Topic 4.3b: Simply Supported Beam - Example 2
For the loaded, simply supported beam shown, determine expressions for the internal shear forces and bending moments in each section of the beam, and draw shear force and bending moment diagrams for the beam.
Solution: We first need to determine the external support reaction by applying our standard static equilibrium conditions and procedure..
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: Resolve any forces not already in x and y direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum Fx = 0 (no external x-forces acting on structure.)
Sum Fy = (-800 lbs/ft)(8 ft) - (1,200 lbs/ft)(8 ft) + Ay + Cy = 0
Sum TA = (Cy)(12 ft) - (800 lbs/ft)(8 ft)(4 ft) - (1,200 lbs/ft)(8 ft)(12 ft) = 0
Solving for the unknowns: Cy = 11,700 lbs; Ay = 4,270 lbs
Part B: Determine the Shear Force and Bending Moment 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: We note that the loading of the beam (800 lb./ft) remains uniform until 8 feet, where it changes to 1200 lb./ft. As a result, for our first beam section, we cut the beam at an arbitary position x, where 0 <>
1. Draw a FBD of the beam section showing and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram - Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V1 and M1, as this is section 1 of the beam.
2. Resolve all forces into x & y components (yes)
3. Apply translational equilibrium conditions (forces only) to the section 1 of the beam:
Sum Fx = 0 (no net external x- forces)
Sum Fy = 4,300 lbs - 800 lbs/ft(x)ft - V1 = 0;
Solving for the shear force: V1 = [4,300 - 800x] lbs
4. We can find the bending moment from static equilibrium principles; summing torque about the left end of the beam. Referring to the free body diagram for beam section 1, we can write:
Sum Torque left end = -800 lb/ft * (x) * (x/2) - V1 (x) + M1 = 0
To make sure we understand this equation, let's examine each term. The first term is the torque due to the uniformly distributed load - 800 lb./ft * (x) ft (this is the load) times (x/2) which is the perpendicular distance, since the uniform load may be considered to act in the center, which is x/2 from the left end. Then we have the shear force V1 times x feet to the left end, and finally we have the bending moment M1 (which needs no distance since it is already a torque).
Next we substitute the expression for V1 (V1 = [-4,300 - 800x] lb.) from our sum of forces result above into the torque equation to get:
Sum Torque left end = -1000 lb/ft * (x) * (x/2) -[-4,300 - 800x] (x) + M1 = 0 ;
and solving for M1 = [-400x2 + 4,300x] ft-lbs for 0 < x < 8 ft.
We will now also find the bending moment expression by integration of the shear force equation. Integration
, solving M1 = -400x2 + 4,300x + C1
Our boundary condition to find the integration constant, C1, is at x = 0, M = 0 (since this is a simply support beam end.)
Applying the boundary condition: 0 = -400(0)2 + 4,300(0) + C1, and solving gives us: C1 = 0.
Therefore the bending moment expression for section 1 of the beam is:
M1 = [-400x2 + 4,300x] ft-lbs for 0 < x < 8 ft.
(The shear force and bending moment diagrams are shown at the bottom of this example page.)
Section 2: We continue in the same manner with beam section 2. We note that the loading changes once more at 12 ft, due to the upward support force acting at that point. So for beam section 2, we cut the beam at location x, where 8 <>
1. Draw a FBD of the beam section showing and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) We have labeled them as V2 and M2, as this is section 2 of the beam.
2. Resolve all forces into x & y components (yes).
3. Apply translational equilibrium conditions (forces only):
Sum Fx = 0 (no net external x- forces)
Sum Fy = 4,300 lbs - 800 lbs/ft*(8 ft) - 1,200lbs/ft *(x - 8)ft - V2 = 0; and solving for the shear force expression: V2 = [7,500 - 1,200x] lbs
4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. We will do it both ways for this section.
Rotational Equilibrium:
Sum of Toque left end = - (800 lb./ft * 8 ft) * 4 ft. -1200 lb/ft * (x - 8')*[(x-8')/2+8'] - V2 * x + M2 = 0;
Please notice the second term. In that term the quantity (1200 * (x-8')) is load due to the 1200 lb/ft acting over the distance (x-8). However we still need to multiply the force expression times the distance to obtain the torque. The uniform load of 1200 lb/ft acts at the center of its distance (x-8'), so the lever arm to point A would be [(x-8')/2 + 8'] (See diagram.)
We next substitute the value for V2 (V2 = [7,500 - 1,200x] lb.) from above and obtain:
- (800 lb./ft * 8 ft) * 4 ft. -1200 * (x - 8')*[(x-8')/2+8'] - [7,500 - 1,200x]* x + M2 = 0;
and then solving for M2 we find:
M2 = [-600x2 + 7,500x - 12,800] ft-lbs for 8 < x < 12
Next we find the bending moment, M2, from integration of shear force expression,V2.
Integration: 
, and solving, M2 = -600x2 + 7,500x + C2
We obtain our boundary condition for beam section 2 by remembering that the bending moment must be continuous along the beam. This means that value of the bending moment at the end of section 1 (at x = 8 ft.) must also be the value of the bending moment at the beginning of section 2 (at x = 8 ft.). Thus our boundary condition to find C2 is: at x = 8 ft M = 8,560 ft-lbs (from equation M1). Now applying the boundary condition and solving for the integration constant, C2, we have:
8560 ft-lbs = -600(8)2 +7500(8) + C2, and solving: C2 = -13,000 ft-lbs
Therefore our bending moment expression is:
M2 = [-600x2 + 7,500x - 12,800] ft-lbs for 8 < x < 12
(The shear force and bending moment diagrams are shown at the bottom of this example page.)
Section 3: Finally, we continue with the last section of the beam, cutting the beam at location x, where 12 <>
1. Draw a FBD of the beam section showing and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) We have labeled them as V3 and M3, as this is section 3 of the beam.
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 = 4,300 lbs - 800 lbs/ft(8 ft) + 11,700 lbs - 1,200lbs/ft (x - 8)ft - V3 = 0
Solving for the shear force expressing: V3 = [-1,200x + 19,200] lbs
4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section.
Rotational Equilibrium:
Sum of Toque left end = - (800 lb./ft * 8 ft) * 4 ft. - 1200 lb. *(x- 8 ft.)*[(x-8')/2 +8'] + 11,700 lb. * 12 ft -V3 * x + M3 = 0;
Once again the distance from end A at which the effective load due to the uniform load of 1200 lb/ft [1200 lb/ft * (x-8')]may be considered to act must be determined carefully. That distance [(x-8')/2 + 8')] is shown at the top of the adjacent diagram.
We next substitute the value for V3 (V3 = [-1,200x + 19,200] lb.) from above and obtain:
- (800 lb./ft * 8 ft) * 4 ft. - 1200 lb. *(x- 8 ft.)*[(x-8')/2 +8'] + 11,700 lb. * 12 ft - [-1,200x + 19,200] lb* x + M3 = 0; and then solving for M3 we find: M3 = [-600x2 + 19,200x - 153,000] ft-lbs for 12 <>.
Determine the bending moment by Integration: 
,
or, M3 = -600x2 + 19,200x + C3
We find our boundary condition for beam section 3, by realizing at the end of the beam (a free end) the bending moment must go to zero, so our boundary condition to find C3 is: at x = 16 ft M = 0 ft-lb.
Appling the boundary condition, and solving for the integration constant C3, we have:
BC: 0 = -600(16)2 + 19,200(16) + C3; and then C3 = -153,000 ft-lbs
So the final expressionn for the bending moment on section 3 will be:
M3 = [-600x2 + 19,200x - 153,000] ft-lbs for 12 <>
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 = -800x + 4,300 lb.; V2 = -1,200x + 7,500 lb.; V3 = -1,200x + 19,200 lb
M1 = -400x2 + 4,300x ft-lb.; M2 = -600x2 + 7,500 - 12,800 ft-lb.; M3 = -600x2 + 19,200x - 153,000ft-lb
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