## Chapter 3

## Q. 3.12

## Q. 3.12

In Fig. 1 a cylindrical aluminum core (2) is surrounded by a titanium sleeve (1), and both are attached at each end to a rigid end-plate. Assume that the core and the outer sleeve are both stress-free at the reference temperature.

Set up the fundamental equations and solve for the following: (a) the stress induced in each member if the entire composite bar is heated by 100°F, and (b) the resulting elongation of the composite bar.

## Step-by-Step

## Verified Solution

**Plan the Solution** This two-element problem is simple enough so that we can reason that, because α_{\text{alum}} > α_{\text{titan}}, the aluminum will tend to expand more than the titanium and will therefore be in compression if both are heated the same amount. However, in a more complicated situation we would essentially have to solve the entire problem just to determine which elements are in compression and which are in tension. Fortunately, we do not have to do this! We just \underline{\text{assume tension in each element}} and let the final solution tell us which element is in tension and which is in compression.

(a) Solve for the stress in each component.

Equilibrium: The basic question to ask in setting up the equilibrium equation is: What free-body diagram can I use that will relate the internal element forces to each other and to the external loads? (In this example, there are no external loads.) The answer is, of course, one of the end-plates to which both the core and the sleeve are attached, as shown in Fig. 2.

\underrightarrow{+} \sum{F_x}=0: -F_1 – F_2 = 0 Equilibrium (1)

Element Force-Temperature-Deformation Behavior: We use the force-temperature-deformation (F-T-D) format of Eq. 3.27, which involves flexibility coefficients, f_i:

e = fF = αLΔT, f = \frac{L}{AE} (3.27)

e_1 = f_1F_1 + α_1L_1ΔT_1 Element Force-Temperature-Deformation Behavior (2a,b)

e_2 = f_2F_2 + α_2L_2ΔT_2where

f_1 = \left(\frac{ L} {AE}\right)_1 =\frac{(40 in.)}{ (1.0 in^2)(16 × 10^3 ksi)} = 2.50(10^{-3}) in./kip

f_2 = \left(\frac{ L} {AE}\right)_2 =\frac{(40 in.)}{ (1.0 in^2)(10 × 10^3 ksi)} = 4.00(10^{-3}) in./kip

α_1L_1ΔT_1 = (5 × 10^{-6}/°F)(40 in.)(100°F) = 0.020 in.

α_2L_2ΔT_2 = (13 × 10^{-6}/°F)(40 in.)(100°F) = 0.052 in.

Geometry of Deformation: Let δ be the displacement of the right-hand end-plate, as shown in the deformation diagram, Fig. 3, and recall that e_1 and e_2 are the \underline{\text{total elongations}} of the respective elements. Then, the appropriate **compatibility equation** is

e_1 = e_2 Geometry of Deformation (3)

Solution for Element Forces: In Eqs. (1) through (3) we have four equations in four unknowns. Since we want to solve for the member stresses, we can eliminate the e’s by substituting Eqs. (2) into Eq. (3), getting

f_1F_1 + α_1L_1ΔT_1 = f_2F_2 + α_2L_2ΔT_2 Compatibility in Terms of Element Forces (4)

This compatibility equation written in terms of forces can now be solved simultaneously with the equilibrium equation, Eq. (1), to give

F_1 = -F_2 = \frac{α_2LΔT – α_1LΔT}{ f_1 + f_2}= \frac{0.052 in. – 0.020 in.}{2.50(10^{-3}) in./kip + 4.00(10^{-3}) in./kip}

so, the forces in the titanium sleeve and aluminum core, respectively, are

F_1 = 4.92 kips, F_2 = -4.92 kips (5)

The stresses are given by

σ_1 = \frac{F_1} {A_1}, σ_2 = \frac{F_2} {A_2}Therefore,

σ_1 = -σ_2 = 4.92 ksi Ans. (6)

(b) Solve for the elongation of the member. From the deformation diagram in Fig. 3, we see that

δ = e_1 = e_2and we can use either Eq. (2a) or Eq. (2b) to solve for an elongation.

Then,

= [2.50(10^{-3}) in/kip](4.92 kips) = 0.020 in

= 0.0323 in.

or

δ = 0.0323 in. Ans. (b)

**Review the Solution** The signs of member stresses σ_1 and σ_2 agree with our “Plan the Solution” discussion, and the value of the elongation, δ, is between the free-expansion values for aluminum and titanium, which is what we would expect. That is, the final elongation is a compromise between the amount that the aluminum “wants” to expand and the smaller amount that the titanium “wants” to expand.