EM 304 Exam Problem: Failure Temperature of the Space Shuttle Columbia

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In this video, I solve Problem 1 of the EM 304 test that Dr. Judy Wood gave her students on February 6, 2003, five days after the Space Shuttle Columbia disaster. If a thermal tile is lost, how hot can an aluminum tube held between two rigid support plates get before something fails? I check three candidates: compression in the aluminum tube (442.2 °F), bearing on the plate at the tube end (515.9 °F), and shear in the plate around the tube (205.03 °F). The shear stress in the plate arises due to the compressive stress in the tube that the plate is connected to. Since the temperature of 205 °F is the lowest, the failure occurs first as shear in the plate, even though the corresponding compressive stress in the tube (27.8 ksi) is far less than its strength limit of 60 ksi.

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Timestamps:

0:00 Problem 1 of EM 304 on February 6, 2003: Possible cause of failure is the loss of a thermal tile
1:01 Problem assumptions: Aluminum tube is heated uniformly while the remaining structure is at 0° F
3:26 Given three volume elements to examine the corresponding stresses and temperatures at failure
6:20 Material properties for the support plates and aluminum tube
9:33 Three questions: Determine maximum temperature of the aluminum tube before failure, first failure location, stress states at the three volume elements, and draw these stress state diagrams.
11:17 Location descriptions: Point A is inside the top plate, B is the surface between the plate and tube, Point C is the outer surface of the aluminum tube.
12:02 Solution: I used Dr. Judy Wood's solution key and Grok AI's assistance in this solution
12:20 Thermal expansion properties: Support plates are idealized as perfectly rigid and non-expanding, while the aluminum tube uses the 2014-T6 aluminum alloy properties.
12:38 Strength limits of the plates and tube. Note that ksi = kips per square inch = 1000 psi
14:52 Solution to (a): Maximum temperature without failure
17:33 The aluminum tube is assumed to be of constant length, which means the thermal expansion displacement is equal to the resulting compressive displacement.
20:17 Equation for thermal displacement involving the coefficient of linear thermal expansion
22:21 Equation for Young's Modulus and compressive stress
24:30 Rearrange to obtain our temperature increase formula (and force that develops in the tube)
27:16 Candidate 1: Aluminum tube in compression (point C)
29:07 Maximum temperature at failure at point C is 442.2 °F
32:04 Candidate 2: Plate bearing at the tube end (point B)
34:40 Plates are idealized as rigid and non-expanding, so we have to scale up the temperature of the aluminum tube by the ratio of the plate to tube compressive strength limits and obtain failure temperature of 515.9 °F at point B.
35:47 Candidate 3: Shear at the plate connection (point A)
37:19 Free body diagram of the shear stress in the plate due to the compressive stress in the tube
42:27 "Unrolling" the circular shear stress to obtain a rectangle corresponding to the shear area
45:40 By equilibrium, we obtain a formula for the compressive stress in the tube as a function of the shear stress in the plate and the ratio of the shear area to compressive area
49:00 Temperature at failure at point A is 205.03 °F
53:30 Solution to (b): It fails at point A first, shear in the plate, because its failure temperature is the lowest
54:33 Solution to part (c): Stress state at the failure temperature
55:46 Obtain compressive stress in the aluminum from the shear stress in the plate using our earlier derived formula, and get 27.8 ksi, which is also the uniaxial compressive stress at point C and bearing stress at point B. This stress is well below the plate and tube allowable compressive stresses
59:02 Solution to (d): Volume-element sketches
59:38 Note: The "1 + 2" are actually points Dr. Judy Wood gives when grading tests
1:03:30 Shear stress diagram needs to have arrows be in equilibrium aka not rotating
1:06:02 Summary of the Engineering Logic: Compressive stress in the tube causes failure in shear at the plate, even though the compressive stress is far below its strength limit