C= (%. 2 f ) N k \n ” ,C ) ;
15 Ay = -W - By ; //N
16 Az = -C - Bz ; //N
17
18
19
printf ( ” R e a c t i o n At A i s A= (%. 0 f ) N j +(%. 1 f N) k \
n ” ,Ay , Az ) ;
28
Scilab code Exa 4.8 Reaction and direction
1
2
3
4
5
6
7
8
9
10
11
12
13
14
clc ;
W = -1200; //N, j Weight
BD =[ -2.4 ,1.2 , -2.4]; //m, V e c t o r BD
EC =[ -1.8 ,0.9 ,0.6]; //m, V e c t o r EC
//T BD=norm ( T BD ) ∗BD/ norm (BD) ; / / m, v e c t o r o f
t e n s i o n i n BD
//T EC=norm ( T EC ) ∗EC/ norm (EC) ; / / m, v e c t o r o f
t e n s i o n i n EC
// A p p l y i n g e q u i l l i b r i u m c o n d i t i o n s we g e t
// Sum F=0 , and Sum (M A) =0 and s e t t i n g co− e f f i c i e n t
equal to zero
A =[0.8 ,0.771;1.6 , -0.514]; // MAtrix o f co− e f f i c i e n t
b =[ -1440;0]; // m a t r i x b
x = linsolve (A , b ) ; // s o l u t i o n m a t r i x
T_BD = x (1) ; // N, T e n s i o n i n BD
T_EC = x (2) ; //N, T e n s i o n i n EC
printf ( ”T BD= (%. 0 f N) and T EC= (%. 0 f N) \n ” ,x (1) ,x
(2) ) ;
15
16 Ax =2/3* T_BD +6/7* T_EC ; //N, x component o f
r e a c t i o n at
A
17 Ay = -(1/3* T_BD +3/7* T_EC + W ) ; //N, Y component o f
r e c t i o n at A
18 Az =2/3* T_BD -2/7* T_EC ; //N, z component o f
r e a c t i o n at
A
19
20
printf ( ” R e a c t i o n a t A i s A=(%. 0 f N) i +(%. 0 f N) j +(%
. 1 f N) k \n ” ,Ax , Ay , Az ) ;
21 // Answe i n Newton i n s t e a d o f l b s
22 // 1 l b s =4.44N
29
Scilab code Exa 4.9 Tension in vector form
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
clc ;
// p a g e 198
// F r e e body d i a g r a m
m =30 // i n kg
g =9.81 // i n m/ s 2
w = - m * g // i n J
DC =[ -480 240 -160] // i n mm
X = norm ( DC )
T = DC / X
disp ( ” T e n s i o n i n t h e v e c t o r form=” )
disp ( T )
// E q u i l i b r i u m e q u a t i o n s
// From e q u a t i o n 2 , s e t t i n g u n i t v e c t o r =0
Ax =49 // i n N
Ay =73.5 // i n N
A =[ Ax Ay ]
y = norm ( A )
disp ( ” T e n s i o n i n t h e v e c t o r form i n N=” )
disp ( y )
Scilab code Exa 4.10 coordinates
1
2
3
4
5
6
7
8
clc ;
// p a g e 197
Tmin =300 // l b
AC =[12 12 0]
w =[0; -450;0]
x1 = AC * w
disp ( x1 )
x =[0 0 x1 ]
30
9 lambda =[2/3 2/3 -1/3]*[0;0; - x1 ]
10 y = x * lambda
11 disp ( y )
12
13 // L o c a t i o n o f G
14 //EG and Tmin a r e h a v i n g same d i r e c t i o n , s o
their
component s h o u l d be i n p r o p o r t i o n
15 x = -1.8/ Tmin (3) * Tmin (1) +1.8; //m, X co−o r d i n a t e o f G
16 y = -1.8/ Tmin (3) * Tmin (2) +3.6; //m, Y co−o r d i n a t e o f G
17 printf ( ”Co−o r d i n a t e s o f G a r e x=%. 0 f m and y= %. 1 f m
” ,x , y ) ;
31
Chapter 5
Distrubuted forces centroids
and centers of gravity
Scilab code Exa 5.1 centroid
1 clc ;
2 // p a g e 228
3 n =4; // no o f component
4 A =[120*80 ,120*60/2 , %pi *60*60/2 , - %pi *40*40]; //mmˆ 2 ,
A r e a s o f R e c t a n g l e , t r i a n g l e , S e m i c i r c l e , and
Circle respectively
5 x =[60 ,40 ,60 ,60]; //mm, x c o m po n e n t s o f c e n t r o i d s o f
R e c t a n g l e , t r i a n g l e , S e m i c i r c l e , and C i r c l e
respectively
6 y =[40 , -20 ,105.46 ,80]; //mm, y c o m po n e n t s o f c e n t r o i d s
o f R e c t a n g l e , t r i a n g l e , S e m i c i r c l e , and C i r c l e
respectively
7
8 sumA =0;
9 sumxA =0;
10 sumyA =0;
11
12 for ( i =1: n )
13
sumA = sumA + A ( i ) ;
32
14
sumxA = sumxA + x ( i ) * A ( i ) ;
15
sumyA = sumyA + y ( i ) * A ( i ) ;
16
17 end
18
19 // F i r s t Moment o f a r e a
20 Qx = sumyA ; // About X a x i s
21 Qy = sumxA ; // About Y a x i s
22 printf ( ” F i r s t moments o f t h e a r e a a r e Qx= %. 0 f mmˆ3
and Qy=%. 0 f mmˆ3 \n ” ,Qx , Qy ) ;
23
24 // L o c a t i o n o f c e n t r o i d
25 X = sumxA / sumA ; // X co−o r d i n a t e
26 Y = sumyA / sumA ; // Y c o=o r d i n a t e
27 printf ( ”Co−o r d i n a t e s o f c e n t r o i d
a r e X= %. 1 f mm and
Y= %. 1 f mm \n ” ,X , Y ) ;
Scilab code Exa 5.2 Coordinates of centroid
1 clc ;
2 // p a g e 229
3 n =3; // no o f s e g m e n t
4 L =[600 ,650 ,250]; //mm, L e n g t h s o f s e g m e n t AB , BC and
CA r e s p e c t i v e l y
5 x =[300 ,300 ,0]; //mm, x c o m po n e n t s o f
centroids of
s e g m e n t AB , BC and CA r e s p e c t i v e l y
6 y =[0 ,125 ,125]; //mm, y c o m po n e n t s o f c e n t r o i d s o f
s e g m e n t AB , BC and CA r e s p e c t i v e l y
7
8 sumL =0;
9 sumxL =0;
10 sumyL =0;
11
12 for ( i =1: n )
13
sumL = sumL + L ( i ) ;
33
14
sumxL = sumxL + x ( i ) * L ( i ) ;
15
sumyL = sumyL + y ( i ) * L ( i ) ;
16
17 end
18
19
20
21 // L o c a t i o n o f c e n t r e o f g r a v i t y
22 X = sumxL / sumL ; // X co−o r d i n a t e
23 Y = sumyL / sumL ; // Y c o=o r d i n a t e
24 printf ( ”Co−o r d i n a t e s o f c e n t r o i d
25
a r e X= %. 0 f mm and
Y= %. 0 f mm \n ” ,X , Y ) ;
// There i s v a r i a t i o n b e c a u s e o f r o u n d o f f
Scilab code Exa 5.7 Mass of steel
1
2
3
4
5
clc ;
// p a g e 242
p =7850; // kg /mˆ 3 , d e n s i t y o f s t e e l rim
n =2; // no o f component
A =[(20+60+20) *(30+20) , -60*30]; //mmˆ 2 , C r o s s s e c t i o n
A r e a s o f r e c t a n g l e I and I I
6
7 y =[375 ,365]; //mm, y c o m p on e n t s o f
centroids of
R e c t a n g l e s I and I I r e s p e c t i v e l y
8
9
10 sumV =0;
11
12 for ( i =1: n )
13
C ( i ) =2* %pi * y ( i ) ; //mm, D i s t a n c e t r a v e l l e d by C
14
V ( i ) = A ( i ) * C ( i ) ; //mmˆ 3 , Volume o f 1 component
15
sumV = sumV + V ( i ) ; // mmˆ 3 , T o t a l volume o f rim
16
17 end
34
18 sumV = sumV *10^( -9) ; // C o n v e r s i o n i n t o mˆ3
19 g =9.81; //m/ s ˆ 2 , a c c e l e r a t i o n due t o g r a v i t y
20 m = p * sumV ; // kg , mass
21 W = m * g ; //N, Weight
22 printf ( ” mass o f s t e e l i s m= %. 0 f kg and Wight i s W=
%. 0 f N\n ” ,m , W ) ;
Scilab code Exa 5.9 equivalent concentrated mass
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
clc ;
// p a g e 250
n =2; // no o f t r i a n g l e
A =[4.5 ,13.5]; //kN , l o a d s
x =[2 ,4]; //mm, d i s t a n c e s o f c e n t r o i d from p o i n t A
sumA =0;
sumxA =0;
for ( i =1: n )
sumA = sumA + A ( i ) ;
sumxA = sumxA + x ( i ) * A ( i ) ;
end
// L o c a t i o n o f c e n t r o i d
X = sumxA / sumA ; // X co−o r d i n a t e
W = sumA ; //kN , C o n c e n t r a t e d l o a d
printf ( ” The e q u i v a l e n t c o n c e n t r a t e d mass i s W= %. 0 f
kN and i t s l i n e o f a c t i o n i s l o c a t e d a t a
d i s t a n c e X= %. 1 f m t o t h e r i g h t o f A \n ” ,W , X ) ;
20
21 // R e a c t i o n s
22 // A p p l y i n g sum ( F x ) =0
23 Bx =0; //N
24 // A p p l y i n g sum (M A) =0
35
25 By = W * X /6; //kN , R e a c t i o n a t B i n Y d i r e c t i o n
26 // A p p l y i n g sum (M B) =0
27 A = W *(6 - X ) /6; //kN , R e a c t i o n a t B i n Y d i r e c t i o n
28
29 printf ( ” The r e c t i o n a t A=%. 1 f kN , At Bx=%. 1 f kN and
By=%. 1 f kN \n ” ,A , Bx , By ) ;
Scilab code Exa 5.10 Reaction and direction
1
2
3
4
5
6
7
8
9
10
11
clc ;
// p a g e 251
t =0.3; //m t h i c k n e s s o f dam
g =9.81; // m/ s ˆ 2 , a c c e l e r a t i o n due t o g r a v i t y
p1 =2400; // kg /mˆ 3 , d e n s i t y o f c o n c r e t e
p2 =1000; // kg /mˆ 3 , d e n s i t y o f w a t e r
W1 =0.5*2.7*6.6* t * p1 * g /1000; //kN , Weight o f c o n c r e t e
component 1
W2 =1.5*6.6* t * p1 * g /1000; //kN , Weight o f c o n c r e t e
component 2
W3 =1/3*3*5.4* t * p1 * g /1000; //kN , Weight o f c o n c r e t e
component 3
W4 =2/3*3*5.4* t * p2 * g /1000; //kN , Weight o f w a t e r
P =0.5*2.7*6.6* t * p1 * g /1000; //kN , p r e s s u r e f o r c e
e x e r t e d by w a t e r
12
13 // A p p l y i n g sum ( F x ) =0
14 H =42.9; //kN , H o r i z o n t a l r e a t i o n a t A
15
16 // A p p l y i n g sum ( Fy ) =0
17 V = W1 + W2 + W3 + W4 ; //kN , V e r t i c a l R e a c t i o n a t A
18
19 printf ( ” The h o r i z o n t a l r e a c t i o n i s H=%. 1 f kN ,
V e r t i c a l r e c t i o n a t A V=%. 1 f kN , \n ” ,H , V ) ;
20 // A p p l y i n g sum (M A) =0
21 M = W1 *1.8+ W2 *3.45+ W3 *5.1+ W4 *6 - P *1.8; //kN . m, Moment a t
36
A
22
23
24
// We can r e p l a c e f o r c e c o u p l e s y s t e m by s i n g l e
f o r c e acting at d i s t a n c e r i g h t to A
25 d = M / V ; // m D i s t a n c e o f r e s u l t a n t f o r c e from A
26
27
printf ( ” The moment a b o u t A i s M=%. 1 f kN .m
a n t i c l o c k w i s e and \n i f we r e p l a c e i t by f o r c e
c o u p l e s y s t e m r e s u l t a n t , s d i s t a n c e from A i s d=
%0 . 2 f m \n ” ,M , d ) ;
28 // D i f f e r e n c e i s b e c a u s e o f round o f f
Scilab code Exa 5.11 Coordinates of centroid
1
2
3
4
5
6
clc ;
// p a g e 263
n =3; // no o f component
r =60; //mm, r a d i u s
l =100; //mm l e n g t h o f c y l i n d e r
V =[0.5*4/3* %pi *( r ) ^3 , %pi * r * r *l , - %pi /3* r * r * l ]; //mmˆ 3 ,
Volumes o f Hemisphere , c y l i n d e r and c o n e
respectively
7 x =[ -3/8* r , l /2 ,3/4* l ]; //mm, x c o m po n e n t s o f c e n t r o i d s
o f Hemisphere , c y l i n d e r and c o n e r e s p e c t i v e l y
8
9 sumV =0;
10 sumxV =0;
11
12 for ( i =1: n )
13
sumV = sumV + V ( i ) ;
14
sumxV = sumxV + x ( i ) * V ( i ) ;
15
16 end
17
37
18
19
20 // L o c a t i o n o f c e n t r e o f g r a v i t y
21 X = sumxV / sumV ; // X co−o r d i n a t e
22
23 printf ( ”Co−o r d i n a t e s o f c e n t r o i d
a r e X= %. 0 f mm \n ” ,
X);
Scilab code Exa 5.12 components of centroids
1
2
3
4
5
6
7
8
9
10
11
12
13
14
clc ;
// p a g e 264
l =4.5; // i n i n
b =2; // i n
h =.5; // i n
a_I = l * b * h
a_II =((1/4) * %pi * b ^2* h )
a_III = - %pi *( h ^2) * h
a_IV = - %pi *( h ^2) * h
V =[ a_I a_II a_III a_IV ]
// d i s p (V)
x =[.25 ,1.3488 ,.25 ,.25]; // i n , x c o mp o n e n t s o f
c e n t r o i d s o f p a r t I , I I , I I I and IV r e s p e c t i v e l y
15 y =[ -1 , -0.8488 , -1 , -1]; // i n , y c o m po n e n t s o f c e n t r o i d s
o f p a r t I , I I , I I I and IV r e s p e c t i v e l y
16 z =[2.25 ,0.25 ,3.5 ,1.5]; // i n , z c o m p o n e n t s o f
c e n t r o i d s o f p a r t I , I I , I I I and IV r e s p e c t i v e l y
17
18
19 for ( i =1:4)
20
temp =0
21
sum_xV =0
22
sum_xV = V ( i ) * x ( i )
38
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
y ( i ) =[ sum_xV ]
end
x = sum ( y )
printf ( ” The sum o f x ∗V=%f
i n ˆ4 \n ” ,x )
for ( i =1:4)
temp =0
sum_zV =0
sum_zV = V ( i ) * z ( i )
y ( i ) =[ sum_zV ]
end
z = sum ( y )
printf ( ” The sum o f z ∗V=%f
i n ˆ4 \n ” ,z )
for ( i =1:4)
temp =0
sum_yV =0
sum_yV = V ( i ) * y ( i )
y ( i ) =[ sum_yV ]
disp ( y ( i ) )
end
s = sum ( y )
printf ( ” The sum o f y ∗V=%f
i n ˆ4 \n ” ,s )
39
Chapter 6
Analysis of structures
Scilab code Exa 6.1 force
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
clc ;
// p a g e 294
// E n t i r e t r u s s
// A p p l y i n g sum (M C) =0
E =(10*12+5*6) /3; //kN
// A p p l y i n g sum Fx=0
Cx =0
// A p p l y i n g sumFy=0
Cy =10+5 - E ; //kN
// At j o i n t A
//By p r o p o r t i o n 10kN/4=F AB/3=F AD/5
F_AB =10/4*3; //kN , f o r c e i n member AB
F_DA =10/4*5; //kN , f o r c e i n member AD
// At j o i n t D
F_DB = F_DA ; //kN , f o r c e i n member DB
F_DE =2*3/5* F_DA ; //kN , f o r c e i n member DE
40
22
23
24
25
26
27
28
29
30
31
32
33
34
// At j o i n t B
// a p p l y i n g sumFy=0
F_BE =5/4*( -5 -4/5* F_DB ) ; //kN , f o r c e i n member BE
// A p p l y i n g sumFx=0
F_BC = F_AB +3/5* F_DB -3/5* F_BE ; //kN , f o r c e i n member BC
// At j o i n t E
// A p p l y i n g sumFx=0
F_EC = -5/3*( F_DE -3/5* F_BE ) ; //kN , F o r c e i n member EC
printf ( ” The f o r c e s i n member o f t r u s s a r e \n F AB= %
. 1 f kN T \n F AD= %. 1 f kN C , \n F DB= %. 1 f kN T ,
\n F DE= %. 0 f kN C \n F BE= %. 2 f kN \n F BC= %. 2
f kN \n F EC= %. 2 f kN ” , F_AB , F_DA , F_DB , F_DE , F_BE ,
F_BC , F_EC ) ;
35 // V a r i a t i o n i n answe b e c a u s e o f round o f f
Scilab code Exa 6.2 force
1
2
3
4
5
6
7
8
9
10
11
12
13
clc ;
// p a g e 306
// E n t i r e t r u s s
v1 =140; // kn , v e r i c a l f o r c e 1
v2 =140; //kN , V e r t i c a l f o r c e 2
h =80; //kN , H o r i z o n t a l f o r c e
// A p p l y i n g sum (M B) =0
J =( v1 *4+ v2 *12+ h *5) /16; //kN
// A p p l y i n g sum Fx=0
Bx = - h ; //kN , n e g a t i v e s i g n shows i t
x axis
// A p p l y i n g sumFy=0
41
i s along negative
14
15 By = v1 + v2 - J ; //kN
16
17 // F o r c e i n member EF
18 // A p p l y i n g sumFy=0
19 F_EF = By - v2 ; //kN , F o r c e i n member EF
20 printf ( ” F o r c e i n member EF i s %. 0 f kN \n N e g a t i v e
s i g n shows member i s i n c o m p r e s s i o n \n ” , F_EF ) ;
21
22
23
24
// F o r c e i n member GI
F_GI =( - J *4 - Bx *5) /5; //kN F o r c e i n member GI
printf ( ” F o r c e i n member GI i s %. 0 f kN \n N e g a t i v e
s i g n shows member i s i n c o m p r e s s i o n \n ” , F_GI ) ;
25 // Answer d i f f e r e n c e i s b e c a u s e o f r o u n d i n g o f f
variables
Scilab code Exa 6.3 Calculation of force
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
clc ;
// p a g e 307
// E n t i r e t r u s s
vB =1; //kN , v e r i c a l
vD =1; //kN , v e r i c a l
vF =1; //kN , v e r i c a l
vH =1; //kN , v e r i c a l
vJ =1; //kN , v e r i c a l
vC =5; //kN , v e r i c a l
vE =5; //kN , v e r i c a l
vG =5; //kN , v e r i c a l
h =8; //m, h e i g h t
v =5; //m, h o r i z o n t a l
force
force
force
force
force
force
force
force
at
at
at
at
at
at
at
at
B
D
F
H
J
C
E
G
d i s t a n c e b e t w e e n s u c c e s s i v e node
A =12.50; //kN , r e a c t i o n a t A
L =7.50; //kN , r e a c t i o n a t L
42
alpha = atan ( h /3/ v ) ; // rad , a n g l e made by i n c l i n e d
members w i t h X a x i s
19 // a l p h a=a l p h a / %pi ∗ 1 8 0 ; / / C o n v e r s i o n o f a n g l e i n t o
degrees
18
20
21
22
23
24
25
// F o r c e i n member GI
// A p p l y i n g sum (M H) =0
F_GI =( L *2* v - vJ * v ) /(2* v * tan ( alpha ) ) ; //kN F o r c e i n
member GI
26 printf ( ” F o r c e i n member GI i s %. 2 f kN \n ” , F_GI ) ;
27
28
29
30
// F o r c e i n member FH
// A p p l y i n g sum (M G) =0
F_FH =( L *3* v - vH *v - vJ *2* v ) /( - h * cos ( alpha ) ) ; //kN , F o r c e
i n member FH
31 printf ( ” F o r c e i n member FH i s %. 2 f kN \n N e g a t i v e
s i g n shows member i s i n c o m p r e s s i o n \n ” , F_FH ) ;
32
33
34 // F o r c e i n member GH
35 be = atan ( v /(2* v * tan ( alpha ) ) ) ; // rad , a s t a n ( be )=GI / HI
36 // A p p l y i n g sum ( M L ) =0
37 F_GH =( - vH *v - vJ *2* v ) /(3* v * cos ( be ) ) ; //kN , F o r c e i n
member FH
38 printf ( ” F o r c e i n member GH i s %. 3 f kN \n N e g a t i v e
s i g n shows member i s i n c o m p r e s s i o n \n ” , F_GH ) ;
Scilab code Exa 6.4 components of force
1 clc ;
2 // p a g e 319
3 // E n t i r e t r u s s
4 // A p p l y i n g sum ( Fy ) =0
43
5 Ay =480; //N, Y component o f r e a c t i o n a t A
6 // A p p l y i n g sum (M A) =0
7 B =480*100/160; //N,
r e a c t i o n at B
8 // A p p l y i n g sum ( Fx ) =0
9 Ax = -300; //N, X component o f r e a c t i o n a t A
10
11 alpha = atan (80/150) ; // r a d i a n
12
13 // F r e e body member BCD
14
15 // A p p l y i n g sum (M C) =0
16 F_DE =( -480*100 - B *60) /( sin ( alpha ) *250) ; //N,
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18
19
20
21
22
Force in
l i n k DE
printf ( ” F o r c e i n l i n k DE i s F DE=%. 0 f N\n N e g a t i v e
s i g n shows f o r c e i s c o m p r e s s i v e \n ” , F_DE ) ;
// A p p l y i n g sum ( Fx ) =0
Cx = F_DE * cos ( alpha ) -B ; //N, X component o f f o r c e
exerted at C
// A p p l y i n g sum ( Fy ) =0
Cy = F_DE * sin ( alpha ) + Ay ; //N, Y component o f f o r c e
exerted at C
printf ( ” Components o f f o r c e e x e r t e d a t C i s Cx=%. 0 f
N and Cy=%. 0 f N \n ” ,Cx , Cy ) ;
Scilab code Exa 6.5 components of force
1
2
3
4
5
6
7
8
9
clc ;
// p a g e 320
P =18; //kN , F o r c e a p p l i e d a t D
AF =3.6; //m, Length AF
EF =2; //m, Length EF
ED =2; //m, Length ED
DC =2; //m, Length DC
// E n t i r e f r a m e
// A p p l y i n g sum ( M F ) =0
44
10 Ay = - P *( EF + ED ) / AF ; //kN , Y component o f r e a c t i o n a t A
11
12 // A p p l y i n g sum ( Fx ) =0
13 Ax = - P ; //kN , X component o f r e a c t i o n a t A
14 // A p p l y i n g sum ( Fy ) =0
15 F = - Ay ; //kN ,
r e a c t i o n at B
16
17
18 printf ( ” Components o f f o r c e e x e r t e d a t A i s Ax=%. 0 f
kN and Ay=%. 0 f kN \n ” ,Ax , Ay ) ;
printf ( ” F o r c e e x e r t e d a t F i s F=%. 0 f kN \n ” ,F ) ;
// F r e e body member BE
// A p p l y i n g sum ( Fx ) =0
//B=E , and a s i t i s 2 f o r c e member
By =0;
Ey =0;
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20
21
22
23
24
25
26 // Member ABC
27 // A p p l y i n g sum ( Fy ) =0
28 Cy = - Ay ; //kN , Y component o f f o r c e e x e r t e d a t C
29 // A p p l y i n g sum (M C) =0
30 B =( Ay * AF - Ax *( DC + ED + EF ) ) /( ED + DC ) ; //kN ,
Force in
link
DE
31 printf ( ” F o r c e e x e r t e d a t B i s B=%. 0 f kN \n ” ,B ) ;
32 // A p p l y i n g sum ( Fx ) =0
33 Cx = - Ax - B ; //kN , X component o f f o r c e e x e r t e d a t C
34
35
36
37
printf ( ” Components o f f o r c e e x e r t e d a t C i s Cx=%. 0 f
kN and Cy=%. 0 f kN \n ” ,Cx , Cy ) ;
printf ( ” N e g a t i v e s i g n s shows f o r c e s a r e i n n e g a t i v e
d i r e c t i o n \n ” )
Scilab code Exa 6.6 Force
45
1
2
3
4
5
6
7
8
9
10
clc ;
P =3; //kN , H o r i z o n t a l F o r c e a p p l i e d a t A
AB =1; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n A
BD =1; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n D
CD =1; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C
FC =1; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C
EF =2.4; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
// E n t i r e f r a m e
// A p p l y i n g sum ( M E ) =0
Fy = P *( AB + BD + CD + FC ) / EF ; //kN , Y component o f
at F
and B
and B
and D
and F
E and F
reaction
11
12
13 // A p p l y i n g sum ( Fy ) =0
14 Ey = - Fy ; //kN , Y component o f
r e a c t i o n at E
15
16 // F r e e body member ACE
17 // A p p l y i n g sum ( Fy ) =0 , and sum ( M E ) =0 we g e t 2
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equation
A =[ - AB / sqrt ( AB ^2+ EF ^2) , CD / sqrt ( CD ^2+ EF ^2) ; - EF / sqrt (
AB ^2+ EF ^2) *( AB + BD + CD + FC ) ,- EF / sqrt ( CD ^2+ EF ^2) ]; //
Matrix o f c o e f f i c i e n t s
B =[ Ey ; - P *( AB + BD + CD + FC ) ]; // M a t r i x B
X = linsolve (A , B ) ; //kN S o l u t i o n m a t r i x
F_AB = X (1) ; //kN , F o r e c inmember AB
F_CD = X (2) ; //kN , F o r e c inmember CD
Ex = -P - EF / sqrt ( AB ^2+ EF ^2) * F_AB - EF / sqrt ( CD ^2+ EF ^2) *
F_CD ; //kN , X component o f f o r c e e x e r t e d a t E
// F r e e body : E n t i r e f r a m e
// A p p l y i n g sum ( F X ) =0
Fx = -P - Ex ; //kN , X component o f f o r c e e x e t e r e d a t F
printf ( ” Components o f f o r c e e x e r t e d a t F i s Fx=%. 1 f
kN and Fy=%. 0 f kN \n ” ,Fx , Fy ) ;
printf ( ” F o r c e i n member AB i s F AB=%. 1 f kN \n ” , F_AB )
;
printf ( ” F o r c e i n member CD i s F CD=%. 1 f kN \n ” , F_CD )
;
printf ( ” Components o f f o r c e e x e r t e d a t E i s Ex=%. 1 f
46
kN and Ey=%. 1 f kN \n ” ,Ex , Ey ) ;
31
32
printf ( ” N e g a t i v e s i g n s shows f o r c e s a r e i n n e g a t i v e
d i r e c t i o n \n ” )
47
Chapter 7
Forces in beams and cable
Scilab code Exa 7.1 free body diagram we
1
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3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
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clc ;
// Page 335
P =2400; //N, V e r t i c a l F o r c e a p p l i e d a t D
AB =2.7; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
BE =2.7; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
BK =1.5; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
AJ =1.2; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
EF =4.8; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
BD =3.6; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n
// For e n t i r e t r u s s
//By f r e e body d i a g r a m we g e t t h e f o r c e a t
A =1800; //N
B =1200; //N
C =3600; //N
alpha = atan ( EF /( AB + BE ) ) ; // r a d
// a . I n t e r n a l f o r c e s a t j
// A p p l y i n g sum ( M J ) =0
M = A * AJ ; //N . m, C o u p l e on member ACF a t J
// A p p l y i n g sum ( Fx ) =0
F = A * cos ( alpha ) ; //N, A x i a l f o r c e a t J
// A p p l y i n g sum ( Fy ) =0
48
A
E
B
A
E
D
and
and
and
and
and
and
B
B
K
J
F
B
A, B , c
22 V = A * sin ( alpha ) ; //N, s h e a r i n g f o r c e a t J
23 printf ( ” Thus , I n t e r n a l f o r c e s a t J a r e e q u i v a l e n t t o
\n C o u p l e M = %. 0 f N .m \n A x i a l f o r c e F= %. 0 f N
\n S h e a r i n g f o r c e V= %. 0 f N\n ” ,M ,F , V ) ;
24
25 // a . I n t e r n a l f o r c e s a t K
26 // A p p l y i n g sum (M K) =0
27 M = B * BK ; //N . m, C o u p l e on f r a m e
28 // A p p l y i n g sum ( Fx ) =0
29 F =0; //N, A x i a l f o r c e a t J
30 // A p p l y i n g sum ( Fy ) =0
31 V = - B ; //N, s h e a r i n g f o r c e a t J
32 printf ( ” Thus , I n t e r n a l f o r c e s a t K a r e
e q u i v a l e n t to
\n C o u p l e M = %. 0 f N .m \n A x i a l f o r c e F= %. 0 f N
\n S h e a r i n g f o r c e V= %. 0 f N\n ” ,M ,F , V ) ;
Scilab code Exa 7.2 free body diagram
1 clc ;
2 // p a g e 344
3 // Drawing o f s h e a r and b e n d i n g moment d i a g r a m
4 printf ( ” Given p r o b l e m i s f o r d r a w i n g diagram , t h i s
5
6
7
8
9
10
11
12
13
14
15
d i a g r a m i s drawn by s t e p by s t e p manner . \ n ” ) ;
F_A = -20; //kN , f o r c e a p p l i e d a t A
F_C = -40; //kN , f o r c e a p p l i e d a t C
AB =2.5; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n A and B
BC =3; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and B
CD =2; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and D
//By f r e e body o f e n t i r e beam
//By sum ( m D ) =0
R_B = -( CD * F_C +( AB + BC + CD ) * F_A ) /( BC + CD ) ; //kN , R e a c t i o n
atB
//By sum ( m A ) =0
R_D = -( BC * F_C -( AB ) * F_A ) /( BC + CD ) ; //kN , R e a c t i o n atB
// For s e c t i o n 1
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// A p p l y i n g sum ( Fy ) =0
V1 = F_A ; //kN
// A p p l y i n g sum (M1) =0
M1 = V1 *0; //kN .m
// For s e c t i o n 2
// A p p l y i n g sum ( Fy ) =0
V2 = F_A ; //kN
// A p p l y i n g sum (M1) =0
M2 = F_A * AB ; //kN .m
// For s e c t i o n 3
// A p p l y i n g sum ( Fy ) =0
V3 = R_B + F_A ; //kN
// A p p l y i n g sum (M1) =0
M3 = F_A * AB ; //kN .m
// For s e c t i o n 4
// A p p l y i n g sum ( Fy ) =0
V4 = R_B + F_A ; //kN
// A p p l y i n g sum (M1) =0
M4 = F_A *( AB + BC ) + R_B * BC //kN .m
// For s e c t i o n 5
// A p p l y i n g sum ( Fy ) =0
V5 = R_B + F_A + F_C ; //kN
// A p p l y i n g sum (M1) =0
M5 = F_A *( AB + BC ) + R_B * BC //kN .m
// For s e c t i o n 6
// A p p l y i n g sum ( Fy ) =0
V6 = R_B + F_A + F_C ; //kN
// A p p l y i n g sum (M1) =0
M6 = V6 *0 //kN .m
X =[0 ,2.5 ,2.5 ,5.5 ,5.5 ,7.5]
V =[ V1 , V2 , V3 , V4 , V5 , V6 ]; // S h e a r m a t r i x
M =[ M1 , M2 , M3 , M4 , M5 , M6 ]; // Bending moment m a t r i x
50
xtitle ( ’ S h e a r and b e n d i n g moment d i a g r a m ’ , ’X a x i s ’
, ’Y a x i s ’ ) ;
55 plot (X , V ) ; // S h e a r d i a g r a m
56 plot (X ,M , ’ r ’ ) ; // Bending moment d i a g r a m
54
Scilab code Exa 7.3 free body diagram
1 clc ;
2 // Drawing o f s h e a r and b e n d i n g moment d i a g r a m
3 // V a l u e s t a k e n i n N and m i n s t e a d o f l b and i n
4 printf ( ” Given p r o b l e m i s f o r d r a w i n g diagram , t h i s
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6
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17
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19
20
21
22
23
24
25
26
d i a g r a m i s drawn by s t e p by s t e p manner . \ n ” ) ;
F_AC =40; // l b / i n , d i s t r i b u t e d l o a d a p p l i e d a t A t o C
F_E =400; // l b , f o r c e a p p l i e d a t E
AC =12; // i n , p e r p e n d i c u l a r d i s t a n c e b e t w e e n A and B
CD =6; // i n , p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and D
DE =04; // i n , p e r p e n d i c u l a r d i s t a n c e b e t w e e n E and D
EB =10; // i n , p e r p e n d i c u l a r d i s t a n c e b e t w e e n E and B
AB =32; // i n , l e n g t h o f beam AB
F = F_AC * AC ; //N, F o r c e due t o d i s t r i c u t e d l o a d a t AC/2
//By f r e e body o f e n t i r e beam
//By sum ( m A ) =0
By =( F *( AC /2) + F_E *( AC + CD + DE ) ) / AB ; //N, Y componet o f
Reaction at B
//By sum ( m B ) =0
// d i s p ( By )
A =( F *( AB - AC /2) + F_E * EB ) / AB ; //N, R e a c t i o n a t A
// by sum ( Fx ) =0
// d i s p (A)
Bx =0; //N, xcomponent o f r e c t i o n a t B
51
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63
64
// Diagrams
// For s e c t i o n A t o C
// A p p l y i n g sum ( Fy ) =0
i =0;
for x =0:2:12
i = i +1;
X(i)=x;
V ( i ) =A - F * x ; //N
// A p p l y i n g sum (M1) =0
M ( i ) = A *x - F /2* x ^2; //N .m
end
// For s e c t i o n Cto D
// A p p l y i n g sum ( Fy ) =0
for x =12:2:18
i = i +1;
X(i)=x;
V ( i ) =A - F ; //N
// A p p l y i n g sum (M1) =0
M ( i ) = A *x - F *( x -0.15) ; //N .m
52
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end
// For s e c t i o n D t o B
for x =18:2:32
i = i +1;
X(i)=x;
// A p p l y i n g sum ( Fy ) =0
V ( i ) =A -F - F_E ; //N
// A p p l y i n g sum (M1) =0
M ( i ) = A *x - F *( x -0.15) + F_E * DE - F_E *( x -0.045) ; //N .m
end
xtitle ( ’ S h e a r and b e n d i n g moment d i a g r a m ’ , ’X a x i s ’
, ’Y a x i s ’ ) ;
92 plot (X ,V , ’ r ’ ) ; // S h e a r d i a g r a m
93
94
plot (X ,M , ’− ’ ) ; // Bending moment d i a g r a m
Scilab code Exa 7.4 free body diagram
1 clc ;
53
2
3
// Drawing o f s h e a r and b e n d i n g moment d i a g r a m
printf ( ” Given p r o b l e m i s f o r d r a w i n g diagram , t h i s
d i a g r a m i s drawn by s t e p by s t e p manner . \ n ” ) ;
F_B =500; //N, f o r c e a p p l i e d a t B
F_C =500; //N, f o r c e a p p l i e d a t C .
F_DE =2400; //N/m, d i s t r i b u t e d l o a d a p p l i e d a t D t o E
AB =0.4; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n A and B
BC =0.4; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and B
CD =0.4; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and D
DE =0.3; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n E and D
F_E = F_DE * DE ; //N, f o r c e e x e r t e d a t DE/2 from E
4
5
6
7
8
9
10
11
12
13 //By f r e e body o f e n t i r e beam
14 //By sum ( m D ) =0
15 A =( CD * F_C +( BC + CD ) * F_B - F_E * DE /2) /( AB + BC + CD ) ; //N,
Reaction at A
//By sum ( Fy ) =0
Dy = F_C + F_B + F_E - A ; //N, Y component o f
Reaction at D
//By sum ( Fx ) =0
Dx =0; //N, Y component o f
Reaction at D
// For s e c t i o n 1
// A p p l y i n g sum ( Fy ) =0
V1 = A ; //N, s h e a r f o r c e from A t o B
16
17
18
19
20
21
22
23
24 // For s e c t i o n 2
25 // A p p l y i n g sum ( Fy ) =0
26 V2 =A - F_B ; //N, s h e a r f o r c e from B t o C
27
28 // For s e c t i o n 3
29 // A p p l y i n g sum ( Fy ) =0
30 V3 =A - F_B - F_C ; //N, s h e a r f o r c e
from C t o D
31
32 // For s e c t i o n 4
33 // A p p l y i n g sum ( Fy ) =0
34 V4 =A - F_B - F_C + Dy ; //N, s h e a r f o r c e At D
35
36 // For s e c t i o n 5
37 // A p p l y i n g sum ( Fy ) =0
54
38 V5 =0; //N, s h e a r f o r c e a t A
39 // Area u n d e r b e n d i n g c u r v e i s
40
41
42
43
44
45
46
47
48
49
50
51
52
53
change in bending
moment o f t h a t 2 p o i n t s
MA =0; //N .m
MB = MA + V1 * AB ; //N .m
MC = MB + V2 * BC ; //N .m
MD = MC + V3 * CD ; //N .m
ME = MD +1/2* V4 * AB ; //N .m
X =[0 ,0.4 ,0.4 ,0.8 ,0.8 ,1.2 ,1.2 ,1.5];
V =[ V1 , V1 , V2 , V2 , V3 , V3 , V4 , V5 ]; // S h e a r m a t r i x ,
plot (X , V ) ; // S h e a r d i a g r a m
X =[0 , AB , AB + BC , AB + BC + CD , AB + BC + CD + DE ];
M =[ MA , MB , MC , MD , ME ]; // Bending moment m a t r i x
plot (X ,M , ’ r ’ ) ; // Bending moment d i a g r a m
Scilab code Exa 7.5 free body diagram
1 clc ;
2 // Drawing o f s h e a r and b e n d i n g moment d i a g r a m
3 printf ( ” Given p r o b l e m i s f o r d r a w i n g diagram , t h i s
d i a g r a m i s drawn by s t e p by s t e p manner . \ n ” ) ;
4
5
6
7
8
9
10
11
12
13
14
w =20; //kN/m, d i s t r i b u t e d l o a d a p p l i e d a t D t o E
AB =6; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n A and B
BC =3; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and B
F_B = w * AB ; //kN , f o r c e e x e r t e d a t AB/2 from A
//By f r e e body o f e n t i r e beam
//By sum ( m C ) =0
RA =( F_B *( AB /2+ BC ) ) /( AB + BC ) ; //kN , R e a c t i o n a t A
55
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//By sum ( m A ) =0
RC =( F_B *( AB /2) /( AB + BC ) ) ; //kN , R e a c t i o n a t C
// For s e c t i o n 1
// A p p l y i n g sum ( Fy ) =0
VA = RA ; //N, s h e a r f o r c e j u s t t o r i g h t t o A
// For s e c t i o n 2
// A p p l y i n g sum ( Fy ) =0
VB = VA - F_B ; //kN , s h e a r f o r c e j u s t
// For s e c t i o n 3
// A p p l y i n g sum ( Fy ) =0
VC = VB ; //kN , s h e a r f o r c e
l e f t to B
from B t o C
// Bending moment a t e a c h end i s z e r o
// Maximum b e n d i n g moment i s a t D where V=0
VD =0; //kN
x = -( VD - VA ) / w ; //m, l o c a t i o n o f maximum b e n d i n g moment
printf ( ”Maximum b e n d i n g moment i s a t D x= %. 0 f m
from A\n ” ,x ) ;
MA =0; //kN .m
MD = MA +1/2* VA * x ; //kN . m, maximum b e n d i n g moment i s a t
D
MB = MD +1/2* VB *( AB - x ) ; //N .m
MC = MB + VB * BC ; //N .m
printf ( ”Maximum b e n d i n g moment i s a t MD= %. 0 fkN . m
from A\n ” , MD ) ;
43 X =[0 , x , AB , AB + BC ]; //m,
44 V =[ VA , VD , VB , VC ]; //kN , S h e a r m a t r i x ,
45
46 plot (X , V ) ; // S h e a r d i a g r a m
47 X =[0 , x , AB , AB + BC ]; //m
48 M =[ MA , MD , MB , MC ]; //kN . m, Bendi ng moment m a t r i x
49 plot (X ,M , ’ r ’ ) ; // Bending moment d i a g r a m
56
Scilab code Exa 7.8 free body diagram
1
2
3
4
5
6
7
8
9
clc ;
F_B =30; //kN , V e r t i c a l F o r c e a p p l i e d a t B
F_C =60; //kN , V e r t i c a l F o r c e a p p l i e d a t C
F_D =20; //kN , V e r t i c a l F o r c e a p p l i e d a t D
AB =6; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n A and B
BC =3; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n C and B
CD =4.5; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n c and D
DE =4.5; //m, p e r p e n d i c u l a r d i s t a n c e b e t w e e n D and E
AE =6; //m, v e r t i c a l p e r p e n d i c u l a r d i s t a n c e b e t w e e n A
and E
10 AC =1.5; //m, v e r t i c a l p e r p e n d i c u l a r d i s t a n c e b e t w e e n
A and C
11 // For e n t i r e c a b l e
12 //Sum ( M E ) =0 , AB∗Ax−Ay ∗ (AB+BC+CD+DE)+F B ∗ (BC+CD+DE)+
F C ∗ (CD+DE)+F D ∗ (DE) =0
13
14 // F r e e body ABC
15 //Sum ( M c ) =0 g i v e s −Ax∗AC−Ay ∗ (AB+BC)+F B ∗BC=0
16 // we g e t 2 e q u a t i o n s i n Ax and Ay
17 A =[ AB , -( AB + BC + CD + DE ) ; - AC , -( AB + BC ) ]; // M a t r i x o f
18
19
20
21
22
23
24
25
26
27
28
coeficients
B =[ -( F_B *( BC + CD + DE ) + F_C *( CD + DE ) + F_D *( DE ) ) ; - F_B * BC ];
X = linsolve (A , - B ) ; //kN , S o l u t i o n m a t r i x
Ax = X (1) ; //kN , X component o f r e a c t i o n a t A
Ay = X (2) ; //kN , Y component o f r e a c t i o n a t A
// a . E l e v a t i o n o f p o i n t s B and D
// F r e e body AB
// sum (M B) =0
yB = - Ay * AB / Ax ; //m, b e l o w A
printf ( ” E l e v a t i o n o f p o i n t B i s %. 2 f m b e l o w A\n ” , yB
57
29
30
31
32
33
34
35
36
37
38
39
);
// f r e e body ABCD
// sum (M D) =0
yD =( Ay *( AB + BC + CD ) - F_B *( BC + CD ) - F_C * CD ) / Ax ; //m, a b o v e
A
printf ( ” E l e v a t i o n o f p o i n t D i s %. 2 f m a b o v e A\n ” , yD
);
//Maximum s l o p e and maximum t e n s i o n
theta = atan (( AE - yD ) / DE ) ; // r a d
Tmax = - Ax / cos ( theta ) ; //kN , maximum t e n s i o n
theta = theta / %pi *180; // d e g r e e
printf ( ”Maximum s l o p e i s t h e t a= %. 1 f d e g r e e and
maximum t e n s i o n i n t h e c a b l e i s Tmax= %. 1 f kN \n ”
, theta , Tmax ) ;
Scilab code Exa 7.9 free body diagram
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
clc ;
yB =0.5; //m, s a g o f t h e c a b l e
m =0.75; // kg /m, mass p e r u n i t l e n g t h
g =9.81; //m/ s ˆ 2 , a c c e l e r a t i o n due t o g r a v i t y
AB =40; //m, d i s t a n c e AB
// a . Load P
w = m * g ; //N/m , Load p e r u n i t l e n g t h
xB = AB /2; //m, d i s t a n c e CB
W = w * xB ; //N, a p p l i e d a t h a l f w a y o f CB
// Summing moments a b o u t B
// sum (M B) =0
To = W * xB /2/ yB ; //N
// from f o r c e t r i a n g l e
TB = sqrt ( To ^2+ W ^2) ; //N, =P , a s t e n s i o n on e a c h s i d e
i s same
58
printf ( ” Magnitude o f l o a d P= %. 0 f N \n ” , TB ) ;
// s l o p e o f c a b l e a t B
theta = atan ( W / To ) ; // r a d
theta = theta *180/ %pi ; // d e g r e e , c o n v e r s i o n t o d e g r e e
printf ( ” S l o p e o f c a b l e a t B i s t h e t a= %. 1 f d e g r e e \n ”
, theta ) ;
21 // l e n g t h o f c a b l e
22 // a p p l y i n g eq . 7 . 1 0
23 sB = xB *(1+2/3*( yB / xB ) ^2) ; //m
16
17
18
19
20
24
25
printf ( ” T o t a l l e n g t h o f c a b l e from A t o B i s Length=
%. 4 f m\n ” ,2* sB ) ;
Scilab code Exa 7.10 free body diagram
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
clc ;
AB =150; //m, d i s t a n c e AB
s =30; //m, s a g o f c a b l e
w =45; //N/m Uniform w e i g t h p e r u n i t l e n g t h o f c a b l e
// E q u a t i o n o f c a b l e , by 7 . 1 6
// C o o r d i n a t e s o f B
xB = AB /2; //m
C =[99 ,105 ,98.4 ,90]; // t r i a l v a l u e s
for i =1:4
if ((30/ C ( i ) +1) - cosh ( xB / C ( i ) ) ) <0.0001 then c = C ( i
);
break ;
end
end
yB = s + c ; //m
//Maximum and minimum v a l u e s o f t e n s i o n
59
Tmin = w * c ; //N, To
Tmax = w * yB ; //N TB
printf ( ”Minimum v a l u e o f t e n s i o n i n c a b l e i s Tmin= %
. 0 f N\n ” , Tmin ) ;
23 printf ( ”Maximum v a l u e o f t e n s i o n i n c a b l e i s Tmax= %
. 0 f N\n ” , Tmax ) ;
24 // Length o f c a b l e
20
21
22
25
26
27
28
29
S_CB = sqrt ( yB ^2 - c ^2) ; //m, one h a l p h l e n g t h by 7 . 1 7
S_AB =2* S_CB ; //m, f u l l l e n g t h o f c a b l e
printf ( ” F u l l l e n g t h o f c a b l e i s s AB= %. 0 f m\n ” , S_AB )
;
60
Chapter 8
Friction
Scilab code Exa 8.1 value of friction force
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
clc ;
// p a g e 396
h =100; // l b , h o r i z o n t a l f o r c e
W =300; // l b , w e i g h t o f b l o c k
us =0.2; // C o e f f i e c i e n t o f s t a t i c f r i c t i o n
uk =0.20; //Co= e f f i c i e n t o f k i n e t i c f r i c t i o n
// A p p l y i n g sumFx =0 , we g e t
F =h -3/5* W ; // l b , F o r c e a l o n g p l a n e
F=-F
// A p p l y i n g sumFy=0 , we g e t
N =4/5* W // l b , Normal f o r c e t o t h e p l a n e
printf ( ” F o r c e F r e q u i r e d t o m a i n t a i n t h e
e q u i l l i b r i u m i s t h u s %. 0 f l b , up and t o r i g h t \n ” ,
F);
19
61
20 // Maximum f r i c t i o n f o r c e
21 Fm = us * N ; // l b , Maximum f r i c t i o n f o r c e
22 printf ( ” \n Maximum f r i c t i o n f o r c e i s %. 2 f
lb i s l e s s
than t h a t o f r e q u i r e d to maintain e q u i l l i b r i u m
t h a t i s %. 2 f l b \n So , e q u i l l i b r i u m w i l l n a t
m a i n t a i n and b l o c k w i l move down\n ” ,Fm , F ) ;
23 // A c t u a l v a l u e o f f r i c t i o n f o r c e
24 Fk =(0.6*300) -( h ) -( Fm ) ; // l b , A c t u a l v a l u e o f f r i c t i o n
force
25 printf ( ” \ n A c t u a l v a l u e o f f r i c t i o n f o r c e i s %. 2 f l b
d i r e c t e d up and t o t h e r i g h t \n ” , Fk ) ;
Scilab code Exa 8.2 Force P to prevent block
1
2
3
4
5
6
7
8
9
10
11
12
clc ;
// p a g e 397
F =800; //N F i r c e i n v e r i c a l d i r e c t i o n
us =0.35; // C o e f f i e c i e n t o f s t a t i c f r i c t i o n
uk =0.25; //Co= e f f i c i e n t o f k i n e t i c f r i c t i o n
theta =25; // d e g r e e , a n g l e o f i n c l i n a t i o n
theta = theta * %pi /180; // rad , C o n v e r s i o n i n t o r a d i a n
// F o r c e P s t a r t b l o c k moving up
// At s t a t i c e q u i l l i b r i u m Tan ( T h e t a s )=u s
theta_s = atan ( us ) ; // r a d
P = F * tan ( theta + theta_s ) ; //N, F o r c e P t o s t a r t b l o c k
moving up
13 printf ( ” F o r c e P t o s t a r t b l o c k moving up i s %. 0 f N\n
” ,P ) ;
14
15
16 // F o r c e P t o k e e p b l o c k moving up
17 // At k i n e t i c e q u i l l i b r i u m Tan ( T h e t a k )=uk
18 theta_k = atan ( uk ) ; // r a d
19 P = F * tan ( theta + theta_k ) ; //N, F o r c e P t o k e e p b l o c k
62
moving up
20 printf ( ” F o r c e P t o k e e p b l o c k moving up i s %. 0 f N\n ”
,P ) ;
21
22
23 // F o r c e P t o p r e v e n t
b l o c k from s l i d i n g down
24
25 theta_s = atan ( us ) ; // r a d
26 P = F * tan ( theta - theta_s ) ; //N, F o r c e P t o p r e v e n t
block
from s l i d i n g down
27 printf ( ” F o r c e P t o p r e v e n t
i s %. 0 f N\n ” ,P ) ;
b l o c k from s l i d i n g down
Scilab code Exa 8.3 Minimum distance
1 clc ;
2 us =0.25; // C o e f f i e c i e n t
3 // A p p l y i n g e q u i l l i b r i u m
of s t a t i c f r i c t i o n
e q u a t i o n we g e t r e l a t i o n i n
x
4 printf ( ” Apply e q u i l l i b r i u m e q u a t i o n s . I t i s
t h e o r i t i c a l p a r t . \n ” ) ;
5 x =12 -(.75*2) +1.5 // i n , D i s t a n c e a t which t h e a p p l i e d
l o a d can be s u p p o r t e d
6 printf ( ”Minimum d i s t a n c e a t which t h e a p p l i e d l o a d
can be s u p p o r t e d i s %. 0 f i n \n ” ,x ) ;
Scilab code Exa 8.4 force required
1
2
3
4
5
clc ;
// p a g e 411
F =400; // l b , f o r c e e x e r t e
us =0.35; // C o e f f i e c i e n t o f s t a t i c f r i c t i o n
phi = atand ( us ) ; // rad , a n g l e o f f r i c t i o n
63
6 // d i s p ( p h i )
7 theta =8; // d e g r e e , a n g l e o f i n c l i n a t i o n
8 theta = theta * %pi /180; // rad , C o n v e r s i o n i n t o r a d i a n
9
10 // U s i n g s i n e r u l e
11 // f o r c e p t o r a i s e b l o c k
12 // f r e e body , b l o c k B
13 R1 = F * sind (109.3) /( sind (43.4) )
14 // f r e e body wedge A
15 P = R1 * sind (46.6) /( sind (70.7) )
16 printf ( ” f o r c e r e q u i r e d t o r a i s e b l o c k i s P=%. 0 f l b \
n ” ,P ) ;
17
18 // f o r c e t o l o w e r b l o c k
19 // f r e e body , b l o c k B
20 R1 = F * sind (70.7) /( sind (98.0) )
21 // f r e e body wedge A
22 P = R1 * sind (30.6) /( sind (70.7) )
23 printf ( ” f o r c e r e q u i r e d t o l o w e r b l o c k
i s P=%. 0 f l b \
n ” ,P ) ;
Scilab code Exa 8.5 Couple required to loosen clamp
1
2
3
4
5
6
7
8
9
10
11
12
clc ;
// p a g e 412
pitch =2; //mm, p i t c h o f s c r e w
d =10; //mm, mean d i a m e t e r o f t h r e a d
r = d /2; //mm, r a d i u s
us =0.30; // C o e f f i e c i e n t o f s t a t i c f r i c t i o n
M =40; //kN .m , Maximum c o u p l e
// F o r c e e x e r t e d by clamp
L =2* pitch ; //mm, a s s c r e w i s d o u b l e t h r e a d e d
theta = atan ( L /(2* %pi * r ) ) ; // rad , a n g l e o f i n c l i n a t i o n
64
13 phi = atan ( us ) ; // rad , a n g l e o f f r i c t i o n
14 Q = M / r *1000; //N, F o r c e a p p l i e d t o b l o c k
15
16
17
18
19
20
21
representing
screw
Q = Q /1000 //kN , C o n v e r s i o n i n t o kN
W = Q / tan ( theta + phi ) ; //kN , Magnitude o f f o r c e e x e r t e d
on t h e p i e c e o f wood
printf ( ” Magnitude o f f o r c e e x e r t e d on t h e p i e c e o f
wood i s W= %. 2 f kN \n ” ,W ) ;
// C o u p l e r e q u i r e d t o l o o s e n clamp
Q = W * tan ( phi - theta ) ; //kN , F o r c e r e q u i r e d t o l o o s e n
clamp
Couple = Q * r ; //N . m, C o u p l e r e q u i r e d t o l o o s e n clamp
printf ( ” C o u p l e r e q u i r e d t o l o o s e n clamp i s %. 2 f N .m
\n ” , Couple ) ;
Scilab code Exa 8.6 force required
1 clc ;
2 clear all
3 // Page 423
4 r =1 // i n i n
5 us =0.20; // C o e f f i e c i e n t
of s t a t i c
f r i c t i o n between
s h a f t and p u l l y
6
7 // V e r t i c a l F o r c e r e q u i r e d t o r a i s e l o a d
8 rf = r * us ; // i n , P e r p e n d i c u l a r d i s t a n c e from t h e c e n t e r
Of p u l l y t o l i n e o f a c t i o n
9 // summing moment a b o u t B
10 P1 =(2.20*500) /1.8 // l b , downward F o r c e r e q u i r e d t o
r a i s e load
11 printf ( ” F o r c e r e q u i r e d t o r a i s e l o a d i s %f l b i n
downward d i r e c t i o n \n ” , P1 ) ;
12
13
14
// V e r t c a l F o r c e r e q u i r e d t o h o l d l o a d
65
15 // summing moment a b o u t C
16 P =(1.80*500) /2.20 // l b , downward F o r c e r e q u i r e d t o
hold load
17 printf ( ” F o r c e r e q u i r e d t o h o l d l o a d i s %. 0 f l b i n
downward d i r e c t i o n \n ” ,P ) ;
18
19 // H o r i z o n t a l f o r c e P t o s t a r t r a i s i n g t h e l o a d
20 OE = rf ; //mm,
21 OD = sqrt (2) *2; // i n , p y t h a g o r u s theorm
22 theta = asin ( OE / OD ) ; // rad ,
23
24 // from f o r c e t r i a n g l e
25 P =500* cotd (40.9) ; // l b , H o r i z o n t a l f o r c e P t o s t a r t
26
r a i s i n g the load
printf ( ” H o r i z o n t a l f o r c e P r e q u i r e d t o s t a r t r a i s i n g
t h e l o a d i s %. 0 f l b \n ” ,P ) ;
Scilab code Exa 8.7 Tension
1
2
3
4
5
6
7
8
9
10
11
clc ;
// p a g e 431
T1 =150; //N, F o r c e on f r e e end o f h a w s e r
T2 =7500; //N, F o r c e on o t h e r end o f h a w s e r
// a , c o e f f i c i e n t o f f r i c t i o n
bta =2*2* %pi ; // rad , a n g l e o f c o n t a c t , 2 t u r n s
//By e q u a t i o n 8 . 1 3
us = log ( T2 / T1 ) / bta ; // Co− e f f i c i e n t o f s t a t i c f r i c t i o n
printf ( ” C o e f f i c i e n t o f s t a t i c f r i c t i o n b e t w e e n
h a w s e r and b a l l a r d i s u s= %0 . 3 f \n ” , us ) ;
12
13 // Number o f wraps when t e n s i o n
14
15 bta =3*2* %pi // i n r a d
66
i n h a w s e r =75 kN
16 // One t u r n = 2∗ p i a n g l e , b t a c o r r e s p o n d s t o
17 ten = T1 * exp ( bta * us )
18 printf ( ” T e n s i o n
i s %f N \n ” , ten ) ;
Scilab code Exa 8.8 Torque
1 clc ;
2 // p a g e 432
3 // Given
4 T2 =600; // l b , T e n s i o n from s i d e 2
5 us =0.25; // C o e f f i e c i e n t o f s t a t i c
f r i c t i o n between
p u l l e y and b e l t
6 bta =(2* %pi ) /3; //Co= e f f i c i e n t o f k i n e t i c f r i c t i o n
b e t w e e n p u l l e y and b e l t
7 r1 =8 // i n i n
8 // P u l l e y B
9
10 T1 = T2 /( exp ( us * bta ) ) //N, T e n s i o n from s i d e 1
11 // d i s p ( T1 )
12
13 // P u l l e y A
14 //Aumming moment a b o u t A
15 MA =( T2 * r1 ) -( T1 * r1 ) ; // l b −f t , C o u p l e MA a p p l i e d t o
p u l l e y which i s e q u a l and o p p o s i t e t o t o r q u e
16
17
printf ( ” The l a r g e s t t o r q u e which can be e x e r t e d by
b e l t on p u l l e y A i s MA= %0 . 0 f l b −i n \n ” , MA ) ;
67
Chapter 9
Distributed forces Moment of
Inertia
Scilab code Exa 9.4 Area of plate
1 clc ;
2 // p a g e 465
3 // Area o f p l a t e
4
5
6 A =9*.75; // i n ˆ2
7 y =1/2*13.84+1/2*.75; // i n , y co−o r d i n a t e
8
9
10
11
12
13
14
15
16
17
18
of centroid
of the p l a t e
// A l l v a l u e s f o r f l a n g e a r e from t a b l e from book
sumA = A +8.85; // i n ˆ2 T o t a l a r e a
sumyA = y * A +0; // i n ˆ3
Y = sumyA / sumA ; // i n
// d i s p (Y)
// Moment o f i n e r t i a
// For w i d e f l a n f e
Ix1 =291+8.85* Y ^2; // i n ˆ4
// f o r p l a t e
Ix2 =1/12*9*(3/4) ^3+6.75*(7.295 -3.156) ^2; // i n ˆ4
// For c o m p o s i t e a r e a
68
19 Ix = Ix1 + Ix2 ; // i n ˆ4
20
21 printf ( ”Moment o f i n e r t i a I x= %. 2 e i n ˆ4 \n ” , Ix ) ;
22
23 // R a d i u s o f g y r a t i o n
24 kx = sqrt ( Ix / sumA ) ; //mm
25 printf ( ” R a d i u s o f g y r a t i o n i s kx= %. 1 f i n \n ” , kx ) ;
Scilab code Exa 9.5 Principle moment of inertia
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
clc ;
// p a g e 466
// Given
r =90; //mm, r a d i u s o f h a l f
b =240; //mm, w i d t h
h =120; //mm, h e i g h t
circle
// Moment o f i n e r t i a o f r e c t a n g l e
Ixr =1/3* b * h ^3; //mmˆ4
// Moment o f i n e r t i a o f h a l f
a =4* r /(3* %pi ) ; //mm
circle
b =h - a ; //mm, D i s t a n c e b from c e n t r o i d c t o X a x i s
I_AA =1/8* %pi * r ^4; //mmˆ 4 , Moment o f i n e r t i a o f h a l f
c i r c l e w i t h r e s p e c t t o AA’
17 A =1/2* %pi * r ^2; //mmˆ 2 , Area o f h a l f c i r c l e
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19 Ix1 = I_AA - A * a ^2; //mmˆ 4 , P a r a l l e l a x i s t h e o r e m
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21 Ixc = Ix1 + A * b ^2; //mmˆ 4 , P a r a l l e l a x i s t h e o r e m
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23 // Moment o f i n e r t i a o f g i v e n a r e a
24 Ix = Ixr - Ixc ; //mmˆ4
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printf ( ”Moment o f i n e r t i a o f a r e a a b o u t X a x i s i s I x
= %2 . 2 e mmˆ4\ n ” , Ix ) ;
Scilab code Exa 9.7 Principle moment of inertia
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clc ;
// p a g e 479
Ix =10.38; // i n ˆ 4 , Moment o f i n e r t i a a b o u t x a x i s
Iy =6.97; // i n ˆ 4 , Moment o f i n e r t i a a b o u t y a x i s
Ixy = -3.28+0 -3.28
disp ( Ixy ) // i n i n ˆ4
// P r i n c i p a l a x e s
tan_2_theta_m = -(2* Ixy ) /( Ix - Iy )
two_theta_m = atand ( tan_2_theta_m )
theta_m = two_theta_m /2
printf ( ” O r i e n t a t i o n o f p r i n c i p l e a x e s o f s e c t i o n
a b o u t O i s Theta m= %. 1 f d e g r e e \n ” , theta_m ) ;
// P r i n c i p l e moment o f i n e r t i a , eqn 9 . 2 7
Imax =( Ix + Iy ) /2+ sqrt ((( Ix - Iy ) /2) ^2+ Ixy ^2) ; //mmˆ4
Imin =( Ix + Iy ) /2 - sqrt ((( Ix - Iy ) /2) ^2+ Ixy ^2) ; //mmˆ4
printf ( ” P r i n c i p l e moment o f i n e r t i a o f s e c t i o n a b o u t
O a r e \n Imax= %. 2 e i n ˆ4 \n Imin= %. 0 e i n ˆ4\ n ” ,
Imax , Imin ) ;
20 // a n s w e r d i f f e r e n c e i s due t o r o u n d o f f
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Chapter 10
Method of virtual work
Scilab code Exa 10.3 Force exerted by each cylinder
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clc ;
m =1000; // kg , mass o f k r a t e
theta =60; // d e g r e e
theta = theta * %pi /180; // r a d i a n s , c o n v e r s i o n i n t o r a d
a =0.70; //m
L =3.20; //m
g =9.81; //m/ s ˆ2
// From t h e o r y we g e t
W = m * g ; //N, Weight
W = W /1000; //kN , c o n v e r s i o n i n t o kN
S = sqrt ( a ^2+ L ^2 -2* a * L * cos ( theta ) ) ; //m
F_DH = W * S / L / tan ( theta ) ; //kN
printf ( ” F o r c e e x e r t e d by e a c h c y l i n d e r i s F DH=%. 2 f
kN” , F_DH ) ;
Scilab code Exa 10.4 Angle
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clc ;
m =10; // kg , mass o f rim
r =300; //mm, r a d i u s o f d i s k
a =0.08; //m
b =0.3; //m
k =4; //kN/m
g =9.81; //m/ s ˆ2 g r a v i t y
// From t h e o r y we g e t
// s i n ( t h e t a )=k ∗ a ˆ2/m/ g / b∗ t h e t a
dif =1;
for theta =0:0.001:1
dif = sin ( theta ) -k * a ^2/ m / g / b * theta ;
if dif <=0.001 then printf ( ” t h e t a= %. 3 f r a d o r %
. 1 f d e g r e e s \n ” , theta , theta / %pi *180) ;
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end
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