WORLD FOR CIVIL

Topic 5.2b: Bending Stress - Example 3

A simply supported WT 8 x 25 T-beam is loaded is shown in Diagram 1. For this beam we will determine the maximum bending stress in the beam. We will also determine the bending stress 4 ft from the left end of the beam and 2 inches above the bottom of the beam.

STEP 1: Apply Static Equilibrium Principles and determine the external support reactions:
1.) FBD of structure (See Diagram 2)

2.) Resolve all forces into x/y components
3.) Apply equilibrium conditions:
Sum F_x = 0 none
Sum F_y = B_y + C_y - 1,000 lbs/ft (4 ft) - 1,500 lbs/ft (4 ft) = 0
Sum T_B = 1,000 lbs/ft (4 ft) (2 ft) - 1,500 lbs/ft (4 ft) (8 ft) + C_y(6 ft) = 0
Solving: B_y = 3,330 lbs; C_y = 6,670 lbs

STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam. We can normally do this for the Shear Force Diagram, reasonably accurately, by simple looking at the loading and the support reactions. That is, for this example, the shear force begins with a value of zero (since there is not point load or reaction at x = 0), and then decreases at a rate of 1000 lb. per foot for the first 4 feet due to the uniform load, resulting in a shear force value of -4000 lb. at x = 4 ft. Then, also at x = 4 ft., there is an upward support reaction of 3,330 lb. which drives the value of the shear force upward (from -4000 lb.) by that amount to a value of -670 lb. Next nothing happens (no loading) for the next 6 ft., so the shear force remains constant. Then at x = 10 ft there is an upward support force of 6,670 lb. which drives the shear force value up this amount, from a value of -670 lb. to +6000 lb. Then from 10 ft. to 14 ft., a uniform load of 1,500 lb./ft. is applied, driving the shear force downward at a rate of 1500 lb. per each foot for the last 4 ft (for total of - 6000 lb), Which brings the shear force value down to zero at x = 14 ft.(See Diagram 3).

The Bending Moment Diagram may be drawn using the Shear Force Diagram, by remembering that (for non-cantilever beams) the value of the bending moment at a given location, x feet from the left end, is equal to the area under the shear force diagram up to that point. Applying this, we obtain the Bending Moment Diagram shown in Diagram 4.

STEP 3: We will now Apply the Flexure Formula to determine the maximum bending stress for the beam. We may use Flexure Formula: M y / I, or a special form of the Flexure Formula: M / s, where s is the section modulus. This special form of the flexure formula can only be used to find the maximum bending stress, and uses the section modulus, where the section modulus is equal to the moment of inertia of the beam cross section divided by the maximum distance from the neutral axis of the beam to an outer edge of the beam.

As an example we apply this form to determine the maximum bending stress in our beam. First we determine the maximum bending moment from our bending moment diagram - which we observe from Diagram 4 is: M_max = 12,000 ft.-lb, and occurs at x = 10 ft. (We will drop the negative sign which simply tells us that the beam is bent concave facing downward at this point. This means the top of the beam is in tension and the bottom of the beam is in compression.) We also then find the x-x axis section modulus of the beam as listed in the beam table, s = 6.77 in³

Designation	Area	of T	Width	thick	thick	-	x-x axis	x-x axis	x-x axis	x-x axis
-	A	d	bf	tf	tw	d/tw	I	S	r	y
-	in2	in	in	in	in	-	in4	in3	in	in
WT8x25	7.36	8.13	7.073	0.628	0.380	21.40	42.20	6.770	2.400	1.890

Now M / s = (12,000 ft-lb)(12 in./ft.)/ 6.77 in³ = 21,270 lb/in². (Notice that we had to convert the bending moment in ft.-lb. to in.-lb. for the units to be consistent) We have thus determined the maximum bending stress (axial stress) in the beam. Please note that this maximum stress is compressive and occurs at the bottom of the stem, since that is the outer edge of the beam which is farthest from the neutral axis. (The stress at the top of the tee is less than that at the bottom of the stem, since the top of the tee is closer to the neutral axis than the bottom of the stem.)

Finally, we will determine the bending stress at 4 ft from the left end of the beam and 2 inches above the bottom of the beam. For this we need to use the flexure formula in the form M y / I., where M = -8,000 ft-lb = -96,000 in-lb (which we determine from the bending moment graph shown in Diagram 4), I = moment of inertia of beam = 42.2 in⁴, and y = distance from the neutral axis to point at which we wish to find the bending stress. The neutral axis is 1.89 inches below the top of the beam (from the beam data table), then the neutral axis is 8.13"-1.89" = 6.24" from the bottom of the beam. Since we wish to find bending stress 2 inches above the bottom of the beam, then y = 6.24" - 2" = 4.24" from the neutral axis to where we wish to determine the bending stress. Thus 96,000 in-lb * 4.24 in. / 42.2 in4. = 9,450 lb/in². Since the bending moment is negative, meaning the beam is bent concave facing downward, and since the location is below the beam centroid, then this stress is compressive.