We actually have already learned a little bit about the internal forces when we studied the simple truss analysis. 
For a truss member with forces acting on it, 
the member could be in tension, 
or compression.
And although the forces only act on the two ends, force must also develop inside the member, 
because we should be able to run an imaginary cut at any place through the member 
and each segment
must also be in force equilibrium according to Newton’s first law.
These forces are internal forces.
They are exerted by the neighboring materials to keep the segments from falling.
This is a very simple example of internal forces and we are about to learn more complicated cases. 
But before doing that let's take a moment and think why we are interested in learning about internal forces.
[PAUSE PLEASE]
You’ve probably come to this answer: 
we want to know the internal force so that we can make sure it does not exceed what the material can withstand. 
We don’t want the members to fail under the loadings. Actually, it is not the force that matters, because we can simply use a thicker,
bigger member, and that seems to fix the problem. 
It is the force over unit area, stress, that can not exceed a certain value. 
The stress associated with internal normal force, like the ones in the truss members, is normal stress, sigma, 
which is calculated as F over the cross sectional area of the member A, 
and it should not be greater than an allowable value. 
As you can probably tell, this is beyond the current Statics class. But it’s always good to know why you are learning what you learn. 
Now let's look at a more general example.
In this example, we have a cantilever beam. Point A is a fixed support. And we are asked to find the internal forces at point B. Since we are 
asked to solve for internal forces, we need to first expose those forces, make them external, and then we can draw the free body diagram, 
and apply rigid body equilibrium analysis that we have already learned. 
In fact, we are not unfamiliar with this approach since we used it to solve for truss forces already. 
And this method is called the method of sections. 
Now let’s look at the proper procedures when applying the method of sections. 
First step, determine the external support reactions and complete the free body diagram of the entire member if necessary. 
Of course, sometimes it is not necessary to do so and you will be able to tell afterwards. But for now, let’s just solve them. 
Since at point A we have a fixed support, 
there are three support reactions: two forces and a couple moment.
And we can solve for all of them by applying the rigid body equilibrium analysis.
Step 2, cut the member at the specified point, in this case, point B.
And we separated the two parts,
and now we need to specify what kinds of unknown reactions we should draw at the cut.
Let's look at the right segment:
It is being held by the left segment, and it is not allowed to move horizontally, or vertically, 
or rotate. That sounds just like a fixed support. 
In fact, don’t forget in this class we assume all members that we study are non-deformable, therefore, any portion of any member can be 
considered to be held by an imaginary fixed support. So let’s draw the same unknown reactions as a fixed support:
two force reactions and one couple moment reaction.
And because of actions and reactions, we can also draw the unknown reactions on the left segment. 
They have the same magnitudes but opposite directions as the ones on the right. 
And therefore we only need to solve for one set of the unknown reactions, either from the left segment or from the right segment. 
So step 3 is to choose one side and solve for the unknowns. I will demonstrate calculations  
using both sides, just to show you that they will lead to the same answers. 
But when you make your own pick, of course you want to choose the side that makes the calculation easier. 
In this case you should pick the right side so you don't need to solve for the support reactions in the first place.
So we put the left segment in an appropriate coordinate system,
write the three two-dimensional rigid body equilibrium equations:
force along the x direction,
force along the y direction and the couple moment about point B,
and we can solve for all three unknowns.
And we can do the same thing for the right segment and get exactly the same answers.
Note that
we want to summarize the total moment about point B because this way the unknowns V_B and N_B will not cause any moment about 
point B therefore they do not show up in your moment equilibrium equation. 
Now let’s look at these names. 
The first one is normal force, because it is normal, or perpendicular to the cross section of the member.
We are already familiar with this one, because all the forces we solved for simple truss analysis are normal forces. 
The second one, V, is shear force, and it is always tangent to the cross section. 
The third one is the bending moment, and as you can tell, it causes a bending tendency in the member. 
Although you can define the positive directions of these internal forces whichever way you want, there are a set of sign conventions. The purpose 
is for better communication between engineers, so other engineers can quickly interpret your calculations. 
According to the sign conventions, for a 2-D problem, 
the normal force is considered to be positive if it is tensile. 
The sheer force is considered to be positive
if it creates a clockwise rotational effect on the segment. 
And the bending moment is considered to be positive 
if it creates a concave upwards bending effect on the segment. 
I would suggest you always draw your unknown internal forces according to the sign convention to avoid confusion. 
Now let's quickly look at another example.
Determine the internal forces at point C. Again, we are going to apply the method of sections. 
First let’s choose the side we want to solve. 
In this example, we inevitably have to solve for some external support reactions first.
But at point A we have a roller support which only has one support reaction, versus at point B we have a pin support 
with two support reactions. 
Therefore we choose the left side with the roller support.
So we solve for force F_A at the roller support using one moment equilibrium equation,
and now we run an imaginary cut through point C 
and analyze the left segment.
We draw the three internal forces according to the sign convention,
the normal force, the sheer force and the bending moment.
We need to replace the distributed load by an equivalent concentrated load.
Now we set up the rigid body equalibrium equations,
and we can solve for all three unknowns.
What if you have a three dimensional problem?
For example, for this beam that’s subjected to the external forces as shown, what are the internal forces at point C in this beam? 
Well we can solve the problem using the same method of sections following the same 
procedures as described for the 2-D problems.
We run an imaginary cut through point C, 
And separate the two parts. Pick one side to solve, 
determine the support external reactions if necessary, 
put the segment in an appropriate coordinate system, and draw the unknown internal reactions the same way for a fixed support, 
and solve for all unknowns by applying the three dimensional rigid body equilibrium equations. 
Keep in mind that for 3-D problems there are no sign conventions like the ones we had for 2-D problems, therefore the signs for the 
internal reactions will depend on how you set up your coordinate system.
Let's say we choose the left segment,
and these are the six internal reactions and you can solve for all of them by applying the six rigid body equilibrium equations.
Now you have one normal force that is along the axial direction and it's perpendicular to the cross section,
you have two shear forces that are tangent to the cross section,
two bending moments,
and also you have one torsional moment that creates a twisting effect of the member.
As you will learn in the class of mechanics of materials,
these different internal forces will cause different types of stresses, and deformations, and even failures in the members
