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

Written notes — mirrored from Exam Problem by Dr. Judy Wood: Space Shuttle Columbia Disaster, which has the full video and the complete solution.

Problem 1 of EM 304 – Test #1 – February 6, 2003

Last Saturday (February 1, 2003), the space shuttle Columbia disintegrated during re-entry through the Earth's atmosphere.

One possible cause could be the loss of thermal tiles…

• Assume tiles were lost in the localized zone shown below, exposing just one of the many aluminum stiffening tubes, connected to both the top and bottom framework of the wing.
• Assume the remaining structure is completely insulated from the heat and remains at 0°F.
• Assume the one exposed tube is heated uniformly.
• At 0°F, the aluminum tube is stress free and there are no gaps.
  • Heat shields insulate the structure from temperature variations, as shown below.
• Assume the top and bottom of the framework is made of a solid plate.

Material properties:

• L = constant
  • Note this is the length of the aluminum tube within the structural support surfaces (support plates) as shown in the earlier figure above.
• Δtmax = 3,000°F (maximum possible)
  • From wiki: During re-entry, the thermal protection system (TPS) experienced temperatures up to 3,000 °F (1,648 °C), but had to keep the orbiter vehicle's aluminum skin temperature below 350 °F (180 °C).

Top and Bottom plates:

• α_plate = 0.00 με/°F
• E_plate = ∞, G = ∞, v = 0 (rigid)
• τ_max = 10.2 ksi
• σ_max = 104 ksi (tension)
• σ_max = 74 ksi (compression)

Aluminum (2014-T6):

• α_al = 12.8 με/°F
• E_al = 10.6 Msi, G = 3.9 Msi, v = 0.34
• τ_max = 19 ksi
• σ_max = 60 ksi (tension)
• σ_max = 60 ksi (compression)

Given the properties above:

(a) Determine the maximum temperature of the aluminum tube without causing failure at point A, B, or C.

• Note: Failure may occur in the plate or the tube.

(b) Where is it most likely to fail first (choose location A, B, or C). Why?

• Justify your answer with calculations.

(c) Using your answer from part (a), calculate the stress state in locations A, B, and C at the failure temperature.

(d) Show these stresses (from part (c)) on each of the volume elements (A, B, and C) provided on above, representing the state of stress at each of the three locations.

Note: Volume elements C, B, and A are provided below.

Location descriptions:

• Point A: zone of material inside top plate
• Point B: surface of plate facing tube end, and surface of tube end facing plate
• Point C: zone of material in outer surface of the aluminum tube.

Solution to Part (a): Maximum temperature without failure

Determine the maximum temperature of the aluminum tube without causing failure at point A, B, or C.

Note: Failure may occur in the plate or the tube.

Let's look at locations A, B, and C in the earlier given diagrams, also shown below.

We are given that the one aluminum tube is exposed and heated uniformly while the top and bottom support plates (and the rest of the structure) stay at 0 °F.

At 0 °F the aluminum tube is stress-free and there are no gaps.

The plates are modeled as rigid (E = ∞, G = ∞, v = 0, α = 0), so the length L of the tube is fixed.

Only the exposed tube wants to expand due to thermal expansion; the rigid plates prevent that expansion, producing compressive force in the tube and equal-and-opposite reactions on the plates.

Failure can occur in the aluminum tube or in the plates (bearing or shear).

Because L is constant:

Candidate 1: Aluminum tube in compression (point C)

If the first failure is aluminum in compression, then the maximum temperature is:

Calculation check:

60,000/(10.6*12.8) = 442.217

Candidate 2: Plate bearing at the tube end (point B)

The support plate's compressive allowable stress is given as σ_max = 70 ksi.

If the point at B fails, then the compressive stress is greater than 70 ksi.

But we have already checked the point at C, which has a compressive strength of 60 ksi in the aluminum tube, thus the aluminum controls as it would fail before the plate does.

Since the support plates are idealized as rigid and non-expanding, we can't just plug in α = 0 and E = ∞ into the earlier ΔT formula to obtain the temperature at this higher stress.

As the aluminum tube expands due to thermal expansion, this expansion imposes stress on the support plate as well.

Thus, we can obtain the maximum temperature that the support plate will fail at by simply scaling up our earlier maximum temperature by the ratio of the compressive strength of the support plate to the aluminum tube.

Note again that the aluminum tube at point C will fail before this temperature is reached at point B in the support plate.

Calculation check:

70/60 * 442.217 = 515.9198

Candidate 3: Shear at the plate connection (point A)

First note that we are told to assume that the "remaining structure is completely insulated from the heat and remains at 0 °F".

The skin of the wing, aka the support plate, is not affected by the temperature, but it is affected by the thermally expanding aluminum tube.

This thermal expansion of the aluminum tube causes shear stress on the support plate, at the point A.

Thus, by equilibrium:

Recall that the maximum allowable compressive stress in the aluminum is 60 ksi, while the maximum allowable shear stress in the support plate is 10.2 ksi.

Thus we can now obtain the corresponding temperature increase of the aluminum tube that would cause the support plate to fail in shear.

Calculation check:

10,200/(10.612.8)(0.3/0.11) = 205.0279

This is the lowest temperature rise!

Therefore the maximum allowable temperature of the tube is 205 °F.

Solution to part (b): Where it fails first

Where is it most likely to fail first (choose location A, B, or C). Why?

Justify your answer with calculations.

The three candidate temperatures are approximately:

• A: 205 °F
• B: 442 °F
• C: 516 °F

Thus the failure occurs first at point A, which is shear in the support plate at the tube connection.

Solution to part (c): Stress state at the failure temperature

(c) Using your answer from part (a), calculate the stress state in locations A, B, and C at the failure temperature.

We can use our earlier equation relating aluminum compressive stress to the plate shear stress:

Note that Dr. Judy Wood used the stress-strain-temperature equation to obtain this corresponding compressive stress, both choices work.

Calculation check:

10.2*(0.3/0.11) = 27.8182

At point C (outer surface of the aluminum tube), we have uniaxial compression of 27.8 ksi.

At point B (interface): the same compressive force acts as bearing stress on the plate and as axial stress on the tube end.

Because the force is limited by shear capacity at A, the bearing stress is well below the plate allowable (70 ksi).

At point A (inside the plate), the stress state is pure shear at the maximum allowable shear stress in the plate, which was given as 10.2 ksi.

Solution to part (d): Volume-element sketches

(d) Show these stresses (from part (c)) on each of the volume elements (A, B, and C) provided below, representing the state of stress at each of the three locations.

Note: Volume elements C, B, and A are provided below.

On the provided volume elements, the stresses at ΔT = 205 °F are shown below:

Note: The "1 + 2" are actually points Dr. Judy Wood gives when grading tests.

Summary of the Engineering Logic

Because the plates are rigid, the aluminum tube develops compressive stress linearly with temperature.

The connection, however, can transmit only a limited shear force before the plate fails in shear.

That force limit is reached at only 205 °F, which is well before the aluminum itself reaches its 60 ksi compressive yield.

Hence the first failure is shear at the plate-to-tube interface (point A).