We were discussing meaning and importance of Shear
force and bending moment and sign conventions for shear force and
bending moment in our recent posts. We have also discussed the basics of shear force and bending moment
diagrams during
our previous posts.

Today we will see here the concept to draw shear
force and bending moment diagrams for a simply supported beam with a point load
acting at midpoint of the loaded beam with the help of this post.

Let us see the following figure, we have one beam AB of length L and beam is resting or supported freely on the supports at its both ends. And therefore, displayed beam indicates the simply supported beam AB and length L.

Let us consider that one point load W is
acting over the beam at midpoint as displayed in following figure. Load W is
acting at point C which is the midpoint of the loaded beam AB.

First of all we will remind the
important points for drawing shear force and bending moment diagram. Now always
remember that we will have to first determine the reaction forces at each
support.

Let us consider that R

_{A}and R_{B}are the reaction forces at end support A and end support B respectively and we will use concept of equilibrium in order to determine the value of these reaction forces.
ƩF

_{X}=0, ƩF_{Y}=0, ƩM =0,
R

_{A}+ R_{B}–W =0
R

_{A}+ R_{B}= W
ƩM =0

R

_{A}* L – W*L/2 = 0
R

_{A}=W/2
R

_{B}= W/2
Now we have values of reaction forces at
end A and end B and it is W/2. Let us determine now the value of shear force
and bending moment at all critical points.

Let us consider one section XX at a
distance x from the end A between A and C as displayed in following figure. Let
us assume that F

_{X}is shear force at section XX and bending moment is M_{X}at section XX.###
**Shear force diagram**

As we have assumed here section XX at a
distance x from end A between A and C and therefore loaded beam AB will be
divided in two portion and let us consider the left portion of the beam. Shear
force at section XX will be equivalent to the resultant of forces acting on
beam to the left side of the section.

F

_{X}= W/2Shear force will be positive here and we can refer the post sign conventions for shear force and bending moment in order to understand the sign of shear force which is determined here.

As we have already discussed that shear force will be constant between two vertical loads or
we can also say that if there is no load between two points then shear force
will be constant and will be represented by a horizontal line.

Therefore we can say here that shear
force between A and C will be constant and it will have positive value of W/2.

Now suppose if we have considered section XX between C and B and at a distance x from the end A. Again we will
have to remind the basics of shear force and bending moment diagrams and we
will be able to calculate shear force at section XX.

F

_{X}= W/2- W
F

_{X}= - W/2
Therefore, we can say here
that shear force between C and B will be constant and it will have negative
value of W/2. We will have to note it here that shear force at point C is
changing from +W/2 to –W/2.

Now we have information here in respect
of shear force at all critical points i.e.

Shear force at point A, F

_{A}= +W/2
Shear force at point B, F

_{B }= -W/2
Shear force at point C, F

_{C }= Shear force changes from +W/2 to –W/2
We can now draw here shear force diagram
and it is displayed here in following figure.

###
**Bending moment diagram**

As we have considered above one section
XX at a distance x from the end A between A and C and hence bending moment at
section XX i.e. M

_{X}will be determined as mentioned here
M

_{X }= R_{A}*x = W*x/2
M

_{X }= W*x/2
Bending moment at point A, x=0

M

_{A }= 0
Bending moment at point C, x=L/2

M

_{C }= WL/4
Now suppose if we have considered
section XX between C and B and at a distance x from the end A. Again we will
have to remind the basics of shear force and bending moment diagrams and we
will be able to calculate bending moment equation for the section XX and it
could be written as mentioned here.

M

_{X }= R_{A}*x –W (x-L/2)
M

_{X }= W*x/2 –W (x-L/2)
M

_{X }= W*x/2 –W*x + WL/2
M

_{X }= - W*x/2 + WL/2
Bending moment at point C, x=L/2

M

_{B }= - W*L/4 + WL/2
M

_{B }= + WL/4
Bending moment at point B, x=L

M

_{B }= R_{A}*L –W (L-L/2)
M

_{B }= WL/2 –W (L/2)
M

_{B }= 0
Therefore we have bending moment for all
critical points and we have also information about the equation followed by
bending moment i.e. bending moment equation stated above is following the
linear equation and therefore we can draw here the BMD i.e. bending moment
diagram and we have drawn it as displayed in above figure.

Do you have any suggestions? Please write in comment
box

###
**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

We will see another important topic i.e. shear force and bending moment diagram for simply supported beam with an eccentric point load in the
category of strength of material.