PH8151 Engineering Physics Chapter 1

38 CHAPTER 1. PROPERTIES OF MATTER

Load

(kg)

Travelling Microscope

Readings

Depression

for load

M= 0.05 kg

(m)

Load

increasing(m)

Load

decreasing(m)

Mean

(m)

L+ 0.05

L

L+ 0.10

L+ 0.15

L+ 0.20

L+ 0.25

Mean Depression (y) =

Figure 1.34: Non-Uniform bending - Measurement of depression at the mid point of the
beam

Self Learning Activity : Young’s Modulus Non-uniform Bending

Using the non-unifrom bending simulator, determine Young’s

moduli of the available materials.

URL: http://vlab.amrita.edu/?sub=1&brch=280&sim=

1509&cnt=1

Worked out Example 1.11.2

A 1 m bar with square cross-section of

side 5 mm is supported horizontally at its

ends and is loaded with 100 g mass at the

middle. The mid point is depressed by

1.96 mm due to the load. Calculate the

Young’s Modulus of the material of the

load.

Solution:

Since the beam is supported at its ends

and loaded at the middle, its bending is

nonuniform. Length between supports (l)

= 1 m. Square cross-section of side 5 mm

impliesb=d=0. 005 m. Depression at

the mid point (y) = 1.96◊ 10 ≠^3 m. Mass

that produces the depression (M) = 0.1

kg. Then, Young’s Modulus is given by,

Y =

Mgl^3

4 bd^3 y

=

(0. 1 kg)(9.8m/s^2 )(1m)^3

4(5◊ 10 ≠^3 m)(5◊ 10 ≠^3 m)^3 (1. 96 ◊ 10 ≠^3 m)

=20.204 GPa

1.12 Stress due to bending in beams

There are no longitudinal stresses or strains at the neutral axisNNÕ (Figure1.35)ofa
bent beam. If radius of curvature of the neutral surface =R, then strain in layer EF at
a distancexfrom the neutral axisNNÕ=x/R. Stress due to bending at layer EF is

‡(x) = (YoungÕs Modulus)◊(Strain in layer EF) =Y◊

(^3) x
R
4
38