WORLD FOR CIVIL

Topic 5.3b: Horizontal Shear Stress - Example 2

A loaded, simply supported beam is shown in Diagram 1. For two different beam cross sections (a WT 8 x 25 T-beam, and a W 10 x 45 beam) we will determine the maximum Horizontal Shear Stress which would develop 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: 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 lb.; C_y = 6,670 lb.

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 3 and 4. This beam is the same beam used in Beams - Bending Stress Example III. Please see that example, if needed, for a more complete explanation of how the shear force and bending moment diagrams were made.

Beam Table for WT 8 x 25

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
WT 8x25	7.36	8.13	7.073	0.628	0.380	21.40	42.20	6.770	2.400	1.890

Step 3. For the WT 8 x 25 T-beam (table above) we will now apply the Horizontal Shear Stress formula:

Shear Stress = Vay'/Ib, to find the maximum shear stress, which occurs at the neutral axis of the beam:
V = maximum shear force = 6,000 lb. (from the shear force diagram)
I = moment of inertia of cross section, from beam table; I = 42.20 in⁴.
b = width of beam where we wish to find shear stress (neutral axis for maximum) from table; b = .38 in.
a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam. In this case we will go to bottom of beam. Then a = (.38" * 6.24" )= 2.37 in².
y' = distance from neutral axis to the centroid of the area "a" ; y' = 3.12 in. (See Diagram 5) Then placing values into our expression we find:
Maximum Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(2.37 in²)*(3.12 in)/ (42.20 in⁴)(.38 in) = 2770 lb/in²

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

We wish to find the shear stress 3 inches above the bottom of the beam cross section, where the shear force is a maximum. (See Diagram 6)\

V = maximum shear force = 6,000 lb. (from the shear force diagram)
I = moment of inertia of cross section, from beam table; I = 42.20 in⁴.
b = width of beam where we wish to find shear stress (3" above bottom of beam) from table; b = .38 in.
a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam. In this case we will go to bottom of beam. Then a = (.38" * 3" )= 1.14 in². (See Diagram 6)
y' = distance from neutral axis to the centroid of the area "a" ; y' = 4.74 in. (See Diagram 6)
Then the horizontal shear stress 3 inches above the bottom of the beam is:
Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(1.14 in²)*(4.74 in)/ (42.2 in⁴)(.38 in)= 2020 lb/in²
Notice, as we expect, the horizontal shear stress value becomes smaller as we move toward an outer edge of the beam cross section.

Part 2: W 10 x 45 beam: We would like to again determine the maximum horizontal shear stress, and the shear stress 3 inches above the bottom of the beam (at the point where the shear force is a maximum), but now find these values for a W 10 x 45 I-beam.

Beam Table for W 10 x 45 I-Beam

-	-	-	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 10 x 45	13.20	10.12	8.022	0.618	0.350	249.0	49.1	4.33	53.20	13.30	2.00

We have already done the statics, and the shear force and bending moment diagrams are shown in the first part of this example above, so we continue at the point where we apply the horizontal shear stress formula to find the values we desire.

For the WT 8 x 25 T-beam we apply the Horizontal Shear Stress formula: Shear Stress = Vay'/Ib, however since we are looking for the maximum shear stress in the I-Beam, we can use the approximate formula for I-beam, Maximum Horizontal Shear Stress = V_max/A_web. This says the approximate maximum shear stress in an I - Beam is equal to the maximum shear force divided by the area of the web of the I-Beam. Applying this we have:

V_max = maximum shear force = 6,000 lb. (from the shear force diagram)
A_max = area of web: A = (.35" * 8.88" )= 3.11 in². (See Diagram 7)

Then Maximum Horizontal Shear Stress = (6000 lb)/(3.11 in²) = 1930 lb/in² As long as this approximate value is reasonably below the allowable shear stress for the beam material there is no need to use the exact formula for the maximum shear stress. Please remember, however, the approximate formula is only for the maximum horizontal shear stress (which occurs are the neutral axis) in an I-Beam. If we need to know the shear stress at any other location, we must use the standard formula – as we will do in the next part.

We now wish to find the shear stress 3 inches above the bottom of the beam cross section, where the shear force is a maximum. (See Diagram 8). To do so, we apply the standard horizontal shear stress formula: Shear Stress = Vay'/Ib

V = maximum shear force = 6,000 lb. (from the shear force diagram)
I = moment of inertia of cross section, from beam table; I = 249.0 in⁴.
b = width of beam where we wish to find shear stress (3" above bottom of beam) from table; b = .35 in.
a = area from point we wish to find shear stress at (3" from the bottom) to an outer edge of beam. In this case we will go to bottom of beam. Notice that the area is composed of the area of the flange (A1) and part of the area of the web (A2). (See Diagrams 8 and 9.) Then a = (A1 + A2)= (.618" x 8.022") + (2.383 in² x .35in²) = 4.96 in² + .834 in²= 5.794 in² (See Diagram 9)
y' = distance from neutral axis to the centroid of the area "a" Notice that in this case, for an I-Beam, this is not a entirely simple matter. The area we wish to find the centroid of is not a simple rectangle, but rather two rectangles. To find the centroid of this compound area we use: y' = (A1 y1 + A2 y2)/(A1 +A2); where A1 and A2 are the two areas, and y1 and y2 are the distances from the neutral axis of the beam to the centroid of each of the respective areas. (Which is simply the distance from the neutral axis to the center of each of the respective areas, since for a rectangle the centroid is at the center.) Using the values shown in Diagram 9, we have:

y' = (A1 y1 + A2 y2)/(A1 +A2) = (4.96 in² x 4.75 in + .834 in² x 2.94 in)/(4.96 in² + .834 in²) = 4.49 in.
Then the horizontal shear stress 3 inches above the bottom of the beam is:
Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(5.794 in²)*(4.49 in)/ (249.0 in⁴)(.35 in)= 1790 lb/in²
Notice, as we expect, the horizontal shear stress value becomes smaller as we move toward an outer edge of the beam cross section.