WORLD FOR CIVIL

Topic 3.1: Stress, Strain & Hooke's Law - I

In the first general topic (Statics) we examined the process of determining both the external support forces acting on a structure and the internal forces acting in members of a structure (particularly axial members). While this is important and in fact indispensable when designing structures or determining the safety of a loaded structure, knowing the force values is not enough. We can see that from a simple example.

In Diagram 1, the structure shown is composed of axial member AC which is pinned to the floor at point A, and cable BC which is pinned to the wall at point B. In addition, a load of 15,000 lb is attached to the structure at point C. If we solve for the forces acting on and in the structure we will find that at point A there is a support force of 14,180 lb. acting at 37^o (along the direction of the member); which is also the internal force in member AC, 14,180 lb (compression). At point B, the external support force of the wall on the cable has a value of 13,090 lb., acting at an angle of 150^o (from +x-axis) This is also the value of the internal tension in the cable, 13,090 lb. Now, we could ask the question; Is this structure safe? Are members BC and AC strong enough to support the load?

We recognize right away that knowing the force in the cable BC is not enough to tell us if the cable is safe or if it will break. Clearly it depends on several other factors in addition to the force in the cable. It depends on the size of the cable. A 1" diameter steel cable will carry more load than a ¼ " diameter steel cable. It also depends on what the cable is made of. A steel cable will clearly support more than an aluminum cable. To address the first consideration, we will turn to the concept of STRESS.

AXIAL STRESS
What is known as Axial (or Normal) Stress, often symbolized by the Greek letter sigma, is defined as the force perpendicular to the cross sectional area of the member divided by the cross sectional area. Or

In diagram 2, a solid rod of length L, is under simple tension due to force F, as shown. If we divide that axial force, F, by the cross sectional area of the rod (A), this quotient would be the axial stress in the member. Axial stress is the equivalent of pressure in a gas or liquid. As you remember, pressure is the force/unit area. So axial stress is really the 'pressure' in a solid member. Now the question becomes, how much 'pressure' can a material bear before it fails.

Well, we will examine that question in some detail in a bit, but to give an example, a normal operating stress for carbon steel might be 30,000 lb/in². Now let's return to our example shown in Diagram 1 (repeated in Diagram 3). In our structure , if we assume both the member and the cable are made of steel, and if the diameter of the cable is .5 inches, and if the cross sectional area of the member is 1.2 in², are the stresses in the cable BC and in member AC within the 'allowable' stress for steel of 30,000 lb/in²?

For the cable BC: Axial Stress = F/A = 13,090 lb./ (p * .25"²) = 66,700 lb/in²
For the member AC: Axial Stress = F/A = 14,180 lb./ (1.2 in²) = 11,820 lb/in²
These are interesting results. We see from the calculations that the stress in member AC (11,820 lb/in²) is well within the allowable stress of 30,000 lb/in², however, we also see clearly that the stress in the cable AC (66,700 lb/in²) is over twice the allowable stress of 30,000 lb/in². This means that the ½ inch diameter cable is much too small to support the load.

Well, what size cable should we use? Another interesting question whose answer we find by simply reversing our process, using the stress equation to find the minimum size cable for the allowable stress of 30,000 lb/in². That is, we set the stress value to the allowable stress of 30,000 lb/in², put in the axial force in the cable, and solve for the cable area needed.

Axial Stress = F/A : 30,000 lb/in² = 13,090 lb./A; solving for A = .436 in². Since the area of cable = 3.14 (r²), we can solve for the radius r = square root (.436 in²/3.14) = .373 inches. So the minimum diameter steel cable which would safely support the load is d = .746 inches ( or ¾ inch diameter cable). This is an important process. We checked the members in the structure, found one was not safe according to the allowable stress for the material, and then calculated the size member needed so that the structure would be safe.

Our next step is to examine an associated property of stress, strain, and to examine the stress and strain in materials in somewhat more detail. Select Topic 3.2: Stress, Strain & Hooke's Law - II