Beams - Bending Stress

BEAMS: BENDING STRESS
Axial Stress (tension and compression) and the Shear Stress (vertical and horizontal) which develop in a loaded beam depend on the values of the Bending Moments and the Shear Forces in the beam. Determining the axial stress - which is often known as the Bending Stress in a beam; and determining the shear stress - often called the Horizontal Shear Stress (for reasons we will discuss) is important in two ways. First, it will enable us to determine if a particular loaded beam is safe under the applied loading. Second, it will enable us to select the best beam (from a table of beams) for a particular loading. Both of these are very important processes for the safety and efficiency of a beam.

The Bending Stress: We will first develop a relationship for the bending stress which develops in a loaded beam. This relationship is known as the Flexure Formula. In Diagram 1 we have shown a simply supported beam loaded at the center. It deflects (or bends) under the load.

In Diagram 2, we have shown the left end section of the beam. As discussed previously, when examining bending moments, horizontal forces act on the cross sectional face of the beam section. We have shown only the horizontal forces along the top and bottom in Diagram 2a, but the forces act across the whole cross section as shown in the side view in Diagram 2b. The horizontal forces decrease from maximum at the outer edges to zero at the neutral axis (an axis running through the centroid of beam cross section).

We will now go through a relatively brief derivation to arrive at the flexure formula. We first write an expression for the bending moment produced by the horizontal forces with respect to the neutral axis (which is a line passing through the centroid of the beam cross sectional area - shown in Diagram 2a). The expression for the bending moment is simply the sum of the forces times the perpendicular distance to the neutral axis, or:
We now note that we can express the force Fx as ; that is, the force acting on any small horizontal strip of area (dA) is the product of axial stress at that point and the amount of area (dA). (This simply comes from the definition of axial stress = Force/Area). We can now rewrite the expression for the bending moment as:
We now will rewrite this expression one more time by noting that the horizontal forces and accompanying stresses increase linearly from zero at the neutral axis to a maximum value at an outer edge. We can then write: , where y is the distance from the neutral axis to area dA, and ymax is the distance from the neutral axis to the outer edge of the beam cross-section, as shown in Diagram 2a. We can then write the stress at an arbitrary y as: . We now substitute this expression into our relationship for the bending moment and obtain
; then rewriting slightly and factoring out theterm from the summation sign (which we may since it is a constant), we find:
; and finally we recognize that the summation term remaining is simply the Moment of Inertia (I) about the centroid of the beam cross section.. We now rewrite one last time, arranging terms and isolating the stress term by itself, we finally obtain:
M ymax / I

That is, the maximum "Bending Stress" at some location along the beam is equal to the bending moment, M, at that location "times" the distance, y, from the neutral axis to the outer edge of the beam "divided" by the moment of inertia, I, of the beam cross sectional area. If this seems somewhat confusing, it will become clearer as we work through several examples.

While the formula above was derived for the maximum stress, it actually holds for the stress at any point in the beam cross section and is known as the Flexure Formula.

Flexure Formula: M y / I

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

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

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 and y components
3.) Apply equilibrium conditions:
Sum Fx = 0 none
Sum Fy = By + Dy - 2,000 lbs/ft (4 ft) - 5,000 lbs = 0
Sum TB = 5,000 lbs (4 ft) - 2,000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0
Solving: By = 9,500 lbs; Dy = 3,500 lbs

STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.

M (ymax / I) = M / (I / ymax) = M / s , we then see that the section modulus is defined as s = I / ymax. 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: Mmax = 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.

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 Fx = 0 none
Sum Fy = By + Cy - 1,000 lbs/ft (4 ft) - 1,500 lbs/ft (4 ft) = 0
Sum TB = 1,000 lbs/ft (4 ft) (2 ft) - 1,500 lbs/ft (4 ft) (8 ft) + Cy(6 ft) = 0
Solving: By = 3,330 lbs; Cy = 6,670 lbs

STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.

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: Mmax = 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.