Bending moment diagram.
Bending moment is an internal force.
It develops in beams and columns
when the structure is subjected to external loads.
An excessively large bending moment in a member
causes structural failure.
Therefore, to design the system properly
we need to locate the critical moment values in each member.
In lecture ST10,
we showed how to express bending moment in beams algebraically.
These equations can be used to locate the critical moment values mathematically.
In practice, however,
we often use a graphical representation of the bending moment
instead of its algebraic representation.
Such a graph, called a moment diagram,
acts as a visual aid for gaining insight into how this internal force varies
and where its critical values are located in the member.
In this lecture,
we present a convenient technique for drawing moment diagrams
without needing to write their underlying moment equations.
Our technique involves using the shear diagram
to construct the moment diagram.
Herein,
I'm going to assume that you know how to draw a shear diagram,
a topic we covered in lecture ST11.
Consider the simply supported beam shown here.
The beam can be easily analyzed
and its shear diagram can be drawn
using the procedure explained in lecture ST11.
Now, we wish to use the shear diagram
to draw the moment diagram for the beam.
The basic technique that we are going to use is simple.
Pick any two points on the beam.
I am going to label the points i and j
The difference between the moment values at i and j
equals the area under the shear diagram between the two points.
That is, if we refer to the moment at point i as Mi,
and call the moment at j, Mj,
then Mj - Mi
is equal to the area under the shear diagram between i and j
which we are denoting as Area_ij
To draw the moment diagram
let's start from the left end of the beam
and work towards its right end.
We know that bending moment at A is zero.
How do we know this?
because no bending moment can develop at a pin or roller support,
and since there is a roller at A,
the starting moment value is zero.
Bending moment at B can be determined
by adding the area under the shear diagram between A and B
to the moment at A.
This area is 10 kN-m
Since the moment at A is zero,
we get 10 kN-m for the moment value at B.
Please keep in mind that the sign associated with each segment
of the shear diagram matters.
Shear in segment AB is positive,
therefore we end up with a positive area under the shear diagram.
which results in a positive moment value at B.
Knowing the moment value at B,
now we can determine the moment value at C
by adding the area under the shear diagram between B and C
to Mb.
The area under the shear diagram between B and C is -10,
therefore, Mc equals 0.
To draw the shape of the moment diagram,
we need to connect these three points
The shape for each segment can be determined
from the shape of the shear diagram above it.
When the shear diagram is a horizontal line,
the moment diagram takes the shape of an inclined line.
So,
we draw an inclined line connecting these two points.
And,
we need another line for connecting these two points.
So,
here is a rule to remember:
a constant shear results in a linear moment.
Lets consider another example:
A beam with an overhang.
Here,
the shear diagram for the beam consists of three rectangular areas.
Let us calculate each area they are:
To draw the moment diagram for the beam,
we start from the left end of the beam
and work toward the right end.
Since there is a pin support at the left end of the beam,
bending moment at A equals zero.
The moment value at the right end of segment AB,
therefore, equals 0+3.67
Or, 3.67 kN-m.
And since the shear force in the segment is constant,
we connect the two end points using a straight line,
like this:
For segment BC,
we know the moment value at B,
and we know the area under the shear diagram
Therefore,
the moment value at C is equal to 3.67-12.67,
Or, negative 9 kN-m.
The moment diagram takes the shape
of an inclined line in this segment.
Why?
Because shear in the segment is constant.
Finally,
to get to the moment value at the right end of segment CD,
we add this area to this value
which yields zero for Md.
And since the shear remains constant in the segment,
we connect these two points using an inclined line.
Now,
let's consider a beam carrying a uniformly distributed load.
Here is the shear diagram for the beam.
Note the shape of the diagram in segment AB.
Shear varies linearly in the segment.
It goes from positive 8 kN at A
to negative 4 kN at B.
Since we need to calculate the area under each segment of the shear diagram,
we must determine the location of the point at which shear crosses the x-axis.
This can be done using the shared angle property
of two similar triangles.
If we label the base of this triangle x
then the base of this triangle becomes 4-x.
Therefore, we can write:
Solving for x, we get 2.67 meters.
That is, the inclined line representing the shear diagram
crosses the x-axis 2.67 meters to the right of point A.
Now,
we are in a position to calculate the area for each segment of the diagram.
To draw the moment diagram
let us start by determining the moment values at four specific points:
the end points of the beam,
and the two interior points at which the diagram changes shape.
Since the beam rests on a pin at its left end
the bending moment at A is zero.
At this point,
the moment value is equal to 0 + 10.7
Or, 10.7 kN-m.
Then we add this area to 10.7 in order to determine the moment value here.
It equals 8 kN-m.
Finally, when we add this area to 8 we get 0,
which is the moment value at the right end of the beam.
Remember
that we can determine the shape of the moment diagram in each segment
using the shape of the corresponding shear diagram.
We know
that when shear is constant moment varies linearly.
So,
the moment diagram takes the shape of an inclined line in segment BC.
But what is the shape of the diagram in this segment of the beam?
The general rule for determining the shape of the moment diagram is as follows:
If we view the shear and moment diagrams as the graphical representation of two polynomials,
then the degree of the moment polynomial
is always one higher than the degree of the shear polynomial.
In this segment,
the degree of the shear polynomial is 0,
since shear is constant.
Therefore,
the moment polynomial has to have a degree of 1.
Here,
the shear polynomial has a degree of 1,
therefore
the corresponding moment polynomial should have a degree of 2,
which means we need to draw a curve,
a quadratic equation in segment AB.
This can be done in two ways:
we can either draw a curve having a decreasing slope,
like this:
or,
we can draw one with an increasing slope,
like this:
To determine the correct shape,
we need to examine the shear diagram right above it.
If the shear is decreasing in value,
then the moment diagram
should have a decreasing slope.
However,
if the shear value is increasing in the segment,
then the slope of the moment diagram should also increase.
In this case,
shear is decreasing in value from left to right,
therefore,
the slope of the moment diagram must decrease from left to right.
Hence, this is the correct curve.
And given that shear is decreasing here,
it is going from zero to a large negative value,
then the moment diagram must have a decreasing slope,
like this:
Here is the complete moment diagram for the beam.
Now it is your turn.
See if you can construct the shear and moment diagrams
for the following beams
without writing the underlying shear and moment equations