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 = (-2,000 lbs/ft)(4 ft) - (1,000 lbs/ft)(6 ft)
+ Ay + Cy = 0
Sum TA = (Cy)(10 ft) - (2,000 lbs/ft)(4
ft)(2 ft) - (1,000 lbs/ft)(6 ft)(13 ft) = 0
Solving for the unknowns:
Cy
= 9,400 lbs; Ay =
4,600 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.
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 = 4,600 lb - 2,000 lb/ft *x - V1 = 0
Solving: V1
= (4,600 - 2,000x) lbs
4. Integration 
M1 = -1000 x2 + 4600x + C1
a)Boundary condition to find
C1: at x = 0; M = 0
Apply BC: 0 = -1000(0)2 + 4600(0) + C1, Solving: C1 =
0
Therefore… M1 =
[-1000x2 + 4600x] ft-lbs for 0 < x < 4 ft.
Section 2: Cut the beam at x, where 4 < x <
10 ft.
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 = 4600 lbs - 2,000 lbs/ft(4 ft) - V2 = 0
Solving: V2
= -3400 lbs
4. Integration 
M2 = -3400x + C2
a)Boundary condition to find
C2: at x=4 ft M=2400 ft-lbs (from equation M1)
Apply BC: 2400 ft lbs = -3400(4) + C2
Solving: C2
= 16,000 ft-lbs
Therefore… M2 =
[-3400x + 16,000] ft-lbs for 4 < x
< 10
Section 3: Cut the beam at x, where 10 < x
< 16 ft.
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 = 4,600 lbs - 2,000 lb/ft(4 ft) + 9,400 lbs -
1,000 lb/ft(x-10)ft - V3 = 0
Solving: V3 = [-1000 x + 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 10 < 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 = 4,600-2,000x lb, V2 = -3,400 lb, V3 = -1,000x+16,000 lb
M1 = -1,000x2+4,600
ft-lb, M2 = -3,400x+16,000 ft-lb, M3 = -500x2+16,000x-128,000
ft-lb
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