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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.

enter image description here

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 ($\delta$).

Which leads to

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

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?

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2 Answers 2

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

enter image description here

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  • $\begingroup$ 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. $\endgroup$
    – Iuri
    Mar 22, 2022 at 1:20
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    $\begingroup$ 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) $\endgroup$
    – r13
    Mar 22, 2022 at 13:23
  • $\begingroup$ 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. $\endgroup$
    – r13
    Mar 22, 2022 at 15:49
  • $\begingroup$ 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! $\endgroup$
    – Iuri
    Mar 22, 2022 at 23:18
  • $\begingroup$ The concept is correct. But the calculation can be tricky. Good luck. $\endgroup$
    – r13
    Mar 23, 2022 at 0:56
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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.

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  • $\begingroup$ Thank you! I will go through the problem again taking into account this information and probably add another answer. $\endgroup$
    – Iuri
    Mar 22, 2022 at 0:41
  • $\begingroup$ @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. $\endgroup$
    – kamran
    Mar 22, 2022 at 1:01
  • $\begingroup$ Oh, I see. Do not worry, you are helping me and I will only go through it again tomorrow! Thank you again. $\endgroup$
    – Iuri
    Mar 22, 2022 at 1:21
  • $\begingroup$ Members AB and BC are identical in all aspects: length, stiffness (EI), load (temperature difference). Therefore, the stiffness of each is identical. $\endgroup$
    – JohnHoltz
    Mar 22, 2022 at 16:15
  • $\begingroup$ -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. $\endgroup$
    – JohnHoltz
    Mar 24, 2022 at 14:01

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