Solutions Manual for Mechanics of Materials An Integrated Learning System 4th Edition Philpot

SOLUTIONS MANUAL FOR
MECHANICS OF MATERIALS AN
INTEGRATED LEARNING SYSTEM
4TH EDITION PHILPOT

,Mechanics of Materials An Integrated Learning System 4th Edition Philpot Solutions Manual


P3.1 At the proportional limit, a 2 in. gage length of a 0.500 in. diameter alloy rod has elongated
0.0035in. and the diameter has been reduced by 0.0003 in. The total tension force on the rod was 5.45
kips. Determine the following properties of the material:
(a) the proportional limit.
(b) the modulus of elasticity.
(c) Poisson’s ratio.

Solution
(a) Proportional Limit: The bar cross-sectional area is
 
A  d 2  (0.500 in.)2  0.196350 in.2
4 4
and thus, the normal stress corresponding to the 5.45 kip force is
5.45 kips
  27.7566 ksi
0.196350 in.2
Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,
 PL  43.0 ksi Ans.

(b) Modulus of Elasticity: The longitudinal strain in the bar at the proportional limit is
 0.0035 in.
 long    0.001750 in./in.
L 2 in.
The modulus of elasticity is therefore
 27.7566 ksi
E   15,860 ksi Ans.
 long 0.001750 in./in.

(c) Poisson’s ratio: The longitudinal strain in the bar was calculated previously as
 long  0.001750 in./in.
The lateral strain can be determined from the reduction of the diameter:
d 0.0003 in.
 lat    0.000600 in./in.
d 0.500 in.
Poisson’s ratio for this specimen is therefore
 0.000600 in./in.
   lat    0.343 Ans.
 long 0.001750 in./in.




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, P3.2 A solid circular rod with a diameter of d = 16 mm
is shown in Figure P3.2. The rod is made of an
aluminum alloy that has an elastic modulus of E = 72
GPa and Poisson’s ratio of  = 0.33. When subjected to
the axial load P, the diameter of the rod decreases by FIGURE P3.2
0.024 mm. Determine the magnitude of load P.

Solution
The lateral strain in the rod is
d 0.024 mm
 lat    1,500  106 mm/mm
d 16 mm
Using Poisson’s ratio, compute the corresponding longitudinal strain:
 1,500  106 mm/mm
 long   lat    4,545.455  106 mm/mm
 0.33
Use Hooke’s law to calculate the stress in the rod:
  E long  (72,000 MPa)(4,545.455  10 6 mm/mm)  327.273 MPa
The cross-sectional area of the rod is:
 
A  d 2  (16 mm) 2  201.062 mm 2
4 4
Consequently, the force P that acts on the rod must be
P   A  (327.273 MPa)(201.062 mm 2 )  65,802.086 N  65.8 kN Ans.




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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 Copyright Act without the permission of the copyright owner is unlawful.