WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, simply supported W T 12 x 50 beam is shown below. For this beam:
A. Determine the maximum bending stress 12 feet from the left end of the beam.
B. Determine the horizontal shear stress at a point 7 inches above the bottom of the beam cross section and 12 feet from the left end of the beam.
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Part A:
STEP 1: Determine the external support reactions:
1.) FBD of structure (See Diagram)
2.) Resolve all forces into x/y components
3.) Apply Equilibrium Conditions:

Sum F_x: none
Sum F_y: A_y + C_y - 2000 lb/ft(4 ft) - 1000 lb/ft(6 ft) = 0
Sum Torque(A) = T_A = -(8000 lb)(2 ft) + C_y (10 ft) - (6000 lb)(13 ft) = 0
Solving: C_y = 9400 lb. A_y = 4600 lb

STEP 2: Determine the shear force and bending moment at x = 12 ft.
1.) Cut Beam at 12 ft. Draw the FBD of left end of beam, showing and labeling all external forces.
2.) Resolve all forces into x/y directions
3.) Apply Equilibrium Conditions:

Sum F_x: none
Sum F_y: 9400 lb - 2000 lb/ft(4 ft) + 4600 - 1000 lb/ft(2 ft) - V_12’ = 0
Sum Torque(A) = T_A = -(8000 lb)(2 ft) + 9400 lb (10 ft) - (2000 lb)(11 ft) - V(12) + M_12’= 0
Solving: V_12’ = 4000 lb. M_12’ = -8000 ft-lb

STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress at 12 ft.

Maximum Bending Stress (MBS) = M_12'/S (Where M_12' is the bending moment at 12 ft., and S is the section modulus for the beam. The section modulus is available from the Beam Tables. The WT 12 X 50 beam is shown in the diagram to the right. The section modulus for this beam from the beam tables is 18.7 in³.)

MBS = -8000 ft-lb. (12in./ft)/18.7 in³ = -5130 lb/in².

The Maximum bending stress occurs at the bottom of the T-beam (since it is the outer edge most distance from the neutral axis) The negative sign means that the beam is bent concave downward at this point, so the bottom of the beam is in compression.

Part B:
STEP 4: To determine the Horizontal Shear Stress (HSS) at 12 ft from the left end of the beam and 7 inches above the bottom of the beam, apply the horizontal shear stress formula. The form we will use is: HSS = Vay'/lb
Where:
V = Shear force 12 ft from the end of the beam
a = cross sectional area from 7 in above the bottom of the beam to bottom of beam
y' = distance from neutral axis to the centroid of area a
I = moment of inertia of the beam (177 in⁴ for W 12 x 50 beam)
b = width of beam a 7 in above the bottom of the beam

HSS = [(4000 lb.)(7" x ..47")(5.97")]/[(177 in⁴)(.47)] = 944 lb/in²