Indiana Jones and the Unbroken Egg – Engineering Thoughts

We all know stress. It’s that feeling when you have too much to do and consequences loom. The load is heavy and failure is not a good option. You reduce stress by removing some of the load or enlisting more people to carry it – a concept few managers seem to get.

To the structural engineer, stress is how a beam or column feels. Who says engineers are not sensitive creatures? Of course they have a mathematical way to describe these feelings.

Eqn. 1                           S = P/A

Here, S is stress in psi, P is load in pounds, A is cross-sectional area in square inches.

An example might make this more clear. Say Indiana Jones dangles from a rope above a snake pit. (Indiana hates snakes.) Engineers cannot put numbers on Indiana’s personal stress, but they can tell you if his 1000 psi-rated rope is likely to break. You may be thinking the guy cannot weigh much over 200 pounds, nothing near 1000 pounds – what’s the problem?

The problem lies in the units. The rope is rated in psi (pounds per square inch or pounds divided by area) and that is stress; you need to use Eqn 1. Take Indiana’s weight, 185 lbs, and divide that by the cross-sectional area of the rope. The cross-sexual what?

Cross sectional area is the area of the cut that the bad guy would make to dispense with Indiana. It is the circle area you would see looking straight down the rope. Engineers are famous for drawing on the backs of envelopes and napkins, so here you go:20131214-210928.jpg

The area of the circle is like the number of people working on that stressful project with the lousy manager. More area (or more people) means less stress. The area of the circle comes from another formula you may remember from high school, A =  π × r2. Suffice to say the area is .1963 square inches. So using Eqn 1, the stress is:

S = 185 lbs/.1963 = 942 pounds per square inch or 942 psi.

Whew! Less than 1000 psi. Are we relieved for Indiana escaping the snake pit or for ourselves somehow passing our last math class?

The stress we just waded through is called tensile stress. We say the rope carried the load in tension. Tensile stress occurs when we pull from both ends. Compressive stress or compression is when we push from both ends. Ropes do not carry compression; however, shortly after the snake pit, Indiana had another stressful situation involving a room with moving walls. As the walls closed in, he was in imminent danger of suffering from excessive compressive stress.

You might be excited to know that the room with moving walls brings up yet another engineering concept: hydrostatic stress. If all sides of the room, including the ceiling and floor were to press equally on a square block placed inside the room, that would be hydrostatic compressive stress. That is, stresses from all three directions are equal.

Here’s an experiment you can do at home: Take someone with a big hand – well, generally people who have one big hand also have a second big hand. But say they use one big hand to completely envelop an egg and crush it. Regardless of how hard the big-handed person presses, as long as the stresses stay equal all around (or “hydrostatic”), the egg will not break. This is true* only if the egg has no cracks in it. Eggs with cracks brings us to Fracture Mechanics; a totally fascinating subject, but ok, I won’t go there.

In theory, objects under hydrostatic stress (all-over tension or compression) never break. However, since most materials have small flaws (e.g., cracks) in them, the cracks eventually cause failure, albeit at a much higher stress than you might expect.

So, how does this all fit together?

First off, let’s remember that to talk about stress and failure, you must take cross-sectional area, as well as load or weight, into account. Ignoring cross-sectional area is like not knowing how many people a project work load is being spread over. Then we need to take out hydrostatic stresses, as they do not cause failure.

Now let’s examine compression.
Say you have a rectangular bar of steel .25″ x .25″ x 4.” (Note: the 4″ height measurement of the bar will not enter any calculations). You vertically push on this bar with 2250 pounds; it is rated to hold 36,000 psi. Watch the units. We need to divide the load (pounds) by the area (square inches). 20131214-195444.jpg

The cross-sectional area, this time, is a square, so its area is length times width: .25″ x .25″ = .0625 square inches. Using Eqn 1 again, the stress is:

S = P/A = 2250 lbs/.0625 sq. in = 36,000 pounds per square inch or 36,000 psi.   Not another failure! We gotta do something.

Enter the Dogbert School of Management. To alleviate stress, apply more pressure. Let’s push with 10,000 psi in the horizontal direction and the into-page direction. Since the 10,000 psi is already given in stress units, we do not need to use Eqn 1.

Here’s what we have now:20131214-202048.jpg

Ends up, Dogbert is right. This bar will not fail if we add more compressive stress. Though it was no doubt unintentional, Dogbert took advantage of the Principal of Superposition. Engineers understand it and use it a lot. Managers, not so much.

The Principal of Superposition says that you can separate out stresses any way you want so long as the resultant stress state adds up to the same one you started with [2].


Here, we separate the stresses into two parts and add them.The first part is hydrostatic compression. Stresses in all three directions are the same.  Since hydrostatic stress does not contribute to failure, for failure analysis, we can ignore it. That leaves the bar “feeling” like it only has 26,000 psi compressive stress in the vertical direction. No failure.

Time to conclude. We have discussed some important engineering concepts: tensile and compressive stress, units (like psi), cross-sectional area, hydrostatic stress and the Principal of Superposition. More impressive still, especially at a party, is the concept that someone with a big hand can squeeze an egg really hard without breaking it.

[1] Timoshenko, S. P. (1953), History of Strength of Materials ISBN 0-07-064725-9, p. 283.
[2] Popov, E. P. (1968), Introduction to Mechanics of Solids ISBN-10: 0134877691, ISBN-13: 978-0134877693, p. 99.

* OK, it is not quite true: Even if the big hand presses evenly all around, the egg has to be completely filled with fluid to keep all stresses equal. The egg has a little air gap, so it would eventually fail where the radius is largest – around it’s equator. Most people are not strong enough – try it!

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