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. 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 Fy = (-5,000 lbs) - (1,000
lbs/ft)(8 ft) + Ay + Cy = 0
Sum TA = (Cy)(8 ft) - (5,000
lbs)(4 ft) - (1,000 lbs/ft)(8 ft)(12 ft) = 0
Solving for the unknowns:
Cy
= 14,500 lbs; Ay =
-1,500 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 < x < 4 ft. Analyze left
hand section.
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 = -1,500 lbs - V1 = 0
Solving: V1 = -1,500 lbs
4. Integration 
M1 = -1,500x + C1
a)Boundary condition to find
C1: at x=0 M=0
Apply BC: 0 = -1,500(0) + C1
Solving: C1
= 0
Therefore… M1
= [-1,500x] ft-lbs for 0 < x < 4
ft.
Section
2: Cut the beam at x, where 4 < x < 8 ft. Analyze left
hand section.
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 = -1,500 lbs - 5,000 lbs - V2 = 0
Solving: V2 =
-6,500 lbs
4. Integration 
M2 = -6,500x + C2
a)Boundary condition to find
C2: at x=4 ft M=-8,000 ft-lbs (from equation M1)
Apply BC: 8000 ft-lbs = -6,500(4) + C2
Solving:
C2 = 20,000 ft-lbs
Therefore… M2
= [-6,500x + 20,000] ft-lbs for 4 <
x < 8
Section
3: Cut the beam at x, where 8 < x < 16 ft. Analyze left
hand section.
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 = -1,500 lbs - 5,000 lbs + 14,500 lbs
-1,000lbs/ft(x-8)ft - V3 = 0
Solving: V3 = [-1,000x + 16,000] lbs
4. Integration 
M3 = -500x2 + 16,000x + C3
a)Boundary condition to find
C3: at x=16 ft M=0 ft-lbs (end of beam, no external torque so
M3=0)
Apply BC: 0 = -500(16)2 + 16,000(16) + C3
Solving: C3
= -128,000 ft-lbs
Therefore… M3
= [-500x2 +16,000x - 128,000] ft-lbs
for 8 < x < 16
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 = -1,500 lb, V2 = -6,500 lb,
V3 = -1,000x+16,000
lb
M1 = -1,500x
ft-lb, M2 = -6,500x+20,000ft-lb, M3 = -500x2+16,000x-128,000
ft-lb
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