As shown in the figure, a beam is subjected to a concentrated force F = 50.0N ,

Question

As shown in the figure, a beam is subjected to a concentrated force F = 50.0N , a distributed load w = 6.00N/m , and a couple moment M = 60.0N?m . (Figure 1) Find Ayand Dy, the reaction forces at both supports. The weight of the beam is negligible when compared to the forces that act on the beam.

Given the same beam, find Vmax, the absolute value of the maximum shear in the beam. The weight of the beam is negligible compared to the forces that act on the beam.

Given the same beam, find Mmax, the absolute value of the maximum moment in the beam, and xmax, the location of the maximum moment in the x direction. The weight of the beam is negligible compared to the forces that act on the beam.

Explanation / Answer

Part A - Basic statics, nothing unusual here. Pick one support (I recommend D), and sum the moments about that point, equal to zero. The only unknown will be Ay. Then sum the forces vertically to get Dy.

Part B - Draw the shear diagram. Start at the left end. Remember the relationship between load and shear (load is the derivative of shear if working in equation form), and make use of it - for example, for the uniformly distributed load portion (the load is constant), the shear should be linear (the derivative of a straight line is a constant). This would mean that for the first portion, starting at the left end, the shear diagram would be a straight downward line. When you come to a concentrated load or reaction, the shear diagram will "jump" at that point by the magnitude of the load. Once you have the diagram drawn with values, you can see where the maximum value is.

Part C - This will work similar to part B. The derivative of the moment equation is the shear equation, so if you have a linear shear, the moment will be quadratic. Wherever there is a maximum or a discontinuity, find the moment at that point explicitly - cut the beam, draw a free body diagram replacing the missing portion of the beam with the internal moment. The applied moment works the same way a point force works for the shear diagram.

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