WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, simply supported beam is shown below. For this beam:
A. Draw a Free Body Diagram of the beam, showing all external loads and support forces (reactions).
B. Determine expressions for the internal shear forces and bending moments in each sections of the beam.
C. Construct the shear force and bending moment diagrams for the beam.

Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Part A: Statics
STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.
STEP 2: Break any forces not already in x and y direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
SF_x = 0
SF_y = A_y + B_y - 1/2(10 ft)(1,000 lb/ft) - 6,000 lb = 0
ST_A = -(5,000 lbs)(6.67 ft) + (B_y)(10 ft) - (6,000 lbs)(14 ft) = 0
Solving for the unknowns:
B_y = 11,735 lbs; A_y = -735 lbs (acts downward)

Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. For this process we will ‘cut’ the beam into sections, and then use Statics - Sum of Forces to determine the Shear Force expressions, and Integration to determine the Bending Moment expressions in each section of the beam.

Section 1: Cut the beam at x, where (0 <>

1. FBD. (Shown in Diagram)
2. All forces in x & y components (yes)
3. Apply translational equilibrium conditions (forces only):
SF_x = 0 (no net external x- forces)
SF_y = -735 lb - 1/2(x ft)[(1,000(lb/ft)/10ft)](x ft) - V1 = 0
Solving: V1 = [-50x² - 735] lbs

4. Integration

M1 = -16.67x³ - 735x + C1

Boundary condition to find C1: at x = 0; M = 0. (simply supported beam)
Apply BC: 0 = -16.67(0)³ - 735(0) + C1, Solving: C1 = 0
Therefore… M1 = [-16.67x³ - 735x] ft-lbs for (0 <>

Section 2: Cut the beam at x, where (10 <>

1. FBD. (Shown in Diagram)
2. All forces in x & y components (yes)
3. Apply translational equilibrium conditions (forces only):
SF_x = 0 (no net external x- forces)
SF_y = -735 lb - 1/2(10 ft)(1,000 lb/ft) + 11,735 lb - V2 = 0
Solving: V2 = +6,000 lbs

4. Integration

M2 = 6,000x + C2

Boundary condition from Section 1 at (x = 14 ft) ; M = 0. (end of simply supported beam):
Apply BC: 0 = 6,000(14 ft) + C2
Solving: C2 = -84,000 ft-lbs
Therefore… M2 = [6,000x - 84,000] ft-lbs.

Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above, we can draw the shear force and bending moment diagrams for our loaded beam.

V1 = [-50x² - 735] lbs, V2 = +6,000 lbs
M1 = [-16.67x³ - 735x] ft-lbs, M2 = [6,000x - 84,000] ft-lbs.