We were
discussing shear force and bending
moment diagrams for a simply supported beam with a point load acting at
midpoint of the loaded beam and shear force and bending moment diagrams for a simply supported beam with uniform distributed load during our previous posts.

We have
also seen shear force and bending
moment diagrams for a simply supported beam with an eccentric point load and shear force and bending moment diagrams for a simply supported beam with uniform varying load in our
recent post.

Today we
will see here the concept to draw shear force and bending moment diagrams for an
overhanging beam with the help of this post.

Let us see the following figure, we have one beam AC of length L and beam is resting or
supported freely on the supports A and B as displayed in following figure. As
we can see from below figure that beam is extended beyond the support B and
such type of beam will be termed as overhanging beam.

We have
already discussed about the various types of beam in strength of materials and
we have studied there that overhanging beam is basically
defined as a beam where end portion of beam is extended beyond the supports. In
such types of beams, one end or also both ends of the beam might be extended
beyond the supports.

Let us consider that distance between
two supports is a and length of overhanging portion is b and therefore L= a + b

Let us
consider that overhanging beam AC is loaded with uniformly distributed
load with rate of loading w per unit length over the entire length as displayed
in following figure.

First of
all we will remind here 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 support B respectively and we will use the concept of equilibrium in order to determine the value of these reaction forces.
Before
going ahead, we must have to understand here the uniformly distributed loading
and we can refer the post different types of load
acting on the beam in
order to understand the uniformly distributed load.

During determination of the total load,
total uniformly distributed load will be converted in to point load by
multiplying the rate of loading i.e. w (N/m) with the span of load distribution
i.e. L and will be acting over the midpoint of the length of the uniformly load
distribution.

ƩF

_{X}=0, ƩF_{Y}=0, ƩM =0,
R

_{A}+ R_{B}– w. L=0
R

_{A}+ R_{B}= w. L
ƩM

_{A}=0
R

_{B}* a – w. L*(L/2) = 0
R

_{B}* a = w. L^{2}/2

*R*_{B}= w L^{2}/2a
R

_{A}= w. L - w L^{2}/2a

*R*_{A}= w. L (1-L/2a)
Now we
have values of reaction forces at support A and support B and it is as
mentioned above. Let us determine now the value of shear force and bending
moment at all critical points.

Let us
consider one section XX between A and B at a distance x from end A. Now we will
have two portion of the beam AB i.e. left portion and right portion. Let us
deal with the left portion here. 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 between A and B at a distance x from end A 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}= R_{A}– w. x
Force
acting to the left of the section and in upward direction will be considered as
positive and force acting to the left of the section but in downward direction
will considered as negative 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.

Now we
have information here in respect of shear force equation which is following
linear equation and therefore we will determine shear force at all critical
points i.e.

**Shear force at point A, x= 0**

F

_{A}= R_{A }
F

_{A}= w. L (1-L/2a) = w. L (a-b)/2a**Shear force at point B, x= a**

F

_{B }= R_{A}– w. a
F

_{B }= w. L (a-b)/2a - w. a
F

_{B }= w. L/2- w. L. b/2a – w. a
After
calculation, we will have negative value for shear force at support B and
therefore we can say that there will be a point between A and B where shear
force will be zero and we can secure the value of that point by following
equation.

Let us think that the point is D, where shear force will be zero or F

_{D}=0
R

_{A}– w. x = 0
x = R

_{A}/ w

*Shear force will be zero at a distance, x = R*_{A}/ w from the end A.
Let us
consider section XX between B and C at a distance x from the end A. let us
determine the value of shear force at support B and C.

Let us
also determine the value of x for zero value of shear force, we can determine
as mentioned here

F

_{X}= R_{A}– w. a + R_{B}– w(x-a)**Shear force at point B, x= a**

F

_{B}= R_{A}– w. a + R_{B}= w. L – w. a
F

_{B}= w. (L – a)
This value
of F

_{B}will be positive as L is greater than a**Shear force at point C, x= L**

F

_{C}= R_{A}– w. a + R_{B}– w (L-a)
F

_{C}= w. L – w. a – w (L-a)
F

_{C}= 0
Now we
have information about shear force at all critical points as determined above
and we can draw shear force diagram as displayed here in following figure.

###
**Bending moment diagram**

As we have
assumed here section XX between A and B at a distance x from end A, therefore
let us determine here the equation for bending moment at any section between A
and B will be as mentioned here.

M

_{X}will be determined as mentioned here.
M

_{X }= R_{A}. x – w.x^{2}/2

*Bending moment at end A, x = 0*
M

_{A }= 0
Bending
moment at point D will be maximum as shear force is zero at this point and we
can determine the value of maximum bending moment by putting the value of x = R

_{A}/w in above equation of bending moment.

*Bending moment at point B, x= a*
M

_{B }= R_{A}. a – w.a^{2}/2
M

_{B }= w. L (a-b)/2– w.a^{2}/2
M

_{B }= w. (a + b) (a-b)/2– w.a^{2}/2
M

_{B }= w. (a^{2}- b^{2}) /2– w.a^{2}/2
M

_{B }= - w. b^{2}/2
Bending moment
at point B will be negative.

As we can
see here from above equation that bending moment at any section between A and B
is following parabolic law and hence we can draw the bending moment diagram as
displayed in above figure.

Let us
consider section XX between B and C at a distance x from the end A. let us
determine the equation for bending moment at any section between B and C.

M

_{X }= R_{A}. x – w.x^{2}/2 + R_{B }(x-a)

*Bending moment at point C, x= L*
M

_{C }= w. L^{2}(a-b)/2a – w.L^{2}/2 + wL^{2}_{ }(L-a)/2a
M

_{C }= (w. L^{2}/2a) [a-b + L-a] – w.L^{2}/2
M

_{C }= (w. L^{2}/2a) a – w.L^{2}/2
M

_{C }= w. L^{2}/2 – w.L^{2}/2
M

_{C }= 0
Now we
have values for bending moment at each critical point and we can see here from
above equation that bending moment at any section between B and C is following
parabolic law and hence we can draw the bending moment diagram as displayed in
above figure.

###
**Point of contraflexure**

Point of
contraflexure is basically defined as the point at which bending moment will be
zero after changing it sign from negative to positive or from positive to
negative.

Let us try
to secure the point of contraflexure and we can easily say that it will be
between A and B as bending moment is changing its sign from positive to
negative between A and B and hence point of contraflexure will be located somewhere
between A and B.

0

_{=}R_{A}. x – w.x^{2}/2
R

_{A}= w. x/2
x/2 = R

_{A}/ w
x = 2 (R

_{A}/ w)
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. in the category of strength of material.

## No comments:

## Post a comment