Topic 4.3a: Simply Supported Beam - Example 1
Example 1
A loaded, simply supported beam is shown. For this beam we would like to determine expressions for the internal shear forces and bending moments in each section of the beam, and to make shear force and bending moment diagrams for the beam.
We will work very slowly and carefully, step by step, through the solution for this example.
Solution: Part A. We first find the support forces acting on the structure. We do this in the normal way, by applying static equilibrium conditions for the beam.
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 = -4,000 lb. - (1,000 lb./ft)(8 ft) - 6,000 lb. + By + Dy = 0
Sum TB = (Dy)(8 ft) - (6,000 lb.)(4 ft) + (1,000 lb./ft)(8 ft)(4 ft) + (4,000 lb.)(8 ft) = 0
Solving for the unknowns: By =23,000 lb.; Dy = -5,000 lb. (The negative sign indicates that Dy acts the opposite of the initial direction we chose.)
Part B: Now we will 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 the translational equilibrium condition for the beam section (Sum of Forces = zero) to determine the Shear Force expressions in each section. Determining the Bending Moment expression for each section of the beam may be done in two ways.
- 1) By applying the rotational equilibrium condition for the beam section (Sum of Torque = zero), and solving for the bending moment.
- 2) By Integration. The value of the bending moment in the beam may be found from
. That is, the bending moment expression is the integral of the shear force expression for the beam section.
We now continue with the example. We begin by starting at the left end of the beam, and cutting the beam a distance "x" from the left end - where x is a distance greater than zero and less the position where the loading of the beam changes in some way. In this problem we see that from zero to eight feet there is a uniformly distributed load of 1000 lb./ft. However this ends at eight feet (the loading changes). Thus for section 1, we will cut the beam at distance x from the left end, where x is greater than zero and less then eight feet.
Section 1: Cut the beam at x, where 0 < x < 8 ft., and analyze left hand section.
1. Draw a FBD of the beam section shown 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. We check that we have all forces in x & y components (yes)
3. Apply translational equilibrium conditions to determine the shear force expression.
Sum Fx = 0 (no net external x- forces)
Sum Fy = -4,000 lb. - 1,000 lb./ft *(x) ft - V1 = 0 ; and solving: V1 = [-4,000 - 1,000x] lb.
This expression gives us the values of the internal shear force in the beam between 0 and 8 ft. Notice as x nears zero, the shear force value in the beam goes to - 4000 lb., and as x approaches 8 ft., the shear force value becomes -12,000 lb., and that is negative everywhere between 0 and 8 ft. Let's think for a moment what this negative sign tells us. Since we found the shear force (V) by static equilibrium conditions, the negative sign tells us that we choose the incorrect direction for the shear force - that the shear force acts in the opposite direction. However, we choose the positive direction of the shear force (by its definition) and so the negative sign also tells us we have a negative shear force.
To try to simplify a somewhat confusing sign situation we may say this: As long as we work from the left end of the beam, and choose the initial direction of the shear force and bending moment in the positive direction (by their definition), then when we solve for the shear force and bending moment, the sign which results is the correct sign as applies to the shear force and bending moment values.
If we graph the shear force expression above, we obtain the graph shown of the internal shearing force in the beam for the first eight feet. We next will determine the bending moment expression for this first beam section.
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 = -1000 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 - 1000 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 = [-4,000 - 1,000x] lb.) from our sum of forces result above into the torque equation to get:
Sum Torque left end = -1000 lb/ft * (x) * (x/2) -[-4,000 - 1,000x] (x) + M1 = 0 ;
and solving for M1 = [-500x2 - 4,000x] ft-lb.
This is our expression for the internal torque inside the load beam for section 1, the first eight feet, which is graphed in the diagram below.
5. Finally, we may also obtain the expression for the bending moment by integration of the shear force expression. The integrals we will be using are basic types.
For a simple, brief review and/or introduction to basic calculus concepts,
Please Select: Simple Derivatives/Integrals.
Continuing with our example:
Integration:
= 1000(1/2 x2) - 4000 (x) + C1; so M1 = -500x2 - 4,000x + C1
As we the results above show, when we do an indefinite integral, the result include an arbitrary constant, in this case called C1. To determine the correct value for C1 for our problem we must apply a boundary condition: That is, we must know the value of the bending moment at some point on our interval into to find the constant.
For simply supported beams (with no external torque applied to the beam) the value of the bending moment will be zero at the ends of the beam.
(There are many ways to explain why this must be so. One of the easiest explanations is to remember that the bending moment value at a point in a simply supported beam is equal to the total area under the shear force diagram up to that point. However, at the left end, as x goes to zero, the area under the shear force diagram would also go to zero, and thus so would the bending moment value.)
So we have for our "boundary condition" that at x = 0, M1 = 0. We put these values into our expression for the bending moment (M1 = -500x2 - 4,000x + C1), and solve for the value of the integration constant, C1; that is:
0 = -500(0)2 - 4,000(0) + C1, and solving: C1 = 0
Therefore: M1 = [-500x2 - 4,000x] ft-lb. for 0 < x < 8 ft., is our final expression for the bending moment over the first section. (Note, it is the same as found above by summing torque for the beam section.)
We now continue with the next section of the beam. Referring to the beam diagram, we see that at a location just greater than 8 ft., there is no loading, and that this continues until 12 ft. where there is a point load of 6,000 lb. So for our second section, we cut the beam at a location "x", where x is greater than 8 ft., and less than 12 ft - and then analyze the entire left hand section of the beam.
Section 2: We cut the beam at x, where 8 < x <12 ft., and analyze the entire section left of where we cut the beam.
1. Draw a FBD of the beam section shown 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 2) 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 V2 and M2, as this is section 2 of the beam.
2. We check that we have all forces in x & y components (yes)
3. Apply translational equilibrium conditions (forces only):
Sum Fx = 0 (no net external x- forces)
Sum Fy = -4,000 lb. - 1,000 lb./ft(8 ft) + 23,000 lb. - V2 = 0, Solving: V2 = 11,000 lb.
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 = - (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - V2 * x + M2 = 0; then we substitute the value for V2 (V2 = 11,000 lb) from above and obtain:
- (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - (11,000 lb.) * x + M2 = 0; and then solving for M2 we find: M2 = [11,000x - 152,000] ft-lb.
From integration of the shear force, we find:
Integration:
, or M2 = 11,000x + C2
We get our boundary condition from another characteristic of the bending moment expression - which is that the bending moment must be continuous. That is, the value of the bending moment at the end of the first beam section, and the value of the bending moment at the beginning of the second beam section must agree - they must be equal. We determine the value of the bending moment from our M1 equation as x approaches 8 ft. (M1 = [-500 (8)2 - 4,000(8)] = -64,000 ft-lb.)
Then our boundary condition to find C2 is: at x=8 ft M=-64,000 ft-lb. We apply our boundary condition to find C2.
Apply BC: -64,000 ft-lb. = 11,000 lb. (8) + C2, Solving: C2 = -152,000 ft-lb.
Therefore: M2 = [11,000x - 152,000] ft-lb. for 8 < x < 12
In like manner we proceed with section 3 of the beam, cutting the beam at a location greater than 12 ft. and less 16 ft., and then analyzing the entire section left of where we cut the beam.
Section 3: Cut the beam at x, where 12 < x < 16 ft. Analyze left hand section.
1. FBD. (See Diagram Section 3)
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 = -4,000 lb. - 1,000 lb./ft(8 ft) + 23,000 lb. -6,000 lb. - V3 = 0, and Solving:
V3 = 5,000 lb.
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 = - (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - 6,000 lb. * 12 ft -V3 * x + M3 = 0; then we substitute the value for V3 (V3 = 5,000 lb) from above and obtain:
- (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - 6,000 lb. * 12 ft - (5,000 lb.) * x + M3 = 0; and then solving for M3 we find: M3 = [5,000x - 80,000] ft-lb.
We will find the bending moment expression for this section using integration only.
Integration 
, or M3 = 5,000x + C3
We obtain a boundary condition for section 3 by remembering that at a free end or simply supported (no external torque) end, the bending moment must go to zero, thus we have the boundary condition to find C3: at x = 16 ft., M = 0 ft-lb.
Apply BC: 0 = 5,000(16) + C3, and Solving: C3 = -80,000 ft-lb.
Therefore: M3 = [5,000x - 80,000] ft-lb. for 12 < x < 16
We now have our expressions for the shear forces and bending moments in each section of the loaded beam (summarized below). Additional, we have shown the shear force and bending moment diagrams for the entire beam - which is a visual representation of the internal shear forces and internal torque in the beam due to the loading.
Part C: Shear Force and Bending Moment Diagrams: Using the expressions found above, we can draw the shear force and bending moment diagrams for our loaded beam.
V1 = -1,000x+4,000 lb.; V2 = 11,000 lb.; V3 = 5,000 lb
M1 =-500x2+4,000x ft-lb.; M2 = 11,000x -152,000 ft-lb.; M3 = 5,000x-80,000 ft-lb
No comments:
Post a Comment