DEFLECTION OF BEAMS

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The proper performance of machine parts and structures depends on rigidly. deflection of machine spindle or cutting tools have an adverse effect on accuracy and surface finish of the component the floors of buildings must have sufficient rigidity to carry expected loads design of machine parts and structures are based on deflection of beams the lateral displacement of abeam under the load is termed as deflection.
The determination of deflection of beam is essential since it is often required that the maximum defection of the beam should not be greater than the specified value. This chapter confined to the deflection of cantilevers and simply supported beams with point loads and uniformly distributed load.
Different equation of deflected beam:
Consider small length of the beam dx over which the bending moment may be assumed to be constant and equal.
The beam bent due to bending as and small length of the beam ‘ds’ subtend an angle dθ at the center of the curvature (o) of the beam the deflection is assumed to be so small and therefore ds’= dx;
 If(x, y) are the coordinates of a point on the beam the slope of the beam is given as   tan θ = dy/dx
Since ‘θ’ is very small
 θ = d²y/dx²
If R is the radius of curvature over a length dx.
dx = R.dθ (or) 1/R =dθ/dx
But from bending equation
1/R =M/EI
Substituting the value of (1/R) from
M/EI = dθ/dx
From equations
M/EI = d²y/dx² (or) M =EI .d²y/dx²
This is known as differential equation of the profile of the deflected beams the profile of the deflected beam is often referred as elastic curve. The product EI is called the flexural rigidity of the beam and is usually constant along the beam.
Standard cases o beam deflection:
Slope and deflection of cantilever and simply supported beam under various types constant along the beam.
Cantilever with a point load at free end:
Consider a cantilever of length L fixed at A and carrying a point load at free end B.consider section XX at a distance ‘x’from fixed end A.
 BM at section XX,Mx = -W(L-x)
BM at any section is given as
M =EI d²y/dx²
Equating EI d²y/dx² = - W (L-x)
Integrating the above equation
EI dy/dx = -W (Lx-x²/2)+C1
At fixed end A the slope is zero and x=0 so that c1=0
EI dy/dx = -W (Lx-x²/2)….slope equation
Again integrating slope equation
EI.y =-W (Lx²/2-x²/6) +C2
But when x=0, y=0 C2=0
EI.y =-W (Lx²/2-x²/6)….deflection equation
When x=L at free end B the slope and deflection are maximum.
(dy/dx) max = (dy/dx) B =WL²/2EI
And Ymax =YB= WL³/3EI.

Cantilever with a point load at a distance ‘a’ from fixed end:
Consider a cantilever of length L and carrying a point load W at a distance ‘a’ from end A as there is no bending moment over the portion CB and hence the portion CB will be straight and portion AC will acts as a cantilever length a
          Slope of AC, (dy/dx)c =Wa²/2EI
And deflection at C, yc = Wa³/3EI
Deflection at free end, B, yB = Wa³/3EI + Wa³/3EI (L-a)
The slope at any point between CB is constant and is equal to (dy/dx)c;

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