WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A simply supported rectangular 2 x 12 beam is loaded as shown below. For this beam:
A. Determine the maximum bending stress 8 feet from the left end of the beam.
B. Determine the horizontal shear stress at a point 3 inches above the bottom of the beam cross section and 8 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 + C_y - 1,000 lbs/ft (4 ft) - 1,500 lbs/ft (4 ft) = 0
Sum T_B = 1,000 lbs/ft (4 ft) (2 ft) - 1,500 lbs/ft (4 ft) (8 ft) + C_y(6 ft) = 0
Solving: B_y = 3,330 lbs; C_y = 6,670 lbs

STEP 2: Determine the shear force and bending moment at x=8 ft.
1.) Cut beam at 8 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 = -1,000 lbs/ft (4 ft) + 3,330 lbs - V₈ = 0
Sum T_A = -1,000 lbs/ft (4 ft) (2 ft) + 3,330 lbs (4 ft) + M₈ = 0
Solving: V₈ = 667 lbs; M₈ = 10,670 ft-lbs

STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 8'.

MBS = My/I (Where M_8' is the bending moment at 8 ft, and S is the section modulus for the beam. The section modulus is available from the Beam Tables. This is a 2 x 12 beam.)
MBS = -10,670 ft-lbs(12 in/ft)(6 in) /[(1/12)(2 in)(12 in)³] = 2,667 lbs/in²

Part B:
STEP 4: To determine the Horizontal Shear Stress (HSS) at 8 ft from the end of the beam and 3 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 8 ft from the end of the beam
a = cross sectional area from 3 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 (288 in⁴ for 2 x 12 beam)
b = width of beam a 3 in above the bottom of the beam

HSS = [(667 lbs)(6 in²)(4.5 in)]/[(288 in⁴)(2 in)] = 31.3 psi