Machanical Design Shigleys\' Chapter 3 solutions PDF

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Chapter 3

3-1 M o  0 18RB  6(100)  0 RB  33.3 lbf Ans. Fy  0

Ro  RB  100  0 Ro  66.7 lbf Ans. RC  RB  33.3 lbf Ans . ______________________________________________________________________________ 3-2 Body AB: Fx  0

R Ax  RBx

Fy  0

R Ay  RBy

M B  0

R Ay (10)  R Ax (10)  0

RA x  RAy

Body OAC: M O  0

R Ay (10)  100(30)  0

RAy  300 lbf

Ans .

Fx  0

ROx   RAx   300 lbf

Fy  0

ROy  RAy  100  0

ROy   200 lbf

Ans .

Ans .

______________________________________________________________________________

Chapter 3 - Rev. A, Page 1/100

3-3

0.8  1.39 kN Ans . tan 30 0.8  1.6 kN Ans . RA  sin 30 RO 

______________________________________________________________________________ 3-4 Step 1: Find R A & R E 4.5  7.794 m tan 30 M A  0 h

9 RE  7.794(400cos30 )  4.5(400sin 30 )  0 R E  400 N Ans .

F

x

0

R Ax  400cos30  0 RAx  346.4 N

F

y

0

RAy  400  400sin 30  0 RAy  200 N

RA  346.4 2  2002  400 N

Ans.

Step 2: Find components of R C on link 4 and R D

M

C

0

400(4.5)   7.794  1.9  R D  0 R D  305.4 N

F F

Ans .

x

0



 RCx 4  305.4 N

y

0



(RCy )4  400 N

Chapter 3 - Rev. A, Page 2/100

Step 3: Find components of R C on link 2  Fx  0

 RCx  2  305.4  346.4  0  RCx  2  41 N

F

y

0

R  Cy

2

 200 N

____________________________________________________________________________________________________________________ _

Chapter 3 - Rev. A, Page 3/100

3-5 M C  0 1500 R1  300(5)  1200(9)  0 R 1  8.2 kN Ans . Fy  0

8.2  9  5  R2  0 R2  5.8 kN

M 1  8.2(300)  2460 N  m

Ans.

Ans.

M 2  2460  0.8(900)  1740 N  m Ans. M 3  1740  5.8(300)  0 checks! _____________________________________________________________________________ 3-6 Fy  0

R 0  500  40(6)  740 lbf M 0  0

Ans.

M 0  500(8)  40(6)(17)  8080 lbf  in

Ans .

M 1  8080  740(8)   2160 lbf  in Ans. M 2  2160  240(6)  720 lbf  in Ans. 1 M 3   720  (240)(6)  0 checks! 2

______________________________________________________________________________

Chapter 3 - Rev. A, Page 4/100

3-7 M B  0 2.2R1  1(2)  1(4)  0 R1   0.91 kN Ans . Fy  0

0.91  2  R2  4  0 R2  6.91 kN Ans.

M 1   0.91(1.2)   1.09 kN m M 2  1.09  2.91(1)  4 kN  m M 3  4  4(1)  0 checks!

Ans . Ans.

______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, R1  VB  200 lbf

Ans.

Beam BD: M D  0 200(12)  R2 (10)  40(10)(5)  0 R2  440 lbf Ans. Fy  0

200  440  40(10)  R3  0 R3  160 lbf Ans.

Chapter 3 - Rev. A, Page 5/100

M 1  200(4)  800 lbf  in

Ans.

M 2  800  200(4)  0 checks at hinge M 3  800  200(6)  400 lbf  in Ans. 1 M 4  400  (240)(6)  320 lbf  in Ans. 2 1 M 5  320  (160)(4)  0 checks! 2

______________________________________________________________________________ 3-9 q  R1 x

1

 9 x  300 0

1

 5 x 1200

1

 R2 x  1500

0

V  R1  9 x  300  5 x  1200  R2 x  1500 1

1

0

M  R1 x  9 x  300  5 x  1200  R2 x  1500

1

(1) 1

(2)

At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), R1  9  5  R2  0  R1  R2  14 1500R1  9(1500  300)  5(1500  1200)  0 R2  14  8.2  5.8 kN Ans. 0  x  300 :



R1  8.2 kN

Ans.

V  8.2 kN, M  8.2x N m

300  x  1200 : V  8.2 9  0.8 kN M  8.2x  9(x  300)   0.8x  2700 N m 1200  x  1500 : V  8.2 9  5   5.8 kN M  8.2x  9(x  300)  5( x 1200)   5.8 x 8700 N m Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________

Chapter 3 - Rev. A, Page 6/100

3-10 q  R0 x

1

 M0 x

V  R0  M 0 x

1

2

 500 x  8

1

0

 40 x  14  40 x  20

0

1

 500 x  8  40 x  14  40 x  20 1

2

M  R0 x  M 0  500 x  8  20 x 14  20 x  20

0

1

(1)

2

(2)



at x  20 in, V  M  0, Eqs. (1) and (2) give R0  500  40  20  14   0



R0  740 lbf

R0 (20)  M0  500(20  8)  20(20  14)  0



2

M0  8080 lbf  in

Ans. Ans .

0  x  8 : V  740 lbf, M  740x  8080 lbf in 8  x  14 : V  740 500  240 lbf M  740x  8080 500(x  8) 240x  4080 lbf in 14  x  20 : V  740  500  40( x  14)  40 x  800 lbf M  740x  8080 500( x  8)  20( x  14)2  20 x2  800 x  8000 lbf  in Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________

3-11 q  R1 x

1

 2 x  1.2

1

 R2 x  2.2

0

1

 4 x  3.2

0

V  R1  2 x  1.2  R2 x  2.2  4 x  3.2 1

1

1

0

(1) 1

M  R1 x  2 x 1.2  R2 x  2.2  4 x  3.2 at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2),

R1  2  R2  4  0



R1  R2  6

3.2 R1  2(2)  R2 (1)  0  3.2 R1  R2  4 Solving Eqs. (3) and (4) simultaneously, R 1 = -0.91 kN, R 2 = 6.91 kN Ans. 0  x  1.2 :

(2)

(3) (4)

V   0.91 kN, M   0.91x kN m

1.2  x  2.2 : V   0.91 2   2.91 kN M   0.91x  2(x  1.2)   2.91x  2.4 kN m 2.2  x  3.2 : V  0.91  2  6.91  4 kN M   0.91x  2(x 1.2)  6.91( x  2.2)  4 x  12.8 kN  m Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________

Chapter 3 - Rev. A, Page 7/100

3-12 q  R1 x

1

 400 x  4 0

1

 R2 x  10

1

0

0

 40 x  10  40 x  20  R3 x  20

0

1

1

V  R1  400 x  4  R2 x  10  40 x  10  40 x  20  R 3 x  20 1

1

M  R1 x  400 x  4  R2 x 10  20 x 10

2

0

2

 20 x  20  R3 x  20

M  0 at x  8 in 8 R1  400(8  4)  0  R1  200 lbf + at x = 20 , V =M = 0. Applying Eqs. (1) and (2), R2  R3  600 200  400  R2  40(10)  R3  0 

1

(1) 1

(2)

Ans.

Ans. 200(20)  400(16)  R2 (10)  20(10)2  0  R2  440 lbf R3  600  440  160 lbf Ans .

0  x  4 : V  200 lbf, M  200 x lbf  in 4  x  10 : V  200  400   200 lbf, M  200 x  400(x  4)  200x  1600 lbf  in 10  x  20 : V  200  400  440  40( x  10)  640  40 x lbf M  200x  400(x  4)  440( x  10)  20  x  10    20 x2  640 x  4800 lbf  in Plots of V and M are the same as in Prob. 3-8. ______________________________________________________________________________ 2

3-13 Solution depends upon the beam selected. ______________________________________________________________________________ 3-14 (a) Moment at center, l  2 a  xc  2 2 wl  l   wl  l  M c    l  2a        a 2  2 2 4   2  2 At reaction, Mr  wa 2 a = 2.25, l = 10 in, w = 100 lbf/in Mc 

100(10)  10    2.25  125 lbf  in 2  4  100 2.25

2

Mr 

  253 lbf  in

2 (b) Optimal occurs when M c  Mr

Ans.

Chapter 3 - Rev. A, Page 8/100

2 wl  l  wa 2 2 a    a  al  0.25 l  0  2  4 2 

Taking the positive root





l 1 l  l 2  4  0.25l 2    2  1  0.207 l  2 2  for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in Mmin 100 2 2.072  214 lbf  in a

A ns .

______________________________________________________________________________ 3-15 (a) 20  10  5 kpsi 2 20  10 CD   15 kpsi 2

C

R  15 2  8 2  17 kpsi  1  5  17  22 kpsi

 2  5 17  12 kpsi  p  tan 1    14.04 cw 2  15   1  R  17 kpsi 1

8

 s  45  14.04  30.96 ccw

(b) 9  16  12.5 kpsi 2 16  9 CD   3.5 kpsi 2

C

R  5 2  3.5 2  6.10 kpsi  1  12.5  6.1  18.6 kpsi

 2  12.5  6.1  6.4 kpsi 1  5   tan 1    27.5 ccw 2 3.5    1  R  6.10 kpsi

p 

 s  45  27.5  17.5 cw

Chapter 3 - Rev. A, Page 9/100

(c) 24  10  17 kpsi 2 24  10  7 kpsi CD  2 C

R  7 2  6 2  9.22 kpsi  1  17  9.22  26.22 kpsi

 2  17  9.22  7.78 kpsi

1

 7 

 p   90   tan 1    69.7  ccw 2 6     1  R  9.22 kpsi

 s  69.7  45  24.7 ccw

(d) 12  22  5 kpsi 2 12  22 17 kpsi CD  2 C

R  17 2 12 2  20.81 kpsi  1  5  20.81  25.81 kpsi

 2  5  20.81   15.81 kpsi

1

17   

 p   90   tan 1    72.39  cw 2 12 

Cha

 1  R  20.81 kpsi  s  72.39  45  27.39 cw

______________________________________________________________________________

Chapter 3 - Rev. A, Page 11/100

3-16 (a) 8  7  0.5 MPa 2 8 7 CD   7.5 MPa 2 C

R  7.5 2  6 2  9.60 MPa  1  9.60  0.5  9.10 MPa

 2  0.5  9.6  10.1 Mpa 1

 7.5  

  p   90  tan1    70.67 cw 2  6    1  R  9.60 MPa  s  70.67   45   25.67 cw

(b) 96  1.5 MPa 2 9 6 CD   7.5 MPa 2 C

R  7.52  32  8.078 MPa  1  1.5  8.078  9.58 MPa

 2  1.5  8.078  6.58 MPa

 p  tan 1    10.9 cw 2  7.5   1  R  8.078 MPa 1

3

 s  45  10.9  34.1 ccw

Chapter 3 - Rev. A, Page 12/100

(c) 12  4  4 MPa 2 12  4  8 MPa CD  2 C

R  8 2  7 2  10.63 MPa  1  4  10.63  14.63 MPa

 2  4  10.63  6.63 MPa 1

 8 



 

 p   90   tan 1    69.4  ccw 2 7  1  R  10.63 MPa  s  69.4  45  24.4 ccw

(d) 65  0.5 MPa 2 6 5  5.5 MPa CD  2 C

R  5.52  82  9.71 MPa  1  0.5  9.71  10.21 MPa

 2  0.5  9.71  9.21 MPa

 p  tan 1    27.75 ccw 2  5.5   1  R  9.71 MPa 1

8

 s  45  27.75  17.25 cw

______________________________________________________________________________

Chapter 3 - Rev. A, Page 13/100

3-17 (a) 12  6  9 kpsi 2 12  6 CD   3 kpsi 2 C

R  3 2  4 2  5 kpsi  1  5  9  14 kpsi

 2  9  5  4 kpsi 1 1 4 tan     26.6 ccw  3 2  1  R  5 kpsi

p 

 s  45  26.6  18.4 ccw

(b) 30  10  10 kpsi 2 30  10 CD   20 kpsi 2 C

R  202  102  22.36 kpsi  1  10  22.36  32.36 kpsi

 2  10  22.36  12.36 kpsi

1  10  tan1    13.28 ccw 2  20   1  R  22.36 kpsi

p 

 s  45  13.28  31.72 cw

Chapter 3 - Rev. A, Page 14/100

(c) 10  18  4 kpsi 2 10  18 CD   14 kpsi 2 C

R  142  92  16.64 kpsi  1  4  16.64  20.64 kpsi

 2  4  16.64  12.64 kpsi 1

14 

   p   90   tan 1    73.63  cw 2  9   1  R  16.64 kpsi  s  73.63  45  28.63 cw

(d) 9 19  14 kpsi 2 19  9 CD   5 kpsi 2 C

R  5 2  8 2  9.434 kpsi  1  14  9.43  23.43 kpsi

 2  14  9.43  4.57 kpsi 1

5 

   p   90   tan 1    61.0  cw 2 8     1  R  9.34 kpsi

 s  61  45   16  cw

______________________________________________________________________________

Chapter 3 - Rev. A, Page 15/100

3-18 (a) 80  30  55 MPa 2 80  30 CD   25 MPa 2 C

R  25  20  32.02 MPa  1  0 MPa 2

2

 2   55  32.02   22.98   23.0 MPa  3  55  32.0  87.0 MPa 1 2 

23  11.5 MPa, 2

 2 3  32.0 MPa,

1 3 

87  43.5 MPa 2

(b) 30  60   15 MPa 2 60  30 CD   45 MPa 2 C

2 2 R  45  30  54.1 MPa  1  15  54.1  39.1 MPa

 2  0 MPa  3  15  54.1   69.1 MPa 39.1  69.1  54.1 MPa 2 39.1  19.6 MPa 1 2  2 69.1  34.6 MPa 23 2

1 3 

Chapter 3 - Rev. A, Page 16/100

(c) 40  0  20 MPa 2 40  0 CD   20 MPa 2 C

R  202  202  28.3 MPa  1  20  28.3  48.3 MPa

 2  20  28.3  8.3 MPa  3   z  30 MPa 1 3 

48.3  30  39.1 MPa, 2

 1 2  28.3 MPa,

2 3 

30  8.3  10.9 MPa 2

(d) 50  25 MPa 2 50  25 MPa CD  2 C

R  252  302  39.1 MPa  1  25  39.1  64.1 MPa

 2  25  39.1  14.1 MPa  3   z  20 MPa 64.1  20 20  14.1  42.1 MPa,  1 2  39.1 MPa,  2 3   2.95 MPa 2 2 ______________________________________________________________________________

1 3 

3-19 (a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses.  1   x  10 kpsi

 2  0 kpsi  3   y  4 kpsi 10  ( 4)  7 kpsi 2 10  1 2   5 kpsi 2 0  ( 4)  2 kpsi 23 2

1 3 

Chapter 3 - Rev. A, Page 17/100

(b) 0  10  5 kpsi 2 10  0  5 kpsi CD  2 C

R  52  42  6.40 kpsi  1  5  6.40  11.40 kpsi

 2  0 kpsi,  3  5  6.40  1.40 kpsi  1 3  R  6.40 kpsi,

1 2 

11.40  5.70 kpsi, 2

3 

1.40  0.70 kpsi 2

(c) 2 8  5 kpsi 2 8 2 CD   3 kpsi 2 C

R  3 2  4 2  5 kpsi  1  5  5  0 kpsi,  2  0 kpsi

 3  5  5   10 kpsi 1 3 

10  5 kpsi, 2

 1 2  0 kpsi,

 2 3  5 kpsi

(d) 10  30  10 kpsi 2 10  30 CD   20 kpsi 2 C

R  20 2 10 2  22.36 kpsi  1  10  22.36  12.36 kpsi

 2  0 kpsi  3  10  22.36  32.36 kpsi 12.36 32.36  6.18 kpsi,  2 3   16.18 kpsi 2 2 ______________________________________________________________________________

 1 3  22.36 kpsi,

1 2 

Chapter 3 - Rev. A, Page 18/100

3-20

From Eq. (3-15),

 3  ( 6 18  12) 2  6(18)  (6)(12)  18( 12)  92  62  ( 15) 2   6(18)( 12)  2(9)(6)( 15)  ( 6)(6) 2  18( 15)2  ( 12)(9) 2   0

 3  594  3186  0 Roots are: 21.04, 5.67, –26.71 kpsi Ans. 21.04  5.67  7.69 kpsi 1 2  2 5.67  26.71  16.19 kpsi 23 2 21.04  26.71  23.88 kpsi  max  1 3  Ans. 2

_____________________________________________________________________________ 3-21 From Eq. (3-15)





2  3  (20  0  20) 2   20(0)  20(20)  0(20)  402  20 2  0 2  













2    20(0)(20)  2(40)  20 2 (0)  20  20 2  0(0) 2  20(40) 2  0  

 3  40 2  2 000  48 000  0 Roots are: 60, 20, –40 kpsi 60  20  20 kpsi 2 20  40 23  30 kpsi 2 60  40  50 kpsi  max  1 3  2

Ans.

1 2 

Ans.

_____________________________________________________________________________

Chapter 3 - Rev. A, Page 19/100

3-22 From Eq. (3-15) 2 2  3  (10  40  40) 2   10(40) 10(40) 40(40) 202    40    20  

  2 2 2  10(40)(40)  2(20)( 40)( 20) 10( 40)  40( 20)  40(20)  0  3  90 2  0 Roots are: 90, 0, 0 MPa Ans.

230  1 2   1 3   max 

90  45 MPa 2

A ns.

_____________________________________________________________________________ 3-23

 

F 15000   33 950 psi  34.0 kpsi A  4   0.75 2 



FL L 60    33 950  0.0679 in AE E 30 106 



0.0679  1130 106   1130 1   L 60 From Table A-5, v = 0.292

 2   v 1   0.292(1130)   330 d   2d  330 10

6

Ans .

Ans.



Ans.

Ans .

(0.75)  248 10  in 6

An s.

_____________________________________________________________________________ 3-24 F 3000  6790 psi  6.79 kpsi Ans.    A  4 0.752



FL L 60    6790  0.0392 in AE E 10.4 106 



0.0392 6  65310   653 L 60 From Table A-5, v = 0.333

1 



 2   v 1   0.333(653)   217

A ns.

Ans.

Ans .

d   2d  217 10 6  (0.75)  163 10 6  in

Ans.

Chapter 3 - Rev. A, Page 20/100

_____________________________________________________________________________ 3-25  d 0.0001 d   0.0001 2  d d From Table A-5, v = 0.326, E = 119 GPa  0.0001  306.7 10 6  1  2  v 0.326 FL F and   , so  AE A E  1E  306.7 106  (119) 109   36.5 MPa = L F   A  36.5 106 

  0.03

2

An s .  25 800 N  25.8 kN 4 S y = 70 MPa >  , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26



FL L 8(12)    20 000  0.185 in AE E 10.4  106 

Ans .

_____________________________________________________________________________ 3-27



FL L 3    140 106   0.00586 m  5.86 mm Ans . AE E 71.7  109 

_____________________________________________________________________________ 3-28

 

FL L 10(12)    15 000  0.173 in AE E 10.4  106 

Ans .

_____________________________________________________________________________ 3-29 With  z  0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule,