TEST BANK FOR Mechanics of Materials An Integrated Learning System 3rd Edition By Timothy A. Philpot

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 2
1 (1.75 in.) 2.4053 in.
4
A

 
and thus, the normal stress in rod (1) is
1
1 2
1
15 kips
6.23627 ksi
2.4053 in
6.24 ksi )
.
(C
F
A


     Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
2 2
2 (2.50 in.) 4.9087 in.
4
A

 
Accordingly, the normal stress in rod (2) is
2
2 2
2
75 kips
15.2789 ksi
2.4053 in.
15.28 ksi (C