Hello guys. Let us continue with yesterday’s post by taking a look at the question.

Determine the max load that a 0.50 in diameter steel rod can support if the normal stress in the rod must not exceed 24,000 psi.

Before I go on babbling about this question, let me introduce some concepts.

The first thing to know about is the concept of stress. Not the kind of stress you get when you are about to have a job interview. Rather, stress in this case is how much load per unit area of material is experiencing. There are two kinds of stress. The most basic kind is axial stress, “defined as a force acting on a perpendicular cross-sectional area.” It is symbolized by the symbol σ. Shear stress, on the other hand, is symbolized by the Greek letter Tau, and is defined as the force parallel to the cross-sectional area of the member divided by the cross-sectional area itself. The basic formula for stress is σ= F/A, where F is the force/load, while A is the cross-sectional area.

The shear stress refer to the picture labeled shear stress. The other two pictures are axial stresses.

not this kind of stress…

Also, another important thing to remember is that under axial stress (either tension or compression) the material will always undergo deformation, symbolized by delta (). This is total deformation, also known as elongation.

While stress is the load per unit area, strain is the deformation per unit length. It’s symbol is an epsilon, or, and its formula is elongation/ the original length of the member.

We’ve gone through stress and strain. But what relates these two things? It turns out that stress = E times strain. But what is E? E is the modulus of elasticity, a modulus established by the famous Thomas Young. It is the constant of proportionality between stress and strain. This equation I mentioned also happens to be called Hooke’s law, perhaps the most famous equation in materials science.

From top: the formulas for strain, stress, and Hooke’s law

Refer back to the formulas of stress and strain. If I were to combine them algebraically, I would get this equation: = FL/AE, where L= length of member, E=modulus of elasticity, F=load, and A=area. This is in fact another version of Hooke’s law. However, it is very useful if you do not know the measurements for stress and/or strain.

For a formula sheet on what’ve we done so far, here is a formula sheet from PLTW: PLTW Material Testing Formula Sheet

Let’s go back to the question I mentioned earlier. First, we need to identify the knowns and unknowns. From the problem, we know that the steel rod has a 0.50 in diameter. From this we can get the area (note it is the cross-sectional area), which, by using the circle’s area formula, is 0.2 square inches. We also know the stress: 24,000 psi, or lbs per square inch. Our only unknown here is the load, or F.

Once we got the knowns and unknowns, we select a formula to use. In this case, we use the stress formula, because it contains all the elements we have and the ones we want to solve. So using the equation, we plug in what we know. It becomes this:

24,000 psi = F/ 0.2 in^2

Multiplying 0.2 in^2 yields:

F= 24000 psi x 0.2 in^2 = approximately 4800 lbs. Wala, that is the answer!

Here are some extra harder questions that I will give the answers to for tomorrow.

1) A 35 ft long solid steel rod is subjected to a 8000 lb load. This load causes the rod to stretch 0.266 in. The modulus of elasticity of the steel is 30000000 psi. Determine diameter of rod.

2) A 9000 lb load is suspended from the roof of a mall with a 16 ft long solid aluminum rod. The modulus of elasticity of the Al is 10,000,000 psi. If the max rod elongation must be limited to 0.50 in and max stress must be limited to 30,000 psi, find the minimum area that may be used for the rod.

3) Two 40. ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of material H and has a diameter of 3/8 in. Wire (2) is made of material K and has a diameter of 3/16 in. When a load of 225 lb is applied to its lower end, wire (1) stretches 0.10 in. When the same 225 lb load is applied to the lower end of wire (2), wire (2) stretches 0.25 in. Compare materials H and K. Which is the stiffer material; in other words, which material has the greater modulus of elasticity?

Well, there’s the questions. For more in-depth learning, download this pdf file from PLTW on materials testing: Principle of Engineering Material Testing

For tomorrow, I will post my analysis for the Checkmate Poem. Refer to my 1st and 2nd posts.