Shear influence line for statically determinate beams.
Suppose we are asked to design hinge C in this bridge. The bridge is subjected to a moving load.
To do so, we need to determine the maximum shear force that the hinge has to withstand.
We can easily determine this force by drawing what is known as an influence line.
An influence line is a diagram.
It shows how a specific force in the structure changes due to a moving unit load.
Here is the influence line for shear at C.
Before I explained how the diagram is drawn, let me explain how to interpret it.
The diagram shows shear values at C as a unit load moves across the bridge.
For example, when the unit load is to the left of C, shear at C is 0.
When the unit load moves past Point C, when the load is just to the right of C, shear at C jumps from 0 to 1
Shear at C then decreases linearly as the load continues to move toward point D.
When the load reaches D, shear at C becomes 0.
Shear continues to decrease linearly as the load continues to move toward E.
When the unit load reaches E, shear at C reaches its maximum value: 3/2.
Once the moving load crosses over point E, the shear force starts to increase toward 0,
until it reaches 0, when the unit load reaches Point F.
Since the influence line is drawn for a moving unit load,
we must multiply the values given by the diagram by the actual magnitude of the load in order to obtain the correct shear values.
For example, if the moving vehicle exerts a force of 6 kN on the bridge,
then maximum shear at C equals 6*3/2, or 9 kN.
So the hinge at C needs to be designed for a maximum shear force of 9 kN.
Okay, now let's turn our attention to the process of drawing shear influence lines for statically determinate beams.
For simply supported beams without any internal hinges drawing a shear influence line is rather straightforward.
Here is a simply supported beam.
Suppose we wish to draw the influence line for shear at C as a unit load moves across the beam.
If we place the unit load at various locations on the beam and calculate shear at C for each load position,
then graph shear as a function of load position, we would get this diagram.
The diagram tells us that shear at C decreases linearly, from 0 to a negative value as the unit load moves from A to C.
At C there is a jump in shear value from negative to positive.
This means that when the unit load is just to the left of C, shear at C is negative.
And when the unit load moves over just to the right of C, shear at C becomes positive.
Then, shear at C decreases linearly as the load moves from C to B.
Note that the diagram consists of two straight segments separated at C,
as if the beam is cut at C,
then the left end of the cut is pushed down and the right end of the cut is pushed up, as points A and B rotate to accommodate the deformation.
The idea of pushing the beam segments up or down and drawing the shape that results is a convenient way for drawing shear influence lines qualitatively.
To better conceptualize this qualitative approach, we can imagine a pair of vertical rollers placed at C, permitting vertical movement at that point.
We then place a positive shear force at the cut point.
A positive shear actually means a pair of forces:
a downward force at the left roller, and an upward force at the right roller.
These forces are assumed to have the ability to move the two rollers in opposite directions.
The left force can cause a downward movement of the left roller,
and the right force enables the right roller to move up.
The deformed shape that results is the shear influence line.
Let's put this qualitative approach to use.
Here is a simply supported beam with an overhang.
Let's draw the influence line for shear at C.
First, place a pair of vertical rollers at C.
Now subject the rollers to a positive shear, like this:
How does the beam deform? Or, what does the influence line for shear at C look like?
The rollers divide the beam into two rigid segments.
Let's call them bars. We have bar AC, and bar CE.
Here is something very important to remember.
Such bars can move up or down or rotate, if permitted by their boundary conditions.
But they cannot bend.
Simply put, this is not an acceptable deformation for a bar.
Let's see if we can figure out how bar AC deforms.
C is being pushed down, and there is a pin at A permitting the bar to rotate.
Therefore, the bar is going to have a clockwise rotation like this:
What about bar CE? How is it going to deform?
C is being pushed up, forcing E to rotate clockwise, pushing the free end of the bar down.
This is a permissible deformation since there is a roller at B.
Rollers and pins permit rotation.
So this is the entire influence line for shear at C.
Now let's draw the influence line for shear at D.
First, place a pair of vertical rollers at D.
Then subject the rollers to a positive shear force.
This means placing a downward force at the left roller, and an upward force at the right roller.
Now draw the deformed shape of the beam.
Again, we have two bars here: AD and DE.
Bar DE is going to move up vertically.
This is so because both D and E are free to move up.
Bar AD however is not going to have any deformation.
True, point D is being pushed down. But the point actually moves down if the bar can rotate about B.
For the bar to rotate at B, point A has to move up.
But A is pinned, it cannot move up. Therefore, B cannot rotate either.
This means bar AD remains un-deformed.
Here is the entire influence line for shear at D.
Now let's consider a cantilever beam.
Draw the influence line for shear at C.
First, place a pair of vertical rollers at C.
Then subject the rollers to a pair of vertical forces like this:
Now draw the displaced shape that results.
The beam is divided into two bars: AC and CB.
Bar CB moves up since both C and B are free to move vertically.
Bar AC remains un-deformed since A is fixed and cannot move or rotate.
So this is the complete influence line for shear at C.
Now let's consider beams with one or more internal hinges.
Here is a statically determinate beam with an internal hinge at D.
We wish to draw shear influence line for point E.
Place a pair of vertical rollers at E and subject them to a positive shear force.
Now draw the resulting deformed shape.
Here, because of the hinge the beam is divided into three bars: AD, DE, and EC.
Bar EC is going to be pushed up at E.
The bar therefore is going to turn clockwise, since the roller at C permits the point to rotate.
For bar DE, we have a force pushing down at E.
This means the beam has to rotate clockwise at B, which means D needs to move up.
But can D move up?
The hinge at D does permit vertical movement at that point.
But does the point actually move up?
It all depends on the adjacent segment.
D moves up only if bar AD can rotate counterclockwise.
That is, if A can rotate counterclockwise.
In this case, A can rotate, since there is a pin at the point.
Therefore D does move up which causes bar DE to rotate clockwise.
So here is the influence line for shear at E.
Can you draw the influence line for shear at D? Pause the video and give it a try.
I will show you the solution at the end of the lecture.
Let's look at an example involving two hinges.
What does the influence line for shear at E look like?
Start by placing a pair of vertical rollers at E,
and subject the point to a positive shear.
Now draw the deformed shape that results.
Here, the beam is divided into four bars: AF, FE, EG, and GD.
Bar EG wants to rotate clockwise.
But it can do so only if G is permitted to move down in segment GD.
Since there is a roller at D, the rotation is allowed.
So here is the deformed shape of bars EG and GD.
Similarly, bar FE wants to turn clockwise, since E is being pushed down.
This bar can rotate only if the hinge at F is permitted to move up by the adjacent bar, bar AF.
Since there is a pin at A, point F does move up, like this:
So here is the influence line for shear at E.
All right, we now know a thing or two about drawing shear influence lines qualitatively.
Now let's talk about determining shear magnitudes in such influence lines.
Here is a simply supported beam.
Say we want to draw the influence line for shear at C.
A pair of rollers ...
a pair of shear forces...
the deformed shape...
Here is the influence line.
But what are these values?
We know the influence line is drawn based on a moving unit load.
So it should come as no surprise that the overall height at C is a unit force.
The height at the negative and positive sides are then fractions of 1.
The negative height is the ratio of the length of the left segment to the overall length, or 1/4.
The positive height is the ratio of the length of the right segment to the overall length, or 3L/4L, or 3/4.
Notice that the two Heights add up to 1.
Consider this beam.
Here is the influence line for shear at D.
So the height on the negative side here is 2/6, or 1/3.
The height at the positive side is 4/6, or 2/3.
What about the height here?
We can determine this height, let's call it h, using similar triangles.
That is, 1/3 over 2 equals h/4. This gives: h = 2/3.
Here is another example.
Here is the shear influence line for point C.
What is the height of the influence line?
We know that the overall height at the vertical rollers is always 1.
Since in this case there is only one height (on the positive side), then it must be 1.
Let's end the lecture by drawing the influence line for shear at D for this beam.
As usual, we place a pair of vertical rollers at D.
Then subject them to a pair of shear forces, like this:
Bar AD turns clockwise like this.
The rest of the beam remains un-deformed.
This is so because neither B nor C can move vertically.
The overall height of the diagram at D is 1.
Here are a few exercise problems.