WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, cantilever WT 8 x 29 beam is shown below. For this beam:
A. Determine the maximum bending stress 5 feet from the left end of the beam.
B. Determine the horizontal shear stress at a point 6 inches above the bottom of the beam cross section and 5 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 = A_x = 0
Sum F_y = A_y - 2,000 lbs/ft (6 ft) - 1,000 lbs/ft (2 ft) = 0
Sum T_A = 2,000 lbs/ft (6 ft) (3 ft) - 1,000 lbs/ft (2 ft) (11 ft) + M_ext = 0
Solving: A_y = 14,000 lbs; M_ext = 58,000 ft-lbs

STEP 2: Determine the shear force and bending moment at x=5 ft.
1.) Cut beam at 5 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 = 0 none
Sum F_y = 14,000 lbs - 2,000 lbs/ft (5 ft) - V₅ = 0
Sum T_A = -2,000 lbs/ft (5 ft) (2.5 ft)+ 58,000 ft-lbs -V₅(5 ft) + M₅ = 0
Solving: V₅ = 4,000 lbs; M₅ = -13,000 ft-lbs

STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 5'.

MBS = M_5'/S (Where M_5' is the bending moment at 5 ft, and S is the section modulus for the beam. The section modulus is available from the Beam Tables. The WT 8 x 29 beam has a section modulus for the beam from the beam tables is 7.0 in³.)

MBS = 13,000 ft-lbs(12 in/ft)/7.0 in³ = 21,900 lbs/in²

Part B:
STEP 4: To determine the Horizontal Shear Stress (HSS) at 5 ft from the end of the beam and 6 inches above the bottom of the beam, apply the horizontal shear stress formula.
The form we will use is: HSS = Vay'/Ib
Where:
V = Shear force 5 ft from the end of the beam
a = cross sectional area from 6 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 (44 in⁴ for WT 8 x 29 beam)
b = width of beam a 6 in above the bottom of the beam

HSS = [(4,000 lbs)(2.46 in²)(3.19 in)]/[(44 in⁴)(.41 in)] = 7,740 psi