A simply supported beam rests on two supports(one end pinned and one end on roller support) and is free to move horizontally. The deflection and slope of any beam(not particularly a simply supported one) primary depend on the load case it is subjected upon. If the load case varies, its deflection, slope, shear force and bending moment get changed.
This article will help you find the deflection and slope developed at any point of a simply supported beam, subjected to any load.
What is a Simply Supported Beam?
The simply supported beam is one of the most modest structures. The configuration of a simply supported beam is so simple having one hinge support at an end and roller support at the other end. With this setup the beam can only rotate horizontally, any vertical moment is restained.
This roller support also helps the beam expand or contract axially, although the free horizontal movement is prevented by the other support.
This is a determinant structure, which means that if an internal hinge is inserted or any of these supports(pin or roller) is removed, the beam can not carry the load anymore. In this case, the beam will freely move under loading.
Finding Deflection and Slope
There are multiple methods like double integration method, Macaulay's method, Conjugate beam method, Castigliano's theorem, Principle of superposition which help us find the deflection and slope of a beam.
Here we will use the double integration method, which is a simple, effective and straight forward method, that can be used to solve any type of question.
From the Euler-Bernoulli bending theory, at a point along a beam, we have:
1/R = M/EI
Where: R is the radius of curvature of the point
M is the bending moment at that point
EI is the flexural rigidity of the member, E is Young's modulus of the beam material, I is the second moment of area.
We also have dx = R dθ and so 1/R = dθ/dx. Again for small displacements, θ ≃ tan θ ≃ dy/dx and so:
1 / R = d²y / dx² = M / EI
The formula used to find slope and deflection of the beam
The bending moment at any point of the beam section can be found using the double integration formula, that is given below.
M is the Bending Moment at a particular section, which is a function of x
EI is the flexural rigidity of the member
y represents the vertical deflection of the beam and x is the lateral direction.
dy/dx represents the slope of the beam at that particular point.
Using this relation, upon integrating, the function for dy/dx(slope) can be found. Then, putting the appropriate value of x we can find the slope.
Upon integrating again and putting the value of x, the deflection(y) can be determined.
Seems confusing? Let’s work out an example, you can be an expert then.
Example
Find the maximum deflection and slope at both the ends of the beam as shown.
Our initial step would be drawing a rough deflection diagram.
Here in the deflection diagram θA and θB are the slope at point A and B respectively due to the load P. The midpoint deflection is δc and the maximum deflection due to point load P is δmax.
Step 1: Finding the reactions at support(s)
We can use the horizontal equilibrium (ΣH = 0), vertical equilibrium (ΣV = 0) and moment equation (ΣMx = 0) for the reaction calculation.
The reaction at A is Pb/L and B is Pa/L.
Step 2: Writing bending moment pattern at different sections of the beam
In this case, the bending moment from L = 0 to L = a follows one pattern and the bending moment from L = a to L = a + b follows another pattern.
Step 3: Putting the bending moment value in double integration formula
Now putting the double integration formula and then integrating twice.
Step 4: Solving the integration constants using boundary conditions
In this step we will apply the boundary conditions (at x = 0, y = 0 and at x = L, y = 0).
Step 5: Find deflection at any point
Now we can find deflection at any point(x) on the beam using the final equation given in step 4.
If we are interested in finding the deflection at half of the span, then putting x = 0.5 L,
Step 6: Finding maximum deflection
The maximum deflection will not result at the centre. At the point of maximum deflection, the slop should be zero. Equating the slope to zero, we can find the value of x.
Step 7: Finding slope
To find the rotation or slope, the equation of dy/dx(Refer step 3 for the equation) should be written again. Now putting the value of x (depending on the point where the slope is to be found) the slope value can be determined.
In this case, to find the slope of point A, x = 0 and for the slope of point B, x = L.
Beam Deflection and Slope Formula
It is always advised to remember some of the slope and deflection formula for some standard load cases. This will save you from a lot more calculation, so confusion in examinations.
Simply supported beam with a concentric load P
Maximum moment, M = PL/4
Deflection at midspan= Maximum deflection = PL^3/48EI
Slope at both ends = maximum slope = PL^2/16EI
Simply supported beam with an eccentric load P, distance 'a' from left support and 'b' from the right support
Maximum moment, M = Pab/L
Maximum deflection = Pb(L^2 - b^2)^1.5 / 9√3LEI at x=sqrt((L^2 - b^2)/3)
Slope at left support = -Fab(L+b)/6LEI
Slope at right support = Fab(L+a)/6LEI
Simply supported beam with a udl ‘w’ on the entire span
Maximum moment, M = wL^2/8
Deflection at midspan= Maximum deflection = 5wL^4/384EI
Slope at both ends = maximum slope = wL^3/24EI
Simply supported beam with end moments ‘M’ at both supports (one anticlockwise and one anticlockwise)
Deflection at midspan= Maximum deflection = ML^2/8EI
Slope at both ends = maximum slope = ML/2EI
Simply supported beam with central anticlockwise moment M
Deflection under the moment = 0
Slope at both the ends = ML/24EI
Simply supported beam with clockwise moment M at the left support
Deflection under midspan = ML^2/16EI
Maximum deflection = ML^2/9√3EI
The slope at left support = ML/3EI
The slope at right support = ML/6EI
You can download the Beam Deflection Formula Table PDF below and keep it handy.