Topic 4.3c: Cantilever Beam - Example 3
In this example we have a loaded, cantilever beam, as 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 draw the shear force and bending moment diagrams for the beam.
Solution:
Part A: Our first step will be to determine the support reactions and external torque acting on the loaded beam. For a cantilevered beam (with one end embedded or rigidly fixed at the wall), the wall may exert horizontal and vertical forces and an external torque (which we will call an external moment, and label Mext) acting on the beam - as we have shown in the free body diagram of 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 static equilibrium conditions.
Sum Fx = Ax = 0
Sum Fy = -4,000 lbs - 3,000 lbs - (2,000 lbs/ft)(6 ft) - 2,000 lbs + Ay = 0
Sum TA = (-4,000 lbs)(4 ft) - (3,000 lbs)(8 ft) -(2,000 lbs/ft)(6 ft)(11 ft) - (2,000 lbs)(14 ft) +Mext = 0
Solving for the unknowns: Ay = 21,000 lbs; Mext = 200,000 ft-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 static equilibrium condition - 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 an arbitrary location 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 for the shear force and bending moment 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 determine the shear force expression:
Sum Fx = 0 (no net external x- forces)
Sum Fy = 21,000 lbs - V1 = 0 ; Solving for shear force: V1 = 21,000 lbs
4. We now find the bending moment expression by summing torque about the left end. That equation would be as follows:
Sum of Torque left end = -V1 * x + 200,000 ft-lb + M1 = 0;
when we substitute in the value for V1 = 21,000 lb. we obtain the equation:
Sum of Torque left end = -21,000 lb * x + 200,000 ft-lb + M1 = 0.
And solving for M1 = 21,000 x -200,000 ft-lb.
[This is the same result we will find by integration. In general, particularly for non-point loads, integration is the faster method.]
Now we find the bending moment equation by integration of the shear force expression.
Integration
, and then M1 = 21,000x + C1
The boundary condition, which we will use to determine the integration constant C1, is different for the cantilevered end of a beam as compared with the end of a simply supported beam (or a free end). Since we have an external moment acting at the end of the cantilevered beam, the value of the bending moment must become equal to the negative of the external moment at the cantilevered end of the beam, in order for the beam to be in rotational equilibrium (sum of torque = 0). Thus our boundary condition to determine C1 is: at x = 0, M = -200,000 ft-lb (That is, for a cantilever beam, the value of the bending moment at the wall is equal to the negative of the external moment.)
Applying the boundary condition and solving for the integration constant:
-200,000 = 21000(0) + C1; and so C1 = -200,000
Then the bending moment expression is: M1 = [21,000x - 200,000] ft-lb for 0 < x < 4 ft.
Section 2: Since the loading changes at 8 ft, due to the point load and the beginning of the uniformly distributed load, for section 2 we cut the beam at location x, where 4 <>
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 label 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 = 21,000 lbs - 4,000 lbs - V2 = 0 ; and solving: V2 = 17,000 lb
4. We now find the bending moment expression by summing torque about the left end. That equation would be as follows:
Sum of Torque left end = -4000 lb.* 4 ft. -V2 * x + 200,000 ft-lb + M2 = 0;
when we substitute in the value for V2 = 17,000 lb. we obtain the equation:
Sum of Torque left end = -4000 lb.* 4 ft. -17,000 lb. * x + 200,000 ft-lb + M2 = 0
And solving for M2 = [ 17,000x - 184,000] ft-lb for 4 < size="3">.
Next we also determine the bending moment expression by integration of the shear force equation from above.
Integration: 
, and so M2 = 17,000x + 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 = 4 ft.) must also be the value of the bending moment at the beginning of section 2 (at x = 4 ft.). Thus our boundary condition to find the integration constant C2 is: at x = 4 ft., M = -116,000 ft-lb. (from equation M1).
Now applying the boundary condition and solving for the integration constant, C2, we have:
-116,000 ft-lbs = 17,000(4) + C2, and so C2 = -184,000 ft-lb., and our expression for the bending moment on beam section 2 is: M2 = [ 17,000x - 184,000] ft-lb for 4 <>
We continue in like manner for the last section of the beam
Section 3: Cut the beam at 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 label them as V3 and M3, as this is section 3 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 = 21,000 lbs - 4,000 lbs - 3,000 lbs - (2,000 lbs/ft)(x-8)ft - V3 = 0
Solving for the bending moment: V3 = [-2,000x + 30,000] lb
4. We now find the bending moment expression by summing torque about the left end. That equation would be as follows:
Sum of Torque left end = -4000 lb.* 4 ft. -3000 lb.* 8 ft. - 2000 lb/ft *(x-8')*[(x-8)/2 + 8'] -V3 * x + 200,000 ft-lb + M3 = 0; [ If you are unsure how the third term in the equation was obtained, please see example 2.}
When we substitute in the value for V3 = [-2,000x + 30,000] lb. we obtain the equation:
Sum of Torque= -4000 lb.* 4 ft. -3000 lb.* 8 ft. - 2000 lb/ft *(x-8')*[(x-8)/2 + 8'] - [-2,000x + 30,000] lb* x + 200,000 ft-lb + M2 = 0
And solving for M3 = [-1,000x2+30,000x - 224,000] ft-lb.
By Integration: 
, and then M3 =-1,000x2 + 30,000x + C3
Since the right hand end of the beam is a free end, the bending moment must go to zero as x goes approaches 14 ft. So our boundary condition to find the integration constant, C3, is: at x = 14 ft M = 0 ft-lb
Applying the boundary condition: 0 = -1,000(14)2 + 30,000(14) + C3 , and solving C3 = -224,000 ft-lbs
So M3 = [-1,000x2 +30,000x - 224,000] 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 = 21,000 lb., V2 = 17,000 lb., V3 = [-2,000x+30,000] lb.
M1 =[21,000x-200,000] ft-lb., M2 = [17,000x-184,000] ft-lb., M3 = [-1,000x2+30,000x - 224,000] ft-lb.
No comments:
Post a Comment