Shigley's Mechanical Engineering Design 10th Edition Budynas Solutions Manual

Shigley's Mechanical Engineering Design 10th Edition Budynas Solutions Manual Completed download: https://testbankreal.com/download/shigleys-mechanical-engineering-design10th-edition-solutions-manual-budynas-nisbett/

Chapter 4 For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For a cantilever, kl = F/ , = F/kl. For the assembly, k = F/y, or, y = F/k = l +  Thus F Fl 2 F y   k kT kl Solving for k kk 1 k 2  2l T Ans. l 1 kl l  kT  kT kl ______________________________________________________________________________ 4-1

For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For each cantilever, kl = F/l, l = F/kl, and,L = F/kL. For the assembly, k = F/y, or, y = F/k = l + l +L. Thus F Fl 2 F F y    k kT kl k L Solving for k k L kl kT 1 k 2  Ans. 2 l 1 1 kl k Ll  kT k L  kT kl   kT kl k L ______________________________________________________________________________ 4-2

4-3

(a) For a torsion bar, k =T/ =GJ/l. Two springs in parallel, with J =di 4/32, and d1 = d2 = d, J G J G   d4 d4  k  1  2  G 1  2  x l  x 32  x l  x 



1  1 Gd 4    32 x lx Deflection equation, 

Shigley’s MED, 10th edition

Ans.

(1)

Chapter 2 Solutions, Page 1/22



T1 x T2  l  x   JG JG

T2  l  x  (2) x From statics, T1 + T2 = T = 1500. Substitute Eq. (2) results in

T1 

lx T2    T2  1500  x 



x l

T2  1500

Ans.

(3)

lx Ans. (4) l  1  1 3 (b) From Eq. (1), k   0.54 11.5 106      28.2 10  lbf  in/rad 32 5 10  5   10  5 From Eq. (4), T1  1500  750 lbf  in Ans. 10 5 From Eq. (3), T2  1500  750 lbf  in Ans. 10 16Ti 16 1500  From either section,     30.6 103 psi  30.6 kpsi Ans.  di3  0.53 T1  1500

Substitute into Eq. (2) resulting in



Ans.

 



______________________________________________________________________________ 4-4

Deflection to be the same as Prob. 4-3 where T1 = 750 lbfin, l1 = l / 2 = 5 in, and d1 = 0.5 in 1=2=

T1  4 



32 Or,

4 1

d G



T2  6 



32

d G

T1  15 103  d14

T2  10 103  d 24

Equal stress,  1   2 

4 2





750  5

32

 0.5  G 4



4T1 6T2  4  60 103  d14 d2

(1)

(2)

(3)

16T1 16T2 T T   13  23 3 3  d1  d 2 d1 d 2

(4)

Divide Eq. (4) by the first two equations of Eq.(1) results in T1 T2 3 d1 d3  2  d 2  1.5d1 (5) 4T1 6T2 d14 d 24

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 2/22

Statics, T1 + T2 = 1500

(6)

Substitute in Eqs. (2) and (3), with Eq. (5) gives

15 103  d14  10 103  1.5d1   1500 4

Solving for d1 and substituting it back into Eq. (5) gives d1 = 0.388 8 in, d2 = 0.583 2 in Ans. From Eqs. (2) and (3), T1 = 15(103)(0.388 8)4 = 343 lbfin T2 = 10(103)(0.583 2)4 = 1 157 lbfin

Ans. Ans.

Deflection of T is

1 

343  4  T1l1   0.053 18 rad J1G  / 32   0.388 84 11.5 106 

Spring constant is

k

T

The stress in d1 is

1 

1



1500  28.2 103  lbf  in 0.053 18

Ans.

16  343 16T1   29.7 103  psi  29.7 kpsi 3 3  d1   0.388 8

Ans.

16 1 157  16T2   29.7 103  psi  29.7 kpsi Ans. 3 3  d 2   0.583 2  ______________________________________________________________________________ The stress in d1 is

4-5

2 

(a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan , and the area is A =  (r1 + x tan )2. The deflection of the tapered portion is l

l

l

F F dx F 1   dx   2  AE  E 0  r1  x tan    E  r1  x tan   tan  0 

 1 1 F  1 1 F        E  r1 tan  tan   r1  l tan     E tan   r1 r2 



r2  r1 F F l tan  Fl    E tan  r1r2  E tan  r1r2  r1r2 E



4 Fl  d1d 2 E

0

Ans.

(b) For section 1,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 3/22

Fl 4 Fl 4(1000)(2)    3.40(104 ) in 2 2 6 AE  d1 E  (0.5 )(30)(10 )

1 

Ans.

For the tapered section, 4 Fl 4 1000(2)    2.26(104 ) in Ans.  d1d 2 E  (0.5)(0.75)(30)(106 ) For section 2, Fl 4 Fl 4(1000)(2) 2     1.51(104 ) in Ans. 2 2 6 AE  d1 E  (0.75 )(30)(10 ) ______________________________________________________________________________ 4-6

(a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan , and the polar second area moment is J = ( /2) (r1 + x tan )4. The angular deflection of the tapered portion is l

l

l

T 2T dx 1 2T 1    dx   4  GJ  G 0  r1  x tan   3  G  r1  x tan  3 tan  0 

0

 2 2 T  1 1 T 1 1    3    3 3 G  r1 tan  tan   r1  l tan    3 G tan   r13 r23 

2 2 r23  r13 2 T 2 T  l  r23  r13 2 Tl  r1  r1r2  r2       3 G tan  r13r23 3 G  r2  r1  r13r23 3 G r13r23 2 2 32 Tl  d1  d1d 2  d 2   3 G d13d 23

Ans.

(b) The deflections, in degrees, are: Section 1,

1 

Tl  180  32Tl  180  32(1500)(2)  180         2.44 deg 4 GJ     d1 G     (0.54 )11.5(106 )   

Ans.

Tapered section, 32 Tl (d12  d1d 2  d 2 2 )  180     3 Gd13 d 23    2 2 32 (1500)(2) 0.5  (0.5)(0.75)  0.75   180      1.14 deg 3 11.5(106 )(0.53 )(.753 )   

Ans.

Section 2,

2 

Tl  180  32Tl  180  32(1500)(2)  180         0.481 deg 4 GJ     d 2 G     (0.754 )11.5(106 )   

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 4/22

______________________________________________________________________________ 4-7

The area and the elastic modulus remain constant. However, the force changes with respect to x. From Table A-5, the unit weight of steel is  = 0.282 lbf/in3 and the elastic modulus is E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top). F = (A)(lx) Fdx w l   1   l 2 0.282 500(12)    (l  x)dx   lx  x 2     0.169 in AE E 0 E 2  0 2E 2(30)106 2

l

c  

l

o

From the weight at the bottom of the cable, 4(5000) 500(12)  Wl 4Wl    5.093 in 2 AE  d E  (0.52 )30(106 )    c  W  0.169  5.093  5.262 in Ans.

W 

The percentage of total elongation due to the cable’s own weight 0.169 (100)  3.21% Ans. 5.262 ______________________________________________________________________________

4-8

Fy = 0 = R 1  F  R 1 = F MA = 0 = M1  Fa  M1 = Fa VAB = F, MAB =F (x  a ), VBC = MBC = 0 Section AB:  1 F  x2  AB  F  x  a  dx    ax   C1  EI EI  2  AB = 0 at x = 0  C1 = 0

y AB 

 F  x2 F  x3 x2   ax dx   a      C2 EI   2 EI  6 2 

(1)

(2)

yAB = 0 at x = 0  C2 = 0, and Fx 2 y AB  Ans.  x  3a  6 EI Section BC:

 BC 

1 EI

  0  dx  0  C

Shigley’s MED, 10th edition

3

Chapter 2 Solutions, Page 5/22

From Eq. (1), at x = a (with C1 = 0),  

 F  a2 Fa 2  a ( a )   = C3. Thus,   EI  2 2 EI 

Fa 2 2 EI Fa 2 Fa 2  dx   x  C4 2 EI  2 EI

 BC   yBC

(3)

F  a3 a2  Fa 3 From Eq. (2), at x = a (with C2 = 0), y  . Thus, from Eq. (3)  a    EI  6 2  3EI 

Fa 2 Fa 3 a  C4   2 EI 3EI

yBC  



C4 

Fa 3 6 EI

Fa 2 Fa 3 Fa 2 x   a  3x  2 EI 6 EI 6 EI

Substitute into Eq. (3), obtaining

Ans.

The maximum deflection occurs at x= l, Fa 2 ymax  Ans.  a  3l  6 EI ______________________________________________________________________________

4-9

MC = 0 = F (l /2)  R1 l  R1 = F /2 Fy = 0 = F /2 + R 2  F 

R 2 = F /2

Break at 0  x  l /2: VAB = R 1 = F /2,

MAB = R 1 x = Fx /2

Break at l /2  x  l : VBC = R 1  F =  R 2 =  F /2,

MBC = R 1 x  F ( x  l / 2) = F (l  x) /2

Section AB:

 AB 

1 EI

Fx F x2 dx   C1 2 EI 4 2

l F   2 C  0 From symmetry, AB = 0 at x = l /2  1 4 EI Shigley’s MED, 10th edition



C1  

Fl 2 . Thus, 16 EI

Chapter 2 Solutions, Page 6/22

F x 2 Fl 2 F   4x2  l 2   EI 4 16 EI 16 EI

 AB 

y AB 

F 16 EI

yAB = 0 at x = 0 y AB 

2 2   4 x  l  dx 

(1)

F  4 x3 2   l x   C2  16 EI  3 

 C2 = 0, and,

Fx 4 x 2  3l 2   48EI

(2)

yBC is not given, because with symmetry, Eq. (2) can be used in this region. The maximum deflection occurs at x =l /2,

ymax

l F  2  Fl 3 2  l     4    3l 2    48 EI   2  48 EI 

Ans.

______________________________________________________________________________ 4-10 From Table A-6, for each angle, I1-1 = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4 From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam 3 with w = 1 N/mm and l = 3000 mm. Fa 2 wl4 (a  3l )  6 EI 8 EI 2500(2000) 2 (1)(3000) 4   2000  3(3000)  6(207)103 (4.14)106 8(207)(103 )(4.14)(106 )

ymax 

 25.4 mm

Ans.

M O   Fa  (wl 2 / 2) =  2500(2000)  [1(30002)/2] =  9.5(106) Nmm

From Table A-6, from centroid to upper surface is y = 29 mm. From centroid to bottom surface is y = 29.0  100=  71 mm. The maximum stress is compressive at the bottom of the beam at the wall. This stress is My 9.5(106 )(71)  max     163 MPa Ans. I 4.14(106 ) ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 7/22

4-11 14 10 (450)  (300)  465 lbf 20 20 6 10 RC  (450)  (300)  285 lbf 20 20

RO 

M1 = 465(6)12 = 33.48(103) lbfin M2 = 33.48(103) +15(4)12 = 34.20(103) lbfin M max 34.2  15  Z  2.28 in 3 Z Z For deflections, use beams 5 and 6 of Table A-9 2 F1a[l  (l / 2)]  l  l  F2l 3 2 y x 10ft     a  2l   6 EIl 2  48 EI  2  450(72)(120) 300(2403 ) 2 2 2 0.5  120  72  240    48(30)(106 ) I 6(30)(106 ) I (240)

 max 

I  12.60 in 4  I / 2  6.30 in 4

Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) = 6.00 in3 12.60  1  ymidspan      0.421 in 14.98  2  34.2  max   5.70 kpsi 6.00 ______________________________________________________________________________ 4-12

I



(1.54 )  0.2485 in 4

64 From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and l = 39 in Fba 2 wa y [a  b 2  l 2 ]  (2la 2  a 3  l 3 ) 6 EIl 24 EI

yA 

340(24)15 152  242  392  6(30)106 (0.2485)39  (150 /12)(15)  2(39)(152 )  153  393   0.0978 in  6 24(30)10 (0.2485)

Ans.

At x = l /2 = 19.5 in 2 2 3  Fa[l  (l / 2)]  l  l  w(l / 2)   l   l  2 3 y    a  2l    2l       l  6 EIl 2  24 EI   2   2   2  

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 8/22

y

340(15)(19.5) 19.52  152  392  6 6(30)(10 )(0.2485)(39) (150 /12)(19.5)  2(39)(19.52 )  19.53  393   0.1027 in  24(30)(106 )(0.2485) 

Ans.

0.1027  0.0978 (100)  5.01% Ans. 0.0978 ______________________________________________________________________________ % difference 

4-13 I 

 

1 (6)(323 )  16.384 103 mm 4 12

From Table A-9-10, beam 10 Fa 2 yC   (l  a ) 3EI Fax 2 y AB  l  x2   6 EIl dy AB Fa 2  (l  3 x 2 ) dx 6 EIl dy AB  A dx Fal 2 Fal A   6 EIl 6 EI Fa 2l yO   A a   6 EI

At x = 0,

With both loads, Fa 2l Fa 2  (l  a ) 6 EI 3EI Fa 2 400(3002 )  (3l  2a )   3(500)  2(300)  3.72 mm Ans. 6 EI 6(207)103 (16.384)103 At midspan, 2 2 Fa(l / 2)  2  l   3 Fal 2 3 400(300)(5002 ) yE    1.11 mm Ans. l      6 EIl  24 207 103 16.384 103  2   24 EI _____________________________________________________________________________  4-14 I  (24  1.54 )  0.5369 in 4 64 yO  

 

 

From Table A-5, E = 10.4 Mpsi From Table A-9, beams 1 and 2, by superposition

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 9/22

200  4(12) 300  2(12) FB l 3 FA a 2 yB    (a  3l )    2(12)  3(4)(12) 3EI 6 EI 3(10.4)106 (0.5369) 6(10.4)106 (0.5369) yB  1.94 in Ans. ______________________________________________________________________________ 3

2

4-15 From Table A-7, I = 2(1.85) = 3.70 in4 From Table A-5, E = 30.0 Mpsi From Table A-9, beams 1 and 3, by superposition

5  2(5 /12) (60 )  0.182 in Ans. Fl 3 ( w  wc )l 4 150(603 ) yA      6 3EI 8EI 3(30)10 (3.70) 8(30)106 (3.70) ______________________________________________________________________________  4 4-16 I  d 64 From Table A-5, E  207(103 ) MPa From Table A-9, beams 5 and 9, with FC = FA = F, by superposition FB l 3 Fa 1   FBl 3  2 Fa(4a 2  3l 2 )  yB    (4a 2  3l 2 )  I  48EI 24 EI 48EyB 1 I 550(10003 )  2  375  (250) 4(250 2 )  3(1000 2 )  3 48(207)10  2  4





 

 53.624 103 mm 4

d

4

64

I 

4

64

(53.624)103  32.3 mm

Ans.

  ______________________________________________________________________________ 4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by superposition









MA 3 Fax 2 x  3lx 2  2l 2 x  l  x2 6 EIl 6 EIl 1   M A x3  3lx 2  2l 2 x  Fax l 2  x 2  Ans.  6 EIl d M F (x  l)  yBC    A x3  3lx 2  2l 2 x   ( x  l )  [( x  l )2  a(3x  l )] 6EI   x l  dx  6 EIl M l F (x  l)   A (x  l)  [( x  l ) 2  a(3x  l )] 6 EI 6 EI (x  l)   M Al  F ( x  l ) 2  a (3x  l )  Ans. 6 EI ______________________________________________________________________________ y AB 











Shigley’s MED, 10th edition







Chapter 2 Solutions, Page 10/22

4-18 Note to the instructor: Beams with discontinuous loading are better solved using singularity functions. This eliminates matching the slopes and displacements at the discontinuity as is done in this solution. a wa   M C  0  R1l  wa  l  a  2   R1  2l  2l  a  Ans. wa wa 2  Fy  0  2l  2l  a   R2  wa  R2  2l Ans. wa w VAB  R1  wx =  2l  a   wx = 2l  a  x   a 2  Ans. 2l 2l 2 wa VBC   R2   Ans. 2l  w  x2  M AB   VAB dx   2l  ax    a 2 x   C1 2l   2  wx  2al  a 2  lx  Ans. M AB  0 at x  0  C1  0  M AB  2l  wa 2 wa 2 M BC   VBC dx    dx   x  C2 2l 2l wa 2 wa 2 M BC  0 at x  l  C2   M BC  (l  x) Ans. 2 2l M 1 wx 1 w 2 1 2 2 1 3   AB   AB dx  2al  a 2  lx  dx    alx  a x  lx   C3    EI EI 2l EI  2l  2 3   y AB    AB dx  

1 w 2 1 2 2 1 3   alx  a x  lx   C3  dx   EI  2l  2 3  

1 w 1 3 1 2 3 1 4    alx  a x  lx   C3 x  C4   EI  2l  3 6 12  

y AB  0 at x  0  C4  0

 BC

M 1   BC dx  EI EI

 wa 2 1  wa 2  1 2  2l (l  x) dx  EI  2l  lx  2 x   C5 

 AB   BC at x  a  1 EI

 w 2 1 4 1 3  1  wa 2  1 2 wa 3 ala  a  la  C  la  a  C  C   C5 3 5 3      2l  2 3  2  6    EI  2l  

Shigley’s MED, 10th edition

(1)

Chapter 2 Solutions, Page 11/22

  1  wa 2  1 2 1  wa 2  1 2 1 3  lx  x  C dx  5    lx  x   C5 x  C6     EI  2l  2  EI  2l  2 6    2 2 wa l  0 at x  l  C6    C5l 6  1  wa 2  1 2 1 3 1 3    lx  x  l   C5 ( x  l )   EI  2l  2 6 3  

yBC    BC dx  yBC yBC

y AB  yBC at x  a  w 1 1 5 1 4 wa 2  1 2 1 3 1 3  3 ala  a  la  C a  3    la  a  l   C5 (a  l ) 2l  3 6 12 2l  2 6 3   wa 2 C3 a  (2)  3la 2  4l 3   C5 (a  l ) 24l wa 2  a 2  4l 2  . Substituting this back into (2) gives Substituting (1) into (2) yields C5   24l 2 wa C3  4al  a 2  4l 2  . Thus,  24l w y AB  4alx 3  2a 2 x 3  lx 4  4a 3lx  a 4 x  4a 2l 2 x   24 EIl wx  2  y AB  2ax 2 (2l  a)  lx 3  a 2  2l  a   Ans.  24 EIl  w yBC  6a 2lx 2  2a 2 x3  a 4 x  4a 2l 2 x  a 4l  Ans.  24 EIl This result is sufficient for yBC. However, this can be shown to be equivalent to w w yBC  4alx3  2a 2 x3  lx 4  4a 2l 2 x  4a 3lx  a 4 x   ( x  a)4  24 EIl 24 EI w yBC  y AB  ( x  a)4 Ans. 24 EI by expanding this or by solving the problem using singularity functions. ______________________________________________________________________________

4-19

The beam can be broken up into a uniform load w downward from points A to C and a uniform load w upward from points A to B. Using the results of Prob. 4-18, with b = a for A to C and a = a for A to B, results in wx  wx  2 2 y AB  2bx 2 (2l  b)  lx3  b 2  2l  b    2ax 2 (2l  a)  lx 3  a 2  2l  a    24 EIl   24 EIl  wx  2 2 2bx 2 (2l  b)  b 2  2l  b   2ax 2 (2l  a)  a 2  2l  a    24 EIl  w  2  2bx3 (2l  b)  lx 4  b 2 x  2l  b   24 EIl

 yBC

Ans.

  4alx 3  2a 2 x 3  lx 4  4a 2l 2 x  4a 3lx  a 4 x   l ( x  a ) 4  Ans.

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 12/22

w  4blx 3  2b 2 x 3  lx 4  4b 2l 2 x  4b3lx  b 4 x  l ( x  b ) 4  24 EIl w  4alx 3  2a 2 x 3  lx 4  4a 2l 2 x  4a 3lx  a 4 x  l ( x  a ) 4   24 EIl w ( x  b) 4  ( x  a ) 4   y AB  Ans. 24 EI ______________________________________________________________________________ yCD 

4-20

Note to the instructor: See the note in the solution for Problem 4-18. wa 2 wa F  0  R   wa  RB   2l  a  Ans.  y B 2l 2l For region BC, isolate right-hand element of length (l + a  x) wa 2 VAB   RA   , VBC  w  l  a  x  Ans. 2l wa 2 w 2 M AB   RA x   x, M BC    l  a  x  Ans. 2l 2 wa 2 2 EI AB   M AB dx   x  C1 4l wa 2 3 EIy AB   x  C1 x  C2 12l wa 2 3 x  C1 x yAB = 0 at x = 0  C2 = 0  EIy AB   12l w a 2l yAB = 0 at x = l  C1   12 w a 2 3 w a 2l wa 2 x 2 wa 2 x 2 2 EIy AB   x  x l  x  y  l  x 2  Ans.    AB 12l 12 12l 12 EIl w 3 EI BC   M BC dx    l  a  x   C3 6 w 4 EIyBC    l  a  x   C3 x  C4 24 wa 4 wa 4  C3l  C4  0  C4   C3l yBC = 0 at x = l   (1) 24 24 wa 2l wa 2l wa 3 wa 2    C3  C3   AB = BC at x = l   l  a  4 12 6 6 wa 2 2  a  4l  l  a   . Substitute back into yBC Substitute C3 into Eq. (1) gives C4  24   1  w wa 2 wa 4 wa 2l 4 yBC   l  a  x  x l  a       l  a   EI  24 6 24 6  

w  4 l  a  x   4a 2  l  x  l  a   a 4    24 EI 

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 13/22

4-21

Table A-9, beam 7, w l 100(10) R1  R2    500 lbf  2 2 wx 100 x  2(10) x 2  x 3  103  y AB  2lx 2  x 3  l 3    6 24 EI 24  30 10  0.05   2.7778 106  x  20 x 2  x 3  1000 

Slope:

 AB 

d y AB w  6lx 2  4 x3  l 3   dx 24 EI

w wl 3 2 3 3 6 l l  4 l  l    x l 24 EI 24 EI 100 103  w l3 x l  x l   x  10   2.7778 103   x  10  6 x l 24 EI 24(30)10 (0.05)

At x = l,  AB

yBC   AB



From Prob. 4-20, 2 wa 2 100  4  RA    80 lbf  2l 2(10) y AB

RB 

100  4  wa  2l  a    2(10)  4  480 lbf  2l 2(10)

100  42  x wa 2 x 2 2  l x  102  x 2   8.8889 10 6  x 100  x 2    6 12 EIl 12  30 10  0.05 

w  4  l  a  x   4a 2  l  x  l  a   a 4   24 EI 100 10  4  x 4  4  42  10  x 10  4   4 4   6  24  30 10  0.05  

yBC  

4  2.7778 106  14  x   896 x  9216    Superposition, RA  500  80  420 lbf  RB  500  480  980 lbf 

y AB  2.7778 10

6

yBC  2.7778 103

 x  20 x  x  1000  8.8889 10  x 100  x    x  10   2.7778 10  14  x   896 x  9216  2

6

3

6

4

2

Ans. Ans.

Ans.

The deflection equations can be simplified further. However, they are sufficient for plotting. Using a spreadsheet, x y

0 0.5 1 1.5 2 2.5 3 3.5 0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027

x y

4 4.5 5 5.5 6 6.5 7 7.5 -0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 14/22

x y x y

8 8.5 9 9.5 10 10.5 11 11.5 -0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036 12 12.5 0.001244 0.001419

13 0.001575

13.5 14 0.001722 0.001867

0.002

Beam Deflection, Prob. 4-21

0.001 0.000 -0.001 y (in) -0.002 -0.003 -0.004 -0.005 -0.006 0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

x (in)

______________________________________________________________________________ 4-22

(a) Useful relations

k

F 48 EI  3 y l

3 kl 3 1800  36  I   0.05832 in 4 6 48E 48(30)10

From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or, h

4

6 I 4 6(0.05832)   0.514 in 5 5

h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically changes the spring rate, so changing the base will make finding a close solution easier. Trial and error was applied to find the combination of values from Table A-17 that yielded the closet desired spring rate. h (in) 1/2 1/2 1/2 9/16 9/16

Shigley’s MED, 10th edition

b (in) 5 5½ 5¾ 5 4

b/h 10 11 11.5 8.89 7.11

k (lbf/in) 1608 1768 1849 2289 1831

Chapter 2 Solutions, Page 15/22

h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is still reasonably close to 10. (b) I  5.5(0.5)3 /12  0.05729 in 4 Mc ( Fl / 4)c 4 I 4(60)103 (0.05729)    F   1528 lbf I I lc  36  (0.25)

(1528)  36  Fl 3 y   0.864 in Ans. 48 EI 48(30)106 (0.05729) ______________________________________________________________________________ 3

4-23 From the solutions to Prob. 3-68, T1  60 lbf and T2  400 lbf I

d4 64



 (1.25) 4 64

 0.1198 in 4

From Table A-9, beam 6,

Fb x  Fb x  z A   1 1 ( x 2  b12  l 2 )  2 2 ( x 2  b2 2  l 2 )  6 EIl  6 EIl  x 10in (575)(30)(10)  102  302  402   6 6(30)10 (0.1198)(40) 460(12)(10)  102  122  402   0.0332 in  6 6(30)10 (0.1198)(40)

Ans.

 d Fb x Fb x dz      1 1 ( x 2  b12  l 2 )  2 2 ( x 2  b2 2  l 2 )    6 EIl  dx  x 10in   x 10in  dx  6 EIl Fb  Fb     1 1 (3 x 2  b12  l 2 )  2 2 (3 x 2  b2 2  l 2 )  6 EIl  6 EIl  x 10in

 A  y   



(575)(30) 3 102   302  402  6  6(30)10 (0.1198)(40)  460(12) 3 102   122  402    6(30)106 (0.1198)(40) 

 6.02(104 ) rad Ans. ______________________________________________________________________________

4-24 From the solutions to Prob. 3-69, T1  2880 N and T2  432 N I

d4 64



Shigley’s MED, 10th edition

 (30) 4 64

 

 39.76 103 mm 4

Chapter 2 Solutions, Page 16/22

The load in between the supports supplies an angle to the overhanging end of the beam. That angle is found by taking the derivative of the deflection from that load. From Table A-9, beams 6 (subscript 1) and 10 (subscript 2), y A   BC

C

 a2  beam 6   y A beam10

(1)

 d  F1a1  l  x  2    F1a1  2  x  a  2 lx 6lx  3x 2  a12  2l 2           1 C  x l  dx  6 EIl   x l  6 EIl Fa  1 1  l 2  a12  6 EIl Equation (1) is thus

 BC

F1a1 2 Fa2 l  a12  a2  2 2 (l  a2 )  6 EIl 3EI 3312(230) 2070(3002 ) 2 2  510  230 300      3(207)103 (39.76)103  510  300  6(207)103 (39.76)103 (510)  7.99 mm Ans.

yA 

The slope at A, relative to the z axis is

 A  z 

F1a1 2  d  F (x  l)  ( x  l ) 2  a2 (3 x  l )    (l  a12 )    2 6 EIl   x l  a2  dx  6 EI

F1a1 2 F l  a12   2 3( x  l ) 2  3a2 ( x  l )  a2 (3x  l )   x l  a2 6 EIl 6 EI Fa F  1 1 (l 2  a12 )  2  3a2 2  2la2  6 EIl 6 EI 3312(230)  5102  2302   3 3 6(207)10 (39.76)10 (510) 2070 3(3002 )  2(510)(300)   3 3  6(207)10 (39.76)10  0.0304 rad Ans. ______________________________________________________________________________ 

4-25 From the solutions to Prob. 3-70, T1  392.16 lbf and T2  58.82 lbf I

d4



64 From Table A-9, beam 6,

Shigley’s MED, 10th edition

 (1) 4 64

 0.049 09 in 4

Chapter 2 Solutions, Page 17/22

(350)(14)(8) Fb x  y A   1 1  x 2  b12  l 2    82  142  222   0.0452 in Ans.  6  6 EIl  x 8in 6(30)10 (0.049 09)(22)

(450.98)(6)(8) F b x  z A   2 2 ( x 2  b2 2  l 2 )   82  62  222   0.0428 in Ans.  6 6 EIl 6(30)10 (0.049 09)(22)   x 8in

The displacement magnitude is   y A2  z A2  0.04522  0.04282  0.0622 in

Ans.

d y  d Fb x Fb     1 1  x 2  b12  l 2     1 1 (3a12  b12  l 2 )    x  a1 6 EIl  d x  x  a1  dx  6 EIl

 A  z   

(350)(14) 3  82   142  222   0.00242 rad 6  6(30)10 (0.04909)(22) 

Ans.

 dz  d  F b x Fb      2 2 ( x 2  b22  l 2 )    2 2  3a12  b22  l 2     x  a1 6 EIl  dx  6 EIl  d x  x  a1

 A  y    

(450.98)(6) 3  82   62  222   0.00356 rad 6  6(30)10 (0.04909)(22) 

Ans.

The slope magnitude is  A  0.002422   0.00356   0.00430 rad Ans. 2

______________________________________________________________________________ 4-26 From the solutions to Prob. 3-71, T1  250 N and T2  37.5 N I

d4 64



 (20) 4 64

 7 854 mm 4

345sin 45o  (550)(300)   F1 y b1 x 2 2 2  yA   ( x  b1  l )   3002  5502  8502   3 6(207)10 (7 854)(850)  6 EIl  x 300mm  1.60 mm Ans. Fb x F b x  z A   1z 1 ( x 2  b12  l 2 )  2 2 ( x 2  b2 2  l 2 )  6 EIl  6 EIl  x 300mm

 345cos 45  (550)(300) o



 300 6(207)10 (7 854)(850)

2

3



 5502  8502 

287.5(150)(300) 3002  1502  8502   0.650 mm  3 6(207)10 (7 854)(850)

The displacement magnitude is  

Shigley’s MED, 10th edition

Ans.

y A2  z A2  1.602   0.650   1.73 mm 2

Ans.

Chapter 2 Solutions, Page 18/22

 d  F1 y b1 x 2 2 2   F1 y b1 d y   x  b1  l     (3a12  b12  l 2 )    d x  x  a1  dx  6 EIl   x a1 6 EIl

 A  z   

  345sin 45o  (550)

3  3002   5502  8502   0.00243 rad  6(207)10 (7 854)(850) 

Ans.

3

dz  d F b x Fb x      1z 1  x 2  b12  l 2   2 2  x 2  b2 2  l 2     6 EIl   x  a1  dx  6 EIl  d x  x  a1

 A  y   

F1z b1 Fb 3a12  b12  l 2   2 2  3a12  b22  l 2   6 EIl 6 EIl o  345cos 45  (550) 3 3002  5502  8502      6(207)103 (7 854)(850) 





287.5(150) 3  3002   1502  8502   1.91104 rad  6(207)103 (7 854)(850) 

Ans.

The slope magnitude is  A  0.002432  0.0001912  0.00244 rad Ans. ______________________________________________________________________________ 4-27 From the solutions to Prob. 3-72, FB  750 lbf I

d4 64



 (1.25) 4 64

 0.1198 in 4

From Table A-9, beams 6 (subscript 1) and 10 (subscript 2) F2 y a2 x 2  F1 y b1 x 2  yA   x  b12  l 2   l  x 2    6 EIl  6 EIl  x 16in

 300 cos 20  (14)(16)  o

6(30)106 (0.119 8)(30)

 0.0805 in

16

2

 14  30   2

 750sin 20  (9)(16) o

2

6(30)106 (0.119 8)(30)

 30

2

 162 

Ans.

F ax F b x  z A   1z 1  x 2  b12  l 2   2 z 2  l 2  x 2   6 EIl  6 EIl  x 16in

 300sin 20  (14)(16) o



16 6(30)10 (0.119 8)(30) 6

 0.1169 in

2

 14  30

2



 750 cos 20  (9)(16)  o

 30 6(30)10 (0.119 8)(30) 6

2

 162 

Ans.

The displacement magnitude is  

Shigley’s MED, 10th edition

2

y A2  z A2  0.08052   0.1169   0.142 in 2

Ans.

Chapter 2 Solutions, Page 19/22

 d  F1 y b1 x 2 F2 y a2 x 2   d y   x  b12  l 2   l  x 2      6 EIl  d x  x  a1  dx  6 EIl   x  a1

 A  z  

 3a  b  l   6EIl  l  3a  6 EIl  300 cos 20  (14) 3 16  14  30      6(30)10 (0.119 8)(30) 



F1 y b1

2 1

2 1

2

F2 y a2

2

2 1

o

2

2

2

6

 750sin 20  (9) o



302  3 162    8.06 105  rad  6(30)10 (0.119 8)(30)  6

Ans.

dz  d F b x F ax      1z 1  x 2  b12  l 2   2 z 2  l 2  x 2     6 EIl   x  a1  dx  6 EIl  d x  x  a1

 A  y   

F1z b1 F a 3a12  b12  l 2   2 z 2  l 2  3a12   6 EIl 6 EIl o 300sin 20  (14) 750 cos 20o  (9)   2 2 2   302  3 16 2    3 16   14  30   6 6   6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30) 



 0.00115 rad

Ans.

The slope magnitude is  A  8.06 105   0.001152  0.00115 rad Ans. ______________________________________________________________________________ 2

4-28 From the solutions to Prob. 3-73, FB = 22.8 (103) N 4  d 4   50  I   306.8 103  mm4 64 64 From Table A-9, beam 6, F2 y b2 x 2  F1 y b1 x 2  yA   ( x  b12  l 2 )  ( x  b2 2  l 2 )  6 EIl  6 EIl  x  400mm

11103  sin 20o  (650)(400)   4002  6502  10502   3 3 6(207)10 (306.8)10 (1050)  22.8 103  sin 25o  (300)(400)    4002  3002  10502  6(207)103 (306.8)103 (1050)  3.735 mm Ans.

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 20/22

F bx F b x  z A   1z 1 ( x 2  b12  l 2 )  2 z 2 ( x 2  b2 2  l 2 )  6 EIl  6 EIl  x  400mm

11103  cos 20o  (650)(400)    4002  6502  10502  6(207)103 (306.8)103 (1050)  22.8 103  cos 25o  (300)(400)   4002  3002  10502   1.791 mm  3 3 6(207)10 (306.8)10 (1050)

The displacement magnitude is  

y A2  z A2 

 3.735 

2

 1.7912  4.14 mm

Ans.

Ans.

d y  d F b x F bx     1z 1  x 2  b12  l 2   2 z 2  x 2  b2 2  l 2     6 EIl   x  a1  d x  x  a1  dx  6 EIl

 A  z  

 3a  b  l   6 EIl  3a  b  l  6 EIl 1110  sin 20  (650)  3  400   650   6(207)10 (306.8)10 (1050) 



F1 y b1

2 1

3

2 1

F2 y b2

2

2 1

2 2

o

2

3

2

3

2

 1050 2 

 22.8 103  sin 25o  (300)  3  4002   3002  1050 2    3 3  6(207)10 (306.8)10 (1050)   0.00507 rad Ans. dz  d F b x F bx      1z 1  x 2  b12  l 2   2 z 2  x 2  b2 2  l 2     6 EIl   x  a1  dx  6 EIl  d x  x  a1

 A  y   

F1z b1 F b 3a12  b12  l 2   2 z 2  3a12  b22  l 2   6 EIl 6 EIl 3 o 1110  cos 20  (650)  3  4002   6502  1050 2     6(207)103 (306.8)103 (1050) 



 22.8 103  cos 25o  (300)  3  4002   3002  1050 2    3 3  6(207)10 (306.8)10 (1050)   0.00489 rad Ans.

The slope magnitude is  A 

 0.00507    0.00489  2

2

 0.00704 rad Ans.

______________________________________________________________________________ 4-29 From the solutions to Prob. 3-68, T1 = 60 lbf and T2 = 400 lbf , and Prob. 4-23, I = 0.119 8 in4. From Table A-9, beam 6,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 21/22

dz  d  F1z b1 x 2 F bx  x  b12  l 2   2 z 2  x 2  b22  l 2         6 EIl   x 0  dx  6 EIl  d x  x 0 F b F b 575(30)   1z 1  b12  l 2   2 z 2  b2 2  l 2     302  402  6 EIl 6 EIl 6(30)106 (0.119 8)(40)

O  y   



460(12) 122  402   0.00468 rad  6 6(30)10 (0.119 8)(40)

Ans.

 d  F1z a1  l  x  2 F a l  x  2   dz x  a12  2lx   2 z 2 x  a22  2lx          6 EIl  dx  6 EIl  d x  x l   x l F a F a     1z 1  6lx  2l 2  3x 2  a12   2 z 2  6lx  2l 2  3x 2  a22   6 EIl  6 EIl  x l

C  y   

F1z a1 2 F a l  a12   2 z 2  l 2  a22   6 EIl 6 EIl 2 575(10)  40  102  460(28)  402  282     0.00219 rad Ans. 6(30)106 (0.119 8)(40) 6(30)106 (0.119 8)(40) ______________________________________________________________________________ 

4-30 From the solutions to Prob. 3-69, T1 = 2 880 N and T2 = 432 N, and Prob. 4-24, I = 39.76 (103) mm4. From Table A-9, beams 6 and 10 d y  d Fb x Fa x  O  z       1 1 ( x 2  b12  l 2 )  2 2 (l 2  x 2 )   6 EIl   x 0  d x  x 0  dx  6 EIl Fa Fb Fal  Fb    1 1 (3 x 2  b12  l 2 )  2 2 (l 2  3 x 2 )   1 1 (b12  l 2 )  2 2 6 EIl 6 EI  6 EIl  x 0 6 EIl 3 312(280) 2 070(300)(510)  2802  510 2    3 3 6(207)10 (39.76)10 (510) 6(207)103 (39.76)103  0.0131 rad Ans. d y  d  F1a1 (l  x) 2 Fa x  ( x  a12  2lx)  2 2 (l 2  x 2 )      6 EIl   x l  d x  x l  dx  6 EIl

C  z  

Fa Fa Fal  Fa    1 1 (6lx  2l 2  3 x 2  a12 )  2 2 (l 2  3 x 2 )   1 1 (l 2  a12 )  2 2 6 EIl 3EI  6 EIl  x l 6 EIl 3 312(230) 2 070(300)(510)  (5102  2302 )  3 3 6(207)10 (39.76)10 (510) 3(207)103 (39.76)103  0.0191 rad Ans. ______________________________________________________________________________

4-31 From the solutions to Prob. 3-70, T1 = 392.19 lbf and T2 = 58.82 lbf , and Prob. 4-25, I = 0.0490 9 in4. From Table A-9, beam 6

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 22/22

 d  F1 y b1 x 2 F1 y b1 2 2   d y x  b12  l 2     (b1  l )      d x  x 0  dx  6 EIl   x 0 6 EIl 350(14)  142  222   0.00726 rad Ans.  6 6(30)10 (0.04909)(22)

O  z  

dz  d  F2 z b2 x 2 F b  x  b22  l 2      2 z 2  b22  l 2       6 EIl   x 0  dx  6 EIl  d x  x 0 450.98(6)   62  222  6(30)106 (0.04909)(22)

O  y   

 0.00624 rad

Ans.

The slope magnitude is O  0.007262   0.00624   0.00957 rad Ans. 2

 d  F1 y a1 (l  x) 2   d y x  a12  2lx         d x  x l  dx  6 EIl   x l F1 y a1 2  F1 y a1   6lx  2l 2  3 x 2  a12    (l  a12 )   6 EIl  x l 6 EIl 350(8)  222  82   0.00605 rad Ans.  6 6(30)10 (0.0491)(22)

C  z  

dz  d  F2 z a2 (l  x) 2  x  a22  2lx           x l  dx  6 EIl  d x  x l

C  y   

F a F a     2 z 2  6lx  2l 2  3x 2  a22     2 z 2  l 2  a22  6 EIl  6 EIl  x l 450.98(16)  222  162   0.00846 rad  6 6(30)10 (0.04909)(22)

The slope magnitude is C 

 0.00605 

2

Ans.

 0.00846 2  0.0104 rad Ans.

______________________________________________________________________________ 4-32 From the solutions to Prob. 3-71, T1 =250 N and T1 =37.5 N, and Prob. 4-26, I = 7 854 mm4. From Table A-9, beam 6

F1 y b1 2 2   d y  d  F1 y b1 x 2 x  b12  l 2     (b1  l )      d x  x 0  dx  6 EIl   x 0 6 EIl

O  z  

 345sin 45o  (550)  5502  8502   0.00680 rad  3 6(207)10 (7 854)(850)

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 23/22

dz  d  F1z b1 x 2 F bx  x  b12  l 2   2 z 2  x 2  b22  l 2         6 EIl   x 0  dx  6 EIl  d x  x 0

O  y   

345cos 45o  (550) F1z b1 2 2 F2 z b2 2 2  b1  l   b2  l    5502  8502     3 6 EIl 6 EIl 6(207)10 (7 854)(850) 

287.5(150) 1502  8502   0.00316 rad 6(207)103 (7 854)(850)

Ans.

The slope magnitude is O  0.006802  0.003162  0.00750 rad Ans.

   F1 y a1  d y  d  F1 y a1 (l  x) 2 x  a12  2lx      6lx  2l 2  3x 2  a12        d x  x l  dx  6 EIl  x l  6 EIl  x l

C  z  

 345sin 45o  (300)  (l  a )  8502  3002   0.00558 rad  3 6 EIl 6(207)10 (7 854)(850) F1 y a1

2

2 1

Ans.

dz  d  F1z a1 (l  x) 2 F a (l  x) 2  x  a12  2lx   2 z 2 x  a22  2lx          6 EIl   x l  dx  6 EIl  d x  x l

C  y   

345cos 45o  (300) F1z a1 2 F2 z a2 2 2 2   l  a1   6EIl  l  a2    6(207)103 (7 854)(850) 8502  3002  6 EIl 

287.5(700) 8502  7002   6.04 105  rad 6(207)103 (7 854)(850)

The slope magnitude is C 

 0.00558

2

Ans.

 6.04 105   0.00558 rad Ans. 2

________________________________________________________________________ 4-33 From the solutions to Prob. 3-72, FB = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From Table A-9, beams 6 and 10  d F b x F ax  d y O  z       1 y 1  x 2  b12  l 2   2 y 2  l 2  x 2   6 EIl  d x  x 0  dx  6 EIl   x 0 F a F b F al F b    1 y 1  3 x 2  b12  l 2   2 y 2  l 2  3x 2    1 y 1  b12  l 2   2 y 2 6 EIl 6 EI  6 EIl  x 0 6 EIl  300 cos 20o  (14) 750sin 20o  (9)(30) 2 2  14  30   6(30)106 (0.119 8)  0.00751 rad 6(30)106 (0.119 8)(30)

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 24/22

dz  d  F1z b1 x 2 F ax  x  b12  l 2   2 z 2  l 2  x 2         6 EIl   x 0  dx  6 EIl  d x  x 0

O  y   

F a F b F al F b     1z 1  3 x 2  b12  l 2   2 z 2  l 2  3x 2     1z 1  b12  l 2   2 z 2 6 EIl 6 EIl 6 EI  6 EIl  x 0 300sin 20o  (14)  750 cos 20o  (9)(30) 2 2  14  30   6(30)106 (0.119 8)  0.0104 rad 6(30)106 (0.119 8)(30)

Ans.

The slope magnitude is O  0.007512  0.01042  0.0128 rad Ans.  d  F1 y a1 (l  x) 2 F ax    dy  x  a12  2lx   2 y 2  l 2  x 2        6 EIl  dx  x l  dx  6 EIl   x l F a F a F al F a    1 y 1  6lx  2l 2  3 x 2  a12   2 y 2  l 2  3x 2    1 y 1 (l 2  a12 )  2 y 2 6 EIl 3EI  6 EIl  x l 6 EIl

C  z  

 300 cos 20o  (16) 750sin 20o  (9)(30) 2 2   30  16   3(30)106 (0.119 8)  0.0109 rad 6(30)106 (0.119 8)(30)

Ans.

dz  d  F1z a1 (l  x ) 2 F ax  x  a12  2lx   2 z 2  l 2  x 2         6 EIl   x l  dx  6 EIl  d x  x l

C  y   

F a F a F al F a     1z 1  6lx  2l 2  3 x 2  a12   2 z 2  l 2  3 x 2     1z 1  l 2  a12   2 z 2 6 EIl 6 EIl 3EI  6 EIl  x l 

300sin 20o  (16)  750 cos 20o  (9)(30) 2 2 30  1 6    3(30)106 (0.119 8)  0.0193 rad 6(30)106 (0.119 8)(30)

 0.0109    0.0193

The slope magnitude is C 

2

2

Ans.

 0.0222 rad Ans.

______________________________________________________________________________ 4-34 From the solutions to Prob. 3-73, FB = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4. From Table A-9, beam 6

 d  F1 y b1 x 2 F bx   d y x  b12  l 2   2 y 2  x 2  b22  l 2        6 EIl  d x  x 0  dx  6 EIl   x 0

O  z  

11103  sin 20o  (650)   b l  b l    6502  10502     3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050) F1 y b1

2 1

2

F2 y b2

2 2

2

 22.8 103  sin 25o  (300)    3002  10502   0.0115 rad  3 3 6(207)10 (306.8)10 (1050)

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 25/22

dz  d  F1z b1 x 2 F bx  x  b12  l 2   2 z 2  x 2  b22  l 2         6 EIl   x 0  dx  6 EIl  d x  x 0 F b F b   1z 1  b12  l 2   2 z 2  b22  l 2  6 EIl 6 EIl 11103  cos 20o  (650)     6502  10502  6(207)103 (306.8)103 (1050)

O  y   

 22.8 103  cos 25o  (300)   3002  10502   0.00427 rad  3 3 6(207)10 (306.8)10 (1050)

 0.0115    0.00427 

The slope magnitude is O 

2

2

Ans.

 0.0123 rad Ans.

 d  F1 y a1 (l  x) 2 F a (l  x) 2   d y x  a12  2lx   2 y 2 x  a22  2lx         6 EIl  d x  x l  dx  6 EIl   x l F2 y a2  F1 y a1   (6lx  2l 2  3 x 2  a12 )  6lx  2l 2  3x 2  a22    6 EIl  6 EIl  x l

C  z  

11103  sin 20o  (400)     l  a   6EIl  l  a   6(207)10 10502  4002  3 6 EIl (306.8)103 (1050) F1 y a1

2

2 1

F2 y a2

2

2 2

 22.8 103  sin 25o  (750)    10502  7502   0.0133 rad  3 3 6(207)10 (306.8)10 (1050)

Ans.

dz  d  F1z a1 (l  x) 2 F a (l  x) 2  x  a12  2lx   2 z 2 x  a22  2lx          6 EIl   x l  dx  6 EIl  d x  x l

C  y   

F a F a     1z 1  6lx  2l 2  3x 2  a12   2 z 2  6lx  2l 2  3x 2  a22   6 EIl  6 EIl  x l

11103  cos 20o  (400) F1z a1 2 F2 z a2 2  2 2  l  a1   l  a2     10502  4002     3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050)  22.8 103  cos 25o  (750)   10502  7502   0.0112 rad  3 3 6(207)10 (306.8)10 (1050)

Ans.

The slope magnitude is C  0.01332  0.01122  0.0174 rad Ans. ______________________________________________________________________________ 4-35 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-29, I = 0.119 8 in4, and it was found that the greater angle occurs at the bearing at O where (O)y =  0.00468 rad. Since is inversely proportional to I,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 26/22

 new Inew =  old Iold



4 Inew =  dnew /64 =  old Iold / new

1/4

or,

d new

 64  old    I old     new 

The absolute sign is used as the old slope may be negative. 1/4

 64 0.00468  d new   0.119 8   1.82 in Ans.   0.00105  ______________________________________________________________________________ 4-36 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the bearing at C where (C)y =  0.0191 rad. See the solution to Prob. 4-35 for the development of the equation 1/4

d new

 64  old    I old     new 

1/4

 64 0.0191  d new   39.76 103    62.0 mm Ans.   0.00105  ______________________________________________________________________________ 4-37 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-31, I = 0.0491 in4, and the maximum slope is C = 0.0104 rad. See the solution to Prob. 4-35 for the development of the equation 1/4

d new

 64  old    I old     new 

1/4

 64 0.0104  d new   0.0491  1.77 in Ans.   0.00105  ______________________________________________________________________________ 4-38 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-32, I = 7 854 mm4, and the maximum slope is O = 0.00750 rad. See the solution to Prob. 4-35 for the development of the equation

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 27/22

1/4

d new

 64  old    I old     new 

1/4

 64 0.00750  d new   7 854   32.7 mm Ans.   0.00105  ______________________________________________________________________________ 4-39 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-33, I = 0.119 8 in4, and the maximum slope  = 0.0222 rad. See the solution to Prob. 4-35 for the development of the equation 1/4

d new

 64  old    I old     new 

1/4

 64 0.0222  d new   0.119 8   2.68 in Ans.   0.00105  ______________________________________________________________________________ 4-40 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is C = 0.0174 rad. See the solution to Prob. 4-35 for the development of the equation 1/4

d new

 64  old    I old     new 

1/4

 64 0.0174  d new   306.8 103    100.9 mm Ans.   0.00105  ______________________________________________________________________________ 4-41 IAB =  14/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = (0.25)(1.5)3/12 = 0.07031 in4, ICD =  (3/4)4/64 = 0.01553 in4. For Eq. (3-41), p. 116, b/c = 1.5/0.25 = 6   = 0.299. The deflection can be broken down into several parts 1. The vertical deflection of B due to force and moment acting on B (y1). 2. The vertical deflection due to the slope at B, B1, due to the force and moment acting on B (y2 = CD B1 = 2B1).

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 28/22

3. The vertical deflection due to the rotation at B, B2, due to the torsion acting at B (y3 = BC B1 = 5B1). 4. The vertical deflection of C due to the force acting on C (y4). 5. The rotation at C, C, due to the torsion acting at C (y3 = CD C = 2C). 6. The vertical deflection of D due to the force acting on D (y5). 1. From Table A-9, beams 1 and 4 with F =  200 lbf and MB = 2(200) = 400 lbfin 200  63  400  62  y1     0.01467 in 3  30 106  0.04909  2  30 106  0.04909  2. From Table A-9, beams 1 and 4  d  Fx 2 M x2   M x Fx  B1     x  3l   B      3x  6l   B  2 EI   x l  6 EI EI  x l  dx  6 EI

6  l     200  6   2  400    0.004074 rad   Fl  2M B   6  2 EI  2  30 10  0.04909  y 2 = 2(0.004072) = 0.00815 in 3. The torsion at B is TB = 5(200) = 1000 lbfin. From Eq. (4-5)  TL 

1000  6 

B2    0.005314 rad   6  JG  AB 0.09818 11.510 y 3 = 5(0.005314) = 0.02657 in 4. For bending of BC, from Table A-9, beam 1 y4  

200  53 

3  30 106  0.07031

 0.00395 in

5. For twist of BC, from Eq. (3-41), p. 116, with T = 2(200) = 400 lbfin

C 

400  5  0.02482 rad 0.299 1.5 0.253 11.5106

y 5 = 2(0.02482) = 0.04964 in 6. For bending of CD, from Table A-9, beam 1 y6  

200  23 

3  30 106  0.01553

Shigley’s MED, 10th edition

 0.00114 in

Chapter 2 Solutions, Page 29/22

Summing the deflections results in 6

yD   yi  0.01467  0.00815  0.02657  0.00395  0.04964  0.00114  0.1041 in Ans. i 1

This problem is solved more easily using Castigliano’s theorem. See Prob. 4-71. ______________________________________________________________________________ 4-42 The deflection of D in the x direction due to Fz is from: 1. The deflection due to the slope at B, B1, due to the force and moment acting on B (x1 = BC B1 = 5B1). 2. The deflection due to the moment acting on C (x2). 1. For AB, IAB =  14/64 = 0.04909 in4. From Table A-9, beams 1 and 4

 d  Fx 2 M B x 2   M x  Fx x  3 l     3x  6l   B      2 EI   x l  6 EI EI  x l  dx  6 EI

 B1  

6  l    100  6   2  200    0.002037 rad   Fl  2M B   6  2 EI  2  30 10  0.04909  x 1 = 5( 0.002037) =  0.01019 in 2. For BC, IBC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4 2  100  5 M Cl 2 x2    0.04267 in 2 EI 2  30 106  0.001953

The deflection of D in the x direction due to Fx is from: 3. The elongation of AB due to the tension. For AB, the area is A =  12/4 = 0.7854 in2 150  6   Fl  x3    3.82 105  in   6  AE  AB 0.7854  30 10

4. The deflection due to the slope at B, B2, due to the moment acting on B (x1 = BC B2 = 5B2). With IAB = 0.04907 in4,

B2 

5  150  6 M Bl   0.003056 rad EI 30 106  0.04909

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 30/22

x4 = 5( 0.003056) =  0.01528 in 5. The deflection at C due to the bending force acting on C. With IBC = 0.001953 in4 150  53   Fl 3  x5       0.10667 in  3  30 106  0.001953  3EI  BC

6. The elongation of CD due to the tension. For CD, the area is A =  (0.752)/4 = 0.4418 in2 150  2   Fl  x6    2.26 105  in   6  AE CD 0.4418  30 10

Summing the deflections results in xD   xi  0.01019  0.04267  3.82 105  6

i 1

 0.01528  0.10667  2.26 105   0.1749 in Ans.

______________________________________________________________________________ 4-43

JOA = JBC =  (1.54)/32 = 0.4970 in4, JAB =  (14)/32 = 0.09817 in4, IAB =  (14)/64 = 0.04909 in4, and ICD =  (0.754)/64 = 0.01553 in4. T  lOA l AB lBC   Tl   Tl   Tl               GJ OA  GJ  AB  GJ  BC G  J OA J AB J BC 

250(12)  2 9 2      0.0260 rad 6  11.5(10 )  0.4970 0.09817 0.4970  Simplified Tl 250(12)(13) s   GJ 11.5 106   0.09817  

Ans.

 s  0.0345 rad

Ans. Simplified is 0.0345/0.0260 = 1.33 times greater Ans. yD 

Fy lOC 3 3EI AB

  s  lCD  

Fy lCD 3 3EI CD



250 133 

3(30)106  0.04909 

 0.0345(12) 

250 123 

3(30)106  0.01553 

yD  0.847 in Ans. ______________________________________________________________________________

4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 31/22

y

 3000 /12  x 2 25 x 2  x3  25 12 3 wx 2lx 2  x3  l 3          24 EI 24  30 106  485 



 7.159 1010  x  27 106   600 x 2  x3 



Ans.

The maximum height occurs at x = 25(12)/2 = 150 in ymax  7.159 1010 150  27 106   600 1502   1503   1.812 in

Ans.

______________________________________________________________________________ 4-45 From Table A-9, beam 6, Fbx 2 yL  x  b2  l 2   6 EIl Fb 3 yL  x  b2 x  l 2 x   6 EIl dyL Fb   3x 2  b 2  l 2  dx 6 EIl

dyL dx Let  

dy L dx



Fb  b 2  l 2  6 EIl

x 0

and set I  x 0

dL 

 d L4 64

. Thus,

32 Fb  b 2  l 2 

1/4

Ans.

3 El

For the other end view, observe beam 6 of Table A-9 from the back of the page, noting that a and b interchange as do x and –x

dR 

32 Fa  l 2  a 2  3 El

1/4

Ans.

For a uniform diameter shaft the necessary diameter is the larger of d L and d R . ______________________________________________________________________________ 4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the solution for d L of Prob. 4-45,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 32/22

 32nFb  l 2  b 2    d  3 El  

1/4

d

4

32(1.28)(3000)(200)  3002  2002  3 (207)103 (300)(0.001)

d  38.1 mm

I

  38.14 

Ans.

 103.4 103  mm 4

64 From Table A-9, beam 6, the maximum deflection will occur in BC where dyBC /dx = 0

 d  Fa  l  x  2 x  a 2  2lx    0  3x 2  6lx   a 2  2l 2   0   dx  6 EIl  3 x 2  6  300  x  1002  2  3002    0  x 2  600 x  63333  0 x

1 600  6002  4(1)63 333   463.3, 136.7 mm  2

x = 136.7 mm is acceptable.  Fa  l  x  2  ymax   x  a 2  2lx     6 EIl  x 136.7 mm 3 103 100  300  136.7 

136.7 2  1002  2  300 136.7   0.0678 mm Ans. 6  207 10 103.4 10  300  ______________________________________________________________________________ 

3

3

4-47 I =  (1.254)/64 = 0.1198 in4. From Table A-9, beam 6

 F a (l  x) 2  F b x    1 1 ( x  a12  2lx)    2 2 ( x 2  b2 2  l 2 )   6 EIl   6 EIl  2

2

2    150(5)(20  8) 2 2   8  5  2(20)(8)  6    6(30)10  0.1198 (20)

  250(10)(8)  82  102  202    6  6(30)10  0.1198 (20) 

2

1/2

  

 0.0120 in Ans. ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 33/22

4-48 I =  (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9. For F1 = 150 lbf: 0x5 150 15  x Fb x y  1 1  x 2  b12  l 2    x 2  152  202  6 EIl 6  30 106  0.1198  20   5.217 10 6  x  x 2  175 

5  x  20 y

(1)

F1a1  l  x  2 150  5  20  x   x 2  52  2  20  x  x  a12  2lx    6 6 EIl 6  30 10  0.1198  20 

 1.739 106   20  x   x 2  40 x  25 

(2)

For F2 = 250 lbf: 0  x  10 250 10  x Fb x z  2 2  x 2  b22  l 2   x 2  102  202   6 6 EIl 6  30 10  0.1198  20   5.797 106  x  x 2  300 

(3)

10  x  20 F a l  x  2 250 10  20  x   x 2  102  2  20  x  z 2 2 x  a22  2lx    6 6 EIl 6  30 10  0.1198  20   5.797 106   20  x   x 2  40 x  100 

(4)

Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too numerous to tabulate here but the plot is shown below, where the maximum deflection of  = 0.01255 in occurs at x = 9.9 in. Ans. 0.015 0.01 0.005 Displacement (in)

y (in)

0

z (in)

-0.005

Total (in)

-0.01 -0.015 0

5

10

15

20

x (in)

______________________________________________________________________________ Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 34/22

4-49 The larger slope will occur at the left end. From Table A-9, beam 8 MBx 2 ( x  3a 2  6al  2l 2 ) 6 EIl dy AB M B  (3x 2  3a 2  6al  2l 2 ) dx 6 EIl y AB 

With I =  d 4/64, the slope at the left bearing is

dy AB dx

 A  x 0

MB (3a 2  6al  2l 2 ) 4 6 E  d / 64  l

Solving for d 32M B 32(1000) 3(42 )  6(4)(10)  2 102   d4 3a 2  6al  2l 2   4   3 E Al 3 (30)106 (0.002)(10)   0.461 in Ans. ______________________________________________________________________________

4-50 From Table A-5, E = 10.4 Mpsi MO = 0 = 18 FBC  6(100)  FBC = 33.33 lbf The cross sectional area of rod BC is A =  (0.52)/4 = 0.1963 in2. The deflection at point B will be equal to the elongation of the rod BC.

33.33(12)  FL  yB    6.79 105  in   6  AE  BC  0.1963 30 10 

Ans.

______________________________________________________________________________ 4-51 MO = 0 = 6 FAC  11(100) 

FAC = 183.3 lbf

The deflection at point A in the negative y direction is equal to the elongation of the rod AC. From Table A-5, Es = 30 Mpsi.

183.3 12   FL  yA     3.735 104  in   2 6 AE     AC   0.5  / 4 30 10  By similar triangles the deflection at B due to the elongation of the rod AC is y A yB1  6 18



yB1  3 y A  3(3.735)104  0.00112 in

From Table A-5, Ea = 10.4 Mpsi The bar can then be treated as a simply supported beam with an overhang AB. From Table A-9, beam 10,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 35/22

 dy  Fa 2  d  F (x  l) Fa 2  2   yB 2  BD  BC  ( l  a )  7 ( x  l )  a (3 x  l )  (l  a )         x l  a 3EI  dx  6 EI  dx x l  a  3EI F Fa 2 7 Fa Fa 2 3( x  l ) 2  3a( x  l )  a (3 x  l )  |x l  a  7 (l  a )   (2l  3a )  (l  a ) 6 EI 3EI 6 EI 3EI 100  52  7 100  5  (6  5)  2(6)  3(5)  6(10.4)106  0.25(23 ) / 12  3(10.4)106  0.25(23 ) /12 

 

 0.01438 in

yB = yB1 + yB2 =  0.00112  0.01438 =  0.0155 in Ans. ______________________________________________________________________________ 4-52 From Table A-5, E = 207 GPa, and G = 79.3 GPa. FlOC l AB 2 Fl AC l AB 2 Fl AB 3 Fl AB 3  Tl   Tl  yB       l AB    l AB  3EI AB G  dOC 4 / 32  G  d AC 4 / 32  3E  d 34 / 64   GJ OC  GJ  AC 

l 32 Fl AB 2  lOC 2l AB   AC 4   4 4   GdOC Gd AC 3Ed AB 

The spring rate is k = F/ yB. Thus  32l 2  l l 2l AB   k   AB  OC 4  AC 4  4     GdOC Gd AC 3Ed AB  

1

1

 32  2002     2  200  200 200      3 4 3 4 3 4  79.3 10 18 79.3 10 12 3 207 10 8               8.10 N/mm Ans. _____________________________________________________________________________

4-53 For the beam deflection, use beam 5 of Table A-9. F R1  R2  2 F F 1  , and  2  2k1 2k 2 y AB  1 

1   2 l

x

Fx (4 x 2  3l 3 ) 48 EI

 1 k2  k1  x y AB  F    x (4 x 2  3l 3 )  48EI  2k1 2k1k2l 

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 36/22

For BC, since Table A-9 does not have an equation (because of symmetry) an equation will need to be developed as the problem is no longer symmetric. This can be done easily using beam 6 of Table A-9 with a = l /2 F  l / 2  l  x   2 l 2   F Fk2  Fk1 yBC   x  x   2lx  2k1 2k1k2l EIl 4    1 k2  k1  l  x  4 x 2  l 2  8lx  Ans.  F   x   48EI  2k1 2k1k2l  ______________________________________________________________________________

4-54 Fa F , and R2  (l  a ) l l Fa F 1  , and  2  (l  a) lk1 lk2 R1 

y AB  1 

1   2 l

x

Fax 2 (l  x 2 ) 6 EIl

 a  x ax 2 y AB  F   (l  x 2 )  Ans.  k a  k1  l  a    2  2 6 EIl  k1l k1k2l    F (x  l) ( x  l ) 2  a (3x  l )  yBC  1  1 2 x  l 6 EI   a  x (x  l) 2   yBC  F   k a  k l  a  ( x  l )  a (3 x  l ) Ans.     2 1 2   6 EI    k1l k1k2l  ______________________________________________________________________________

4-55 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB (Table A-9, beam 6) y AB 

Fbx 2 x  b2  l 2   6 EIl

dy AB Fb   3x 2  b2  l 2   0 dx 6 EIl

x



l 2  b2 l2 , xmax   0.577l 3 3

3x 2  b 2  l 2  0

Ans.

For x  l/2, xmin  l  0.577l  0.423l Ans. ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 37/22

4-56

M O  1(3000)(1500)  2500(2000)  9.5 106  N·mm

RO  1(3000)  2500  5 500 N

I  4.14(106 ) mm4

From Prob. 4-10,

x2 1  2500 x - 2 000 2 dy x3 2 6 2 EI  9.5 10  x  2 750 x   1250 x  2000  C1 dx 6

M  9.5 106   5500 x 

dy  0 at x  0  C1  0 dx dy x3 2 EI  9.5 106 x  2 750 x 2   1250 x  2000 dx 6 x4 3 EIy  4.75 106 x 2  916.67 x3   416.67 x  2000  C2 24 y  0 at x  0  C2  0 , and therefore

 

 

y

1  3 114 106  x 2  22 103  x3  x 4  10 103  x  2000    24 EI

yB  

1 114 106  30002  22 103  30003 3 6  24  207 10  4.14 10 3 30004  10 103   3000  2000   

 25.4 mm

Ans.

MO = 9.5 (106) Nm. The maximum stress is compressive at the bottom of the beam where y = 29.0  100 =  71 mm

9.5 106  (71) My  max     163 106  Pa  163MPa Ans. I 4.14(106 ) The solutions are the same as Prob. 4-10. ______________________________________________________________________________ 4-57 See Prob. 4-11 for reactions: RO = 465 lbf and RC = 285 lbf. Using lbf and inch units M = 465 x  450 x  721  300 x  1201

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 38/22

dy 2 2  232.5 x 2  225 x  72  150 x  120  C1 dx EIy = 77.5 x3  75 x  723  50 x  1203  C1x EI

y = 0 at x = 0  C2 = 0 y = 0 at x = 240 in 0 = 77.5(2403)  75(240 72)3  50(240  120)3 + C1 x  and, EIy = 77.5 x3  75 x  723  50 x  1203 2.622(106) x

C1 =  2.622(106) lbfin2

Substituting y =  0.5 in at x = 120 in gives 30(106) I ( 0.5) = 77.5 (1203)  75(120  72)3  50(120  120)3 2.622(106)(120) I = 12.60 in4 Select two 5 in  6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4 12.60  1  Ans.     0.421 in 14.98  2  The maximum moment occurs at x = 120 in where Mmax = 34.2(103) lbfin ymidspan 

Mc 34.2(103 )(2.5)   5 710 psi O.K. I 14.98 The solutions are the same as Prob. 4-17. ______________________________________________________________________________

 max 

4-58 I =  (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in. 1 24 RO  12.5  39  (340)  453.0 lbf 2 39 12.5 2 1 M  453.0 x  x  340 x  15 2 dy 12.5 3 2 EI  226.5 x 2  x  170 x  15  C1 dx 6 3 EIy  75.5 x3  0.5208 x 4  56.67 x  15  C1 x  C2 y  0at x  0  C2  0 y  0 at x  39 in y



C1  6.385(104 ) lbf  in 2 Thus,

1  3 75.5 x3  0.5208 x 4  56.67 x  15  6.385 10 4  x    EI

Evaluating at x = 15 in,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 39/22

1 75.5 153   0.5208 154   56.67 15  15 3  6.385 10 4  (15)   30(10 )(0.2485)   0.0978 in Ans.

yA 

6

1 75.5 19.53   0.5208 19.54   56.67 19.5  15 3  6.385 10 4  (19.5)   30(10 )(0.2485)    0.1027 in Ans.

ymidspan 

6

5 % difference

Ans.

The solutions are the same as Prob. 4-12. ______________________________________________________________________________ 3 14 100 7 14 100  420 lbf  and RB   980 lbf  4-59 I = 0.05 in4, RA  10 10 M = 420 x  50 x2 + 980  x  10 1 EI

dy 2  210 x 2  16.667 x 3  490 x  10  C1 dx

EIy  70 x3  4.167 x 4  163.3 x  10  C1 x  C2 3

y = 0 at x = 0  C2 = 0 y = 0 at x = 10 in  C1 =  2 833 lbfin2. Thus, y

1 70 x3  4.167 x 4  163.3 x  10 3  2833 x  6  30 10  0.05 

3  6.667 107  70 x3  4.167 x 4  163.3 x  10  2833 x  Ans.   The tabular results and plot are exactly the same as Prob. 4-21. ______________________________________________________________________________

4-60 RA = RB = 400 N, and I = 6(323) /12 = 16 384 mm4. First half of beam, M =  400 x + 400  x  300 1 dy 2 EI  200 x 2  200 x  300  C1 dx From symmetry, dy/dx = 0 at x = 550 mm 



0 =  200(5502) + 200(550 – 300) 2 + C1

C1 = 48(106) N·mm2

EIy =  66.67 x3 + 66.67  x  300 3 + 48(106) x + C2

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 40/22

y = 0 at x = 300 mm



C2 =  12.60(109) N·mm3.

The term (EI)1 = [207(103)16 384] 1 = 2.949 (1010 ) Thus y = 2.949 (1010) [ 66.67 x3 + 66.67  x  300 3 + 48(106) x  12.60(109)] yO =  3.72 mm

Ans.

yx = 550 mm =2.949 (1010) [ 66.67 (5503) + 66.67 (550  300)3 + 48(106) 550  12.60(109)] = 1.11 mm Ans. The solutions are the same as Prob. 4-13. ______________________________________________________________________________ 4-61 1  M A  Fa  l 1  M A  0  M A  R2l  F (l  a)  R2  l  Fl  Fa  M A 

M

B

 0  R1l  Fa  M A

M  R1 x  M A  R2 x  l

 R1 

1

dy 1 1 2  R1 x 2  M A x  R2 x  l  C1 dx 2 2 1 1 1 3 EIy  R1 x3  M A x 2  R2 x  l  C1 x  C2 6 2 6

EI

y = 0 at x = 0



C2 = 0

y = 0 at x = l



1 1 C1   R1l 2  M Al . Thus, 6 2

EIy 

y

1 1 1 1 3  1  R1 x3  M A x 2  R2 x  l    R1l 2  M Al  x 6 2 6 2  6 

1  3  M A  Fa  x3  3M A x 2l   Fl  Fa  M A  x  l   Fal 2  2M Al 2  x   6 EIl

Ans.

In regions, 1  M A  Fa  x3  3M A x 2l   Fal 2  2M Al 2  x   6 EIl  x  M A  x 2  3lx  2l 2   Fa  l 2  x 2    Ans.  6 EIl 

y AB 

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 41/22

1  3  M A  Fa  x3  3M A x 2l   Fl  Fa  M A  x  l    Fal 2  2M Al 2  x   6 EIl 1 3 3  M A  x 3  3x 2l   x  l   2 xl 2   F   ax 3   l  a  x  l   axl 2      6 EIl

yBC 

 1  M  x  l  l  Fl  x  l   x  l   a 3x  l  6 EIl  x  l   M l  F  x  l  a 3x  l   Ans.      6 EI



2

2

A

2

A

The solutions reduce to the same as Prob. 4-17. ______________________________________________________________________________ w b  a  1    R1  4-62  M D  0  R1l  w  b  a  l  b   b  a    2l  b  a  2 2l   w w 2 2 M  R1 x  xa  x b 2 2 dy 1 w w 3 3 EI  R1 x 2  xa  x  b  C1 dx 2 6 6 1 w w 4 4 EIy  R1 x3  xa  x  b  C1 x  C2 6 24 24 y = 0 at x = 0 y = 0 at x = l



C2 = 0

1 1 w w 4 4 C1    R1l 3   l  a    l  b   l 6 24 24 

y

1 EI

 1 w b  a  w w 4  2l  b  a  x3  x  a  x  b  2l 24 24 6

4

1  1 w b  a  w w 4 4  x   2l  b  a  l 3   l  a    l  b    l 6 2l 24 24   



w 4 2  b  a  2l  b  a  x 3  l x  a  l x  b 24 EIl

4



4 4  x  2  b  a  2l  b  a  l 2   l  a    l  b    

Ans.

The above answer is sufficient. In regions,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 42/22



w 4 4 2  b  a  2l  b  a  x 3  x  2  b  a  2l  b  a  l 2   l  a    l  b      24 EIl wx  4 4  2  b  a  2l  b  a  x 2  2  b  a  2l  b  a  l 2   l  a    l  b     24 EIl

y AB 

yBC 



w 4 2  b  a  2l  b  a  x 3  l  x  a  24 EIl



4 4  x  2  b  a  2l  b  a  l 2   l  a    l  b    

yCD 



w 4 4 2  b  a  2l  b  a  x 3  l  x  a   l  x  b  24 EIl



4 4  x  2  b  a  2l  b  a  l 2   l  a    l  b    

These equations can be shown to be equivalent to the results found in Prob. 4-19. ______________________________________________________________________________ 4-63 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4, R1 = 0.5(180)(10) = 900 lbf Since the loading and geometry are symmetric, we will only write the equations for the first half of the beam 2 For 0  x  8 in M  900 x  90 x  3 At x = 3, M = 2700 lbfin Writing an equation for M / I, as seen in the figure, the magnitude and slope reduce since I 2 > I 1. To reduce the magnitude at x = 3 in, we add the term,  2700(1/I 1  1/ I 2) x  3 0. The slope of 900 at x = 3 in is also reduced. We account for this with a ramp function,  x  31 . Thus, 1 1 1 1 M 900 x 90 0 1   2700    x  3  900    x  3  x3 I I1 I2  I1 I 2   I1 I 2   5128 x  9520 x  3  3173 x  3  195.5 x  3 0

E

1

2

2

dy 1 2 3  2564 x 2  9520 x  3  1587 x  3  65.17 x  3  C1 dx

Boundary Condition:

Shigley’s MED, 10th edition

dy  0 at x  8 in dx

Chapter 2 Solutions, Page 43/22

0  2564  8   9520  8  3  1587  8  3  65.17  8  3  C1  2

2

3

C1 =  68.67 (103) lbf/in2 Ey  854.7 x3  4760 x  3  529 x  3  16.29 x  3  68.67(103 ) x  C2 2



y = 0 at x = 0

3

4

C2 = 0

Thus, for 0  x  8 in 1  2 3 4 y 854.7 x 3  4760 x  3  529 x  3  16.29 x  3  68.7(103 ) x  6   30(10 )

Ans.

Using a spreadsheet, the following graph represents the deflection equation found above

Beam Deflection 0 0

1

2

3

4

5

6

7

8

-0.002 -0.004 y (in)

-0.006 -0.008 -0.01 -0.012

x (in)

The maximum is ymax  0.0102 in at x  8 in Ans. ______________________________________________________________________________ 4-64 The force and moment reactions at the left support are F and Fl respectively. The bending moment equation is M = Fx  Fl Plots for M and M /I are shown. M /I can be expressed using singularity functions

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 44/22

M F Fl Fl l  x  x I 2 I1 2 I1 4 I1 2

0



F l x 2 I1 2

1

where the step down and increase in slope at x = l /2 are given by the last two terms. Integrate

dy F 2 Fl Fl l E  x  x x dx 4 I1 2 I1 4 I1 2 dy/dx = 0 at x = 0  C1 = 0

1

F l  x 4 I1 2

2

 C1

2

3

F 3 Fl 2 Fl l F l x  x  x  x  C2 12 I1 4 I1 8I1 2 12 I1 2 y = 0 at x = 0  C2 = 0 2 3 F  3 l l  2 y 2 x  2 x  6lx  3l x   24 EI1  2 2  3 2  F  l 5Fl 3 l y x l /2  Ans.  2    6l    3l (0)  2(0)    24 EI1   2  96 EI1 2  2 3 F  l  3Fl 3 3  l  2 y x l  2 l  6 l l  3 l l   2 x    Ans.    2   2   16EI    24 EI1   1 Ey 

The answers are identical to Ex. 4-10. ______________________________________________________________________________ 4-65 Place a dummy force, Q, at the center. The reaction, R1 = wl / 2 + Q / 2 wx2 M x  wl Q  M   x  2 Q 2  2 2 Integrating for half the beam and doubling the results

 1 ymax   2  EI

l /2

 0

 M M  Q

  2  dx    Q  0 EI

l /2

 0

 wl  wx 2   x  x   2    2 dx 2     

Note, after differentiating with respect to Q, it can be set to zero

ymax

w  2 EI

l /2

 0

l /2

w  x3l x 4  5w x  l  x dx      2 EI  3 4 0 384 EI 2

Ans.

______________________________________________________________________________ 4-66 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at the free end and positive to the left

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 45/22

M  Qx 

ymax

wx2 2

M  x Q

l l  1 l  M   1  wx 2  w   M  x 3dx     x  dx   dx     2 EI 0  EI 0   Q   Q 0 EI 0  2 

wl 4 Ans. 8 EI ______________________________________________________________________________ 

4-67 From Table A-7, I1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4 First treat the end force as a variable, F. Adding weight of channels of 2(5)/12 = 0.833 lbf/in. Using the variable x as shown in the figure M  F x 

5.833 2 x   F x  2.917 x 2 2

M  x F

A 

1 EI



60

0

M

M 1 dx F EI



60

0

( F x  2.917 x 2 )( x ) d x 

60 1 ( Fx 3 / 3  2.917 x 4 / 4) 0 EI

(150)(603 ) / 3  (2.917)(604 ) / 4  0.182 in in the direction of the 150 lbf force 30(106 )(3.70)  y A   0.182 in Ans. ______________________________________________________________________________ 

4-68 The energy includes torsion in AC, torsion in CO, and bending in AB. Neglecting transverse shear in AB M  Fx,

M x F

In AC and CO, T T  Fl AB ,  l AB F The total energy is

 T 2l   T 2l  U        2GJ  AC  2GJ CO

Shigley’s MED, 10th edition

l AB

 0

M2 dx 2 EI AB

Chapter 2 Solutions, Page 46/22

The deflection at the tip is



 

k

U Tl AC T TlCO T    F GJ AC F GJ CO F

l AB

 0

Tl l Tl l M M 1 dx  AC AB  CO AB  EI 3 F GJ AC GJ CO EI AB

l AB

 Fx dx 2

0

2 2 3 3 Tl AC l AB TlCO l AB Fl AC l AB FlCO l AB Fl AB Fl AB      4 4 4 GJ AC GJ CO 3EI AB G  d AC / 32  G  d CO / 32  3E  d AB / 64  2  l AC l 32 Fl AB 2l AB   CO4   4 4    Gd AC GdCO 3Ed AB 

F





 2 32l AB

 l AC l 2l AB   CO4   4 4   Gd AC GdCO 3Ed AB 

1

1

  2  200  200 200    8.10 N/mm Ans.    32  2002   79.3 103 184 79.3 103 12 4 3  207 103 84   ______________________________________________________________________________



4-69 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4 Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in R1 

1 1 (10)180  Q  900  0.5Q 2 2

For 0  x  3 in M   900  0.5Q  x

M  0.5 x Q M  0.5 x Q

For 3  x  13 in M   900  0.5Q  x  90( x  3) 2

By symmetry it is equivalent to use twice the integral from 0 to 8 3 8  8 M M  1 1  2 2    2 dx   900 x dx  900 x  90  x  3  x dx     EI 2 3  0 EI Q Q 0 EI1 0 3

300 x 3 1   EI1 0 EI 2

8

1 4 9 2   3 3 300 x  90( 4 x  2 x  2 x )  3

120.2 103  8100 1 8100 3 3 145.5 10   25.3110       30 106  0.1755  30 106  0.4604 EI1 EI 2 

 0.0102 in Ans. ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 47/22

4-70 I =  (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) in4, A = (0.52)/4 = 0.1963 in2. Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB. Resolve the force F into components in the x and y directions obtaining 0.6 F in the horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the dummy variable x to originate at the end where the loads are applied on each segment, 0.6 F: AB

OA

M  0.6 x F

M  0.6 F x

M  4.2 F

Fa  0.6 F

M  4.2 F

Fa  0.6 F

0.8 F: AB

M  0.8F x

M  0.8 x F

OA

M  0.8F x

M  0.8 x F

T  5.6 F Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection of B in the direction of F is* T  5.6 F

 B  F

U  Fa L  Fa  TL  T 1 M    M dx     F  AE OA F  JG OA F EI F 0.6 15 15 5.6 15 15  0.6    5.6  6  0.1963  30 10 6.136 103 11.5 106  

7 15 15  4.22  15 2  dx  0.6 x  d x  30 106  3.068 103  0 30 106  3.068 10 3  0 7



15

15 15 2 2  0.8 x  d x   0.8 x  d x 6 3  6 3  30 10  3.068 10  0 30 10  3.068 10  0

 1.38 105   0.1000  6.71103   0.0431  0.0119  0.1173  0.279 in

Shigley’s MED, 10th edition

Ans.

Chapter 2 Solutions, Page 48/22

*Note. Be careful, this is not the actual deflection of point B. For this, dummy forces must be placed on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and then substitute the values of Fx = 9 lbf, Fy =  12 lbf, and Fz = 0. This can be done separately and then added by vector addition. The actual deflections of B are found to be B = 0.0831 i  0.2862 j  0.00770 k in From this, the deflection of B in the direction of F is

 B F  0.6  0.0831  0.8  0.2862  0.279 in which agrees with our result. ______________________________________________________________________________ 4-71 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending. IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = 0.25(1.53)/12 = 0.07031 in4, ICD =  (0.754)/64 = 0.01553 in4. For the torsion of bar BC, Eq. (3-41) is in the form of  =TL/(JG), where the equivalent of J is Jeq = bc 3. With b/c = 1.5/0.25 = 6, JBC = bc 3 = 0.299(1.5)0.253 = 7.008 (103) in4. Use the dummy variable x to originate at the end where the loads are applied on each segment, M AB: Bending M  F x  2 F  x 2 F T Torsion T  5F 5 F M x BC: Bending M  F x F T Torsion T  2 F 2 F M CD: Bending M  F x x F U Tl T 1 M   M dx F JG F EI F 6 5F  6  2F  5 1 2  5  2  F  x  2 d x 6    3 6 6 0.09818 11.5 10 7.008 10 11.5 10  30 10  0.04909 0

D 

5

2

1 1  F x 2d x  F x 2d x   6 6 30 10  0.07031 0 30 10  0.01553 0  1.329 104  F  2.482 104  F  1.14110 4  F  1.98 10 5  F  5.72 10 6  F  5.207 104  F  5.207 104  200  0.104 in

Ans.

______________________________________________________________________________ Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 49/22

4-72 AAB =  (12)/4 = 0.7854 in2, IAB =  (14)/64 = 0.04909 in4, IBC = 1.5 (0.253)/12 = 1.953 (103) in4, ACD =  (0.752)/4 = 0.4418 in2, IAB =  (0.754)/64 = 0.01553 in4. For (D )x let F = Fx =  150 lbf and Fz =  100 lbf . Use the dummy variable x to originate at the end where the loads are applied on each segment, M y M y  Fz x 0 CD: F Fa Fa  F 1 F M y M y  F x  2 Fz x BC: F Fa Fa  Fz 0 F M y M y  5 F  2 Fz  Fz x 5 AB: F Fa Fa  F 1 F 5 Fa U FL 1   D  x      F x  2 Fz  x d x F  AE CD F EI BC 0 1  EI AB 

6

  5F  2 F

z

0

 FL  Fa  Fz x  5  d x     AE  AB F

F  2 1 3 F 1   5  Fz  52  6   6 3  0.4418  30 10 30 10 1.953 10   3  

F  6 F 1   25F  6   10 Fz  6   z  62  5  1 6    2 30 10  0.04909   0.7854  30 10 6

 1.509 107  F  7.112 104  F  4.267 104  Fz  1.019 104  F  1.019 104  Fz  2.546 107  F  8.135 104  F  5.286 104  Fz

Substituting F = Fx =  150 lbf and Fz =  100 lbf gives

 D  x  8.135 104   150   5.286 104   100   0.1749 in

Ans.

______________________________________________________________________________ 4-73 IOA = IBC =  (1.54)/64 = 0.2485 in4, JOA = JBC = 2 IOA = 0.4970 in4, IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, ICD =  (0.754)/64 = 0.01553 in4 Let Fy = F, and use the dummy variable x to originate at the end where the loads are applied on each segment,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 50/22

OC:

M Fx

M  x, F

DC:

M Fx

M x F

T  12 F

T  12 F

U 1 M  TL  T    M dx  F EI F  JG OC F The terms involving the torsion and bending moments in OC must be split up because of the changing second-area moments.

 D  y 

 D  y

2 12 F  4  12 F  9  1  12   12   F x 2d x 6  6   6 0.4970 11.5 10 0.09818 11.5 10 30 10  0.2485 0 11



13

12

1 1 1 F x 2d x  F x 2d x  F x 2d x    6 6 6 30 10  0.04909 2 30 10  0.2485 11 30 10  0.01553 0

 1.008 104  F  1.148 103  F  3.58 107  F  2.994 104  F  3.872 105  F  1.2363 103  F  2.824 103  F  2.824 103  250  0.706 in

Ans.

For the simplified shaft OC,

 B  y 

13 12 12 F 13 1 1 2 12  F x d x  F x 2d x  6    6 6 0.09818 11.5 10 30 10  0.04909 0 30 10  0.01553 0

 1.6580 103  F  4.973 104  F  1.2363 103  F  3.392 103  F  3.392 103  250  0.848 in

Ans.

Simplified is 0.848/0.706 = 1.20 times greater Ans. ______________________________________________________________________________ 4-74 Place a dummy force Q pointing downwards at point B. The reaction at C is RC = Q + (6/18)100 = Q + 33.33 This is the axial force in member BC. Isolating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energy. Thus, only energy in the member BC needs to be considered. Let the axial force in BC be F, where

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 51/22

F  Q  33.33

F 1 Q

 FL  F   0  33.3312   1  6.79 105  in Ans.    2 6 AE  Q  BC   Q 0   0.5  / 4  30 10  Q 0 ______________________________________________________________________________

B 

4-75

U Q

IOB = 0.25(23)/12 = 0.1667 in4 AAC =  (0.52)/4 = 0.1963 in2 MO = 0 = 6 RC  11(100)  18 Q RC = 3Q + 183.3 MA = 0 = 6 RO  5(100)  12 Q

 RO = 2Q + 83.33

Bending in OB. BD: Bending in BD is only due to Q which when set to zero after differentiation gives no contribution. AD: Using the variable x as shown in the figure above M   7  x  Q OA: Using the variable x as shown in the figure above M  100 x  Q  7  x 

M    2Q  83.33 x

Axial in AC: F  3Q  183.3

Shigley’s MED, 10th edition

M  2 x Q F 3 Q

Chapter 2 Solutions, Page 52/22

 U 

 FL  F 

 1

M



B    M Q dx         Q Q 0  AE  Q  Q 0  EI Q  0 183.3 12  1  3  6   0.1963  30 10 EI

5

 100 x  7  x  d x   2 83.33 x dx 6

2

0

0

5 6    1.12110   100 x 7  x d x  166.7 x 2 dx       6 0 10.4 10  0.1667  0 

1

3

 1.121103   5.768 107  100 129.2   166.7  72    0.0155 in

Ans.

______________________________________________________________________________ 4-76

There is no bending in AB. Using the variable, rotating counterclockwise from B M  R sin  P Fr  cos  P F  sin  P

M  PR sin  Fr  P cos 

F  P sin  MF  2 PR sin 2  P

A  6(4)  24 mm2 , ro  40  12 (6)  43 mm, ri  40  12 (6)  37 mm, From Table 3-4, p. 135, for a rectangular cross section 6 rn   39.92489 mm ln(43 / 37)

From Eq. (4-33), the eccentricity is e = R  rn =40  39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38)     M  M  1   MF  2 2 F R  F  2 2 CFr R  Fr   d   d   d           0 AeE 0 AE 0 AE 0 P AG  P  P   P 

  d 

2    P  R sin   2 PR  sin   2 2 PR sin  2 CPR  cos    d   d   d   d 0 0 0 0 AeE AE AE AG  PR  R EC   (10)(40)  40 (207 103 )(1.2)    1  2    1  2      4 AE  e G  4(24)(207 103 )  0.07511 79.3 103  

2

2

2

2

  0.0338 mm Ans. ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 53/22

4-77

Place a dummy force Q pointing downwards at point A. Bending in AB is only due to Q which when set to zero after differentiation gives no contribution. For section BC use the variable, rotating counterclockwise from B M M  PR sin   Q  R  R sin    R 1  sin   Q Fr Fr   P  Q  cos   cos  Q F F   P  Q  sin   sin  Q MF   PR sin   QR 1  sin     P  Q  sin  MF  PR sin 2    PR sin  1  sin    2QR sin  1  sin   Q

But after differentiation, we can set Q = 0. Thus, MF  PR sin  1  2sin   Q A  6(4)  24 mm2 , ro  40  12 (6)  43 mm, ri  40  12 (6)  37 mm, From Table 3-4, p.135, for a rectangular cross section 6 rn   39.92489 mm ln(43 / 37)

From Eq. (4-33), the eccentricity is e = R  rn =40  39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38),     1   MF  2 F R  F  2 2 CFr R  Fr  d   d   d        d    0 0 AE 0 AE 0 Q AG  Q    Q  PR 2 2 PR 2 2 PR 2  sin  1  sin  d   sin  d   sin  1  2sin   d   AeE 0 AE 0 AE 0 CPR 2  cos 2  d  0 AG 2  CE    PR  PR    PR  CPR PR   R    1     2     1  2   4 G   4  AeE 4 AE  4  AE 4 AG AE  4  e

 



2

M  M  AeE  Q

3    1.2  207 10   40    1  2    24  207 103  4  0.07511 4 79.3 103      0.0766 mm Ans. ______________________________________________________________________________



10  40 

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 54/22

4-78

A = 3(2.25) 2.25(1.5) = 3.375 in2 (1  1.5)(3)(2.25)  (1  0.75  1.125)(1.5)(2.25) R  2.125 in 3.375 Section is equivalent to the “T” section of Table 3-4, p. 135, 2.25(0.75)  0.75(2.25)  1.7960 in 2.25ln[(1  0.75) /1]  0.75ln[(1  3) / (1  0.75)] e  R  rn  2.125  1.7960  0.329 in rn 

For the straight section 1 I z  (2.25)  33   2.25(3)(1.5  1.125) 2 12 2 1 2.25      (1.5)  2.253   1.5(2.25)  0.75   1.125   2    12  2.689 in 4

For 0  x  4 in M   Fx

M   x, F

V F

V 1 F

For    /2 Fr  F cos 

Fr  cos  , F

F  F sin 

F  sin  F

M  (4  2.125sin  ) F MF MF  F (4  2.125sin  ) F sin   2 F (4  2.365sin  )sin  F

M  F (4  2.125sin  )

Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the curved part, integrating from 0 to π/2, and double the results

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 55/22



 /2 2 1 4 2 F (4)(1) (4  2.125sin  ) 2  F d  0 Fx dx  E I 3.375(G / E ) 0 3.375(0.329)  /2 2 F (4  2.125sin  ) sin  F sin 2  (2.125) d   d 0 0 3.375 3.375 2  /2 (1) F cos  (2.125)  d 0 3.375(G / E )



 /2

Substitute I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi 2  6700   43 4 1          16    17(1)  4.516     6  30 10   3  2.689  3.375(11.5 / 30) 3.375(0.329)   2   4  

2.125    2  2.125         4 1  2.125          3.375  4  3.375   4   3.375 11.5 / 30   4  

 0.0226 in Ans. ______________________________________________________________________________

4-79 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to  , and double the results M M  FR 1  cos   R 1  cos  F Fr Fr  F sin   sin  F F F  F cos   cos  F MF  F 2 Rcos 1  cos    MF   2 FRcos 1  cos  F

From Eq. (4-38),  FR 2



FR



  2 (1  cos  ) 2 d  cos 2  d   0 0 AeE AE   

2 FR  1.2 FR  2  cos  1  cos   d  sin  d    0 0 AE AG 

2 FR  3 R 3 E   0.6   AE  2 e 2 G

A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4, p. 135,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 56/22

h 4.5   34.95173 mm ro 37.25 ln ln 32.75 ri and e = R  rn = 35  34.95173 = 0.04827 mm. Thus, 2 F  35  3 35 3 207     0.6   0.08583F 3  13.5  207 10  2 0.04827 2 79.3  1 where F is in N. For  = 1 mm, F   11.65 N Ans. 0.08583 Note: The first term in the equation for  dominates and this is from the bending moment. Try Eq. (4-41), and compare the results. ______________________________________________________________________________ rn 

4-80 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal reaction to be applied at B and subject to the constraint ( B ) H  0.

M

M   R sin  H

FR (1  cos  )  HR sin  2

0  

 2

By symmetry, we may consider only half of the wire form and use twice the strain energy Eq. (4-41) then becomes, U 2 ( B ) H   H EI



 /2

0

 /2



M 

  M H  Rd  0 0

 FR   2 (1  cos  )  HR sin   ( R sin  ) R d  0 F F  F 30   H 0  H    9.55 N 2 4 4  

Ans.

The reaction at A is the same where H goes to the left. Substituting H into the moment equation we get, M

FR  (1  cos  )  2sin   2

Shigley’s MED, 10th edition

M R  [ (1  cos  )  2sin  ] F 2

0  

 2

Chapter 2 Solutions, Page 57/22

U 2  M  2  /2 FR 2 P   [ (1  cos  )  2sin  ]2 R d M Rd  2  0 P EI  F  EI 4 3  /2 FR  ( 2   2 cos 2   4sin 2   2 2 cos   4 sin   4 sin  cos  ) d 2  0 2 EI FR 3  2               2    4    2 2  4  2   2 2 EI   2  4 4  (3 2  8  4) FR 3 (3 2  8  4) (30)(403 )   0.224 mm Ans. 8 EI 8 207 103    24  / 64  ______________________________________________________________________________ 

4-81 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the curved beam portion. The shear and axial components will be negligible compared to bending. Place a fictitious force Q pointing to the left at point A. M M  PR sin   Q ( R sin   l )  R sin   l Q Note that the strain energy in the straight portion is zero since there is no real force in that section. From Eq. (4-41), 

 /2



0

  

PR 2  EI

 1  M  1 M Rd   EI  Q   Q 0 EI



 /2

0



 /2

0

PR sin   R sin   l Rd

PR 2   1(52 )     R sin   l sin   d  EI  4 R  l    (5)  4  6 4  30 10    0.125  / 64   4 2

 0.551 in Ans. ______________________________________________________________________________

4-82 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. M AB x P

Straight portion:

M AB  Px

Curved portion:

M BC  P  R(1  cos  )  l 

M BC   R(1  cos  )  l  P

From Eq. (4-41) with the addition of the bending strain energy in the straight portion of the wire,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 58/22

 /2 1  M BC  M AB  1  M AB dx      M BC Rd 0 EI 0 P  EI  P   P l 2 PR  /2 2  x dx  R (1  cos  )  l  d    EI 0 EI 0 3 Pl PR  /2 2  R (1  2 cos   cos 2  )  2 Rl (1  cos  )  l 2 d    0 3EI EI Pl 3 PR  /2 2  R cos 2    2 R 2  2 Rl  cos   ( R  l ) 2 d    3EI EI 0  Pl 3 PR   2     R   2 R 2  2 Rl   ( R  l ) 2   3EI EI  4 2 

 

l

P  EI 

l3  3  2 2  3  4 R  R  2 R  2 Rl   2 R ( R  l )   

 43  3 1  2 2  3  4 (5 )  5  2(5 )  2(5)(4)   2  5  5  4   6 4 30 10    0.125  / 64  

 0.850 in Ans. ______________________________________________________________________________

4-83 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Place a dummy force, Q, at A vertically downward. The only load in the straight section is the axial force, Q. Since this will be zero, there is no contribution. In the curved section M  R 1  cos   Q

M  PR sin   QR 1  cos  

From Eq. (4-41)   /2 1  M   1    M Rd     0 EI  Q    Q 0 EI

PR 3  EI 



 /2

0



 /2

0

PR sin   R 1  cos   Rd

PR 3  1  PR 3  sin   sin  cos   d  1    EI  2  2 EI 1 53 

2  30 106   0.1254  / 64 

 0.174 in

Ans.

______________________________________________________________________________ 4-84 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion.

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 59/22

Place a dummy force, Q, at A vertically downward. The load in the straight section is the axial force, Q, whereas the bending moment is only a function of P and is not a function of Q. When setting Q = 0, there is no axial or bending contribution. In the curved section M  P  R 1  cos    l   QR sin 

M   R sin  Q

From Eq. (4-41) 

 /2

   0 

 

 1  M  1 M Rd   EI  Q   Q 0 EI

PR 2 EI

 /2



 /2

0

P  R 1  cos    l    R sin   Rd

  R sin   R sin  cos   l sin   d   0

1  52 

2  30 106   0.1254  / 64 

PR 2  1  PR 2 R  l  R    R  2l    EI  2  2 EI

5  2  4    0.452 in

Since the deflection is negative,  is in the opposite direction of Q. Thus the deflection is

  0.452 in 

Ans. ______________________________________________________________________________

4-85 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is tension FAB FAB  F 1 F For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no bending in section DE. For section BCD, let  be counterclockwise originating at D M M  FR sin   R sin  0   F Using Eqs. (4-29) and (4-41) 3  1   FR M  Fl  Fl  FAB    sin 2  d 1  0  M Rd  0 EI  F  AE EI  AE  AB F

  403   Fl  FR3 F  l  R 3  9.81  80         AE 2 EI E  A 2 I  207 103     22  / 4  2   24  / 64        6.067 mm Ans. ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 60/22

4-86 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC =  (0.54)/64 = 3.068 (10-3) in4 Applying a force F at point B, using statics (AC is a two-force member), the reaction forces at O and C are as shown. FOA OA: Axial FOA  3F 3 F M OA  2 x F

Bending M OA  2 Fx

AB: Bending

M AB  x F

M AB   F x

AC: Isolating the upper curved section

M AC  3R  sin   cos   1 F 10 20 1 1  Fl  FOA 2    4 Fx dx  F x 2d x     EI OAB 0  AE OA F  EI OAB 0 M AC  3FR  sin   cos   1

 

9 FR 3  EI  AC

 /2

  sin   cos   1

3F 10 

0

0.5 10.4 106 

 3 

9 F 103 

d

4 F 103 

3 10.4 10 6  0.1667 

30 10  3.068 10 6

2

 /2

3

 sin 

2



F  203 

3 10.4 10 6  0.1667 

  2sin  cos   2sin   cos 2   2 cos   1 d

0

    1.73110 5  F  7.69110 4  F  1.538 10 3  F  0.09778 F   1  2   2   4 2 4  0.0162 F  0.0162 100   1.62 in Ans. _____________________________________________________________________________ 4-87 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC =  (0.54)/64 = 3.068 (10-3) in4 Applying a vertical dummy force, Q, at A, from statics the reactions are as shown. The dummy force is transmitted through section OA and member AC.

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 61/22

FOA 1 Q

OA: FOA  3F  Q

M AC   3F  Q  R sin    3F  Q  R 1  cos  

AC:

M AC  R  sin   cos   1 Q

 Fl   FOA   1   /2  M AC Rd       M AC   Q  AE OA  Q   EI  AC 0  Q 0

    

3FlOA 3FR 3   AE OA  EI  AC 3 100 10

10.4 10  0.5 6



 /2

  sin   cos   1

2

d

0

3 100 103

     1  2   2    0.462 in 4 2 30 10  3.068 10   4 6

3

Ans.

______________________________________________________________________________ 4-88

I =  (64)/64 = 63.62 mm4 0 /2

M  R sin  F T T  FR(1  cos  )  R(1  cos  ) F According to Castigliano’s theorem, a positive  U/ F will yield a deflection of A in the negative y direction. Thus the deflection in the positive y direction is M  FR sin 

( A ) y  

U 1   F  EI



 /2

0

F ( R sin  ) 2 R d 

1 GJ



 /2

0

 F [ R(1  cos  )]2 R d  

Integrating and substituting J  2 I and G  E /  2 1     FR 3  3   4  (1  )  4  2      4  8  (3  8)  4 EI    (250)(80)3  [4  8  (3  8)(0.29)]   12.5 mm Ans. 4(200)103  63.62  ______________________________________________________________________________ FR 3 ( A ) y   EI

4-89 The force applied to the copper and steel wire assembly is Fc  Fs  400 lbf (1) Since the deflections are equal,  c   s

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 62/22

 Fl   Fl       AE c  AE  s Fc l Fs l  2 6 3( / 4)(0.1019) (17.2)10 ( / 4)(0.1055) 2 (30)106

Yields, Fc  1.6046 Fs . Substituting this into Eq. (1) gives 1.604 Fs  Fs  2.6046Fs  400



Fs  153.6 lbf

Fc  1.6046 Fs  246.5 lbf F 246.5 c  c   10 075 psi  10.1 kpsi Ans. Ac 3( / 4)(0.1019)2 F 153.6 s  s   17 571 psi  17.6 kpsi Ans. As ( / 4)(0.10552 ) 153.6(100)(12)  Fl     0.703 in Ans.   2 6  AE  s ( / 4)(0.1055) (30)10 ______________________________________________________________________________

4-90 (a) Bolt stress

 b  0.75(65)  48.8 kpsi

Ans.

  Fb  6 b Ab  6(48.8)   (0.52 )  57.5 kips 4 F 57.43 Cylinder stress c   b   13.9 kpsi Ans. Ac ( / 4)(5.52  52 ) (b) Force from pressure  D2  (52 ) P p (500)  9817 lbf  9.82 kip 4 4 Fx = 0

Total bolt force

Pb + Pc = 9.82

(1)

Since  c   b , Pc l Pbl  2 2 ( / 4)(5.5  5 ) E 6( / 4)(0.52 ) E Pc = 3.5 Pb

(2)

Substituting this into Eq. (1) Pb + 3.5 Pb = 4.5 Pb = 9.82  Pb = 2.182 kip. From Eq. (2), Pc = 7.638 kip Using the results of (a) above, the total bolt and cylinder stresses are 2.182  b  48.8   50.7 kpsi Ans. 6( / 4)(0.52 )

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 63/22

7.638  12.0 kpsi Ans. ( / 4)(5.52  52 ) ______________________________________________________________________________

 c  13.9 

4-91

Tc + Ts = T

c = s

(1)

Tcl Tl  s  JG c  JG s



Substitute this into Eq. (1)  JG c T T  T  JG s s s



 Tc 

Ts 

 JG c T  JG s s

(2)

 JG s T  JG s   JG c

The percentage of the total torque carried by the shell is

% Torque 

100  JG s

 JG s   JG c

Ans.

______________________________________________________________________________ 4-92 RO + RB = W (1) OA = AB  Fl   Fl       AE OA  AE  AB 400 RO 600 RB 3   RO  RB AE AE 2 Substitute this unto Eq. (1) 3 RB  RB  4 2



RB  1.6 kN

(2)

Ans.

3 RO  1.6  2.4 kN Ans. 2 2 400(400)  Fl  A    0.0223 mm Ans.   3  AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________

From Eq. (2)

4-93 See figure in Prob. 4-92 solution. Procedure 1: 1. Let RB be the redundant reaction. Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 64/22

 RO = 4 000  RB

2. Statics. RO + RB = 4 000 N 3. Deflection of point B.  B 

(1)

RB  600   RB  4000  400   0 AE AE

(2)

4. From Eq. (2), AE cancels and RB = 1 600 N Ans. and from Eq. (1), RO = 4 000  1 600 = 2 400 N Ans.  Fl 

2 400(400)

A    0.0223 mm Ans.   3  AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________ 4-94 (a) Without the right-hand wall, the deflection of point C would be 5 103  8 2 103  5 Fl C     AE  / 4  0.752 10.4 106  / 4  0.52 10.4 106  0.01360 in  0.005 in  Hits wall

Ans.

(b) Let RC be the reaction of the wall at C acting to the left (). Thus, the deflection of point C is now

5 103   RC  8  2 103   RC  5    C    2 6 2  / 4 0.75 10.4 10  / 4 0.5 10.4        106  0.01360 

4 RC 5   8  2   0.005 6  2  10.4 10  0.75 0.5 

or, 0.01360  4.190 106  RC  0.005



RC  2 053 lbf  2.05 kip  Ans.

 Fx = 5 000 + RA  2 053 = 0  RA = 2 947 lbf,  RA = 2.95 kip  Ans. Deflection. AB is 2 947 lbf in tension. Thus RA  8 2947 8   5.13 103  in  Ans. AAB E  / 4  0.752 10.4 106 ______________________________________________________________________________

 B   AB 

4-95 Since OA = AB, TOA (4) TAB (6)  JG JG Shigley’s MED, 10th edition



3 TOA  TAB 2

(1)

Chapter 2 Solutions, Page 65/22

Statics. TOA + TAB = 200 (2) Substitute Eq. (1) into Eq. (2), 3 5 TAB  TAB  TAB  200  2 2 From Eq. (1)

TAB  80 lbf  in

Ans.

3 3 TOA  TAB  80  120 lbf  in Ans. 2 2 80  6  180  TL  A    0.3900   4 6 JG  / 32 0.5 11.5 10        AB

 max 

16T d3

 AB 

 16  80 

  0.53 

 OA 

16 120 

  0.53 

Ans.

 4890 psi  4.89 kpsi

 3260 psi  3.26 kpsi

Ans.

Ans.

______________________________________________________________________________ 4-96 Since OA = AB, TOA (4) TAB (6)  4  / 32  0.5 G  / 32  0.754 G



TOA  0.2963 TAB

(1)

Statics. TOA + TAB = 200 (2) Substitute Eq. (1) into Eq. (2), 0.2963TAB  TAB  1.2963TAB  200

From Eq. (1)



TAB  154.3 lbf  in

TOA  0.2963TAB  0.2963 154.3  45.7 lbf  in

A 

154.3  6  180  0.1480 4 6  / 32 0.75 11.5 10     

 max 

16T d3

 AB 



 OA 

16 154.3

  0.753 

16  45.7 

  0.53 

Ans.

Ans.

Ans.

 1862 psi 1.86 kpsi

 1862 psi 1.86 kpsi

Ans.

Ans.

______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 66/22

4-97 Procedure 1 1. Arbitrarily, choose RC as a redundant reaction. 2. Statics. Fx = 0, 12(103)  6(103)  RO  RC = 0 RO = 6(103)  RC (1) 3. The deflection of point C. 12(103 )  6(103 )  RC  (20) 6(103 )  RC  (10) R (15) C    C 0 AE AE AE 4. The deflection equation simplifies to

 45 RC + 60(103) = 0

 RC = 1 333 lbf = 1.33 kip

RO = 6(103)  1 333 = 4 667 lbf = 4.67 kip

From Eq. (1),

Ans. Ans.

FAB = FB + RC = 6 +1.333 = 7.333 kips compression FAB 7.333   14.7 kpsi Ans. A (0.5)(1) Deflection of A. Since OA is in tension, R l 4 667(20)  A   OA  O OA   0.00622 in Ans. AE (0.5)(1)(30)106 ______________________________________________________________________________

 AB 

4-98 Procedure 1 1. Choose RB as redundant reaction. 2. Statics. RC = wl  RB

(1)

1 w l 2  RB  l  a  (2) 2 3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9, MC 

R l  a  w l  a   4l  l  a    l  a 2  6l 2   0 yB  B   3EI 24 EI  3

2

4. Solving for RB. w  2 2 RB  6l  4l  l  a    l  a     8 l  a  

w  3l 2  2al  a 2  8 l  a 

Ans.

Substituting this into Eqs. (1) and (2) gives

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 67/22

RC  wl  RB 

w 5l 2  10al  a 2   8 l  a 

Ans.

1 2 w wl  RB  l  a    l 2  2al  a 2  Ans. 2 8 ______________________________________________________________________________ MC 

4-99 See figure in Prob. 4-98 solution. Procedure 1 1. Choose RB as redundant reaction. 2. Statics. RC = wl  RB

(1)

1 M C  wl 2  RB  l  a  (2) 2 3. Deflection equation for point B. Let the variable x start at point A and to the right. Using singularity functions, the bending moment as a function of x is

1 M   wx 2  RB x  a 2

1

M  xa RB

1

U 1 M  M dx  RB EI 0 RB l

yB 

1 1 1  1    w x 2  0  dx   w x 2  RB  x  a    x  a  dx  0    EI 0 2 EI a  2  l

l

or, 1 1 a 3 3  R  w   l 4  a 4    l 3  a 3    B  l  a    a  a    0  2 4 3  3  Solving for RB gives

RB 

w 8 l  a 

3

w 2 2 3  l 4  a 4   4a  l 3  a3     8  l  a   3l  2al  a 

Ans.

From Eqs. (1) and (2) RC  wl  RB 

w 5l 2  10al  a 2   8 l  a 

Ans.

1 2 w wl  RB  l  a    l 2  2al  a 2  Ans. 2 8 ______________________________________________________________________________ MC 

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 68/22

4-100 Note: When setting up the equations for this problem, no rounding of numbers was made in the calculations. It turns out that the deflection equation is very sensitive to rounding. Procedure 2 1. Statics.

R1 + R2 = wl

1 2 wl 2 2. Bending moment equation. R2l  M 1 

(1) (2)

1 M  R1 x  w x 2  M 1 2 dy 1 1 EI  R1 x 2  w x 3  M 1 x  C1 dx 2 6 1 1 1 EIy  R1 x3  w x 4  M 1 x 2  C1 x  C2 6 24 2

(3) (4)

EI = 30(106)(0.85) = 25.5(106) lbfin2. 3. Boundary condition 1. At x = 0, y =  R1/k1 =  R1/[1.5(106)]. Substitute into Eq. (4) with value of EI yields C2 =  17 R1. Boundary condition 2. At x = 0, dy /dx =  M1/k2 =  M1/[2.5(106)]. Substitute into Eq. (3) with value of EI yields C1 =  10.2 M1. Boundary condition 3. At x = l, y =  R2/k3 =  R1/[2.0(106)]. Substitute into Eq. (4) with value of EI yields 1 3 1 1 R1l  wl 4  M 1l 2  10.2M 1l  17 R1 (5) 6 24 2 Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in, are 12.75 R2 

1 0   1   24 1   0  2287 12.75 532.8   

 R1   1      3  R2    12  10       M 1  576 

Solving, the simultaneous equations yields R1 = 554.59 lbf, R2 = 445.41.59 lbf, M1 = 1310.1 lbfin

Ans.

For the deflection at x = l /2 = 12 in, Eq. (4) gives

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 69/22

y x 12in 

1 1 500 4 1 1 12  1310.112 2  554.59 123  6  6 24 12 2 25.5 10   10.2 1310.112  17  554.59  

 5.51103  in

Ans.

______________________________________________________________________________ 4-101 Cable area, A 

 4

(0.52 )  0.1963 in 2

Procedure 2 1. Statics. RA + FBE + FDF = 5(103) 3 FDF + FBE = 10(103)

(1) (2)

2. Bending moment equation. M  RA x  FBE x  16  5000 x  32 1

1

dy 1 1 2 2  RA x 2  FBE x  16  2500 x  32  C1 dx 2 2 1 1 2500 3 3 EIy  RA x3  FBE x  16  x  32  C1 x  C2 6 6 3

EI

3. B.C. 1: At x = 0, y = 0



(3) (4)

C2 = 0

B.C. 2: At x = 16 in, FBE (38)  Fl  yB     6.453(106 ) FBE   6 0.1963(30)10  AE  BE

Substituting into Eq. (4) and evaluating at x = 16 in 1 EIyB  30(106 )(1.2)(6.453)(106 ) FBE  RA 163   C1 (16) 6 Simplifying gives 682.7 RA + 232.3 FBE + 16 C1 = 0 (5) B.C. 2: At x = 48 in, FDF (38)  Fl  yD     6.453(106 ) FDF   6 0.1963(30)10  AE  DF Substituting into Eq. (4) and evaluating at x = 48 in,

Simplifying gives Shigley’s MED, 10th edition

 

1 1 2500 RA 483  FBE (48  16)3  (48  32)3  48C1 6 6 3 18 432 RA + 5 461 FBE + 232.3 FDF + 48 C1 = 3.413(106)

EIyD  232.3FDF 

(6)

Chapter 2 Solutions, Page 70/22

Equations (1), (2), (5) and (6) in matrix form are 1 1 0   RA   5000   1     1 3 0   FBE   10 000   0    0  682.7 232.3 0 16   FDF      6    3.413 10 C 18 432 5 461 232.3 48     1   

Solve simultaneously or use software. The results are RA =  970.5 lbf, FBE = 3956 lbf, FDF = 2015 lbf, and C1 =  16 020 lbfin2. 3956 2015  BE   20.2 kpsi,  DF   10.3 kpsi Ans. 0.1963 0.1963 EI = 30(106)(1.2) = 36(106) lbfin2 y 

1 2500 3 3  970.5 3 3956   x  x  16  x  32  16 020 x  6  6 6 3 36 10  

 

161.8 x  659.3 x  16  

1 36 106

B: x = 16 in,

3

3

3

 833.3 x  32  16 020 x



yB 

1  161.8 163   16 020 16    0.0255 in 6   36 10 

yC 

1  161.8  323   659.3  32  16 3  16 020  32   6   36 10 

Ans.

C: x = 32 in,

 0.0865 in

Ans.

D: x = 48 in, yD 

 

1  161.8 483  659.3  48  16 3  833.3  48  32 3  16 020  48   6   36 10

 

 0.0131 in Ans. ______________________________________________________________________________

4-102 Beam: EI = 207(103)21(103) = 4.347(109) Nmm2. Rods: A = ( /4)82 = 50.27 mm2. Procedure 2 1. Statics. Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 71/22

RC + FBE  FDF = 2 000

(1)

RC + 2FBE = 6 000

(2)

2. Bending moment equation. M =  2 000 x + FBE x  75 1 + RC x  150 1 dy 1 1 2 2  1000 x 2  FBE x  75  RC x  150  C1 dx 2 2 1000 3 1 1 3 3 EIy   x  FBE x  75  RC x  150  C1 x  C2 3 6 6

EI

(3) (4)

3. B.C 1. At x = 75 mm, FBE  50   Fl  yB     4.805 106  FBE   3 50.27  207 10  AE  BE

Substituting into Eq. (4) at x = 75 mm, 4.347 109   4.805 106  FBE   

1000 753   C1  75   C2  3

Simplifying gives 20.89 103  FBE  75C1  C2  140.6 106 

(5)

B.C 2. At x = 150 mm, y = 0. From Eq. (4), 

or,

1000 1 3 1503   FBE 150  75   C1 150   C2  0  3 6

70.31103  FBE  150C1  C2  1.125 109 

(6)

B.C 3. At x = 225 mm, FDF  65  Fl  yD    6.246 106  FDF   3  AE  DF 50.27  207 10

Substituting into Eq. (4) at x = 225 mm,

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 72/22

4.347 109  6.246 106  FDF   

1000 1 3 2253   FBE  225  75  3 6 1 3  RC  225  150   C1  225  C2 6

Simplifying gives 70.31103  RC  562.5 103  FBE  27.15 103  FDF  225C1  C2  3.797 109 

(7)

Equations (1), (2), (5), (6), and (7) in matrix form are  2 103   1 1 1 0 0      RC   3 1 2 0 0 0  6 10      FBE       0 20.89 103  0 75 1   F   140.6 106       DF   3       9 0 70.3110  0 150 1 C1  1.125 10       70.31103  562.5 103  27.15 103  225 1   C2   9   3.797 10  

Solve simultaneously or use software. The results are RC =  2378 N, FBE = 4189 N, FDF =  189.2 N Ans. 7 2 8 3 and C1 = 1.036 (10 ) Nmm , C2 =  7.243 (10 ) Nmm . The bolt stresses are BE = 4189/50.27 = 83.3 MPa, DF =  189/50.27=  3.8 MPa Ans. The deflections are From Eq. (4)

yA 

1  7.243 108   0.167 mm 9   4.347 10 

Ans.

For points B and D use the axial deflection equations*. 4189  50   Fl  yB     0.0201 mm   50.27  207 103  AE  BE

Ans.

189  65  Fl  yD    1.18 103  mm Ans.   3  AE  DF 50.27  207 10 *Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for calculating the very small deflections, especially for point D. ______________________________________________________________________________

4-103 (a) The cross section at A does not rotate. Thus, for a single quadrant we have

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 73/22

U 0 M A The bending moment at an angle  to the x axis is FR M M  MA  1 1  cos   2 M A The rotation at A is  /2 U 1 M A   M Rd  0  M A EI 0 M A

Thus,

1 EI

 /2



  M

A



0

FR 1  cos  1 Rd  0 2 



FR   FR  0 MA    2 2 2 

or, FR  2  1   2   Substituting this into the equation for M gives FR  2 M  cos    (1) 2   The maximum occurs at B where  =  /2 MA 

M max  M B  

FR

Ans.

 (b) Assume B is supported on a knife edge. The deflection of point D is  U/ F. We will deal with the quarter-ring segment and multiply the results by 4. From Eq. (1) M R  2   cos    F 2   Thus, U 4 D   F EI

 /2

 0

M FR 3 M Rd  F EI

 /2

 0

2

2 FR 3   2   cos   d         EI  4   

3

FR  2  8  Ans.  4 EI ______________________________________________________________________________ 

4-104

Pcr  I

C 2 EI l2



 D4  d 4  

 D4

1  K 

64 64 2 4  C E   D Pcr  2  1  K 4   l  64 

Shigley’s MED, 10th edition

4

where K 

d D

Chapter 2 Solutions, Page 74/22

1/4

  64 Pcr l 2   D 3 Ans. 4   CE 1  K   ______________________________________________________________________________



D 2 1  K 2  ,



D 4 1  K 4  



D 4 1  K 2 1  K 2  , where K = d / D. 4 64 64 The radius of gyration, k, is given by I D2 2 k   1 K 2   A 16 From Eq. (4-46) S y2l 2 S y2l 2 Pcr  Sy  2 2  Sy  2 2 4 k CE 4  D /16 1  K 2  CE  / 4  D 2 1  K 2 

4-105 A 

I

4 Pcr   D 1  K 2

2

S

y



 D 2 1  K 2  S y  4 Pcr 

4 S y2l 2 D 2 1  K 2 

 2 D 2 1  K 2  CE

4 S y2l 2 1  K 2 

 1  K 2  CE

  4 S y2l 2 1  K 2  4 Pcr  D  2 2 2   S y 1  K   1  K  CE 1  K  S y 

1/2

1/2

  S yl 2 Pcr   2  Ans. 2 2 2   S y 1  K   CE 1  K   ______________________________________________________________________________

0.9

4-106 (a) M A  0, (0.75)(800) 

0.92  0.52

FBO (0.5)  0  FBO  1373 N

Using nd = 4, design for Fcr = nd FBO = 4(1373) = 5492 N

l  0.92  0.52  1.03 m, S y  165 MPa In-plane: 1/2

 bh3 /12  I k      A  bh  l 1.03   142.7 k 0.007218 1/2

 0.2887h  0.2887(0.025)  0.007 218 m,

C  1.0

1/2

2 9  l   2 (207)(10 )       6  k 1  165(10 ) 

Shigley’s MED, 10th edition

 157.4

Chapter 2 Solutions, Page 75/22

Since (l / k )1  (l / k ) use Johnson formula. Try 25 mm x 12 mm, 2   165 106   1   6 Pcr  0.025(0.012) 165 10    (142.7)   29.1 kN 9  2  1(207)10     This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane. Out-of-plane: k  0.2887(0.012)  0.003 464 in, l 1.03   297.3 k 0.003 464 Since (l / k )1  (l / k ) use Euler equation. Pcr  0.025(0.012)

C  1.2

1.2 2  207 109

 8321 N 297.32 This is greater than the design load of 5492 N found earlier. It is also significantly less than the in-plane Pcr found earlier, so the out-of-plane condition will dominate. Iterate the process to find the minimum h that gives Pcr greater than the design load.

With h = 0.010, Pcr = 4815 N (too small) h = 0.011, Pcr = 6409 N (acceptable) Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans.

 

P 1373   10.4 106 Pa  10.4 MPa dh 0.012(0.011) No, bearing stress is not significant. Ans. ______________________________________________________________________________

(b)  b  

4-107 This is an open-ended design problem with no one distinct solution. ______________________________________________________________________________ F = 1500( /4)22 = 4712 lbf. From Table A-20, Sy = 37.5 kpsi Pcr = nd F = 2.5(4712) = 11 780 lbf

4-108

(a) Assume Euler with C = 1 1/4

1/4  64 11790  502   64 Pcr l 2    d  3   3 6   1 30 10     CE  Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in



Pcr l 2 I d  64 C 2 E 4

Shigley’s MED, 10th edition

 1.193 in

Chapter 2 Solutions, Page 76/22

l 50   160 k 0.3125  2 2 (1)30 106     126   37.5 103     2 6 4   30 10  / 64 1.25 Pcr   14194 lbf 502 1/2

1/2

2  l   2 CE        k 1  S y 

Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory.

 use Euler

Ans.

1/4

 64 11780 162    0.675 in, so use d = 0.750 in d  3 6   1 30 10   k = 0.750/4 = 0.1875 in l 16   85.33 use Johnson k 0.1875

(b)

2    37.5 103   1   3 Pcr   0.750  37.5 10    85.33  12 748 lbf 6 4  2  1 30 10   



2

Use d = 0.75 in. (c)

n( a ) 

14194  3.01 4 712

Ans.

12 748  2.71 Ans. 4 712 ______________________________________________________________________________ n(b ) 

4-109 From Table A-20, Sy = 180 MPa 4F sin = 2 943 735.8 sin  In range of operation, F is maximum when  = 15 735.8 Fmax   2843 N per bar sin15o F

Pcr = ndFmax = 3.50 (2 843) = 9 951 N l = 350 mm, h = 30 mm

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 77/22

Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm l 350   242.6 k 1.443 2 9  l   2 1.4  207 10       180 106   k 1  

1/2

Pcr  A

C 2 E

l / k 

2

 5(30)

 178.3

1.4 2  207 103

 242.6 

2

 use Euler  7 290 N

Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm l 350   202.1 k 1.732 1.4 2  207 103 C 2 E Pcr  A  6(30)  12 605 N 2 2 l / k   202.1 O.K. Use 25  6 mm bars

Ans. The factor of safety is

12605  4.43 Ans. 2843 ______________________________________________________________________________ n

4-110 P = 1 500 + 9 000 = 10 500 lbf

Ans.

MA = 10 500 (4.5/2)  9 000 (4.5) +M = 0 M = 16 874 lbfin e = M / P = 16 874/10 500 = 1.607 in

Ans.

From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq. (4-55) k2 

I 2.059   0.953 in 2 A 2.160

P  ec  10500  1.607  3 / 2   1    17157 psi  17.16 kpsi Ans. 1  2    A k  2.160  0.953  ______________________________________________________________________________

c  

4-111 This is a design problem which has no single distinct solution. ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 2 Solutions, Page 78/22

4-112 Loss of potential energy of weight = W (h + ) 1 Increase in potential energy of spring = k 2 2 1 W (h + ) = k 2 2 2 W 2 W or,  2   h  0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields k k

 2  0.6   1.2 = 0 Taking the positive root (see discussion on p. 206 of text) 1  max  0.6  (0.6) 2  4(1.2)   1.436 in  2

Ans.

Fmax = k  max = 100 (1.436) = 143.6 lbf Ans. ______________________________________________________________________________ 4-113 The drop of weight W1 converts potential energy, W1 h, to kinetic energy

1 W1 2 v1 . 2 g

Equating these provides the velocity of W1 at impact with W2. 1 W1 2 v1  v1  2 gh (1) 2 g Since the collision is inelastic, momentum is conserved. That is, (m1 + m2) v2 = m1 v1, where v2 is the velocity of W1 + W2 after impact. Thus W1h 

W1  W2 W v2  1 v1 g g



v2 

W1 W1 v1  2 gh W1  W2 W1  W2

(2)

The kinetic and potential energies of W1 + W2 are then converted to potential energy of the spring. Thus, 1 W1  W2 2 1 v2  W1  W2    k 2 2 g 2 Substituting in Eq. (1) and rearranging results in

W1  W2 W12 h  2  2 0 (3) k W1  W2 k Solving for the positive root (see discussion on p. 206) 2

2 W1  W2  W12 h  1  W1  W2     2  4  8  2 k k  W1  W2 k    

Shigley’s MED, 10th edition

(4)

Chapter 2 Solutions, Page 79/22

W1 = 40 N, W2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm. 1

2 40  400  402 200   40  400    29.06 mm  4  8    32  40  400 32    32    

   2  2

Ans.

Fmax = k = 32(29.06) = 930 N Ans. ______________________________________________________________________________ 1 4-114 The initial potential energy of the k1 spring is Vi = k1a 2 . The movement of the weight 2 1 1 2 W the distance y gives a final potential of Vf = k1  a  y   k2 y 2 . Equating the two 2 2 energies give 1 2 1 1 2 k1a  k1  a  y   k2 y 2 2 2 2

Simplifying gives

 k1  k2  y 2  2ak1 y  0

2k1a . Without damping the weight will vibrate between k1  k2 2k1a these two limits. The maximum displacement is thus y max = Ans. k1  k2 With W = 5 lbf, k1 = 10 lbf/in, k2 = 20 lbf/in, and a = 0.25 in

This has two roots, y = 0,

2  0.25 10  0.1667 in Ans. 10  20 ____________________________________________________________________________ ymax 

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Chapter 2 Solutions, Page 80/22

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Chapter 2 Solutions, Page 81/22