Solutions Manual Mechanics of Materials An Integrated Learning System 3rd Edition By Timothy A. Philpot

Solutions Manual
Mechanics of Materials An Integrated Learning System 3rd Edition
By
Timothy A. Philpot



( All Chapters Included - 100% Verified Solutions )




1

,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
 
A  ( D 2  d 2 )  [(60 mm) 2  (50 mm) 2 ]  863.938 mm 2
4 4
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
Pmax   allow A  (200 N/mm2 )(863.938 mm2 )  172,788 N  172.8 kN Ans.




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,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
P P 27 kips
  Amin    1.500 in.2
A  18 ksi
The cross-sectional area of the aluminum tube is given by

A  (D2  d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d

[(2.50 in.) 2  d 2 ]  1.500 in.2
4
4
(2.50 in.) 2  d 2  (1.500 in.2 )

4
(2.50 in.) 2  (1.500 in.2 )  d 2

 d max  2.08330 in.

The outside diameter D, the inside diameter d, and the wall thickness t are related by
D  d  2t
Therefore, the minimum wall thickness required for the aluminum tube is
D  d 2.50 in.  2.08330 in.
tmin    0.20835 in.  0.208 in. Ans.
2 2




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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 Copyright3Act 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,
Fy   F1  15 kips  0
 F1  15 kips  15 kips (C)

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):
Fy   F2  30 kips  30 kips  15 kips  0
 F2  75 kips  75 kips (C)

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
F 15 kips
A1,min  1   0.375 in.2
 40 ksi
The minimum rod diameter is therefore

A1,min  d12  0.375 in.2  d1  0.69099 in.  0.691 in. Ans.
4
Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of
F 75 kips
A2,min  2   1.875 in.2
 40 ksi
The minimum diameter for rod (2) is therefore

A2,min  d 22  1.875 in.2  d 2  1.545097 in.  1.545 in. Ans.
4




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