In previous lectures
we talked about shear force in beams.
how to calculate them,
and how to represent them using mathematical equations.
Please review lectures ST8 through ST10 for the background content.
This lecture deals with shear diagrams,
the graphical representation of the shear force in beams.
Consider the simply supported beam.
It is subjected to two concentrated loads:
one here, and one here.
These applied forces result in the development of an internal shear force
that varies across the length of the beam.
For example,
the shear force at this point has a different value than the shear force here.
A graph that shows how shear varies in the beam is a useful visual design tool
for determining the locations at which the force attains its maximum and minimum values.
For example,
the shear diagram for this beam reveals that the internal force attains its maximum positive value
in the right segment, and it reaches its maximum negative value in the middle segment.
Let's start the conversation
by going through the process of drawing the shear diagram for this beam.
Suppose these two crates exert a total force of 1000 N on the beam,
and the force exerted by the end crate is 600 N
The beam is 8 meters in length.
The distance between the two supports is 6 meters.
And the crates are placed on the beam 5 meters apart.
Let's start by taking a closer look at the beam.
Here is a two-dimensional representation of it.
The free-body diagram of the beam can be drawn this way:
which enables us to write the static equilibrium equations
and then solve them for the unknown support reactions.
Once all the unknown forces are determined
the free-body diagram can be used to formulate the necessary
algebraic equations for representing shear in the beam.
We talked about shear equations in lecture ST10.
If you're not sure how to write them, please review that lecture.
Once we have the shear equations
we can then graph them in order to come up with the shear diagram for the entire beam.
This is the equation for the shear in the left segment of the beam.
The equation tells us that shear is constant throughout the segment.
So it's graph looks like this:
The second equation covers the shear in the middle segment of the beam.
it can be graphed this way.
And the third equation translates into this line.
So, here's the shear diagram for the entire beam.
The process of drawing the shear diagram from the shear equation is rather
straightforward and does not need much discussion.
Instead, in this lecture,
we're going to talk about the construction of the shear diagram
directly from the free-body diagram of the beam,
without using the shear equation.
Before we start,
let's make an important observation.
According to these equations,
shear is not defined at the points of application of the concentrated loads.
The equations give us the shear values within each segment of the beam, but not at its boundaries.
Why is that?
Why can't we define shear at the location of a concentrated load?
To see why, let's isolate points a B, C and D.
At each point , we have a concentrated force.
We say that these forces are external to the beam.
They are acting on the beam,
and are not inside of it.
In contrast, a shear force is internal to the beam,
it does not show up on the free-body diagram unless we cut the beam.
Let's take a closer look at point A.
Since the point is dimensionless,
we cannot break it up into smaller segments.
For static equilibrium to be maintained,
for the sum of the forces in the y-direction to be zero,
there must be an opposing force just to the right of A.
This force which must develop inside the beam is called a shear force.
The location of the force is one dimensionless point to the right of point A.
Let's label that point A+.
So, a shear force must develop at A+ for point A to remain in equilibrium.
The magnitude of this force is 300 N.
We say shear at point A+ is 300 N.
The same idea applies to point B.
Since the sum of the forces in the y-direction must be zero,
then a shear force on either side of B must develop.
We refer to these points as B- and B+.
In this case, shear at B- is 300 N,
at B+, we have a shear force of -700 N.
Using the same line of reasoning,
we can conclude that shear just to the left of C is -700 N,
but at C+, the force becomes 600 N.
And, shear to the left of is D 600 N.
Let us devise a technique for constructing the shear diagram directly from the free-body diagram
by examining the relationship between the two diagrams.
Starting from the left end, and working toward the right end of the beam,
we can correlate the diagrams in the following way:
There is an upward force of 300 N at A.
This corresponds to a jump in shear value.
It jumps from 0 to 300 N.
So we can say, an upward external force on the beam causes an increase in shear value.
The amount of increase is equal to the magnitude of the upward force.
The left beam segment is not subjected to any loads within it.
That is, there is no load on the beam from point A+ to point B-,
therefore, no change in shear takes place.
As you can see, shear in this segment of the beam has remained unchanged.
The value of shear from A+ all the way to B- remains 300 N.
There is a downward force of 1,000 N at B.
This causes a drop of 1,000 units in shear.
As you can, see shear has decreased from positive 300 N at B-
to -700 N at point B+.
Then, shear remains constant between B+ and C-,
since the segment is not subjected to any external loads.
The upward reaction force of 1300 N at C causes an increase of 1,300 units in shear.
shear goes from -700 N at C- to 600 N at C+.
Shear remain constant between C+and D-.
Then at D, shear drops by 600 N to zero.
Let us go through another example,
showing how to draw the shear diagram directly from the free-body diagram.
Consider this beam.
It is subjected to a concentrated load of 10 kN at B.
First, we draw the free-body for the beam.
Then, we use the static equilibrium equations to calculate the support reactions.
They are:
Let's start from the left end of the beam and work our way toward the right end.
Since there is no force present at A,
then the starting value for shear is 0.
Further, since there's no load being applied within segment AB,
shear remains zero throughout the segment.
The downward force at B causes a drop in the shear value.
Shear drops from 0 to -10 kN.
So, shear just to the right of B is - 10 kN.
Since there is no load applied to segment BC,
shear remains constant throughout the segment.
Then at C, shear jumps up by 15 units.
It goes from -10 at the left side of C, to +5 at the right side of C.
Shear remains constant within CD.
Then it drops by 5 units at the very end of the segment.
So, this is the shear diagram for the beam.
In the left segment shear is zero.
In the middle segment shear is negative.
And, in the right segment we have a positive shear.
Now, let's look at a case in which the beam is subjected to a uniformly distributed load.
We start by drawing the beam's free-body diagram.
We then calculate the unknown support reactions
Notice how the load divides the beam into two segments.
The left segment is subjected to a uniformly distributed load
and the right segment is load free.
We start from the left end of the beam.
The upward force at A causes a positive shear force to develop at the left end of segment AB.
The magnitude of the force is 8 kN.
Since the segment is subjected to a load, shear changes value within it.
That is, shear at the right end has a different value than the shear at the left end of the segment.
The change in shear between the segment ends
is equal to the total area under the load diagram within the segment.
The area associated with the distributed load is that of this rectangle.
Here, it equals 3 times 4, or 12.
Since the load is downward,
we consider this to be a negative change
which means shear at the right end of the segment equals 8 -12, or -4 kN.
Further, since the load is constant shear is going to change linearly.
This means we need to connect these end points using a straight line.
The rule is as follows:
When there is no load applied to the segment, shear within the segment does not change value.
When there is a constant load applied to the segment shear within it changes linearly.
What if the load was triangular?
Then the shear diagram within this segment becomes a nonlinear curve representing a quadratic equation.
Couple of exercise problems at the end of the lecture deal with such nonlinear diagrams.
Back to our problem.
Shear in the right segment of the beam remains constant, since BC is not subjected to any loads.
So, within the segment shear remains -4 kN.
At C, since there is an upward force the diagram returns to zero.
So, here is a complete shear diagram for the beam.
See if you can draw the shear diagram for each of the following beams.