WORLD FOR CIVIL

Topic 5.2a: Bending Stress - Example 2

A loaded, simply supported W 10 x 45 beam is shown in Diagram 1. For this beam we will first determine the maximum bending moment (and where it occurs in the beam). Then we will determine the maximum bending stress at that location, and also the bending stress at that location along the beam and 8 inches from the bottom of the beam cross section.

STEP 1: Apply Static Equilibrium Principles and determine the external support reactions:
1.) Draw Free Body Diagram 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 + D_y - 2,000 lbs/ft (4 ft) - 5,000 lbs = 0
Sum T_B = 5,000 lbs (4 ft) - 2,000 lbs/ft (4 ft) (6 ft) + D_y(8 ft) = 0
Solving: B_y = 9,500 lbs; D_y = 3,500 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 -5000 lb. (at x = 0') due to the downward acting load of 5000 lb. at x= 0'. Next nothing happens (no loading) for the next 4 ft., so the shear force remains constant. Then at x = 4 ft there is an upward support force of 9,500 lb. which drives the shear force value up this amount, from a value of -5000 lb. to + 4,500 lb. Again there is no change in loading for the next 4 ft., so the shear force remains constant until x = 8 ft. At that point, a uniform load of 2,000 lb./ft. is applied, driving the shear force downward at a rate of 2000 lb. per each foot for the 4 ft (at total of - 8000 lb.), until at x = 12 ft. the value of shear force is -3,500 lb. Also at x = 12 ft. is the upward support reaction of +3,500 lb., which can be considered to bring the shear force back to zero.(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 what is known as the section modulus. If we rewrite the standard flexure formula several times as follows for the maximum stress:
M (y_max / I) = M / (I / y_max) = M / s , we then see that the section modulus is defined as s = I / y_max. That is, 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 = 20,000 ft.-lb., and occurs at x = 4 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 below, s = 49.1 in³

-	-	-	Flange	Flange	Web	Cross	Section	Info.	Cross	Section	Info.
Designation	Area	Depth	Width	thick	thick	x-x axis	x-x axis	x-x axis	y-y axis	y-y axis	y-y axis
-	A	d	bf	tf	tw	I	S	r	I	S	r
-	in2	in	in	in	in	in4	in3	in	in4	in3	in
W 10x45	13.20	10.12	8.022	0.618	0.350	249.0	49.1	4.33	53.20	13.30	2.00

Now M / s = (20,000 ft-lb.)(12 in./ft.)/ 49.1 in³ = 4, 890 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. Since the beam is symmetric about the neutral axis, the stress at the top of the beam and at the bottom of the beam are equal in value (4,890 psi.) with the top in tension and the bottom in compression.

Finally, we will determine the bending stress at 4 ft (where the maximum bending moment occurs), and 8 inches above the bottom of the beam. For this we need to use the flexure formula in the form M y / I, where M = 20,000 ft-lb. = 240,000 in-lb., I = moment of inertia of beam = 249 in⁴, and y = distance from the neutral axis to point at which we wish to find the bending stress. Since we wish to find the bending stress 8 inches above the bottom of the beam, and since the neutral axis is 5.06 inches above the bottom of the beam (at the beam center), then y = 8 - 5.06 = 2.94 inches. Then 240,000 in-lb. * 2.94 in. / 249 in4. = 2830 lb./in². And since the location is above the beam centroid (and the bending moment is positive), this is a tensile stress.