WORLD FOR CIVIL

Topic 5.3a: Horizontal Shear 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 Horizontal Shear Stress which develops in the beam due to the loading. We will also determine the Horizontal Shear Stress 3 inches above the bottom of the beam at the position in the beam where the shear force is a maximum.

Step 1: Out first step in solving this problem is 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 bending moment diagram, but will include it for completeness. We have shown the shear force and bending moment graphs in Diagram 3a and 3b. This beam is the same beam used in Topic 4.7b:Bending Stress - Example 1 Please see that example, if needed, for a more complete explanation of how the shear force and bending moment diagrams were made.

Step 3. We will now apply the Horizontal Shear Stress formula: Shear Stress = Vay'/Ib
We wish to find the maximum shear stress, which occurs at the neutral axis of the beam:

V = maximum shear force = 5,000 ft-lb. (from the shear force diagram)
I = moment of inertia of cross section; for rectangle
I = (1/12) bd³ = 1/12 (2" * 4"³) = 10.67 in⁴.
b = width of beam section where we wish to find shear stress at; b= 2 in.
a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam
a= (2" x 2")= 4 in².
y' = distance from neutral axis to the centroid of the area "a" which we used; y'= 1 in.
(See Diagram 4)

Placing the values into the equation, we find:
Maximum Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(4 in²)*(1 in)/ (10.67 in⁴)(2 in)= 937 lb/in²

This is the correct value; we notice it is not very large. The beam is clearly able to carry the load without failing in shear.

Part II We now would also like to determine the Horizontal Shear Stress 3 inches above the bottom of the beam at the position in the beam where the shear force is a maximum (which is actually through out the beam, since the value of shear force is either +5000 lb., or - 5000 lb. through out the beam.)

We again apply the Horizontal Shear Stress formula: Horizontal Shear Stress = Vay'/Ib

We wish to find the shear stress 3 inches above the bottom of the beam cross section. (See Diagram 5)
V = shear force = 5,000 ft-lb. (from the shear force diagram)

I = moment of inertia of cross section; for rectangle I = (1/12) bd³ = 1/12 (2" * 4"³) = 10.67 in⁴.
b = width of beam section where we wish to find shear stress at; b = 2 in.
a = area from point we wish to find shear stress at (3" above bottom of the beam) to an outer edge of beam. We will go to the top edge of the beam, then a = (2" x 1")= 2 in².
y' = distance from neutral axis to the centroid of the area "a" which we used; y'= 1.5 in. (See Diagram 5)
Then the horizontal shear stress 3 inches above the bottom of the beam is:
Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(2 in²)*(1.5 in)/ (10.67 in⁴)(2 in)= 703 lb/in²

Notice, as we expect, the horizontal shear stress value becomes smaller as we move toward the outer edge of the beam cross section.