재료역학 솔루션 3판 - jaelyoyeoghag sollusyeon 3pan

CCHHAAPPTTEERR

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( a ) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical steel shaft shown. ( b ) Determine the maximum shearing stress caused by the same torque T in a solid cylindrical shaft of the same cross-sectional area.

SOLUTION

( a ) Given shaft: () 4421

44 64 64

66 3 3

(45 30 ) 5 10 mm 5 10 m 2

(5 10 )(45 10 ) 5 10 N m 45 10

Jcc

Tc J T Jc

τ τ

−

− −

=−

=−=× =×

×× ==×⋅ × T =⋅5 kN m 

( b ) Solid shaft of same area:

()

22 2 2 3 2 21

2 4 3 3 6 3

(45 30 ) 3 10 mm

or 33 mm

2 , 2 (2)(5 10 ) 87 10 Pa (0)

Acc

A cA c

Tc T Jc J c

ππ

π π π τ π τ π

=−= −= ×

===

×

==×

τ=87 MPa 

Knowing that the internal diameter of the hollow shaft shown is d = 22 mm, determine the maximum shearing stress caused by a torque of magnitude T = 900 N ⋅ m.

SOLUTION

()

44 4 4 4 21

max 12

11 (40) 20 mm 22 0 m 11 (22) 11 mm 22

(20 11 ) 228329 mm 22 (900)(0) 78 MPa 228329 10

Jcc

Tc J

ππ

τ −

 == =  =  == = 

=−= −=

== = ×



null

( a ) For the 75 mm diameter solid cylinder and loading shown, determine the maximum shearing stress. ( b ) Determine the inner diameter of the hollow cylinder, of 100 mm outer diameter, for which the maximum stress is the same as in part a.

SOLUTION

( a ) Solid shaft: 11 (75) 37 mm 22

c == d =

max 336

2 (2)(4520)

54 10 Pa (i.) 54 MPa (0)

Tc T J c

τ ππ

== = = × 

( b ) Hollow shaft: 2 11 (100) 50 mm 22

c == d =

()

44 2 21 22max 44 2 4 94 126 max

2 (2)(4520)(0)

0 3614 10 m (54 10 )

JTcc cc Tc cc

πτ π

−

==

=− = − = ×

×

c 11 ==0 m dc 21 =87 mm =43 mm 

( a ) Determine the torque that can be applied to a solid shaft of 20-mm diameter without exceeding an allowable shearing stress of 80 MPa. ( b ) Solve Part a , assuming that the solid shaft has been replaced by a hollow shaft of the same cross-sectional area and with an inner diameter equal to half of its own outer diameter.

SOLUTION

( a ) Solid shaft:

44 94

11

(0) 0 m 22

(0) 15 10 m 22

ππ −

== =

=− = ×

96 max (15 10 )(80 10 ) 125. 0.

J T c

τ ××− == = T =⋅125 N m 

( b ) Hollow shaft: Same area as solid shaft.

()

2 22 2 2 2 21 2 2 2

44 4 4 94 21

13 24

22 (0) 0 m 33 1 0 m 2

(0 0 ) 26 10 m 22

Acc c c cc

c c

Jcc

ππ ππ

ππ −



=−=−= =



== =

=−= − = ×

69 max 2

(8010)(26) 181. 0.

J T c

τ ××− == = T =⋅181 N m 

The solid spindle AB has a diameter ds = 38 mm and is made of a steel with an allowable shearing stress of 84 MPa, while sleeve CD is made of a brass with an allowable shearing stress of 50 MPa. Determine the largest torque T that can be applied at A.

SOLUTION

Solid spindle AB : 11 (38) 19 mm 22

c == = ds

44 9

max 96 all

(0) 204 10

22 (204 10 )(84 10 )

905 N m. 0. AB

Tc J J T c

ππ

−

== = ×

=

××

== = ⋅



Sleeve CD : 22 11 (75) 37 mm 22

c == = d

()

12 44 4 4 64 21 66 all 2

37 6 31 mm

(0 0 ) 1 10 m 22 (1 )(5010) 2080 N m CD 0.

cct

Jcc

J T c

ππ

−

=−= −=

=−= − =×

×× == = ⋅

Allowable value of torque T is the smaller. T =⋅2 kN m 

The torques shown are exerted on pulleys A and B. Knowing that both shafts are solid, determine the maximum shearing stress ( a ) in shaft AB , ( b ) in shaft BC.

SOLUTION

( a ) Shaft AB : TdAB =⋅=300 N m, 0 m, c =0 m

max 33

2 (2)(300) (0) 56 10 Pa

Tc T J c

τ ππ

== =

=× τmax=56 MPa

( b ) Shaft BC : 300 400 700 N m

0 m, 0 m

TBC dc

=+= ⋅

max 33

2(2)(700) (0) 36 10 Pa

Tc T J c

τ ππ

== =

=× τmax =36 MPa 

Knowing that each portion of the shafts AB , BC , and CD consist of a solid circular rod, determine ( a ) the shaft in which the maximum shearing stress occurs, ( b ) the magnitude of that stress.

SOLUTION

Shaft AB :

max 3

48 N m 1 7 mm 0 m 2 2

(2) (48)

72 MPa (0)

T

Tc T J c

τ π

=⋅

== =

==

Shaft BC : T =− 48 + 144 =96 N m⋅

max 33

12 (2)(96)

9 mm 83 MPa 2 (0)

Tc T cd J c

τ ππ

== == = =

Shaft CD : T =−48 144+ + 60 =156 N m⋅

max 33

12 (2156)

10 mm 85 MPa 2 (0)

Tc T cd J c

τ ππ

×

== == = =

Answers: ( a ) shaft CD ( b ) 85 MPa 

Knowing that an 8-mm-diameter hole has been drilled through each of the shafts AB, BC , and CD, determine ( a ) the shaft in which the maximum shearing stress occurs, ( b ) the magnitude of that stress.

SOLUTION

Hole: 11

1 4mm 2

cd ==

Shaft AB :

()

44 4 4 94 21

48 N m 1 7 mm 2

(0 0 ) 4 10 m 22

Jcc

ππ −

=⋅

=−= − = ×

max 29 (48)(0) 78 MPa 4 10

Tc J

τ == − = ×

Shaft BC : T =− 48 + 144 =96 N m⋅ 22 1 9mm 2

cd ==

() 44 4 4 94 21

2 max 9

(0 0 ) 9 10 m 22 (96)(0) 87 MPa 9 10

Jcc

Tc J

ππ

−

=−= − =×

== = ×

Shaft CD : T =−48 144+ + 60 =156 N m⋅ 22 1 10 mm 2

cd ==

() 2144 (0 4 0 ) 4 18 10 94 m 22

Jcc ππ − =−= − =×

max 29

(156)(0) 87 MPa 18 10

Tc J

τ == − = ×

Answers: ( a ) shaft CD ( b ) 87 MPa 

In order to reduce the total mass of the assembly of Prob. 3, a new design is being considered in which the diameter of shaft BC will be smaller. Determine the smallest diameter of shaft BC for which the maximum value of the shearing stress in the assembly will not be increased.

PROBLEM 3 Under normal operating conditions, the electric motor exerts a torque of 2 kN ⋅ m on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress in ( a ) shaft AB , ( b ) shaft BC , ( c ) shaft CD.

SOLUTION

See the solution to Problem 3 for figure and for maximum shearing stresses in shafts AB , BC , and CD. The largest shearing occurs in shaft AB. Its magnitude is 81 MPa. Adjust the diameter of shaft BC so that its maximum shearing stress is 81 MPa.

3 3 3 3 6

81 10 Pa 1 10 N m 2

2(2)(1)

22 10 m 22 mm (81 10 )

BC BC

T

Tc T J c

T c

τ π

πτ π

−

=× =× ⋅

==

×

== =×=

×

dc == 2 44 mm 

The allowable shearing stress is 100 MPa in the steel rod AB and 60 MPa in the brass rod BC. Knowing that a torque of magnitude T = 900 N ⋅ m is applied at A and neglecting the effect of stress concentrations, determine the required diameter of ( a ) rod AB , ( b ) rod BC.

SOLUTION

max 43 max

2 ,,

2

Tc T Jc c J

π τ πτ

== =

Shaft AB : max 6

3 3 6

100 MPa 100 10 Pa

(2)(900) 17 10 m 17 mm (1 0 0 1 0 )

−

==×

==×=

×

dcAB ==235 mm 

Shaft BC : max 6

3 3 6

60 MPa 60 10 Pa

(2)(900) 21 10 m 21 mm (60 10 )

−

==×

==×=

×

dcBC ==242 mm 

The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum rod BC. Knowing that a torque of magnitude T = 1250 N ⋅ m is applied at A , determine the required diameter of ( a ) rod AB , ( b ) rod BC.

SOLUTION

max 43 max

2

Tc T Jcc J

π τ πτ

== =

( a ) Rod AB : 3663

(2)(1250)

15 10 m (50 10 ) 25 10 m 25 mm

−

==×

×

=× =

dcAB == 2 50 mm

( b ) Rod BC : 3663

(2)(1250)

31 10 m (25 10 )

31 10 m 31 mm

−

==×

×

=× =

dcBC == 2 63 mm

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa. Determine ( a ) the largest inner diameter of rod AB for which the factor of safety is the same for each rod, ( b ) the largest torque that can be applied at A.

SOLUTION

Solid rod BC : 4

6 all

336 all all

2

25 10 Pa 1 0 m 2

(0) (25 10 ) 132 N m 22

Tc Jc J

π τ

ππ τ

==

=×

==

== ×= ⋅

Hollow rod AB :

()

6 all all

1 all 44 all all 2 1 22

50 10 Pa 132 N m 11 (0) 0 m 22 ?

2 T

c J Tcc cc

τπ τ

=×

=⋅

== =

=

== −

1244 all 2 all 494 6

2 (2)(132)(0)

0 3 10 m (50 10 )

Tc cc πτ

−

=−

=− =×

×

( a ) c 1 =× =7 10 m− 3 7 mm dc 11 == 21 5 mm 

( b ) Allowable torque. T all=⋅132 N m 