Let’s talk about internal loadings. When we have a structural member, for example, a rod,
or a beam, we need to make sure that the material it’s made of can resist the loading acting within
the member. The loadings we are concerned with in this video include the sheer forces,
normal forces and bending moments. To figure out these loadings, we can use the method of sections.
Imagine we have a beam like this, and we want to figure out the internal loadings at point C.
To find it, we make a cut through point C like this. Once we split the member, we would have
three loads. The first is the normal force. That acts perpendicular to the cross section.
We represent that with the letter N. Next, we have the shear force. That’s tangent to the
cross section. Usually this is represented with the letter V. Lastly, we have a bending moment,
which we represent with the letter M. The normal force is positive if it creates tension.
A shear force is positive when it causes the beam segment to rotate clockwise,
and the moment is positive when the segment will bend in a concave upward manner.
So whenever we cut a member, we assume these forces to be drawn like this.
If we get a negative answer, all it means is that the its opposite to our assumption.
To figure out these values, you should be familiar with finding moments, distributed loads,
and 2 force members. There are videos on all of these subjects, so if you need a refresh,
please take a look at the description. Let’s start off with a simple example.
Here, we have a beam and 2 forces are applied to it. We need to figure out the normal force, shear
force and the bending moment at point C. Looking at the beam, at A, we have a pin support and at B
we have a roller. We can easily figure out the support at B by writing a moment equation about
point A. Remember a roller only has one reaction and that’s straight upwards. So we have the 2
forces times the perpendicular distance and the reaction at B times the perpendicular distance.
Let’s solve. Now we don’t need to figure out the reactions at A because I am going to cut the
member at C and only focus on the right side. This way, we can forget about the left side.
At C, we have the normal force, shear force and the moment.
To find the normal force, we can write an equation of equilibrium for the x axis forces. Since there
are no x axis forces, the normal force is 0. To find the shear force, we can write an equation
of equilibrium for the y axis forces. So we have the shear force, the 15 kN force applied at the
top and the support reaction at B. Let’s solve. Lastly, we need to find the moment at C. To do
that, let’s write a moment equation about point C. So we have the moment at C,
plus the moment created by the 15kN force and the moment created by the reaction at B. Let’s solve.
And those are our answers. Let’s take a look at this problem where we need
to find our usual forces at point D. Now since we have this member joining here, it’s easier to
focus on the left side, which means we first need to figure out the reactions at pin A. To do that,
we need to express the distributed load as a single resultant force. So we multiply the
length by height, which gives us 1200 N, and we place it at the centroid of the distributed load,
which is the midpoint of the length, so at 4 m. Now if we consider member AB, the force of member
BC would be like this since it’s a 2 force member. The angle can be found using the inverse of tan.
Lastly, at A, we have x and y reactions. We can figure out force BC by writing a moment equation
about pin A. Note that only the y component of force AB creates a moment since the x component’s
line of action goes through point A. Let’s solve. Now we can finally figure out the reactions at A.
First, we will write an equation of equilibrium for the x axis forces.
We only have the x component of force BC. Let’s solve. Next an equation of equilibrium for the
y axis forces. We have our resultant force and the y component of force BC. Let’s solve.
Now that we found the reactions at A, we can cut the member at point D and forget about the right
side. There is something we need to make sure we keep in mind. When we cut the member like this,
we need to recalculate the resultant force created by the distributed load. What we
calculated before was for the whole member, now that we cut it in half, we have this section of
the distributed load to worry about. As before, we multiply the length by height, this time,
giving us 600 N. We place that at the centroid of this distributed load, so that’s at 2 m.
Now looking at where we cut, we have the shear force, the normal force and the moment at D.
To find the normal force, we can write an equation of equilibrium for the x axis forces.
Let’s solve. We got a negative value, which means the normal force actually faces to the left.
Next, we can find the shear force by writing an equation of equilibrium for the y axis forces.
Let’s solve. Lastly, to figure out the moment,
let’s write a moment equation about point D. Let’s solve and those are our answers.
Let’s take a look at one last example where we need to figure out the normal force,
shear force and bending moment at point D. So the first step is to figure out the
reactions at A. To do that, we first need to figure out the force in member BC. Notice that
it’s a two-force member, so we can draw it like this. The angle can be found using the inverse
of tan. Now if we write a moment equation about point A, we can figure out force BC.
Using that, we can start to figure out the reactions at A. First, an equation
of equilibrium for x axis forces. Let’s solve for the x reaction. Next, an equation of equilibrium
for the y axis forces. Let’s solve. Now we can cut the member at D and
forget about the top part. At D, we have the normal force, shear force and the moment.
We can find the shear force by writing an equation of equilibrium for the x axis forces. Let’s solve.
Next, an equation of equilibrium for the y axis forces to figure
out the normal force. Let’s solve. Lastly, we have the moment. For that,
we will write a moment equation about point D. Solving gives us our answers.
That should cover the types of problems you will face when it comes to internal loadings. In
the next video, we will go over how to draw shear force and moment diagrams.
I hope this video helped. Thanks for watching and best of luck with your studies.