In the previous video, we learned how to 
determine the internal forces at a specified 
location on a member. However, we would rather 
know where the maximum internal reaction 
happens in the member, because that’s where 
the member is mostly likely to fail. 
And in most cases we can not easily tell that 
location. Therefore, we need to know how 
exactly internal forces vary with 
different locations, in other words, we want to 
know the internal forces as functions of location 
x on the member. 
You can write the functions for all four types of 
internal forces following a similar approach, 
but in this video we will focus on the shear force 
function and the bending moment function. 
And the graphs of these functions are known as 
the shear force and bending moment diagrams. 
Let’s use this example to demonstrate that. 
Again, the goal is to write the shear force and 
bending moment functions of this beam 
so we can calculate the shear force and bending 
moment at any given location.
Also we want to sketch the diagrams so we can 
easily visualize the variations in the shear force
and bending moment, and be able to tell where 
the maximum shear force or bending moment 
happens. 
We have already learned the method of sections 
therefore we will cut the member at an arbitrary 
location x, and apply the method of sections 
to write the shear force and bending moment as 
expressions of x.
So the first thing we do is to complete the free 
body diagram of this member by determining all 
the external support reactions and marking them 
on the members. You may ask, since when we 
apply the method of sections, we can choose 
either side for our analysis,
so is it really necessary to solve all support 
reactions? Can we just solve for the roller 
support only?
That is true. However there's an advantage of 
solving all the external forces completely 
because this way we can double-check our 
diagrams.
Therefore I would encourage you to always solve 
for all the external support reactions.
Then we want to set up an x axis to represent 
the different location along this member. You 
can set it up whichever way you want, 
from left to right, from right to left, from the 
center, or even setting up multiple x axes, but I 
always set it up this way, from left to right, 
and for one particular problem, I always only use 
one x axis.
On this axis, point A is the origin, 
x equals to zero, and point B is this point right 
here when x equals to 8, and any point between 
A and B 
can be represented by various x values on this 
axis. 
So next thing we want to do is to section the 
member. Before doing that, let’s take a look at 
the loadings on the member.
And you can tell the loadings change. 
Intuitively we might be able to tell we can not 
use a single shear force function or a single
bending moment function to describe the entire 
member. 
Since points C and D are where changes 
happen, for this problem, we are going to cut the 
members three times. First, section 1 between 
points A and C, secondly, section 2 between 
points C and D, and lastly, section 3 between 
points D and B. 
Keep in mind that every time we section the 
member, we will take the entire left segment for 
analysis. Because this way, the length of 
the segment is x, as defined by the x-axis we 
set up. 
So first, we section the member at location 1 
and take the entire left segment for analysis. 
The entire length of this segment is again
an arbitrary x. We draw the internal forces 
according to the sign conventions that we 
learned before. 
On this segment we have a distributed load with 
intensity of 80 pound per foot, 
and we need to replace it by a concentrated 
load first. The magnitude of this concentrated 
force is area of the rectangle, which 
is 80 times x, and it is placed at the centroid 
location of the distributed load, which is half of x 
distance from the section.
And now we can apply the rigid body equilibrium 
to this segment, and write the equilibrium 
equations. 
Since we are not interested in the normal force, 
and it is zero anyway, we write two equations, 
the resultant force along the y direction, 
and the resultant moment about where we cut 
the member, let’s call it point x. 
This way the unknown force V_x has no 
moment about this point x and does not show 
up in the moment equilibrium equation. 
Since x is not known, we can not find the 
definite values for the shear force and bending 
moment.
But we can still solve them as expressions of x,
and these are the shear force and bending 
moment functions of location x for the first part 
of the member. 
And we do the same thing for section 2. Keep in 
mind we are still using the entire left segment  
for the analysis, therefore the length of the 
segment is still x. 
And also be careful when finding the moment 
arm 
for the concentrated force,
and we can solve for two more functions
that apply to the second part of the member.
And we do it one more time. Again the segment 
has a length of x,
again be careful with the moment arms,
and we can solve for the last set of functions
that apply to the third part of the member.
And after we summarize the results, we can see 
that both shear force and bending moment 
functions are piecewise functions, 
because they are each made up of three 
different equations that apply to different parts of 
the member.
And if you know the functions, of course you will 
be able to graph them. You’ve probably learned 
this in pre-calculus class. These are the two 
graphs I made using Microsoft Excel. 
And these are the shear force diagram and the 
bending moment diagram. 
From these two diagrams you can easily 
visualize how shear force
and bending moment change with location along 
the member, where they change direction, from 
positive to negative or vice versa, 
and most importantly, where the absolute 
maximum shear force and bending moment 
occur. 
And those are important information used in the 
design of a structure. 
However, this method is quite time consuming 
and sometimes we want to be able to quickly 
sketch the diagrams that can be less accurate. 
To do that, we’ll learn a graphical method as 
well. 
Let’s look at these two diagrams I made. On the 
top, we have the shear force diagram, 
which is made up of three parts, or, three 
straight lines. 
Two of these lines are horizontal, 
but the first one, let’s look at its slope.
It decreases from V equals to 300 pounds at x 
equals to 0 to V equals to -20 pounds at x 
equals to 4 ft. 
So the slope is -80 pounds per foot.
And if you recall,
for the first 4 feet on the member, there is a 
distributed load with load intensity w that is 
also -80 pounds per foot. It is negative because 
as you can see the force is pointing downwards. 
Well, we know that load intensity is defined as 
force over length so this makes sense.
And there is no distributed load anywhere else 
on the member, therefore everywhere else on the 
shear force diagram the slope is zero,
and you see the two horizontal lines.
You might be wondering what happens here.
Well at this location,
there is a concentrated load applied here, and 
that corresponds to this step change in the 
shear force diagram 
with the magnitude equal to the vertical 
component of this applied force, 346 pounds.
And also, here we have another concentrated 
load applied at point B,
and as you can see on the shear force diagram,
at this location there's another step change,
with the magnitude of 365 pounds that will return 
the curve back to zero.
Another very important relationship is that, 
if you look at the bending moment diagram, the 
slope function is the shear force function. 
Like these two straight lines,
they are both decreasing, with constant negative 
slopes, corresponding to the shear force values 
here,
-18 pounds and -365 pounds.
And this part is mostly increasing,
with slope corresponding to the
positive values on the sheer force diagram.
And you can not quite see it, but there is a 
maximum value in the bending moment diagram 
that corresponds to where shear force is zero.
And lastly, here on the member we have an 
external couple moment applied at point B,
and on the bending moment diagram, you see 
this step change 
with the magnitude of 200 pound-foot
that will return the curve back to zero.
As a summary for the graphical method, if there 
is distributed load on the member, 
then the load intensity function w corresponds to 
the slope in the shear force diagram.
Here the load intensity is pointing upwards, 
hence positive slope, and increasing shear force 
on the diagram. 
You know that load intensity is not always a 
constant.
Here is just a very simple example.
But it’s always true that w(x) equals to dV/dt.
If anywhere on the member there is a 
concentrated load,
then on the shear force diagram there is a 
corresponding step change.
And then, the shear force is the slope function, 
or the derivative function, of the bending moment 
function,
V(x) equals to dM/dx.
The slope anywhere on the bending moment diagram corresponds to the shear force value at that location. 
And lastly, the external couple moment
applied on the member corresponds to a step 
change in the bending moment diagram. 
Now let’s look at another example quickly and use the graphical method to sketch the shear force and bending moment diagrams. 
So for this example, first step,
we need to determine all of the external 
loadings, including all the support reactions, and 
complete the free body diagram of this member. 
Now the free body diagram is complete, 
the 2nd step, is to determine how many 
sections we should have on the diagrams.
We should have a different section whenever the 
loading situation changes. 
And step 3, we start with the shear force diagram. 
I work from left to right. 
First there is the 160 newtons concentrated load 
at point A, so on the diagram there is a step 
change.
And for the next 4 meters there is no change in 
the force, or the load intensity of zero, therefore 
we have this horizontal line with slope of zero.
And then there is the concentrated load of 100 
newtons downwards, so we have a step change 
again for 100 newtons.
Then for the next 3 meters there is a downwards 
distributed load with intensity of 40 newtons per 
meter, 
which corresponds to a decreasing line with 
slope of negative 40. 
For the last 3 meters, there is an upwards 
distributed load with intensity of 20 newtons per 
meter,
which corresponds to an increasing line with 
slope of positive 20, returning the curve back to 
zero.
And this completes the shear force diagram. 
Pay attention to the point where shear force is 
zero, because as you learned from calculus,
this critical point corresponds to either a local maximum or a local minimum bending moment, 
and you know we are always interested to know 
the maximum bending moment in the member.
And step 4, we can use the couple moment 
loading information on the member and the 
shear force diagram 
that we already made to complete the bending 
moment diagram.
As you can see, there are two couple moments 
on the member, which will correspond to step 
changes in the bending moment diagram. 
If you are not sure which direction is the step 
change, you just need to do a quick method of 
sections analysis 
and I will leave it to you to figure that out. 
And finally, we use the shear force diagram that 
has been sketched, 
look at the values of the shear forces because 
they correspond to the slopes of the bending 
moment diagram, take into considerations 
of the step changes, and complete the bending 
moment diagram.
Review calculus if you need to. 
From these diagrams, we can tell the absolute 
maximum shear force in this member is 160 
newton,  
and the absolute maximum bending moment in 
this member is 750 newton meters. 
These information will become very useful in the structural design and material strength analysis. 
