WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, cantilever 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 section of the beam.
C. Make 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:
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.

Sum F_x = A_x = 0
Sum F_y = ( -1,500 lbs/ft)(6 ft) - (1,000 lbs/ft)(2 ft) - (800 lbs/ft)(4 ft) + A_y = 0
Sum T_A = (-1,500 lbs/ft)(6 ft)(3 ft)-(1,000 lbs/ft)(2 ft)(7 ft)-(800 lbs/ft)(4 ft)(10 ft) + M_ext = 0

Solving for the unknowns:
A_y = 14,200 lbs; M_ext = 73,000 ft-lbs

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):

Sum Fx = 0 (no net external x- forces)
Sum Fy = 14,200 lbs - 1,500 lbs/ft(x) - V1 = 0
Solving: V1 = [-1,500x + 14,200] lbs

4. Integration

M1 = -750x² + 14,200x + C1

a)Boundary condition to find C1: at x=0 M=-73,000 ft-lbs (That is, for a cantilever beam, the value of the bending moment at the wall is equal to the negative of the external moment.)
Apply BC: -73,000 = -750(0)² + 14200(0) + C1
Solving: C1 = -73,000
Therefore… M1 = [-750x² + 14,200x - 73,000] ft-lbs for 0 <>

Section 2: Cut the beam at x, where 6 <>

1. FBD. (Shown in Diagram)
2. All forces in x & y components (yes)
3. Apply translational equilibrium conditions (forces only):

Sum Fx = 0 (no net external x- forces)
Sum Fy = 14,200 lbs - (1,500 lbs/ft)(6 ft) - (1,000 lbs/ft)(x - 6)ft - V2 = 0
Solving: V2 = [-1,000x + 11,200] lbs

4. Integration

M2 = -500x² + 11,200x + C2

a)Boundary condition to find C2: at x=6 ft M=-14,800 ft-lbs (from equation M1)
Apply BC: -14,800 ft-lbs = -500(6)² + 11,200(6) + C2
Solving: C2 = -64,000 ft-lbs
Therefore… M2 = [ -500x² + 11,200x - 64,000] ft-lbs for 6 <>

Section 3: Cut the beam at x, where 8 <>

1. FBD. (Shown in Diagram)
2. All forces in x & y components (yes)
3. Apply translational equilibrium conditions (forces only):

Sum Fx = 0 (no net external x- forces)
Sum Fy = 14,200 lbs - (1,500 lbs/ft)(6 ft) - (1,000 lbs/ft)(2 ft) - (800 lbs/ft)(x-8)ft - V3 = 0
Solving: V3 = [-800x + 9,600] lbs

4. Integration

M3 =-400x² + 9,600x + C3

a)Boundary condition to find C3: at x=12 ft M=0 ft-lbs (free end of beam, no external torque so M3=0)
Apply BC: 0 = -400(12)² + 9,600(12) + C3
Solving: C3 = -57,600 ft-lbs
Therefore… M3 = [-400x² +9,600x - 57,600] ft-lbs for 8 <>

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 = [-1,500x+14,200] lb V2 = [-1,000x+11,200] lb V3 = [-800x+9,600] lb
M1 =[-750x²+14,200x-73,000] ft-lb M2 = [-500x²+11,200x-64,000] ft-lb M3 = [-400x²+9,600x - 57,600] ft-lb