Mechanics of Materials an Integrated Learning System 3rd Edition Solutions
- Home
- Documents
- Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Solution Manual for Mechanics of Materials 3rd Edition by Philpot
-
View
2.092 -
Download
360
Embed Size (px)
DESCRIPTION
Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Text of Solution Manual for Mechanics of Materials 3rd Edition by Philpot
-
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as
a compression member. If the axial normal stress in the member must be limited to 200 MPa,
determine the maximum load P that the member can support.
Solution
The cross-sectional area of the stainless steel tube is
2 2 2 2 2( ) [(60 mm) (50 mm) ] 863.938 mm4 4
A D d
The normal stress in the tube can be expressed as
P
A
The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable
normal stress, rearrange this expression to solve for the maximum load P
2 2max allow (200 N/mm )(863.938 mm ) 172,788 172.8 kN NP A Ans.
-
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip
load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness
required for the tube.
Solution
From the definition of normal stress, solve for the minimum area required to support a 27-kip load
without exceeding a stress of 18 ksi
2
min
27 kips1.500 in.
18 ksi
P PA
A
The cross-sectional area of the aluminum tube is given by
2 2( )4
A D d
Set this expression equal to the minimum area and solve for the maximum inside diameter d
2 2 2
2 2 2
2 2 2
max
[(2.50 in.) ] 1.500 in.4
4(2.50 in.) (1.500 in. )
4(2.50 in.) (1.500 in. )
2.08330 in.
d
d
d
d
The outside diameter D, the inside diameter d, and the wall thickness t are related by
2D d t Therefore, the minimum wall thickness required for the aluminum tube is
min
2.50 in. 2.08330 in.0.20835 in. 0.208 in.
2 2
D dt
Ans.
-
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.3 Two solid cylindrical rods (1) and (2)
are joined together at flange B and loaded, as
shown in Figure P1.3/4. If the normal stress
in each rod must be limited to 40 ksi,
determine the minimum diameter required
for each rod.
FIGURE P1.3/4
Solution
Cut a FBD through rod (1). The FBD should include the free end of the rod at A.
As a matter of course, we will assume that the internal force in rod (1) is tension
(even though it obviously will be in compression). From equilibrium,
1
1
15 kips 0
15 kips 15 kips (C)
yF F
F
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again,
we will assume that the internal force in rod (2) is tension. Equilibrium of this
FBD reveals the internal force in rod (2):
2
2
30 kips 30 kips 15 kips 0
75 kips 75 kips (C)
yF F
F
Notice that rods (1) and (2) are in compression. In this situation, we are
concerned only with the stress magnitude; therefore, we will use the force
magnitudes to determine the minimum required cross-sectional areas. If
the normal stress in rod (1) must be limited to 40 ksi, then the minimum
cross-sectional area that can be used for rod (1) is
211,min
15 kips0.375 in.
40 ksi
FA
The minimum rod diameter is therefore
2 21,min 1 10.375 in. 0.6909 0.691 9 i4
inn. .A d d
Ans.
Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of
222,min75 kips
1.875 in.40 ksi
FA
The minimum diameter for rod (2) is therefore
2 22,min 2 21.875 in. 1.54509 1.545 in.7 in.4
A d d
Ans.
-
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.4 Two solid cylindrical rods (1) and (2) are
joined together at flange B and loaded, as shown in
Figure P1.3/4. The diameter of rod (1) is 1.75 in.
and the diameter of rod (2) is 2.50 in. Determine the
normal stresses in rods (1) and (2).
FIGURE P1.3/4
Solution
Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We
will assume that the internal force in rod (1) is tension (even though it obviously will
be in compression). From equilibrium,
1
1
15 kips 0
15 kips 15 kips (C)
yF F
F
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we
will assume that the internal force in rod (2) is tension. Equilibrium of this FBD
reveals the internal force in rod (2):
2
2
30 kips 30 kips 15 kips 0
75 kips 75 kips (C)
yF F
F
From the given diameter of rod (1), the cross-sectional area of rod (1) is
2 21 (1.75 in.) 2.4053 in.
4A
and thus, the normal stress in rod (1) is
11 21
15 kips6.23627 ksi
2.4053 in6.24 ksi )
.(C
F
A
Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
2 22 (2.50 in.) 4.9087 in.
4A
Accordingly, the normal stress in rod (2) is
22 22
75 kips15.2789 ksi
2.4053 in.15.28 ksi (C)
F
A
Ans.
-
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.5 Axial loads are applied with rigid bearing plates to the
solid cylindrical rods shown in Figure P1.5/6. The diameter
of aluminum rod (1) is 2.00 in., the diameter of brass rod (2)
is 1.50 in., and the diameter of steel rod (3) is 3.00 in.
Determine the axial normal stress in each of the three rods.
FIGURE P1.5/6
Solution
Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal
force in rod (1) is tension (even though it obviously will be in compression). From equilibrium,
1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F
FBD through rod (1)
FBD through rod (2)
FBD through rod (3)
Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal
force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):
2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in
rod (3) is:
3
3
8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0
26 kips 26 kips (C)
yF F
F
-
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
From the given diameter of rod (1), the cross-sectional area of rod (1) is
2 21 (2.00 in.) 3.1416 in.
4A
and thus, the normal stress in aluminum rod (1) is
11 21
16 kips5.0930 ksi
3.1416 in5.09 ksi (C)
.
F
A
Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
2 22 (1.50 in.) 1.7671 in.
4A
Accordingly, the normal stress in brass rod (2) is
22 22
14 kips7.9224 ksi
1.7671 in.7.92 ksi (T)
F
A Ans.
Finally, the cross-sectional area of rod (3) is
2 23 (3.00 in.) 7.0686 in.
4A
and the normal stress in the steel rod is
33 23
26 kips3.6782 ksi
7.0686 in3.68 ksi (C)
.
F
A
Mechanics of Materials an Integrated Learning System 3rd Edition Solutions
Source: https://dokumen.tips/documents/solution-manual-for-mechanics-of-materials-3rd-edition-by-philpot.html