Can anyone derive this formula of shagging in a bridge:-

δ=Wl³/4Ybd³

Here Y=young's modulus

l=length

b=breadth of bridge

d=depth

others I think all know or who will attempt this will know!!!

 Hi,

This derivation is a long one. First of all let’s start with the meaning of bending of beam

The neutral filament CD is at a distance ‘r’  from the upper filament.R is the radius of curvature of beam and Ф is the angle subtended at the centre of curvature.

See the following fig,

Stress = change in length / original length

Y(young’s modulus) = stress/strain

Hence stress = Y(r/R)

Let ‘a’ be cross section of the beam then,

Stress = force/area

F = Y (r/R )a

Moment of force about CD = Y(r/R)ar

Let I_g be the bending moment of inertia given as ∑ar² = Ak²(A is the total area)

Hence moment of force = (Y/R)I_g……………..eqn1

Now coming to the 2^nd part of the question,

A beam fixed at 1 end and loaded at the other is called a Cantelever. See the fig below. 

Let EF be the neutral axis. When weight W is loaded , EF^/ becomes the new neutral axis.

Consider a section P at a distance x from fixed end E.

Moment of couple = W*PF^/ = W*(l-x)

Cantilever is in equilibrium hence,

W*(l-x) = (Y/R)I_g

Now I_g = Ak² where k is radius of gyration.

W*(l-x) = (Y/R)Ak²……………………….eqn 2

Consider a section Q very near to P at a distance dx, radius of curvature of Q will be same as P.

PQ = RdӨ = dx

R = dx/dӨ

Substituting in equation2, we get,

W*(l-x) = Yak²dӨ/dx

dӨ = W*(l-x)dx/Yak²……………….eqn 3

tangents meet line FF^/  at C and D

hence the depression dy of Q below P is

dy = (1-x)dӨ………………………….eqn 4

substituting value of dӨ from eqn 3 in the above equation we get,

dy = W(1-x)²dx/(Yak²)…………………………….eqn 5

now integrate this expression from 0 to l to get y,

y = (1/Yak²)(Wl³/3)……………………………….eqn 6

again I_g = ak² therefore eqn 6 becomes,

y = Wl³/(3YI_g)…………………………………..eqn 7

now we come to a special case of this equation which is non uniform bending

see fig below 

W is applied at centre of beam which is supported at ends. Bending is non uniform.

The beam is analogous to two separate cantilevers of length l/2 having weight W/2 at their ends.

Hence in the formula for depression derived before, substitute W/2 in place of weight and l/2 in place of length . from eqn 7

y = (W/2)(l/2)³/(3YI_g)

y = Wl³/(48YI_g)……………………………………..eqn 8

In case of a bridge, it is a rectangular bar, with b as breadth and d as depth hence geometrical moment of inertia will be

I_g = bd³/12

Hence depression produced is putting above value in eqn 8

y = Wl³/(4Ybd³)……………………………..eqn 9

which is the required equation.