# Stress in a L-shaped Beam due to Thermal Expansion

A L-shaped frame $$ABC$$ (right angle at $$B$$) consisting of beams $$AB$$ and $$BC$$ are fixed at ends $$A$$ and $$C$$ to rigid walls. Find the tensile stress at $$A$$ when the temperature of the frame is raised by $$\Delta T$$. Properties: length of $$AB = BC = L$$, flexural rigidity $$EI$$, coef. of thermal expansion $$\alpha$$ and circular cross-section with diameter $$d$$. Also, assume that the right angle at $$B$$ is maintained during deflection.

Initial thoughts: due to $$\Delta T$$, beam $$AB$$ will cause beam $$BC$$ to deflect at $$B$$ rightwards ($$\rightarrow$$), and $$BC$$ will cause $$AB$$ to deflect at $$B$$ upwards ($$\uparrow$$).

1. Since the right-angle remains after the temperature change, the L-shape of the bar is also maintained.
2. The upward deflection does not influence the tensile stress at $$A$$.
3. The thermal expansion of $$AB$$ ($$\delta_T = \alpha \Delta T L$$) is counteracted by a reaction force from $$BC$$, which has the same magnitude of its 'virtual' deflection ($$\delta$$).

$$\alpha \Delta T L - \delta = 0$$

This is equal to zero because the shape of the beams are maintained.

From the integral method for deflection (or energy method),

$$\delta = \frac{F}{A} \frac{L^3 A}{3} = \sigma \frac{L^3 A}{3}$$

Substituting $$\delta$$ and solving for $$\sigma$$

$$\sigma = \frac{3\alpha \Delta T}{AL^2}$$

I am worried that my assumptions are wrong and I cannot think of solving it other way. Could someone check these assumptions? Is there a way that I could do it differently to check the answer?

The frame will be deformed as shown below - a distorted L if so claimed.

• That is right, but I do not understand how the upward deflection will influence the tensile stress at $A$. Do you have any clues? I shall add it to @kamran's tips and try to solve it again.
– Iuri
Mar 22, 2022 at 1:20
• Remember this is a rigid frame, the thermal expansion force of BC will cause joint B lifts upward, and induce a bending moment on member AB (due to the restraint of the rigid joint) and support A. So, at least, the stress at A is the combination of axial stress due to the thermal force of AB plus bending stress caused by the thermal force of BC. I think you can analyze the horizontal and vertical thermal effects separately, then combine the results. (Drawing free body diagram according to the deformed shape will help to visualize the forces)
– r13
Mar 22, 2022 at 13:23
• Simply put, this frame is structurally indeterminate to the 3rd degree, we can't analyze it directly as a single axial load could result in 3 reactions on each support, as well as the internal forces of the members.
– r13
Mar 22, 2022 at 15:49
• Oh, now I see. I can interpret this L-shaped frame as one beam $AB$ with an upward and other rightward force acting at B and, yes, by the Superposition Principle add them together to get the final tensile stress at $A$. That is it, right? Thank you once again!
– Iuri
Mar 22, 2022 at 23:18
• The concept is correct. But the calculation can be tricky. Good luck.
– r13
Mar 23, 2022 at 0:56

you are right your method is not correct because the stiffness of the two members are not rqual.

the stifness of

$$K_{Ab}= EA/L$$

but stiffnes of the cantilever BC is.

$$K_{BC}= 3EI/L^3$$

The thermal extention of ab will be countered by a force $$F = K_{BC} \delta AB$$

Final delta, note we use F/2 because as AB recoils the F decreases to zero.

$$\delta_{final}= F/2* K_{AB}$$

## Edit

After some comments The stiffness of the two members is not identical! because AB is loaded axially and extends longitudinally but BC is a cantilever beam both loaded transversely and deflecting by rotation not by elongation. Also even though there is a rule for springs in series, $$\frac{1}{K_{final}}= \frac{1}{K_1}+\frac{1}{K_2}$$ I don't know if it applies here because usually, the load is the same in a series spring system, here it is not.

• Thank you! I will go through the problem again taking into account this information and probably add another answer.
– Iuri
Mar 22, 2022 at 0:41
• @luri, i now realize that the force of the cantilever is decreasing as the AB recoils. i will modify my answer later. i am flying my Cessna now. but i guess you need to use F/2. Mar 22, 2022 at 1:01
• Oh, I see. Do not worry, you are helping me and I will only go through it again tomorrow! Thank you again.
– Iuri
Mar 22, 2022 at 1:21
• Members AB and BC are identical in all aspects: length, stiffness (EI), load (temperature difference). Therefore, the stiffness of each is identical. Mar 22, 2022 at 16:15
• -1. The two members are the same as shown by the image provided in r13's answer. You did not account for the moment at occurs at B to keep the two members joined at a 90 degree angle. That is, the members are not pinned together. Mar 24, 2022 at 14:01