Topic 3.82c: Mixed Mechanical/Thermal - Example 3
Partially Constrained Thermal Deformation:
In a structure, rather than being completely constrained or completely free, members more often are partially constrained. This means a member may expand (or contract) but not as much as it would if unconstrained.
Example 3
In the structure shown three metal rods (steel, aluminum, brass) are attached to each other and constrained between two rigid walls. The rods are initially unstressed and then experience a temperature increase of 80o F. We would like to determine the stress which develops in each rod, and the amount of deformation of the aluminum rod.
The steel, aluminum and brass rod have areas and lengths respectively of 1.5 in2, 6 ft. , 1 in2, 10 ft., .75 in2, 8 ft. as shown in the diagram. The linear coefficient of expansion for the materials are as follows: Steel = 6.5 x 10-6/oF; Brass = 11 x 10-6/oF; Aluminum = 13 x 10-6/oF. And Young's modulus for the materials are Est = 30 x 10 6 psi; Eal = 10 x 10 6 psi; Ebr = 15 x 10 6 psi.
Solution:
PART I: STATICS
In this problem we first consider the effect of the change in temperature. Even though the total length of the three members is fixed, the members may expand or contract against each other. Since all three members are trying to expand they put each other into compression, and also push outward on the walls. In response an external force due to the wall acts inward on the members at each end as shown in Diagram 2.
With a little thought we realize that the forces in each member are the same and equal to the force exerted by the wall on the structure. This may be seen in Diagram 3, where we have cut the structure through the steel member. Since the structure is in equilibrium, this section of the structure must also be in equilibrium. But we can see that this is only possible if there is an internal force in the steel member which is equal and opposite to the force exerted by the wall.
And, of course, this same argument would hold if we cut the structure through the aluminum member, or through the brass member (as shown in Diagram 4)
So we can write (from static equilibrium) that Fst = Fal = Fbr = F
PART II: DEFORMATION
We now consider the geometry of the problem to obtain an additional equation to use in determining the force in the member.
After a little consideration we see that following deformation relationship must be true: 
st + al + st = 0 That is, the sum of the deformations (some of which may be positive expansions and some of which may be negative contractions) must be zero, since the rods are fixed between two walls. On substituting our expression for the total deformation (thermal +/- mechanical) we obtain:
[
- FL / EA ]st + [ - FL / EA ]al + [ - FL / EA ]br = 0
The sign of the mechanical deformation term is negative (-) for all the rods since they are all in compression.. We now substitute values from our problem into the equation and obtain:
[(6.5 x 10-6/oF)(80oF)(72 in) - F (72 in) / (30x106 lbs/in2 )(1.5 in2)] + [(13 x 10-6/oF)(80oF)(120 in) - F (120 in) / (10x106 lbs/in2 )(1 in2)] + [(11 x 10-6/oF)(80oF)(96 in) - F (96 in) / (15x106 lbs/in2 )(.75 in2)] = 0
Rewriting the equation and combining terms we can write:
.0374 in. + .1248 in. + .0845 in. = [1.6 x 10-6 in/lb (F) + 12 x 10-6 in/lb (F) + 8.53 x 10-6 in/lb (F)], or
.2467 in = [22.13 x 10-6 in/lb]F
Now solving for F = 11,150 lb. This is the force in each rod, and we then calculate the stress in each rod from Stress = Force/Area. Finding: Steel Stress = 11,150 lb/1.5 in2 = 7,430 lb/in2; Aluminum Stress = 11,150 lb/1 in2 = 11, 150 lb/in2; Brass Stress = 11,150 lb/.75 in2 = 14,870 lb/in2.
PART III: DEFORMATION
Now that we have the amount of force in the aluminum member, its deformation may be calculated from [ - FL / EA ], oral = [(13 x 10-6/oF)(80oF)(120 in) - 11,150 lb (120 in) / (10x10-6 * 1 in2] = -.009 in. The negative deformation means that the aluminum is forced to shrink by that amount due to the effects of the steel and brass acting on it.
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