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This says that the shear at some point x2 equals the shear at the starting point x1 plus the area under the distributed load curve between the two points.
dM/dx = V says that the slope of the moment diagram is equal to the shear force.
Also, since the maximum moment occurs where dM/dx = 0 , we know that the maximum moment will be where the shear force is zero.
Integrating dM = V dx , we get
∫ dM = ∫ V dx
M2 - M1 = ∫ V dx or M2 = M1 + ∫ V dx
So, the moment at a point x2 equals the moment at the starting point x1 plus the area under the shear diagram between the two points.
These equations are valid between concentrated forces and moments. We also know that at points of concentrated forces we get a discontinuity or a jump in the shear diagram in the direction of and equal in magnitude of the force. At points of concentrated moments, we get a discontinuity or a jump iin the moment diagram equal in magnitude to the concentrated moment. The direction of the jump is up when the moment is clockwise and down when counterclockwise.
Let’s state a procedure for the summation method.
1. Draw a free body diagram of the beam and use equilibrium to determine the reaction forces. 2. Draw the shear diagram using the following rules. a. Start at the left end at zero b. When you encounter a concentrated force, follow the force. That is jump in the direction of the force the magnitude of the force. c. Between concentrated forces, apply
V2 = V1 + ∫ w dx to find the shear at the end of
the section and dV/dx = w to draw the line over the section. 3. Draw the moment diagram using the following rules. a. Start at the left end at zero. b. When you encounter a concentrated moment, jump up if the moment is CW or down if the moment is CCW a magnitude equal to the magnitude of the moment. c. Between concentrated moments, apply
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Lecture #14, page 2 Back Lecture #14, page 3 |