Unit 1 Engineering Physics

34 CHAPTER 1. PROPERTIES OF MATTER

C D

E

F

O

l /2 l /2

R R

R

(R-y

)

y

EF = y, OF = OC = OD = R, OE = OE - EF = R - y, CE = DE = l /2

A B

Figure 1.30: Uniform bending - circular geometry of neutral filament

Rearranging,

Y=

Wl^2 a

8 Igy

(1.38)

If the beam is of rectangular cross-section, then

Ig=

bd^3

12

(1.39)

wherebis the breath anddis the depth (or thickness) of the beam. If the load is due
the a suspended object of massM, the corresponding weight isW=Mg, wheregis the
acceleration due to gravity. Then the expression (1.38) can be written as:

Y =

3

2

M gal^2

bd^3 y

(1.40)

Experimental Procedure

Support the given beam symmetrically on two knife edges with lengthlbetween knife
edges and and suspend two equal loadsLat equal distancesafrom the knife edges as
shown in Figure1.28. A pin is fixed vertically at the mid point of the beam. Focus a
travelling microscope to the tip of the pin such that the horizontal cross-wire coincides
with the image of pin-tip. Note down the reading on the vertical scale of the microscope.
Increase the load on both sides by adding 50 g each. Raise the microscope such that the
horizontal cross-wire now coincides with the new position of pin tip image and note down
the reading from vertical scale. This is repeated by increasing the loads on both sides in
steps of 50 g tillL+250 g and also for decreasing loads. The readings are tabulated as
shown (Figure1.31)and mean elevation (y) for M = 0.05 kg is calculated. Using a vernier

34