WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, simply supported W 10 x 45 beam is shown below. For this beam:
A. Determine the maximum bending stress 6 feet from the left end of the beam.
B. Determine the horizontal shear stress at a point 4 inches above the bottom of the beam cross section and 6 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 = 0 none
Sum F_y = B_y + D_y - 2,000 lbs/ft (4 ft) - 5,000 lbs = 0
Sum T_B = 5,000 lbs (4 ft) - 2,000 lbs/ft (4 ft) (6 ft) + D_y(8 ft) = 0
Solving: B_y = 9,500 lbs; D_y = 3,500 lbs

STEP 2: Determine the shear force and bending moment at x=6 ft.
1.) Cut beam at 6 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 = -5,000 lbs + 9,500 lbs - V₆ = 0
Sum T_A = 9,500 lbs (4 ft) - 4,500 lbs (6 ft) + M₆ = 0
Solving: V₆ = 4,500 lbs; M₆ = -11,000 ft-lbs

STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 6'.

MBS = M_6'/S (Where M_6' is the bending moment at 6 ft, and S is the section modulus for the beam. The section modulus is available from the Beam Tables. The W 10 x 45 beam has a section modulus for the beam from the beam tables is 49.1 in³.)

MBS = -11,000 ft-lbs(12 in/ft)/49.1 in³ = -2,688 lbs/in²

Part B:
STEP 4: To determine the Horizontal Shear Stress (HSS) at 6 ft from the end of the beam and 4 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 6 ft from the end of the beam
a = cross sectional area from 4 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 (249 in⁴ for W 10 x 45 beam)
b = width of beam a 4 in above the bottom of the beam

HSS = [(4,500 lbs)(6.153 in²)(4.37 in)]/[(249 in⁴)(.35 in)] = 1,388 psi