Shear and Moment Diagrams
Previously we talked about shear and moment in beams, and how to represent them using algebraic equations.
In this lecture, we're going to focus on their graphical representations,
also called shear and moment diagrams.
Such diagrams allow us to quickly and accurately determine the locations
at which shear and bending stresses are maximum or minimum,
and visualize how the stresses vary across the beam,
an ability that plays an important role in structural design.
Before we start explaining how shear and moment diagrams can be drawn,
let's establish a sign convention.
In this example, note that the left region of the shear diagram is labeled positive,
and the diagram's right region is labeled negative.
What exactly does positive or negative shear mean?
That is what the sign convention is for.
Consider a simple beam.
If we cut it, say at point C,
we need to show the internal shear force and bending moment at the cut-point.
The shear force has to be shown as a pair of vertical forces acting in opposite directions,
so the pair has to be drawn either like this,
or like this:
In the first scenario the shear force tends to rotate the beam segments in the clockwise direction.
In the second scenario the beam segments want to rotate in the counterclockwise direction.
By convention, the shear force direction that tends to cause a clockwise rotation is considered positive.
If the shear force tends to cause a counterclockwise rotation, we call it a negative shear.
Therefore this is a positive shear:
And this is a negative shear:
Similarly, when we cut the beam, the internal bending moment can be drawn either like this,
or like this:
In the first scenario, the bending moment tends to cause a concave up deformation, like this:
In the second scenario, the beam segments tend to deform concave down like this:
A positive moment tends to cause a concave up bending in the beam,
a negative moment tends to bend the beam concave down,
therefore this is considered a positive moment:
And this is a negative moment:
Now on to drawing shear and moment diagrams.
One way to draw the diagrams is to graph their corresponding equations.
For example, for this beam we can easily formulate the shear and moment equations.
We talked about the procedure for coming up with these equations in a previous lecture,
so we're not going to talk about that again today.
Let's just assume the equations have been derived already, and we have them, here they are:
Our task is to graph them.
For shear we have two equations.
For the left segment of the beam, where X is between 0 and 5, shear equals to: 375-100X
This represents a line segment, to draw it we need two points,
let's pick the segment's endpoints,
so we need to evaluate the equation at: X=0 and at X=5
When X is zero, shear value is 375.
When x is 5 we get -125.
Now locate the points on the coordinate system, then connect them with a straight line.
Note that the line crosses the X axis somewhere between 0 and 5,
making the left part of the diagram positive, and the right part negative.
For X between 5 and 10, shear is given using equation -125.
In this problem, since shear is not really a function of X,
it remains constant throughout the segment.
Here is the complete shear diagram for the beam:
We can graph the moment equation in a similar manner.
For the first segment of the beam, where X is between 0 and 5,
the moment equation is quadratic.
This means we need more than two points to graph it.
I'm going to use three points for generating the graph,
the segment's end points and the point where moment is maximum.
How do we determine the point of maximum?
We take the derivate of the moment equation with respect to X,
set it to zero, then solve for X.
That is, at the point of maximum the derivative of the moment equation vanishes.
Our moment equation is M=375X-50X^2
Its derivative with respect to X is 375-100X
equate the equation to 0 and solve for X.
This gives us: X=3.75
That is, moment is maximum at X=3.75 meters from the left end of the beam.
Now that we have our three points, let's determine the moment value at each point.
At X=0, M is 0.
At X=3.75, M is 703.
And at X=5, M=625
Here is the graph of the quadratic equation:
Since the diagram is above the X axis,
moment is considered positive in the entire region, where X is between 0 and 5.
Moment in the right segment of the beam where X is between 5 and 10,
is given by equation: 1250-125X
This is a line equation, to graph it, let's evaluate it at its endpoints.
At X=5, moment equals 625.
At X=10, moment is 0.
So moment changes linearly from 625 to 0 in the segment, like this:
In summary, in this case, shear is partly positive and partly negative,
whereas bending moment remains positive throughout the beam.
There is a faster technique for drawing shear and moment diagrams without using equations,
we will discuss and illustrate it in the next lecture.