In this video I go through an example problem of drawing shear and moment diagrams of a beam that has a triangular load on it.
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Quick Tips (I explain these in more depth in the video)
The area under the loading curve = the change in shear (positive area is a positive change and negative area is a negative change)
The area under the shear function/diagram = the change in moment (positive area is a positive change and negative area is a negative change)
Positive slope on the loading function = concave up shear function
Negative slope on the loading function = concave down shear function
Positive slope on the shear function = concave up moment function
Negative slope on the shear function = concave down moment function
Distributed load intensity = the slope of the shear function
Amount of shear = slope of the moment diagram
Where the shear diagram crosses the axis is a max or min on the moment diagram
Drawing shear and moment diagrams process:
1) Establish your coordinate system with the positive X direction being along the length of the beam, starting on the left.
2) Solve for the support reactions on the beam using moment equilibrium equations and force equilibrium equations in the Y direction.
3) Draw the free body diagram of the beam with the solved support reactions written as forces as well as the loading scenario.
4) Draw the shear X axis and the moment X axis vertically aligned below the beam to line up the changes of the loading with the shear and moment diagrams. Then label these diagrams with V for shear and M for moment and label the units for each.
5) Plot the shear diagram starting from the left end. As you encounter forces in the loading scenario, draw the shear diagram as outlined in Note A. Label the distance X along the diagram where the shear function crosses the axis.
6) Plot the moment diagram starting from the left end. As you encounter changes in the shear diagram, draw the moment diagram as outlined in Note B. Label magnitude of the moment at the maximum and minimum points along the function as well as what point along the X axis they occur.
Note A:
A point load (a force at a single point) causes the shear diagram to jump straight up or down; up if the load is pushing up and down if the force is pushing down. This includes reaction forces. Point moments will not directly effect the shear diagram.
Distributed loads cause the shear diagram to decrease or increase in a way that can be modeled by a function. You can model the distributed load as a function and then integrate that function to get the shear function (see the Quick Tips at the top for more specifics on this).
Note B:
The moment reactions at the ends of the beams are zero if the beam is supported by pin, roller or free connections. This means that the moment diagram will start and end at zero. However, if there is a fixed support then there will be a moment reaction which will cause the diagram to not start or end at zero, depending on which side the fixed support is on (usually it is on the left end). If there is a support reaction then it will cause a vertical jump in your moment diagram just as a point moment would. Remember that the support reaction is in the opposite direction of the internal moment. If the support reaction is counter clockwise (positive) then the internal forces which you are drawing are negative and thus will have a negative drop.
A point moment will cause a vertical jump. The direction of the jump being as explained in the previous point.
The integral of the shear function is the moment function. This only applies to sections of the function that are continuous, in other words, there are no vertical jumps up or down
(see the Quick Tips at the top for more specifics on how to draw the diagram).