Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

ExAmpLE 5.7
An aluminum rod (1) [E = 70 GPa; α =
22.5 × 10 −6 /°C] and a brass rod (2) [E =
105 GPa; α = 18.0 × 10 −6 /°C] are connected
to rigid supports, as shown. The crosssectional
areas of rods (1) and (2) are 2,000
mm 2 and 3,000 mm 2 , respectively. The temperature
of the structure will increase.
(a) Determine the temperature increase
that will close the initial 1 mm gap
between the two axial members.
(b) Compute the normal stress in each rod if
the total temperature increase is 60°C.
Plan the Solution
First, we must determine whether the temperature increase will cause sufficient elongation
to close the 1 mm gap. If the two axial members come into contact, the problem becomes
statically indeterminate and the solution will proceed with the five-step procedure outlined
in Section 5.5. To maintain consistency in the force–temperature–deformation relationships,
tension will be assumed in both members (1) and (2), even though it is apparent that both
members will be compressed because of the temperature increase. Accordingly, the values
obtained for the internal axial forces F 1 and F 2 should be negative.
SOLUTION
(a) The axial elongation in the two rods due solely to a temperature increase can be
expressed as
δ = α D TL and δ = α DTL
1, T 1 1 2, T 2 2
If the two rods are to touch at B, the sum of the elongations in the rods must equal 1 mm:
We solve this equation for DT:
δ + δ = α D TL + α D TL = 1mm
1, T 2, T 1 1 2 2
1 mm
−6 −6
(22.5 × 10 /C) ° D T(900 mm) + (18.0 × 10 /C) ° D T(600 mm) = 1mm
∴D T = 32.2° C
Ans.
(b) Given that a temperature increase of 32.2°C closes the 1 mm gap, a larger
temperature increase (i.e., 60°C in this instance) will cause the aluminum and brass
rods to compress each other, since the rods are prevented from expanding freely by
the supports at A and C.
(1) (2)
A B C
900 mm 600 mm
Step 1 — Equilibrium Equations:
Consider a free-body diagram (FBD)
of joint B after the aluminum and
brass rods have come into contact.
The sum of forces in the horizontal
direction consists exclusively of the
internal member forces.
F 1 (1) (2)
F 2
A B C
∑ F = F − F = 0 ∴ F = F
x 2 1 1 2
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