In our previous topics, we have seen some important
concepts such as deflection and slope of a simply supported beam with point load,

*Deflection of beams and its various terms*,*Concepts of direct and bending stresses*,*shear stress distribution diagram*and*basic concept of shear force and bending moment*in our previous posts.
Now we will start here, in this post, another
important topic i.e. deflection and slope of a simply supported beam carrying
uniformly distributed load throughout length of the beam.

We have already seen terminologies
and various terms used
in deflection of beam with the help of recent posts and now we will be
interested here to calculate the deflection and slope of a simply supported
beam carrying uniformly distributed load throughout length of the beam with the
help of this post.

###
**Basic
concepts**

There are basically three important methods by which we can easily determine the deflection and slope at any section of a loaded beam.

Double integration method

Moment area method

Macaulay’s method

Double
integration method and Moment area method are basically used to determine
deflection and slope at any section of a loaded beam when beam will be loaded
with a single load.

While Macaulay’s method is basically used to
determine deflection and slope at any section of a loaded beam when beam will
be loaded with multiple loads.

We will use double integration method here to
determine the deflection and slope of a simply supported beam carrying uniformly distributed load throughout length of the beam.

Differential
equation for elastic curve of a beam will be used in double integration
method to determine the deflection and slope of the loaded beam and hence we
must have to recall here the differential
equation for elastic curve of a beam.

*Differential equation for elastic curve of a beam*

After first integration of differential equation, we
will have value of slope i.e. dy/dx. Similarly after second integration of
differential equation, we will have value of deflection i.e. y.

Let us come to the main subject i.e. determination
of deflection and slope of a simply supported beam carrying uniformly
distribution load throughout length of the beam.

Let us consider a beam AB of length L is simply supported
at A and B and loaded with uniformly distributed load as displayed in following
figure.

We have following information from above figure,

w = Rate of loading in N/m

R

_{A}= Reaction force at support A = w L/2
R

_{B}= Reaction force at support B = w L/2
θ

_{A}= Slope at support A
θ

_{B}= Slope at support B###
*Boundary
condition*

*Boundary condition*

We must be aware with the boundary conditions applicable in such a problem where beam will be simply supported and loaded with uniformly distributed load. We have following boundary condition as mentioned here.

Deflection
at end supports i.e. at support A and at support B will be zero, while slope
will be maximum.

Deflection
will be maximum at the center of the loaded beam

Slope
will be zero at the center of the loaded beam

Let us consider one section XX at a distance x from
end support A, let us calculate the bending moment about this section.

We have taken the concept of sign convention to
provide the suitable sign for above calculated bending moment about section XX.
For more detailed information about the sign convention used for bending
moment, we request you to please find the post “Sign
conventions for bending moment and shear force”.

Let us recall the differential
equation of elastic curve of a beam and we can write the expression for
bending moment at any section of beam as mentioned here in following figure.

Let us consider the bending moment determined
earlier about the section XX and bending moment expression at any section of
beam. We will have following equation as displayed here in following figure.

We will now integrate this equation and also we will
apply the boundary conditions in order to secure the expressions for slope as
well as deflection at a section of the beam and we can write the equations for
slope and deflection for loaded beam as displayed here.

Where, C

_{1}and C_{2}are the constant of integration and we can secure the value of these constant C_{1}and C_{2}by considering and applying the boundary condition.
Let us use the boundary condition
as we have seen above. At x = 0, deflection will be zero and similarly at x = L
deflection will also be zero.

After applying the boundary conditions in above equation of deflection of beam, we will have following values of constant C

_{1}and C

_{2}as mentioned here.

C

_{1 }= - wL^{3}/24
C

_{2 }= 0
Let us insert the values of C

_{1}and C_{2}in slope equation and in deflection equation too and we will have the final equation of slope and also equation of deflection at any section of the loaded beam. We can see the slope equation and deflection equation in following figure.###
**Slope at the end supports**

*At x = 0,*

*θ*_{A}= Slope at support A
Let us use the slope equation and insert the value
of x = 0, we will have value of slope at support A i.e. θ

_{A}*θ*

_{A}= -w.L^{3}/24EI

**θ**_{A}

**= -W.L**^{2}/24EI

*At x = L,*

*θ*_{B}= Slope at support B
Let us use the slope equation and insert the value
of x = L, we will have value of slope at support B i.e. θ

_{B}*θ*

_{B}= -w.L^{3}/24EI

*θ*_{B}

*= -W.L*^{2}/24EI
Negative sign represents that tangent at support A
makes an angle with beam axis AB in anti-clockwise direction.

###
**Maximum
deflection**

As we have seen in boundary conditions that in case
of simply supported beam loaded with uniformly distributed load, deflection will be
maximum at the center of the loaded beam.

At
x = L/2, y

_{M}= Maximum deflection
Let us use the deflection equation and insert the
value of x = L/2, we will have value of deflection at centre of the loaded beam.

*y*_{M}= - (5/384) [w.L^{4}/EI]

*y*_{M}= - (5/384) [W.L^{3}/EI]
W = w. L = Total load on beam due to uniformly
distributed load

Negative sign represents here that deflection in the
loaded beam will be in downward direction.

We will see deflection and slope of a cantilever
beam in our next post.

Please comment your feedback and suggestions in
comment box provided at the end of this post.

###
**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google