Unit 1 Engineering Physics

32 CHAPTER 1. PROPERTIES OF MATTER

Worked out Example 1.10.1

Calculate the Young’s modulus of the following cantilever. The length of the can-

tilever is 1 m which is suspended with a load of 150 gm at the free end. The depression

is found to be 4 cm. The thickness of the beam is 5 mm and breadth of the beam is

3cm[ 6 ].

Solution:

For a rectangular cantilever, the Young’s Modulus is given by Equation (1.34). Sub-

stituting the relevant quantities, we have:

Y=

4 Mgl^3

bd^3 y

=

4(150◊ 10 ≠^3 kg)(9.8m/s^2 )(1m)^3

(3◊ 10 ≠^2 m)(5◊ 10 ≠^3 m)^3 (4◊ 10 ≠^2 m)

=39.2GPa

1.11 Uniform Bending and Nonuniform Bending..................

It can be shown thatthe rate of change of the bending moment with respect to distance
along the beam axis is equal to the shear force[ 7 ]. Therefore, if the shear force is zero in
a region of the beam, then the bending moment is constant in that same region. Pure
bending or uniform bending refers to flexure of a beam under a constant bending moment,
which means that the shear force is zero. Nonuniform bending refers to flexure of beams
in the presence of shear forces, which means that the bending moment changes as we
move along the axis of the beam. For a uniformly bent beam, the radius of curvatureRof
the neutral filament (in the plane of bending) is constant, and for a beam in nonuniform
bending,Rchanges with distance along the axis of the beam.

1.11.1 Uniform Bending: Theory and Experiment

Consider a beam (or bar) AB arranged horizontally on two knife-edges C and D symmet-
rically so that AC = BD =a,as shown in Figure1.28. The beam is loaded with equal
weights W and W at the ends A and B such that any given filament of the bent beam
forms an arc of a single circle. This bending can be shown to be of uniform type. The
forces acting on the neutral filament of the beam are shown in Figure1.29. The reactions
on the knife-edges at C and D are equal to W and W acting vertically upwards. The
moment of external bending couple on the part AC of the beam is=W◊AC=W◊a

Internal bending moment =

YIg

R

(1.35)

whereY - Youngs’ modulus of the material of the bar
Ig- Geometrical moment of inertia of the cross-section of beam
R- Radius of curvature of the neutral filament of the bar.
In the equilibrium position, moment of the external bending couple = internal bending
moment

Wa=

YIg

R

(1.36)

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