WORLD FOR CIVIL

Topic 5.1b: Bending Stress - Example 1

In Diagram 1, we have shown a simply supported 20 ft. beam with a load of 10,000 lb. acting downward at the center of the beam. The beam used is a rectangular 2" by 4" steel beam. We would like to determine the maximum bending (axial) stress which develops in the beam due to the loading.

Step 1: Out first step in solving this problem is, of course, to apply static equilibrium conditions to determine the external support reactions. In this particular example, because of the symmetry of the problem, we will not go through the statics in detail, but point out that the two support forces will support the load at the center equally with forces of 5000 lb. each as shown in Diagram 2.

Step 2: The second step is to draw the shear force and bending moment diagrams for the beam. We really don't need the shear force diagram at this point, except we will use it to make the bending moment diagram. We will normally be able to draw the shear force diagram by simply looking at the load forces and the support reactions. If necessary we will determine the shear force and bending moment expressions and make the shear force and bending moment diagrams from these expressions, as we did in the proceeding topic.

We first draw the shear force diagram. Due to the 5000 lb. support force, the shear force value begins at +5000 lb., and since there is no additional loads for the next 10 feet, the shear force remains constant at 5000 lb. between 0 and 10 feet. At 10 feet, the 10,000 lb. downward load drives the shear force down by that amount, from + 5000 lb. to a value of - 5000 lb. Then as there are no additional loads for the next 10 feet, the shear force will remain constant over the remainder of the beam. Graphing the shear force values produces the result in Diagram 3a.

Then using the fact that for non-cantilevered beams the bending moment values are equal to the area under the shear force diagram, we develop the bending moment graph shown in Diagram 3b.

Step 3. We now apply the flexure formula: Bending Stress = M y / I
We wish to find the maximum bending stress, which occurs at the outer edge of the beam so:
M = maximum bending moment = 50,000 ft-lb. = 600,000 in-lb. (from bending moment diagram)
y = distance from the neutral axis of the cross section to outer edge of beam = 2 inches
I = moment of inertia of cross section; for rectangle I = (1/12) bd³ = 1/12 (2" * 4"³) = 10.67 in⁴.
Then, Maximum Bending Stress = M y / I = (600,000 in-lb)*(2 in)/(10.67 in⁴) = 112, 500 lb/in²
This is the correct value, but it is clearly excessive for normal steel. Thus if we tried to use a rectangular 2"x4" steel beam, it would fail under the load. We will have to use a stronger beam.
Notice that the maximum bending moment does not depend on the type of beam. The values of "y" and "I" in the flexure formula do depend on the beam used. Thus, if we had used a rectangular 2"x6" beam (instead of a 2" x 4" beam), the value of y would be: y = 3", and the value of I would be: I = (1/12)(2" * 6"³) = 36 in⁴. Then the maximum bending stress for this beam would be:
Maximum Bending Stress = M y / I = (600,000 in-lb)*(3 in)/(36in⁴) = 50,000 lb/in² . This value is in a more reasonable range for acceptable axial stresses in steel.