$\frac{fb}{y} = \frac{M}{I} = \frac{E}{R}$

Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 5M

Year: Dec 2013

Consider a portion of a beam, as shown in the diagram. Let the initial length of the portion be ‘dx’. After a sagging moment is applied to this portion, the sides WX change to W’X’ and YZ to Y’Z’, such that they intersect at a point O, which is the centre of curvature for the bending and subtends an angle θ at the centre. Let the Neutral layer ‘NN’ be located at a distance of R from the point O.

Consider a layer CD at a distance y from the Neutral Layer NN. Since this layer is a part of the unwarped portion of the beam, it also has a length of ‘dx’.

Original Length of CD = dx = R θ

New Length of EF = C’D’ = (R + y) θ

Change in the Length of CD = (R + y) θ - Rθ = yθ

Strain in the layer CD = $\frac{change \ \ in \ \ the \ \ length \ \ CD}{original \ \ length \ \ of \ \ CD} = \frac{yθ}{Rθ} = \frac{y}{R}$

Stress in the layer CD, f=E x Strain in CD

$$f = E × \frac{y}{R}$$

$\frac{f}{y} = \frac{E}{R}.............(i)$

Now, to calculate the force on the layer of CD. Let ‘dA’ be the C/S area of the layer.

Force in the layer = Stress in the layer × Area of the layer

$Force \ \ in \ \ the \ \ layer = f × dA = \frac{E}{R}ydA$

Moment of this Force about the Neutral Axis = $(\frac{E}{R}ydA)y$

$= \frac{E}{R}y^2dA.......(ii)$

Total moments in different layers = $\int\frac{E}{R}y^2dA$

The total moment $M = \frac{E}{R}\int y^2dA$

The term $\int y^2dA$ represents the moment of inertia I of the section about the Neutral Axis.

$\frac{M}{I} = \frac{E}{R}........(iii)$

From the equations (I) and (III), we have,

$\frac{f}{y} = \frac{M}{I} = \frac{E}{R}$

Since we’re referring to Bending Stresses in this case, we have,

$\frac{f_b}{y} = \frac{M}{I} = \frac{E}{R}$