Unit – 3
Moment of Inertia
Shape  Area(a)  Moment of Inertia  Radius of gyration (r)  Polar moment of Inertia (J) 
Rectangle 

 
Triangle 


 
Circle  
Semicircle  
Hollow circle  
Hollow rectangle  
Quarter circle 
Assumption: The following assumptions are undertaken in order to derive a differential equation of elastic curve for the loaded beam
1. Stress is proportional to strain i.e. hooks law applies. Thus, the equation is valid only for beams that are not stressed beyond the elastic limit.
2. The curvature is always small.
3. Any deflection resulting from the shear deformation of the material or shear stresses is neglected.
It can be shown that the deflections due to shear deformations are usually small and hence can be ignored.
Fig.
Consider a beam AB which is initially straight and horizontal when unloaded. If under the action of loads the beam deflect to a position A’B’ under load or infact we say that the axis of the beam bends to a shape A’B’. It is customary to call A’B’ the curved axis of the beam as the elastic line or deflection curve.
In the case of a beam bent by transverse loads acting in a plane of symmetry, the bending moment M varies along the length of the beam and we represent the variation of bending moment in B.M diagram. Further, it is assumed that the simple bending theory equation holds good.
If we look at the elastic line or the deflection curve, this is obvious that the curvature at every point is different; hence the slope is different at different points.
To express the deflected shape of the beam in rectangular coordinates let us take two axes x and y, xaxis coincide with the original straight axis of the beam and the y  axis shows the deflection.
Further, let us consider an element ds of the deflected beam. At the ends of this element let us construct the normal which intersect at point O denoting the angle between these two normal.
But for the deflected shape of the beam the slope is at any point C is defined,
Tan i (1) or Assuming tani
Further
Ds = Rdi
However,
Ds = dx [usually for small curvature]
Hence
ds
Or
Or
Substituting the value of , on e get
From the simple bending theory
So the basic differential equation governing the deflection of beams is
This is the differential equation of the elastic line for a beam subjected to bending in the plane of symmetry. Its solution y = f(x) defines the shape of the elastic line or the deflection curve as it is frequently called.
The governing differential equation is defined as
or
On integrating one get,
this equation gives the slope of theloaded beam.
Integrate once again to get the deflection.
Where A and B are constants of integration to be evaluated from the known conditions of slope and deflections for the particular value of x.
Illustrative examples: let us consider few illustrative examples to have a familiarity with the direct integration method
Case 1: Cantilever Beam with Concentrated Load at the end:  A cantilever beam is subjected to a concentrated load W at the free end, it is required to determine the deflection of the beam.
In order to solve this problem, consider any Xsection XX located at a distance x from the left end or the reference, and write down the expressions for the shear force and the bending moment
Therefore
The governing equation
Substituting the value of interms of then integrating the equation one get
Integrating once more,
The constants A and B are required to be found out by utilizing the boundary conditions as defined below
i.e at x= L ; y= 0 ——————– (1)
At x = L ; dy/dx = 0 ——————– (2)
Utilizing the second condition, the value of constant A is obtained as
While employing the first condition yields
Substituting the values of and we get
The slope as well as the deflection would be maximum at the free end hence putting we get,
Case 2: A Cantilever with Uniformly distributed Loads: In this case the cantilever beam is subjected to U.d.l with rate of intensity varying w / length. The same procedure can also be adopted in this case.
Boundary conditions relevant to the problem are as follows:
1. At x = L; y = 0
2. At x= L; dy/dx = 0
The second boundary conditions yields
Whereas the first boundary conditions yields
Thus
So will be at
Case 3: Simply Supported beam with uniformly distributed Loads: In this case a simply supported beam is subjected to a uniformly distributed load whose rate of intensity varies as w / length.
In order to write down the expression for bending moment consider any crosssection at distance of x metre from left end support.
The differential equation which gives the elastic curve for the deflected beam is
Integrating, once more one gets
Boundary conditions which are relevant in this case are that the deflection at each support must be zero.
i.e. at x = 0; y = 0 : at x = l; y = 0
Let us apply these two boundary conditions on equation (1) because the boundary conditions are on y, This yields B = 0.
So the equation which gives the deflection curve is
In this case the maximum deflection will occur at the center of the beam where x = L/2 [ i.e. at the position where the load is being applied]. So, if we substitute the value of x = L/2
Then
Key takeaway:
(i) The value of the slope at the position where the deflection is maximum would be zero.
(ii) The value of maximum deflection would be at the center i.e. at x = L/2.
The final equation which is governs
Deflection for Common Loadings:
1. Concentrated load at the free end of cantilever beam (origin at A):
Fig.
 Maximum Moment, M =−PL
 Slope at end: θ= PL2/2EI
 Maximum deflection: δ=PL3 /3EI
 Deflection Equation (y is positive downward): EIy=(Px2)(3L−x)/6
2. Concentrated load at any point on the span of cantilever beam
Fig.
 Maximum Moment: M= wa
 Slope at end: θ=wa2/2EI
 Maximum deflection: δ = wa3(3L−a)/6EI
 Deflection Equation (y is positive downward),
 EIy=Px2(3a−x)/6 for 0 < x <a
 EIy=Pa2(3x−a)/6 for a < x <L
3. Uniformly distributed load over the entire length of cantilever beam
Fig.
 Maximum Moment: M=−wL2/2
 Slope at end: θ = wL3/6EI
 Maximum deflection: δ=wL4/8EI
 Deflection Equation (y is positive downward): EIy=wx2(6L2−4Lx+x2)/120L
4. Triangular load, full at the fixed end and zero at the free end
Fig.
 Maximum Moment: M=−wL2/6
 Slope at end: θ= wL3/24EI
 Maximum deflection, δ=wL4/30EI
 Deflection Equation (y is positive downward): EIy=wx2(10L3−10L2x+5Lx2−x3)/120L
5. Moment load at the free end of cantilever beam
Fig.
 Maximum Moment: M=−M
 Slope at end: θ=ML/EI
 Maximum deflection: δ=ML2/2EI
 Deflection Equation (y is positive downward): EIy=Mx2/2
6. Concentrated load at the mid span of simple beam
Fig.
 Maximum Moment: M=PL/4
 Slope at end: θA=θB = WL2/16EI
 Maximum deflection: δ=PL3/48EI
 Deflection Equation (y is positive downward): EIy=Px{(3/4)L2−x2)}/12 for 0<x<L/2
7. Uniformly distributed load over the entire span of simple beam
Fig 15
 Maximum Moment: M=wL2/8
 Slope at end: θL=θR=wL3/24EI
 Maximum deflection: δ = 5wL4/384EI
 Deflection Equation (y is positive downward): EIy=wx(L3−2Lx2+x3)/24
8. Triangle load with zero at one support and full at the other support of simple beam
Fig 16
 Maximum Moment: M=woL2/9√3
 Slope at end,
 θL= 7wL3/360EI
 θR= 8wL3/360EI
 Maximum deflection: δ=2.5wL4/384EI at x=0.519L
 Deflection Equation (y is positive downward), EIy=wx(7L4−10L2x+3x)/360L
9. Triangular load with zero at each support and full at the midspan of simple beam
 Maximum Moment: M=wL2/12
 Slope at end, θL=θR=5wL3/192EI
 Maximum deflection: δ=wL4/120EI
 Deflection Equation (y is positive downward): EIy=wox(25L4−40L2x2+16x4)/960L for 0<x<L/2
Cantilever Beams
Cantilever, End Load  Deflection: Slope: Shear: Moment:
 
Cantilever, Intermediate Load  Deflection: Slope: Shear: Moment:
 
Cantilever, uniform Distributed Load  Deflection: Slope: Shear: Moment:
 
Cantilever, Triangular Distributed Load  Deflection: Slope: Shear: Moment:
 
Cantilever, End Moment  Deflection: Shear: Moment:

Simply Supported Beams
Simply supported, Intermediate Load  Deflection: For :Slope: Shear: Moment:
 
Simply supported, Center Load  Deflection: Slope: Shear: Moment:
 
Simply Supported, 2 Loads at Equal Distances from Supports  Deflection: Slope: Shear: Moment:
 
Simply Supported, Moment at Each support  Deflection: Slope: Shear: Moment:  
Simply supported, uniform distributed load  Deflection: Slope: Shear: Moment:  
Simply Supported, Moment at One Support  Deflection: Slope: @ Shear: Moment 
FixedFixed Beams
Fixedfixed, Center Load  Deflection: Shear Moment:
 
FixedFixed Uniform Distributed Load  Deflection: Shear: Moment:

Betti’s theorem is also known as Maxwell – Betti reciprocal work theorems.
It says that the deflection at C due to a unit load at W at point B is the same as the deflection at B if same load W were applied at C.
.
Key takeaways:
Reciprocal theorem: The reciprocal theorem states that the work done by forces acting through displacement of the second system is the same as the work done by the second system of forces acting through the displacements of the first system.
References:
 U.S. Department of Energy, Material Science. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
 Elements of Strength of Materials (Timoshenko)
 Strength of Materials (S. Ramamrutham)
 A Textbook Of Strength Of Materials (R K Bansal)
 Strength of Materials (Dr B.C.Punmia)