Shear and Moment Diagrams
In this lecture, we focus on drawing shear and moment diagrams without the use of any equations.
This is particularly useful for practical structural analysis and design.
More specifically, we're going to learn how to construct the shear diagram directly from the load diagram.
And, how to construct the moment diagram from the shear diagram.
Suppose we wish to construct the shear and moment diagrams
for a simply supported beam subjected to a uniformly distributed load.
Observe that the beam has two upward reactions, each having a magnitude of WL/2.
The basic fact that enables us to draw the shear diagram directly from the load diagram is this:
The change in shear in a beam segment is equal to the total load applied to the segment.
So if we know the shear at A and we know how much load is being applied to segment AB,
then we can easily calculate shear at B.
That is, the shear at B equals to the shear at A plus the total load applied to the segment.
The total load applied to the beam segment here
is the area of the rectangle having L as its base and W as its height.
Therefore the total load is -WL.
The negative sign indicates that the load is acting downward.
To draw the shear diagram for the beam segment, we need to have three pieces of information.
One, the shear at the left end of the segment,
two, the shear at the right end of the segment,
and three, the shape of the curve that connects the two points.
In the case of a simply supported beam,
shear at the left end of the beam always equals to the support reaction at that end.
Since there is an upward reaction of WL/2 at the left end of the beam,
then the shear there is positive WL/2.
The shear at the right end of the beam segment equals to the shear at the left end,
plus the area under the load diagram between the two points.
Therefore, the shear at the right end is WL/2-WL, or -WL/2.
Now that we know the shear force at either end of the segment,
we need to draw the curve representing shear within the segment.
Since here the load is constant throughout the beam segment,
then we draw a straight line connecting the two points.
That is, shear is linear in the beam.
A beam segment subjected to a uniformly distributed load carries a linearly varying shear.
Concerning the shape of the diagram, the underlying rule is this:
If we view the load diagram and the shear diagram as polynomials,
then the degree of the shear polynomial is one higher than that of the load polynomial.
Since the degree of the load polynomial in this example is zero, the load is constant,
then the shear polynomial has a degree of one;
Shear varies linearly in the beam.
Note that the shear diagram consists of a positive region and a negative region,
with the shear being 0 at the midpoint of the beam.
Let's label the positive region R1, and the negative region R2.
We are now in a position to draw the moment diagram using the constructed shear diagram.
To do so, we need to calculate the area under the shear diagram.
Since the diagram consists of two regions, we need to calculate two areas:
a positive area associated with R1, and a negative area associated with R2.
R1 forms a right triangle with a height of WL/2, and a base of L/2.
Therefore its area is: WL^2/8
R2 has an area equal to WL^2/8, but this area is negative since the region is below the x-axis.
We draw the moment diagram segment by segment.
It's customary to do so from left to right.
So first we draw the moment diagram for R1, then we do it for R2.
To draw the moment diagram for R1, we start from the left end of the beam.
We examine the reaction at that end.
Since there is no moment reaction there, then moment at the left end of the segment is zero.
Moment at the right end of the segment equals to the moment at the left end,
plus the area under the shear diagram in the segment.
That is, the moment at the right end of R1 equals to: 0+WL^2/8
Now that we have the moment values at the ends of the segment,
we need to connect them using a curve indicating how bending moment changes within the segment.
Similar to the curve representing the shear diagram, the moment diagram curve is a polynomial.
The degree of this polynomial is one higher than that of the shear polynomial.
In this example, since the shear diagram is linear, the shear polynomial has a degree of 1,
then the moment diagram needs to be drawn using a polynomial of degree 2,
a curve representing a quadratic equation.
But such a curve can be drawn in two ways:
either this way,
or this way:
As you can see, the first curve has a decreasing slope.
The slope of the curve decreases as we move from left to right.
The second curve has an increasing slope. Its slope increases from left to right.
So which one is the correct curve?
Does the moment in this segment of the beam have an increasing slope or a decreasing one?
To answer the question we need to examine the shear diagram.
If shear in the segment is increasing in value, then the slope of the moment curve should increase.
If shear is decreasing, moment should also have a decreasing slope.
Shear in region R1 is positive, but decreasing in value from left to right.
That is the shear value of the left end of the segment is higher,
then the shear value at the right end of the segment.
The shear at the left end of the segment is WL/2, whereas shear at the right end of the segment is 0.
Therefore the moment curve needs to have a decreasing slope.
So this is the correct diagram:
We employ the same technique to draw the moment diagram for the right segment of the beam.
The moment at the left end of R2 is the same as the moment at the right end of R1.
Therefore we start with a moment value of positive WL^2/8.
The moment at the right end of R2 equals to the moment at the left end of the segment,
plus the area under the shear diagram in the segment.
Since this area is -WL^2/8, then the right moment is 0.
Just like before there are two options for connecting the two points:
a curve with a decreasing slope,
or one with an increasing slope.
To decide the correct shape for the diagram, examine the shear diagram in the segment.
Does shear have an increasing value or a decreasing value?
In R2, shear goes from zero at the left end of the segment, to -WL^2 at the right end of the segment.
That is, shear decreases in value from zero to a large negative value.
Therefore the slope of the moment diagram needs to follow the same pattern.
It needs to change from zero to a large negative value.
This curve fits the requirement:
Its slope changes from zero at the left end, to a large negative value at the right end.
Therefore the moment diagram for the entire beam looks like this:
Let's look at another example.
A simply supported beam subjected to two concentrated loads.
We start by calculating the support reactions.
The left reaction is 6 kN, and the right reaction is 8kN.
Note that the concentrated loads divide the beam into three segments:
AB, BC and CD.
We are going to construct the shear diagram segment by segment.
For segment AB, because there is an upward reaction of 6 kN at the left end of the segment,
the shear at that point is positive 6.
The area under the load diagram in this segment is 0.
There is no distributed load applied to the segment,
therefore the shear at the right end of the segment is 6 plus 0, or 6.
That is, the shear at the right end equals to the shear at the left end,
plus the area under the load diagram within the segment.
Therefore shear in the segment remains constant with a magnitude of 6 kN.
At point B, since there is a downward force of 10 kN, there is going to be a drop of 10 units in the shear value.
So shear is going to drop from positive 6 at the immediate left of B,
to -4 at the immediate right of B.
Therefore the shear at the left end of the segment BC is -4.
Since there is no load applied within segment BC, the total load applied to the segment is 0.
Hence the shear at the right end of the segment is -4+0, or -4.
Similar to segment AB, in segment BC shear remains constant.
The shear just to the left of point C is -4.
Given that there is a downward force of 4 kN at C,
then the shear just to the right of the point is -4-4, or -8.
That is, there is a drop of 4 in the shear value right at point C.
Similar to the other two segments, total load applied within segment CD is 0.
Therefore shear remains constant throughout the segment.
Shear remains -8 in segment CD.
To draw the moment diagram, let's start by calculating the area under the shear diagram in each segment.
The shear area for segment AB is positive 24,
for BC the area is negative 8. And for segment CD the area is negative 16.
For segment AB, the moment at the left end of the segment is 0,
since there is no moment at the pin support at A.
Then the moment at the right end of the segment is 0 plus 24, or 24.
Since shear is constant in the segment, moment is linear.
Why?
If we view shear and moment graphs as polynomials,
the degree of the moment polynomial should be one higher than that of the shear polynomial.
Shear is a polynomial of degree 0, since does not change within the segment,
therefore the bending moment has to be represented as a polynomial of degree 1,
which is an equation for a line. So bending moment varies linearly in the segment.
For segment BC, we know the moment at the left end of the segment.
It is the same as the moment at the right end of the previous segment.
So the moment at the left end of segment BC is positive 24.
The moment at the right end of the segment is: 24+(-8), or positive 16.
Similar to the previous segment,
the moment in segment BC varies linearly as shown here.
Finally, we can draw the moment diagram for segment CD
following the same pattern that we used for the other segments.
Here is the complete moment diagram for the beam.