WORLD FOR CIVIL

Topic Exam Solution - Problem 1

A loaded, simply supported W 10 x 29 beam is shown below. For this beam:
A. Determine the maximum bending stress 10 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 10 feet from the left end of the beam.

Solution:

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 = 0 none
Sum F_y = B_y + D_y - 4,000 lbs - 6,000 lbs - 1,000 lbs/ft (8 ft) = 0
Sum T_B = +4,000 lbs (8 ft) + 6,000 lbs (4 ft) - 8,000 lbs (4 ft) + D_y (8 ft) = 0
Solving: B_y = 21,000 lbs; D_y = -3,000 lbs

STEP 2: Determine the shear force and bending moment at x=10 ft.

1.) Cut beam at 10 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 = -4,000 lbs - 6,000 lbs - 1,000 lbs/ft (2 ft) + 21,000 lbs - V₁₀ = 0
Sum T_A = -6,000 lbs (4 ft) + 21,000 lbs (8 ft) -2,000 lbs (9 ft) - V₁₀(10 ft) + M₁₀ = 0
Solving: V₁₀ = 9,000 lbs; M₁₀ = -36,000 ft-lbs

STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 10'.
MBS = M_10'/S (Where M_10' is the bending moment at 10 ft, and S is the section modulus for the beam. The section modulus is available from the Beam Tables. The W 10 x 29 beam has a section modulus for the beam from the beam tables is 30.8 in³.)Therefore:
MBS = 36,000 ft-lbs(12 in/ft)/30.8 in³ = 14,030 lbs/in²

STEP 4: To determine the Horizontal Shear Stress (HSS) at 10 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 10 ft from the end of the beam = 9,000 lb.
a = cross sectional area from 6 in above the bottom of the beam to bottom of beam =(2.9 in² + 1.69 in²)
y' = distance from neutral axis to the centroid of area a = (3.8 in)
I = moment of inertia of the beam (158 in⁴ for W 10 x 29 beam)
b = width of beam a 6 in above the bottom of the beam =(.29 in)
Then putting in values:
HSS = [(9,000 lbs)(2.9 in² + 1.69 in²)(3.8 in)]/[(158 in⁴)(.29 in)] = 3360 psi