WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, cantilever 2 x 10 rectangular beam is shown below. For this beam:
A. Determine the maximum bending stress 4 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 4 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 - 1,000 lbs/ft (8 ft) - 5,000 lbs = 0
Sum T_A = -5,000 lbs (12 ft) - 1,000 lbs/ft (8 ft) (12 ft) + M_ext = 0
Solving: A_y = 13,000 lbs; M_ext = 156,000 ft-lbs

STEP 2: Determine the shear force and bending moment at x=4 ft.
1.) Cut beam at 4 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 = 13,000 lbs - V₄ = 0
Sum T_A = -V₄(4 ft) +156,000 ft-lbs + M₄ = 0
Solving: V₄ = 13,000 lbs; M₄ = 104,000 ft-lbs

STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 4'.

MBS = My/I (Where M_4' is the bending moment at 4 ft, and S is the section modulus for the beam. The section modulus is available from the Beam Tables. This is a 2 x 10 beam.)

MBS = 104,000 ft-lbs(12 in/ft)(5 in) /[(1/12)(2 in)(10 in)³] = 37,400 lbs/in²

Part B:
STEP 4: To determine the Horizontal Shear Stress (HSS) at 4 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 4 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 (167 in⁴ for 2 x 10 beam)
b = width of beam a 6 in above the bottom of the beam

HSS = [(13,000 lbs)(12 in²)(2 in)]/[(167 in⁴)(2 in)] = 934 psi