8va Solutions Shigley PDF

 

41

Chapter 3

(d)

 = 121 (1)(2) = 0.666  = 6667 7 in

100 lbf/in

3

 I   x  O

 A

6"

B

12"

1350 lbf V (lbf)

 = − 600 (6) = −18 1800 00 lb lbf  f · in 2

 M 1

450 lbf  

750

 M 2  x 

O 7.5"  M  (lbf • in)

 = −1800 + 12 750(7.5) = 10 1013 13 lb lbf  f · in

At A, top of beam

450

600

4

 = 1800(1) = 27 2700 00 psi   Ans. 0.6667 At A, y = 0 3 750 = 56 τ   = 563 3 psi   Ans. 2 (2)(1)

 M 2

σ max max

 x 

O  M 1

max max

3-26

wl 2

 M max max

 =

⇒

8

  σ max max

 =

wl 2 c

(a)   l

8 I 

⇒

 I   = 8clσ  I 

  w

2

 = 12(12) = 14  = (1/12)(1.5)(9.5) = 107.2 in 144 4 in in,,  I  = 8(1200)(107.2) w = = 10.4 lb lbf/ f/in in   Ans. 4.75(144 ) in,,  I  =  = (π/64)(2 − 1.25 ) = 0.665 6656 6 in (b)   l = 48 in 8(12)(10 )( )(0 0.6656) w =   = 27.7 lb lbf/ f/in in   Ans. 1(48)  . in,,  I  = (1/12)(2)(3 ) − (1/12)(1.625)(2.625 ) = 2.05 051 1 in (c)   l = 48 in 8(12)(10 )( )(2 2.051)   = 57.0 lb w = lbf/ f/in in   Ans. 1.5(48) 3

4

2

4

4

4

3

2

3

3

3

2

 I 

(d)   l

4 . 2 48in

. 2(1 24)

= 72 in; Table A-6,  =  = c  = 2.158 8(12)(10 )( )(2 2.48) w =   = 21.3 lb lbf/ f/in in   Ans. 2.158(72)  = 3.85in (e)   l = 72 in; Table A-7, I  = 0.842"

"

max

2.158"

3

2

4

2

w

 =

(f)   l

72 in in,,  I 

 =

 =

8(12)(103 )( )(3 3.85) 2(722 )

(1/12)(1)(43 ) w

 =

= 35.6 lb lbf/ f/in in   Ans.

5.33 333 3 in4

 =

8(12)(103 )( )(5 5.333) (2)(72)2

  = 49.4 lb lbf/ f/in in   Ans.

4

 

42 3-27

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

(a) Model (c)

π  = 64  = (0.5 ) = 3.068(10− ) in

500 lbf  500 lbf 

4

 I 

0.4375

 A

1.25 in 500 lbf

500 lbf

 

4

= π4 (0.5 ) = 0.19 1963 63 in 2

 Mcc  M

σ 

V (lbf)

3

2

218.75(0.25)

 =  I  = 3.068(10− ) = 17 82 825 5 ps psii = 17.8 kp kpsi si   Ans.  = 43  AV  = 43 0.500 = 340 3400 0 psi   Ans. 1963 3

500

O

τ max max

500

 M  (lbf • in)

 M max  500(0.4375)    218.75 lbf • in

O

(b) Model (d) 1333 lbf/in 0.25"

1.25" 500 l bf

500 l bf

 = 500(0.25) + 12 (500)(0.375) = 218.75 lb lbf  f · in  = 500 lbf 

 M max max

 

V (lbf) 500

V max max O

Same M and V  500

 M 

 

∴

M max

= 17.8 kpsi   Ans.  = 34 3400 00 ps psii   Ans.

σ 

τ max max

O

3-28 F  l

 p2 b

 p1 a

 = − F  x   x − +  p  x   x  −  − l  −  p +a  p  x   x  −  − l  + ter terms ms fo forr x   x    > l + a V  =  x  −  x  −  x    > l + a  = − F  +  + p  x   − l  −  p 2+a p  x   − l  + teterms rms for x   M  =  = − F x  +  +  p  x   x  −  − l  −  p + p  x   x  −  − l  + ter terms ms for x   x    > l + a q

1

1

1

1

1

0

1

1

2

1

2

2

1

2

2

3

6a

2

At x 

 = (l + a)+, V  =  =  =  M  =  = 0, terms for x   x    > l + a = 0 − F  +  + p a −  p 2+a p a = 0   ⇒   p − p  = 2aF  1

1

2

2

1

2

(1)

 

43

Chapter 3

− F (l + a) +

 p1 a 2

2

 

From (1) and (2)

p1

−  p 6+a p 1

a3

2 p1

=0   ⇒

 = 2aF (3l + 2a),

  p2

2

b

From similar triangles

2

2

(2)

(3)

2

a p2

=  p + p 1

2

 = 2aF (3l + a)

a

 p2

− p  = 6 F (al + a)

⇒   b =  p + p

2

1

(4)

2

 M max max occurs where V 

 = 0  =

F  a  2  2b b

l

 p2  p2

 = l + a − 2b

 x max max

 p2 b

 p1

b

 = − F (l + a − 2b) +  p2 (a − 2b) −  p 6+a p 1

 M max max

Fl

=− −  = − F l

F  a (

−

b 2 )

2

 p1

+

2(

Normally M max max The fractional increase in the magnitude is 

2

a

−

b 2 2 )

1

2

 p1

 p 2

(a

− 6+a

1

1

2

= 1500 lbf, a = 1.2 in, l = 1.5 in

 

p1

 = 2(1500) [3(1.5) + 2(1.2)] = 14 37 375 5 lbf/in 1.2 2

2(1500)  p2

(4)

 

a

(

−

= F (a − 2b) − ( p /2)(a − 2bF) l − [( p + p )/6a](a − 2b)

For example, consider F  (3)

− 2b)

 = 1.2 [3(1.5) + 1.2] = 11 87 875 5 lbf/in b = 1.2(11 87 875) 5)/(14 37 375 5 + 11 875 875)) = 0.5429 in 2

Substituting into (5) yields 

= 0.03 036 6 89

or

3.7 .7% % hig igh her th thaan

−Fl

3

b 3 2 )

3

(5)

 

44

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany M Mechanical echanical Engineering D Design esign

20  = 600(15) + 3000 = 8500 lbf  2 15

 y

 600 lbf/ft

3-29

R1

3000 lbf 

5  = 600(15) − 3000 = 3500 lbf  2 15

 x  5'

15'  R1

 

 R2

R2

 = 3500 = 5.833 ft 600

V (lbf)

a

5500 a O 3000

 x  3500

 M  (lbf • ft)

3500(5.833)  20420

O

 x  15000

 y

(a)

+ 5(12) = 3 in ¯ = 1(12) 24

 y

 

 z

 I  z

 y

At x 

 = 5 ft,  =

 

y

 = −3 in,

 y

At x 

 = 14.17 ft,  =

 

 = 13 [2(5 ) + 6(3 ) − 4(1 )] = 136 in −15000(12)(−3) = −3970 psi σ   = −  x 

3

3

136

 = − −15000(12)5 = 6620 psi 136 20420(12)(−3) y = −3 in,   σ  x  = − = 5405 psi 136

 = 5 in,  

  σ  x 

 = 5 in,

 y

 = − 20420(12)5 = −9010 psi 136

  σ  x 

Max tension 66 6620 20 ps psii   Ans. Max compression 9010 90 10 ps psii   Ans. 5500 lbf  (b)   V max max

 =

= =− 5 in

Q n.a.

 = V I bQ = 5500(25) = 50 506 6 ps psii   Ans. 136(2)

τ max max V

 = |σ 2 | = 9010 = 45 4510 10 ps psii   Ans. 2 max max

3

 =  y¯ A = 2.5(5)(2) = 25 in

 z

(c)   τ max max

3

4

 

45

Chapter 3

3-30  y

 = cl F 

F  a

 R1

c

 x 

l  R1

 

  == cl Fx 

 M 

R2

6 M  σ 

≤ x  ≤  ≤ a

6(c/ l ) Fx 

 = bh  =  =

3-31

0

2

6c Fx 

 =   cl   F  =  = V ,

From Prob. 3-30,  R1

V   = 32 bh = 32 (c/bhl ) F 

  h

  ⇒

bh 2

0

blσ  bl σ max max

≤ x  ≤  ≤ a   Ans.

≤ x  ≤  ≤ a

 = 32 l bFτ c

∴ h

τ max max

0

 =  

 Ans.

max max

From Prob. 3-30

e  x  h

Fc

3

6 F cx 

sub in x 

 = e and equate to h above  =

l bσ max max

6 F ce

2 l bτ max max

h( x )

=

 

l bσ max max

= 

cσ  = 38 Flblbτ  τ 

e

max max

 Ans.

2 max

3-32

 = bl F 

F  a

 R1

b l

 R1

 R2

  == bl Fx 

 M 

32 M  σ max

 =

π d 3

  =  =

d 

32 b

  = π d 

3

32 b Fx  π l σ max max

l Fx 

1/3



0

3-33

≤ x  ≤  ≤ a   Ans.

t 

t 

b

Square:

Round:

 

 

b

Am

2

sq sq

all all

 = (b − t ) T   = 2 Am t τ  τ   = 2(b − t ) t τ  τ  Am = π ( b − t ) /4 2

2

2

all all

 = 2π (b − t ) t t τ τ 

all /4 all

T rd rd  

46

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

Ratio of torques T sq sq T rd rd

Twist per unit length square:

= π (b −−t )t )t τ τt τ τ    /2 = π4 = 1.27 2

2G θ 1 t   L

θ sq sq

 =

Round:

2

2(b

t τ  τall  

all all

C 

 A

 L  A

=  = −

 = C 

θ rd rd

all all

m

 L

 A

m

−

Ratio equals 1, twists are the same.

− −

=

π (b t ) C  π ( b t ) 2 /4

m

4(b t ) C  2 ( b t )

= C 4((bb−−t )t ) 2

Note the weight ratio is W sq sq W rd rd

3-34   l

ρl (b

2

= ρl π (b −− t t )() t ) = bπ−t t  = 19 = 6.04 π = 2.86

thin-walled assumes b

≥ 20t 

with b

= 20t  with b = 10t 

6

 = 0.05  = 40 in in,, τ   = 11500 ps psi, i, G = 11.5(10 ) psi, t  = 050 0 in r m = r i + t  t / /2 = r i + 0.025 for r i   > 0 =0 for r i = 0 π  Am = (1 − 0.05) − 4 r m − r m = 0.95 − (4 − π )r m 4  L m = 4(1 − 0.05 − 2r m + 2π r m /4) = 4[0.95 − (2 − π/ 2)r m ]   T  =  = 2 Am t τ τ  =  = 2(0.05)(11 500 500)) Am = 1150 Am Eq. (3-45): all all



2

2

2



2

2

Eq. (3-46): T L m l 180 T L m (40) 180  = θ  l 180 = = π 4G A t  π 4(11.5)(10 ) A (0.05) π

θ (deg)

1

2

6

m

2

m

= 9.9645(10− ) T ALm 4

2

m

Equations can then be put into a spreadsheet resulting in: r i

0 0.10 0.20 0.30 0.40 0.45

 

r m

0 0.125 0.225 0.325 0.425 0.475

 

Am

0.902 5 0.889 087 0.859 043 0.811 831 0.747 450 0.708 822

 

Lm

3.8 3.585 398 3.413 717 3.242 035 3.070 354 2.984 513

 

r i

0 0.10 0.20 0.30 0.40 0.45

 

T (lbf  in)

·

1037.9 1022.5 987.9 933.6 859.6 815.1

 

r i

0 0.10 0.20 0.30 0.40 0.45

 

θ (deg) (deg)

4.825 4.621 4.553 4.576 4.707 4.825

 

47

Chapter 3

1200

1000

   )   n    i   •    f    b    l    (

      T

800

600

400

200

0

0

0. 1

0.2

0. 3

0.4

0.5

r i (in)

4.85 4.80 4.75    )   g   e    d    (  

4.70

        

4.65 4.60 4.55 4.50

0

0.1

0.2

0. 3

0. 4

0.5

r i (in)

Torque carrying capacity reduces with r i . However, this is based on an assumption of uniform stresses which is not the case for small r i . Also note that weight also goes down with an increase in r i . 3-35

From Eq. (3-47) where θ 1 is the same for each leg.

 = 13 G θ  L c ,   T   = 13 G θ  L c T  =  = T  + T   = 13 G θ   L c + L c = 13 G θ  τ   = G θ  c ,   τ   = G θ  c τ   = G θ  c   Ans. T 1

1

1

1

3 1 1

2

1 1

max max

2

1

2

3 2 2

1



3 1 1

3 2 2

1 2



1



 L i ci3   Ans.

1 max

3-36

 = G θ  c

(a)   τ max max

1 max

G θ 1

τ max max

12 00 000 0

9.6(104 ) psi/in

 = c = 1/8 =  = 13 G θ  ( L Lcc ) /  = 13 (9.6)(10 )()(55/8)(1/16) = 4.88 lbf · in   Ans. max

T 1/16

1

3

1 16

4

3

 

48

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

T 1/8

 = 13 (9.6)(10 )()(55/8)(1/8) = 39.06 lbf · in   Ans.  = 9.6(10 )1/16 = 600 6000 0 psi,   τ  /  = 9.6(10 )1/8 = 12 000 psi   Ans. 4

3

4

τ 1/16

4

1 8

9.6(104 )

 = 12(10 ) = 87(10− ) rad/in = 0.458◦ /in   Ans.

(b)  

θ 1

3

6

3-37   Separate strips: For each 1/16 in thick strip,

 Lcc2 τ   L

T  ∴

(1)(1/16)2 (12 00 000) 0)

  == 3 = = 15.625 3  = 2(15.625) = 31.25 lb lbf  f · in   Ans.

lbf   in

·

 T max max

For each strip, 3(15.625)(12)   ==  L Lc3cT Gl = (1)(1 = 0.192 rad   Ans. /16) (12)(10 ) k t t  = T  T /θ   = 31.25/0.192 = 162.8 lbf · inin /rad /θ  =  /rad   Ans. θ 

3

3

Solid strip: From Eq. (3-47),2  Lcc τ   L T max max 3

 =

6

1(1/8)2 12 00 000 0

=

3

= 62.5 lbf · in   Ans.

12 000 000(12) (12)  = θ  l = Gτ τ llc = 12(10  = = 0.09 0960 60 ra rad d   Ans. )(1 /8)

θ 

1

6

 = 62.5/0.0960 = 65 651 1 lb lbf  f · in/rad   Ans.

k l

 = 60 MP MPa, a, H  35 kW (a)   n = 2000 rpm

3-38   τ all all

9.55(35)103

9.55 H   

Eq. (4-40) 16T 

T 

=

2000

    =  = =

16(167.1)

3

 = 20 200 0 rp rpm m

(b)   n

π τ max max

 

∴

π (60)106

= 167.1 N · m 1/3



= 24.2(10− ) m  24.2 mm   Ans. 3

 T 

  = 1671 N · m

  =  =

d 

3-39   τ all all

1/3

16T 

 = π d    ⇒   d 

τ max max

n

 =  =

16(1671) π (60)106

1/3



= 52.2(10− ) m  52.2 mm   Ans. 3

 = 110 MPa, θ  =  = 30◦,   d = 15 mm, l = ? T    ⇒   T  = π τ    == 16  = 16 τ d  d  π d 

3

τ 

θ 

3

Tl

 =  J G  =

  180 π

 

49

Chapter 3

l

π  J G θ 

 = 180

T 

π

= 180



d 4 G θ 

π

32 ( π/ 16) τ  τ d  d 3

π (0.015)(79.3)(109 )(30)

= 360 3-40   d 

110(106 )

3 in, replaced by 3 in hollow with t 

=

(a)

 

T solid solid

3

 =  =

(b)   W solid solid

360 τ 

= 2.83 m   Ans.

 = π (3 − 2.5 )  = 32 τ  1.5

 = 16 τ (3 )   T 

%T 

π d G θ 

1/4 in 4

π

=

4

hollow hollow

( π/ 16)(33 )

4

4

− (π/32 32)) [( [(3 3 − 2.5 ) /1.5] (100) = 48.2%   Ans. (π/ 16)(33 )

2

2

2

2

 = k d  = k (3 (3 ) ,   W   = k (3 (3 − 2.5 ) (3 ) − k (3 (3 − 2.5 ) %W  =  = k (3 (100) = 69.4%   Ans. k (3 (3 ) hollow hollow 2

2

2

2

3-41   T 

= 5400 N · m,   τ   = 150 MPa d /2) 4.023(10 )  = T J c ⇒ 150(10 ) = (π/32)[5400( =   τ  = (a) d  − (0.75d ) ] d  all all

4

6

4

4.023(104 )

  =  =

3

1/3

= 6.45(10− ) m = 64.5 mm From Table A-17, the next preferred size is d = 80 mm; ID  = 60mm   Ans. π   J  =  = 32 (0.08 − 0.06 ) = 2.749(10− ) mm d 

(b)



4

150(106 )

2

4

τ i

4

6

4

.030)  = 5400(0 = 58.9(10 ) Pa = 58.9 MPa   Ans. 2.749(10 ) 6

6

− 3-42

025 5 H  63 02 025(1) 5(1)   == 63 02 = = 12 60 605 5 lbf · in n 5

(a)   T 

 = π d  ⇒   d C C   =

τ 

1/3

    = =

16T  3

C 

16T  π τ 

16(12 605 605)) π (14 00 000) 0)

1/3



= 1.66 in   Ans.

From Table A-17, select 1 3/4 in 605)) 605  = 16(2)(12 = 23.96(10 ) psi = 23.96 kp kpsi si π (1.75 )

τ start start (b) design activity

3

3

 

50

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

3-43   ω

= 2π n /60 = 2π (8) /60 = 0.8378 rad/s 1000   ==  H  = = 119 194 4 N·m 0.8378 ω

T 

1/3

    = = 16T 

d C  C 

π τ 

1/3

16(1194) π (75)(106 )



= 4.328(10− ) m = 43.3 mm 2

From Table A-17, select 45 mm   Ans. 3-44   s

 =

√ 

 A ,   d 

 =  =

 

4 A  A/π /π

Square: Eq. (3-43) with b

=c

 = 4.c8T 

τ max max

3

 = ( A4.8) T /

( τ max max ) sq

3 2

T  16T  3.545T   = 16  =  =  A/π π d  π (4 A /π ) / ( A) /

( τ max max ) rd

Round:

3

( τ max max ) sq ( τ max max ) rd

3 2

3 2

.8 = 34.545 = 1.354

Square stress is 1.354 times the round stress   Ans. 3-45   s

 =

√ 

   =  =

 A,   d 

 A/π 4 A /π

Square: Eq. (3-44) with b

= c, β = 0.141 Tl Tl θ   = = 0.141c G 0.141( A) / G sq sq

4

4 2

Round:

 =  JTGl = (π/32) (4T A l  A/π /π ) /

θ rd rd θ sq sq θ rd rd

4 2

.141 = 16/.02832 = 1.129

Square has greater θ  by a factor of 1.13   Ans. 3-46

 D z

808 lbf   D x 

 y

 x   E 

 z

 D

92.8 lbf 

3.9 in

4.3 in

C  z Q 2.7 in C  x 

362.8 lbf 

362.8 lbf 

G

= 6( A .2832 Tl ) / G 4 2

 

51

Chapter 3

  =  M  D

 z

7C  x 

 − 4.3(92.8) − 3.9(362.8) = 0

 = 259.1 lbf 

C  x 

  = −  M C  C 

7 D x 

 z

 − 2.7(92.8) + 3.9(362.8) = 0

 D x 

 = 166.3 lbf  4.3  M  D ⇒   C  z = 808 = 496.3 lbf   x  7

     M C  C 

 x 

⇒

  D z

 = 27.7 808 = 311.7 lbf  311.7 lbf 

 y  E 

808 lbf 

496.3 lbf 

Q

C 

362.8 lbf 

166.3 lbf 

 x 

92.8 lbf   D

 x 

Q

 z

C 

 D

259.1 lbf   M  y  M  z 259.1(2.7)  699.6 lbf • in

O

O 166.3(4.3)  715.1

lbf • in

 = 808(3.9) = 31  = 3151 51 lb lbf  f · in

Tor orque que : T   x =4.3+

in

   =  =

Bending Q :  M   x  4.3+ in

=

699.62

2

+ 1340 = 15 1512 12 lb lbf  f · in

Torque:

 = 16  = T  = 16(3151) = 82 8217 17 ps psii π d  π (1.25 )

τ 

3

3

Bending:

 = ± 32(1512) = ±78 7885 85 ps psii π (1.25 )

σ b

3

Axial: σ a

362.8  = − AF  = − (π/4)(1 = −29 296 6 ps psii .25 ) 2

|σ  | = 7885 + 296 = 81 8181 81 ps psii max max

8181 2

2

     = +

τ max max

82172

= 91 9179 79 ps psii   Ans.

311.7(4.3)  1340 lbf • in

 

52

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

σ max max tens.

 =

7885

− 296

2

3-47

  +

7885



2

 x 

 B

 A y 3 in

 A z  y

2

− 296 + 8217 = 12845 ps psii   Ans. 2

 B z

2.6 in

92.8 lbf 

 A

 B y

P 1.3 in

 z  E 

92.8 lbf 

808 lbf  362.8 lbf 

  = −  M  B

+ 1.3(92.8) + 3 A y = 0

5.6(362.8)

 z

 A y

 = 637.0 lbf   M  A = −2.6(362.8) + 1.3(92.8) + 3 B y = 0  z

   =   =   =  B y

 M  B

 M  A

 y

 y

274.2 lbf  0

 ⇒

  A z

0

 ⇒

  B z

 = 53.6 808 = 1508.3 lbf 

 = 23.6 808 = 700.3 lbf 

Torsion:   T 

  == 808(1.3) = 10 1050 50 lb lbf  f · in τ  =  = 16(1050) = 534 5348 8 psi π (1 ) Bending:   M  p = 92.8(1.3) = 120.6 lbf · in  M  A = 3  B y + B z  = 3 274.2 + 700.3 = 22 2256 56 lb lbf  f · in = M  32(2256) = ±22980 ps σ b = ± psii π (1 ) 3

 

2

2

 

2

2

max max

3

.8  = − (π/924)1  =  = −12 120 0 ps psii

Axial:   σ  inAP

2

  −  =

τ max max

σ max max tens

 =



22980

22980

− 120 +

2

2

− 120 + 5348 = 12730 ps psii   Ans. 2

 

2

22980

2



− 120 + 5348 = 24049 ps psii   Ans. 2 2

 

53

Chapter 3

3-48 Gear F

1000 lbf • in

 = 1000 = 40 400 0 lb lbf  f  2.5 F n = 400 tan 20 = 145.6 lbf  Torque at C    T C  C  = 400(5) = 2000 lbf · in

2.5 R

Shaft ABCD  y

F t t 

F t 

F n

 



 R Ay 666.7 lbf   A 3"  z

P

2000 lbf • in

 = 2000 = 666.7lbf  3

 B 145.6 lbf 

 R Az

2000 lbf • in 10"

 R Dy C  C  5"

400 lbf 

 D

 

x 

 R Dz

( M  A ) z

  

( M  A ) y

∴

0

18 R Dy

145.6(13)

666.7(3)

0

  R Dy

216.3 lbf 

 =   ⇒ − −  =   ⇒  =  = 0   ⇒ −18 R Dz + 400(13) = 0   ⇒   R Dz = 288.9 lbf  

F  y

 = 0   ⇒   R Ay + 216.3 − 666.7 − 145.6 = 0   ⇒   R Ay = 596.0 lbf  F   z = 0   ⇒   R Az + 288.9 − 400 = 0   ⇒   R Az  = 111.1 lbf   M  B = 3 596 + 111.1 = 1819 lbf · in  M C  C  = 5 216.3 + 288.9 = 1805 lbf · in

   

2

2

2

2

Maximum stresses occur at B.   Ans. 32 M  B

σ max max

 =

σ  B

2

  = π (1.25 ) = 9486 psi

σ  B

 =

τ  B

 = 16π d T  B  = 16(2000) = 5215 psi π (1.25 )

   + + σ  B

32(1819)

2

2

π d 3

3

3

2

τ  B

 =

3

9486 2

   + +

3-49   r 

 =

σ  B

2

2

2

    = +

τ max max

9486

2

τ  B2

 = 7049 psi   Ans.

d /2

  == 90◦, σ  σ r  [1 − 1 + (1 − 1)(1 − 3)co 3)coss 180 180]] = 0   Ans. r  = 2

(a) For top, θ 

52152

= 11 79 792 2 psi   Ans.

 

54

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

 = σ 2 [1 + 1 − (1 + 3)cos 180 180]] = 3σ   

Ans.

σ θ  θ 

 = − σ 2 (1 − 1)(1 + 3)si 3)sin n 180 = 0   Ans.  = 0◦, For side, θ  =

τ r  r θ 

σ 

σ r  r 

[1

1

(1

1)(1

3)coss 0] 3)co

0   Ans.

 = 2 − + − − = σ  σ θ   = [1 + 1 − (1 + 3)co 3)coss 0] = −σ    Ans. θ  2  = − σ 2 (1 − 1)(1 + 3)s 3)sin in 0 = 0   Ans.

τ r  r θ 

(b)

1

σ θ  θ /σ 

 = 2  = r 

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

+ 1

 

4r 2

3 104

     − + = +

100

σ θ  θ /σ 

3.000 2.071 1.646 1.424 1.297 1.219 1.167 1.132 1.107 1.088 1.074 1.063 1.054 1.048 1.042 1.037

1

16 r 4

cos 180

1 2

2

25 r 2

3 104

 + 16 r 

4



   3.0 2.5

2.0

1.5

1.0

0.5

0

0

5

10 r  (mm)  (mm)

15

20

 

55

Chapter 3

(c)

1

σ θ  θ /σ 

 = 2  = r 

+ 1

 

4r 2

3 104

     − + =

100

1

16 r 4

cos0

σ θ  θ /σ 

3 104

1 25 2 r 2

 − 16 r 

4



   0.2

5 6 7 8 9 10 11 12 13 14 15 16 17

1.000 0.376

1 18 9 20

0 0..0 03 20 7 0.025

0

0.135 0.034

0.2

0.011 0.031 0.039 0.042 0.041 0.039 0.037 0.035 0.032

0.4

0.6

0.8

1.0

0

5

10

15

r  (mm)  (mm)

3-50

  == 11.5 = 1.5

 D/  D /d 

  == 11/8 = 0.125 . K ts ts = 1.39 . K t t  = 1.60

r /d 

Fig. A-15-8:

 

Fig. A-15-9:

 

 Mcc 32 K t t M  32(1.6)(200)(14)    = K t  M =   = = 45 63 630 0 psi  I  π d  π (1 )

σ  A

3

3

 = K t s T J c = 16πK d t s T   = 16(1.39)(200)(15) = 21 24 240 0 psi π (1 )

τ  A

3

σ max max

 =

σ  A

2

   + + σ  A

2

2

3

2

τ  A

 =

45.63 2

   + + 45.63 2

= 54.0 kpsi   Ans. τ max max

45.63 2

2

+  =  

21.242

31.2 kpsi   Ans.

=

2

21.242

20

 

56 3-51

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r  fore, from Eq. (3-50)

 = r i . There =

r i2  pi

σ t t , max

 = r   − r 

2

2

o

i

 

o

3-52

If  pi

2

r 2

o

i

r i2

r o2

 = r   − r  2

+ r 

  Ans.

  + −   

r i2  pi

σ r  r,  max

1

i

r 2

=   pi

r o2

 = 0, Eq. (3-4 (3-49) 9) bec becomes omes

2

r o2

− r  = − pi

1

2

i

  Ans.

i

2

2 2

2

− por o − r i r o po /r  σ t t  = r o − r i 2

2

 por o2 2

1

r i2 2

= − r o − r i r   = r i . So The maximum tangential stress occurs at r  = 2 por o σ t t ,  = −  Ans. r   − r  2

+  2

max

2

2

o

i

For σ r  r , we have

2

2 2

2

− por o + r i r o po/r  σ r  r  = r o − r i 2

2

= r 

r i2

2

o

r 2

 − 1

 −

 = 0 at r  =  = r i . Thus at r  =  = r o

So σ r  r 

2

 po r o

σ r  r,  max

  r i2

 por o2

 = r   − r 

2

2

o

i

  −  r i2

r o2

r o2

= − po   Ans.

3-53 2 av

F  r av t 

 p

 =   p A = π r   p  =

σ 1

F 

 = σ   =  A

F 

2

wall

2 π r av  p

= 2π r  t  =  pr  2t  av av

av av

 Ans.

 

57

Chapter 3

3-54   σ t t   > σ l   > σ r  r 

 = (σ t t −  = r i where σ l is intermediate in value. From Prob. 4-50 σ r    r ) /2 at r  =

τ max max

 = 12 (σ t t ,

τ max max

τ max max

 =

 pi 2

max

max )

− σ r r,  2

r o2 r o2

r i r i2

  + +   −

1

 = 75 mm, r i = 69 mm, and τ   = 25 MPa. This gives

Now solve for  pi using r o  pi 3.84 MP MPaa   Ans.

 =

3-55

max max

Given r o

 = 5in, r i = 4.62 625 5 in and referring to the solution of Prob. 3-54, 350 (5) + (4.625)  + 1 τ   = 2 (5) − (4.625) = 2 42 424 4 ps psii   Ans.



max max

3-56

From Table A-20,  S  y

2

2

2

2



 = 57 kp kpsi si;; also, r o = 0.87 875 5 in and r i = 0.62 625 5 in

From Prob. 3-52

2 por o2

σ t t , max

 = − r   − r  2

o

Rearranging  po

 =

r o2

2

i

r i2 (0.8 S  y )

  − 

2r o2

 = 11 20 200 0 ps psii   Ans.

Solving, gives  po 3-57

390 MP MPaa ; also r 

From Table A-20, S   y

From Prob. 3-51

 =

 =

  +  2

 =  pi

σ t t , max

r o

r o2

i

r i

 = (σ t t )

τ max max

= r i = 0.375 in

 = 0.8S  y

 = 68.5 MP MPaa   Ans.

Since σ t t  and σ r  r  are both positive and σ t  t   > σ r  r 

Eq. (3-55) for r 

 =

therefore   pi

2

 − r i

where σ t t  is max at r i

20mm.

2

solving gives   pi 3-58

25mm,, r  25mm

o

max /2

  −  r o2

r i2

r o2

2

 + r i

 

58

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design



2

 + 

0.282 2π (7200)

(σ t t ) max

 =

386

60

×

2

0.375

3

0.292 8

2

+5 +

(0.3752 )( )(5 52 )



292)   − 3++3(00..292 (0.375 ) = 85 8556 56 ps psii

0.3752

1

2

8556 τ max

 =

Radial stress:

dr 

( σ r  r ) max

 =

σ r  r 



= k 

2 2

2



2

2

2

2



r i2 r o2



√ r  r   =   −   = 2r  = 0   ⇒   r  = i o r  3

2

  + 

0.282 2π (7200) 386



 = k  r i   + r o − r ir  r o  − r 

 

d σ  σr  r   

Maxima:

= 4278 psi   Ans.

2

60

3

0.292

0.3752

8

2

 

+5 −

0.375(5)

 = 1.36 3693 93 in

0.3752 (52 ) 1.36932

2

  − 1.3693



= 36 3656 56 ps psii   Ans. 3-59

= 2π (2069)/60 = 216.7 rad/s, ρ = 3320 kg/m , ν = 0.24, r i = 0.01 0125 25 m, r o = 0.15 m;

 

ω

3

use Eq. (3-55)

 = 3320(216.7)

σ t t 

2

 +  3

0.24

8

(0.0125) 2

2

+ (0.15) + (0.15)

1

− 3++3(00..2424) (0.0125)

2



2

(10) −6

2.85 MPa   Ans.

=

3-60

/16)  = 386(1/16)((6π/ 4)(6 − 1 ) = 5.655(10− ) lb lbf  f · s  /in

ρ

2

4

τ max max is at bore and equals

σ t t 

2

2

4

2

Eq. (3-55) 2π (10 00 000) 0)

( σ t t ) max = 5.655(10−4 )

60



= 4496 psi  = 4496 = 2248 psi   Ans. 2

τ max max

2

3

+ 0.20 8



0.52



+ 3 + 3 − 1 3+ 3(00..2020) (0.5) + 2

2

2



 

59

Chapter 3

3-61

= 2π (3000)/60 = 314.2 ra rad/ d/ss 0.282(1.25)(12)(0.125) m = 386 = 1.370(10− ) lbf  · s  /in

 

ω

3

2

 F 

F  6"

  == m ω r  =  = 1.370(10− )(314.2 )(6) = 811.5lbf   = (1.25 − 0.5)(1/8) = 0.09375in .5  = 0.811 = 8656 psi   Ans. 093 09 3 75 2

F 

3

2

2

 Anom

σ nom nom

Note: No te: Str Stress ess con concen centra tratio tion n Fig Fig.. A-1 A-15-1 5-1 giv gives es K t t 

3-62 to 3-67

=. 2.25 wh whic ich h in incr crea ease sess σ 

max max

and fat fatigu igue. e.

 = 0.292,   E = 30 Mpsi (207 GPa),   r i = 0  R = 0.75 in (20 mm),   r o = 1.5in(40mm) ν

Eq. (3-57)  ppsi

 =

 pPa

 =

30(106 ) δ



0.753

207(109 ) δ 0.0203

(1.52

2

2



− 0.75 )(0.75 − 0) = 1.5(10 )δ 2(1.5 − 0) (0.04 − 0.02 )( )(0 0.02 − 0) = 3.881(10 2(0.04 − 0) 2



2

2

2

2

7



3-62

 = 12 [40.042 − 40.000] = 0.021 mm   Ans.

δmax

 = 12 [40.026 − 40.025] = 0.0005 mm   Ans.

δmin

From (2)

 = 81.5 MP MPa, a,   p  = 1.94 MPa   Ans.

 pmax

min

3-63

 = 12 (1.5016 − 1.5000) = 0.000 0008 8 in   Ans.

δmax

 = 12 (1.5010 − 1.5010) = 0   Ans.  = 12 00 000 0 ps psi, i,   p  = 0   Ans.

δmin

Eq. (1)

 

pmax

min

12

(1)

)δ

(2)

 

60

Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design

3-64

 = 12 (40.059 − 40.000) = 0.029 0295 5 mm   Ans.

δmax

 = 12 (40.043 − 40.025) = 0.009 mm   Ans.

δmin  

Eq. (2)

114.5 MP MPa, a,   pmin

pmax

 =

34.9 MPa   Ans.

 =

3-65

 = 12 (1.5023 − 1.5000) = 0.00 001 1 15in   Ans.

δmax

 = 12 (1.5017 − 1.5010) = 0.00 000 0 35 in   Ans.  = 17250psi   p  = 5250 psi   Ans.

δmin  

Eq. (1)

pmax

min

3-66

1

δ

(40.076

40.000)

0.03 038 8 mm   Ans.

 = 2 −  = 1  = 2 (40.060 − 40.025) = 0.0175 mm   Ans.

max

δmin

 = 147.5 MP MPaa   p  = 67.9 MPa   Ans.

 

Eq. (2)

pmax

min

3-67

 = 12 (1.5030 − 1.500) = 0.00 0015 15 in   Ans.

δmax δmin  

Eq. (1)

 = 12 (1.5024 − 1.5010) = 0.00 0007 07 in   Ans.

pmax

 = 22500psi   p  = 10500 ps psii   Ans. min

3-68

 = 12 (1.002 − 1.000) = 0.00 001 1 in   r i = 0,   R = 0.5in,   r o = 1 in ν = 0.292,   E  =  = 30 Mp Mpsi si

δ

Eq. (3-57)  p

 =

30(106 )( )(0 0.001) 0.53



(12

2

2

Eq. (3-50) for outer member at r i 0.5 in 2 0.5 (2.25)(104 ) ( σ t t ) o 1 12 0.52

 =

 =

−

2



− 0.5 )()(00.5 − 0) = 2.25(10 ) psi   Ans. 2(1 − 0) 12

 + = 0.52

4

37500 ps psii   Ans.