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# BeerMOM_ISM_C03

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```CCHHAAPPTTEERR 33
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
261

T
18 mm

PROBLEM 3.1
Determine the torque T that causes a maximum shearing stress of 70 MPa in the steel
cylindrical shaft shown.

SOLUTION
4max
3
max
3 6
;
2
2
(0.018 m) (70 10 Pa)
2
641.26 N m
\uf070\uf074
\uf070 \uf074
\uf070
\uf03d \uf03d
\uf03d
\uf03d \uf0b4
\uf03d \uf0d7
Tc J c
J
T c

641 N m\uf03d \uf0d7T \uf074\uf020
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
262

T
18 mm

PROBLEM 3.2
For the cylindrical shaft shown, determine the maximum shearing stress caused by
a torque of magnitude T \uf03d 800 N \uf0d7 m.

SOLUTION
4max
max 3
3
6
;
2
2
2(800 N m)
(0.018 m)
87.328 10 Pa
\uf070\uf074
\uf074 \uf070
\uf070
\uf03d \uf03d
\uf03d
\uf0d7\uf03d
\uf03d \uf0b4
Tc J c
J
T
c

max 87.3 MPa\uf074 \uf03d \uf074\uf020

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
263

2.4 m
30 mm
45 mmT

PROBLEM 3.3
(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: \uf028 \uf0294 42 1
4 4 6 4 6 4
6 6
3
3
2
(45 30 ) 5.1689 10 mm 5.1689 10 m
2
(5.1689 10 )(45 10 ) 5.1689 10 N m
45 10
J c c
J
Tc JT
J c
T
\uf070
\uf070
\uf074\uf074
\uf02d
\uf02d
\uf02d
\uf03d \uf02d
\uf03d \uf02d \uf03d \uf0b4 \uf03d \uf0b4
\uf03d \uf03d
\uf0b4 \uf0b4\uf03d \uf03d \uf0b4 \uf0d7\uf0b4

5.17 kN mT \uf03d \uf0d7 \uf074\uf020
(b) Solid shaft of same area:

\uf028 \uf0292 2 2 2 3 22 1
2
4
3
3
6
3
(45 30 ) 3.5343 10 mm
or 33.541 mm
2,
2
(2)(5.1689 10 ) 87.2 10 Pa
(0.033541)
A c c
Ac A c
Tc TJ c
J c
\uf070 \uf070
\uf070 \uf070
\uf070 \uf074 \uf070
\uf074 \uf070
\uf03d \uf02d \uf03d \uf02d \uf03d \uf0b4
\uf03d \uf03d \uf03d
\uf03d \uf03d \uf03d
\uf0b4\uf03d \uf03d \uf0b4

87.2 MPa\uf074 \uf03d \uf074\uf020
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
264

3 in.
4 ft
T

PROBLEM 3.4
(a) Determine the maximum shearing stress caused by a 40-kip \uf0d7 in. torque T in
the 3-in.-diameter solid aluminum shaft shown. (b) Solve part a, assuming that
the solid shaft has been replaced by a hollow shaft of the same outer diameter
and of 1-in. inner diameter.

SOLUTION
(a)
4
(40 kip in.)(1.5 in.)
(1.5 in.)
2
7.5451 ksi
\uf074
\uf070
\uf03d
\uf0d7\uf03d \uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8
\uf03d
Tc
J

7.55 ksi\uf074 \uf03d \uf074\uf020
(b)
4 4
(40 kip in.)(1.5 in.)
[(1.5 in.) (0.5 in.) ]
2
7.6394 ksi
\uf074
\uf070
\uf03d
\uf0d7\uf03d
\uf02d
\uf03d
Tc
J

7.64 ksi\uf074 \uf03d \uf074\uf020

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
265

T = 40 kip · in.
T'
3 in.
T'
T = 40 kip · in.
T
4 in.
(b)
(a)

PROBLEM 3.5
maximum shearing stress. (b) Determine the inner diameter of the 4-in.-
diameter hollow cylinder shown, for which the maximum stress is the
same as in part a.

SOLUTION
(a) Solid shaft:
4
1 1 (3.0 in.) 1.5 in.
2 2

2
\uf070
\uf03d \uf03d \uf03d
\uf03d
c d
J c

max
3
3
2
2(40 kip in.)
(1.5 in.)
7.5451 ksi
\uf074
\uf070
\uf070
\uf03d
\uf03d
\uf0d7\uf03d
\uf03d
Tc
J
T
c

max 7.55 ksi\uf074 \uf03d \uf074\uf020
(b) Hollow shaft: 1 1 (4.0 in.) 2.0 in.
2 2
\uf03d \uf03d \uf03doc d

\uf028 \uf0294 42
max
4 4
max
4
4
2
2(40 kip in.)(2.0 in.)(2.0 in.)
(7.5451 ksi)
9.2500 in 1.74395 in.
and 2 3.4879 in.
\uf070
\uf074
\uf070\uf074
\uf070
\uf02d\uf03d \uf03d
\uf03d \uf02d
\uf0d7\uf03d \uf02d
\uf03d \uf05c \uf03d
\uf03d \uf03d
o i
o o
o
i o
i
i i
c cJ T
c c
Tcc c
c
d c

3.49 in.id \uf03d \uf074\uf020

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
266

60 mm
30 mm
D
200 mmT \ufffd 3 kN · m

PROBLEM 3.6
A torque 3 kN mT \uf03d \uf0d7 is applied to the solid bronze cylinder shown.
Determine (a) the maximum shearing stress, (b) the shearing stress at point D
which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c) the
percent of the torque carried by the portion of the cylinder within the 15-mm

SOLUTION
(a) 3
4 3 4 6 4
3
3 3
6
6
1 30 mm 30 10 m
2
(30 10 ) 1.27235 10 m
2 2
3 kN 3 10 N
(3 10 )(30 10 ) 70.736 10 Pa
1.27235 10m
c d
J c
T
Tc
J
\uf070 \uf070
\uf074
\uf02d
\uf02d \uf02d
\uf02d
\uf02d
\uf03d \uf03d \uf03d \uf0b4
\uf03d \uf03d \uf0b4 \uf03d \uf0b4
\uf03d \uf03d \uf0b4
\uf0b4 \uf0b4\uf03d \uf03d \uf03d \uf0b4\uf0b4

70.7 MPam\uf074 \uf03d \uf074
(b) 315 mm 15 10 mD\uf072 \uf02d\uf03d \uf03d \uf0b4

3 6
3
(15 10 )(70.736 10 )
(30 10 )
D
D c
\uf072\uf074 \uf074
\uf02d \uf02d
\uf02d
\uf0b4 \uf0b4\uf03d \uf03d \uf0b4 35.4 MPaD\uf074 \uf03d \uf074
(c) 3
2
D D D D
D D D D
D D
T JT
J
\uf072 \uf074 \uf070\uf074 \uf072 \uf074\uf072\uf03d \uf03d \uf03d

3 3 6
3
(15 10 ) (35.368 10 ) 187.5 N m
2
187.5100% (100%) 6.25%
3 10
D
D
T
T
T
\uf070 \uf02d\uf03d \uf0b4 \uf0b4 \uf03d \uf0d7
\uf0b4 \uf03d \uf03d\uf0b4 6.25% \uf074\uf020
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
267

4 in.
8 in.
ds
t 5 in.14
3 in.
D
C
A
B
T

PROBLEM 3.7
The solid spindle AB is made of a steel with an allowable shearing stress of
12 ksi, and sleeve CD is made of a brass with an allowable shearing stress
of 7 ksi. Determine (a) the largest torque T that can be applied at A if the
allowable shearing stress is not to be exceeded in sleeve CD, (b) the
corresponding required value of the diameter sd of spindle AB.

SOLUTION
(a) Analysis of sleeve CD:
\uf028 \uf029
2
1 2
4 4 4 4 4
2 1
3
3
2
1 1 (3) 1.5 in.
2 2
1.5 0.25 1.25 in.
= (1.5 1.25 ) 4.1172 in
2 2
(4.1172)(7 10 )= 19.21 10 lb in.
1.5
oc d
c c t
J c c
JT
c
\uf070 \uf070
\uf074
\uf03d \uf03d \uf03d
\uf03d \uf02d \uf03d \uf02d \uf03d
\uf02d \uf03d \uf02d \uf03d
\uf0b4\uf03d \uf03d \uf0b4 \uf0d7

19.21 kip in.T \uf03d \uf0d7 \uf074\uf020
(b) Analysis of solid spindle AB:
3
3 3
3
3
=
19.21 10 1.601 in
2 12 10
(2)(1.601) 1.006 in. 2s
Tc
J
J Tc
c
c d c
\uf074
\uf070
\uf074
\uf070
\uf0b4\uf03d \uf03d \uf03d \uf03d\uf0b4
\uf03d \uf03d \uf03d

2.01 in.\uf03dd \uf074
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
268

4 in.
8 in.
ds
t 5 in.14
3 in.
D
C
A
B
T

PROBLEM 3.8
The solid spindle AB has a diameter ds \uf03d 1.5 in. and is made of a```