Let’s go through how to draw shear force and moment diagrams. Here, I have a beam, and I’ve
drawn both the shear force and moment diagrams for it. We are going to look at 2 methods of drawing
these. The first is the segment method. What we do is figure out where a change occurs on the beam,
so right here on this beam a change occurs because a force is applied here. Then we
cut the beam and find the shear force and the bending moment at that location. Then we do the
same for the next segment, and once we figure out the shear force and the moment for each segment,
we plot it. The second method involves knowing the relationship between a shear force diagram
and a moment diagram. So let’s take look at these 2 graphs here.
Here are the things you should keep in mind. Notice that if the shear force is positive,
the slope of our moment diagram is positive. If the shear force is negative, then the slope of
our moment diagram is negative. The area under the shear force diagram gives us the moment equations,
and if we take the derivative of our moment diagram, we the get the shear force equations.
I will go over both methods with 4 examples, and show more relationships between them, and by the
end, you should have a very good idea of how to draw these diagrams, so let’s get started.
Let’s take a look at our first example. We need to draw the shear and moment diagrams. The first step
is to figure out the reactions at pin A and roller B. If we write a moment equation about point A,
that eliminates the 2 reactions and we can directly solve for the reaction at B.
Remember, B is a roller, which means it’ll only have a single force upwards. Let’s solve.
Next, we can write an equation of equilibrium for the y axis forces,
that will allow us to figure out AY. Let’s solve.
There is no x reaction since there are no horizontal forces being applied to the beam.
Now that we figured out the reactions at the supports, the next step is to figure out where
a change on this beam occurs. We see that our 30kn force is applied 2 m away from point A,
so that’s our first segment, and then from 2 m to 6 m is our second segment.
So we are looking at the beam from 0 m to less than 2 m, and then from greater than 2 m to 6 m.
Let’s draw each segment separately. First the segment from 0 m to 2 m. Now instead
of saying this segment is 2 m in length, we want to represent it as variable. In other words,
x is the distance from point A. This allows us to figure out values as a function of the length
instead of fixed distances. For this section, we have the reaction at A, we have the shear force,
and we have a moment. The 30 kN force is not included, because that’s where we made the cut.
So we’re looking at the beam from 0 m to less than 2 m. This should be familiar to
you from internal loadings, if you need a refresh, please check the description.
Now, if we write an equation of equilibrium for the y axis forces,
we can figure out the shear force. Notice that the length of the beam
didn’t matter to us. Now to find the moment, we do need to consider the length of the beam.
Instead of finding the moment at 2 m, let’s write it as a function of x, that way, we can figure out
the moment on this section at any location between 0 m to 2 m. That’s why we wanted to represent the
length as a variable. So we are finding the moment at the edge of our beam, right here,
and instead of using a fixed length, we are writing it as a function of x.
We have the 20 kn multiplied by a distance of x, that creates a clockwise moment, so it’s positive.
Then we have our moment here, which is counter clockwise, so it’s negative.
Let’s set those aside, and draw the next segment, which is from 2 m to 6 m.
Now if the previous section had a length of x, and the whole beam is 6 m, the length of this section
is simply 6 minus x. That’s how we will represent this length. We have the shear force, the moment,
and at the other end, we have the 10 kN force. Again, the 30 kN force is not included because
we are looking at the beam from greater than 2 m to 6 m. So remember that wherever we make a cut,
we don’t include the force at that cut. Now as before, we will write an equation
of equilibrium for the y axis forces to figure out the shear force. We have the shear force
and the reaction at the end. Let’s solve. Again, notice that we don’t care about the
length of the beam for the shear force. Next, we need to figure out the moment.
Again, we want to write the moment as a function of x. So we have the moment at the beginning, and
then we have the reaction times the perpendicular distance, which is 6-x. Let’s simplify.
Now we can plot our graphs. Let’s start with the shear force diagram.
So from 0 m to 2 m, we have a constant value of 20 kN. Then from 2 m to 6 m, we have a shear force
of --10kN. And finally, at 6 m, it goes back to 0 because of our 10 kn reaction. Now notice the
large jump here, if we calculate the difference, you will get 30 kN which is the force that’s being
applied at that point. This is why when we draw the segments, we don’t include the force at the
cut, because our equations will take care of it. Now a moment diagram. From 0 m to 2 m,
our equation is 20x. So let’s plot it. At 0 m, we have 0, and at 2 m, we get 40 m, since that’s
20 times 2. Now from 2 m to 6 m, we use the other equation and that our moment diagram. You should
notice some interesting things. If we take the slope of our moment diagram, that gives us the
shear diagram. In other words, the derivative of our moment equations gives us the shear values.
This also means that the area under the shear force graph gives us our moment equations.
Also notice that when we our shear force is positive, the slope of our moment diagram is
positive. When our shear force is negative, the slope of our moment diagram is negative. We will
use these techniques in the last 2 examples. Let’s take a look at this problem involving
a distributed load. So as usual, our first step is to figure out the reactions at pin
A and roller B. To do so, we need to express our distributed load as a resultant force.
We can do that by multiplying the load by the length, giving us 320 kN. We place this at
the centroid of our distributed load, so at 4 m. Now if we write a moment equation about point A,
we can directly solve for the reaction at B. Don’t forget to include the moment
applied at the end of the beam. Let’s solve. Now, we can write an equation of equilibrium
for the y axis forces. That allows us to figure out the reaction at A. Let’s solve.
Note that there isn’t an x reaction at A since we have no horizontal forces affecting the beam.
Now we can find the segments on the beam where changes occur. In this case, we have a distributed
load from 0m to 8 m. So that’s one segment. The next segment is from B to C. Let’s draw the first
segment separately. We will say the length of this segment is x. We have the vertical reaction at A,
the distributed load, the shear force and the moment. We don’t care about the reaction at B
since that’s where we made the cut. We are looking at the beam starting from 0 m, to less than 8 m.
We can figure out the shear force by writing an equation of equilibrium for the y axis forces.
Starting off, we have the vertical reaction at A. Then we have the distributed load, but remember,
we don’t have a fixed length, so we multiply the load by the length, which in this case is x,
and then we have the shear force. Let’s isolate for V.
Now to find the moment. Let’s write our moment equation.
So first, we have our moment. Next, we need to think about the moment created by the distributed
load. We can find the resultant force of the distributed load by multiplying the load by the
length, so that’s 40 times x. Then we place the resultant force at the midpoint of our length,
that’s x divided by 2. So the moment created by the distributed load is the multiplication of the
two. Remember, we aren’t using a fixed length, we want it written as a function of x. Lastly,
we have the reaction at A times the perpendicular distance of x. Let’s isolate for the moment.
Now we move onto the next segment. So that’s from greater than 8 m to 11 m. The length of this
section is the total length of the beam minus x. Again, we ignore the vertical reaction at B since
that’s the location of our cut. At the left end, we have the shear force and the moment.
On the right side, we have the 20 kN force along with the moment applied.
Let’s figure out the shear force by writing an equation of equilibrium for the y axis forces.
We have the shear force and the force applied. Let’s solve.
Next, we can write a moment equation. We have our moment plus the 20 kn force
times the perpendicular distance and lastly the moment at the end. Let’s isolate for the moment.
We now have all the equations we need to plot our graphs.
Let’s start with our shear force diagram. Let’s plot our first equation, so at 0m,
we have a shear force of 133.75 kN. Next we plug in 8 m and get our second point.
Now for the next segment. The equation is a fixed value, so let’s draw that as well. Notice the jump
from here to here. If we find the difference, you will see that it’s equal to the reaction at B we
found earlier. Again, this is why we don’t include it in our diagrams; our equations will take care
of it. Next, we can draw the moment diagram. Let’s start with the first segment which is a
parabola. That’s for the segmenting start from 0 m to less than 8 m.
The x coordinate of the vertex of the parabola can be found using this equation. Hopefully you
remember this from your functions class. If we plug this value into our equation,
we get the peak of the parabola which is our maximum bending moment.
Next, the 2nd segment, which is a linear graph. Now, it’s usually a good idea to write down the
locations of interest. You can see where the shear force becomes zero by setting it to 0.
That gives us 3.34 m. You can also see where the moment is 0 by setting the moment to 0.
That gives us 6.69 m. Again, notice how our shear force equations are the derivative of
our moment equations. You should also notice that at the location where our sheer force is 0
is where the maximum bending moment occurs. And those are our answers.
Now we are going to switch to a different method to draw our diagrams. Instead of cutting the
member at specific locations and then considering each segment, we are going to look at the forces
applied to the beam and construct our diagrams. Even with this method, we must still first
figure out the support reactions on our beam. So let’s start off with a moment equation about
point A. That will allow us to solve for the reaction at B. Let’s solve.
Now an equation of equilibrium for the y axis forces to find the reaction at A. Let’s solve.
Note that there is no x react at A since we don’t have any horizontal forcers.
Now we will begin constructing our shear diagram. Starting at x = 0, so at the very left side,
we have a force of 450 N straight upwards. That shear force will remain the same until we come
to this point right here, which is where the 800 N force is applied. So we subtract 800 from 450
giving us -350. The shear force remains the same until we get to this point. Here,
another 600 N is subtracted from our value giving us -950. This continues until we get
to the very end, where the support force at B, which is 950 N is added, giving us zero.
Now let’s draw a moment diagram. Remember that the area under the sheer force diagram gives us
the moment. So starting at 0, we have a moment of zero since the area is 0. Now if we move to 1 m,
the area under the shear graph is 450nm. So that’s the value of our moment at 1 m. Now
it’s a positive slope because our shear force is positive. So it’s a linear increase towards 1 m,
and the max value at this point is 450 N. Then we look at it from 1 m to 2 m. The area under
the graph is 350 N, and that’s negative, which means our moment will be decreasing,
and it’s 450 minus the 350 N. Giving us 100 nm. So right up to 2 m, our moment is 100 nm,
and it’s a negative slope. Now at 2 m, we have a moment applied, that’s a clockwise moment, so we
add that value to our 100 nm. That gives us 1300 nm, and it’s an instant jump since it’s a moment
directly applied at this point. Then we look from 2 m to 3 m. The area under the graph is 350 nm,
and it’s negative, so we subtract 350. So at 3 m, our value is now 950 with a negative slope.
Again, it’s a negative slope because our sheer force diagram between 2 m and 3 m
is negative. Lastly, we look between 3 m and 4 m. The area under the graph is 950 nm.
This is also negative, so we subtract that value, giving us 0 nm. And that is our moment diagram.
Let’s take a look at one last example covering applied moments, forces and distributed forces.
The first step is to figure out the reactions at the supports. To do that, we have to show
the resultant force created by the distributed load. So that’s the load multiplied by the length,
giving us 45 kN. We place that at the centroid, so that’s 1.5 m away from B. Now we can write a
moment equation about point A. This allows us to solve for the reaction at B. Let’s solve.
Next, we can write an equation of equilibrium for the y axis forces. Solving gives us the vertical
reaction at A. Note that we got a negative value, which means the reaction is actually
facing downwards. There is no x reaction at A since we have no horizontal forces.
Now we are going to draw our shear force diagram. Starting at x = 0, we have the support reaction
at A, which gives us a shear force of 14.3 kN. Note that this support reaction is facing down
so it’ll be negative. Now we move forward until we get to this point. This is where our 8kN force is
applied. That’s also downwards, so we just add that to our current value, giving us 22.3 kN.
Now we keep going forward until we get to point B, where we have a support reaction upwards.
Let’s add that, giving us a positive value of 45 kN. From here on, we have a distributed load.
When we have a distributed load, our shear force changes with respect to the length. That means
we will not have a straight line but rather a downward slope line. It’s a negative slope
because our load is downwards. If we keep going to the very right edge, we have a total load of 15
times 3 m, giving us 45 kN. It’s negative since it’s downwards, so adding it to our previous
value of positive 45 brings us to 0. So that’s the distributed load on our shear force diagram.
Now we can draw our moment diagram. So remember, we are calculating the moment for the whole beam
at x = 0. So initially we don’t have a moment since we are at 0 m. Then we move
forward for 2 m. The area under the graph is 28.6 kNm. So our line has a negative slope
since the shear force is negative and we end with a moment of 28.6 kNm. Now at that location,
we have a moment that’s being applied. That’s a clockwise moment, so that’s positive, and we add
that to our value, giving us 8.6 kNm. This created a jump upwards like this. Now from 2 m to 3 m, the
area under the shear force diagram is 14.3 kNm, that’s negative since the shear force is negative.
Let’s add it to our current value, giving us 22.9 kNm. Next, we look at the segment from 3 m to 5 m.
The area under the graph is 44.6 kNm. This is still negative since our shear force is negative.
So let’s add it to our current value, giving us 67.5 kNm. Up until now, all of our lines have
been linear. Now looking at the last section, so from 5 m to 8 m, the area under the shear force
diagram is positive. That area is the area of a triangle, so that’s ½ times the height
times the length, giving us a positive 67.5 kNm. That means at the end, the moment is 0. However,
this is not a linear graph, this will be parabolic and it’s slope will be positive because our shear
force is positive. So remember, distributed loads create a parabolic moment line and that
makes sense. Remember how the derivative of our moment equation gives us the shear force equation,
so the derivative of a parabolic equation gives us a linear equation. And that is our moment diagram.
That should give you a good idea on how to draw these shear force diagrams and moment
diagrams. Doing plenty of questions will help out a lot with this method. I hope this
video helped and if it did, please consider sharing it with your friends and classmates,
they too might find it helpful. Thanks for watching and best of luck with your studies!