Unit 1 Engineering Physics

30 CHAPTER 1. PROPERTIES OF MATTER

A

B’

B

W

C

D

P Q

O

dy

y

x

dx

dθ

dθ

R

R

l-x

l

Figure 1.25:Neutral surface of a bent cantilever

Hence,
1
R

=

1

(l≠x)

dy

dx

(1.29)

The internal bending moment at point P of the neutral filament is given by the expression
(1.28):

bending moment of the beam =

YIg

R

=

YIg

(l≠x)

dy

dx

(1.30)

whereY is the Young’s Modulus of the material of the cantilever and the geometric
moment of inertiaIg=(bd^3 )/ 12 (for a beam of rectangular cross-section).

Moment of the weightWabout the point P =W(l≠x) (1.31)

At equilibrium, since the moment of the external bending couple (equation1.31) is bal-
anced by the internal bending moment (equation1.30),

W(l≠x)=

YIg

(l≠x)

dy

dx

Rearranging,

dy=

W

YIg

(l≠x)^2 dx (1.32)

The total depressionyat the end of the cantilever is obtained by integrating the expression
(1.32) for the entire length (from 0 tol) of the cantilever.

y=

⁄

dy 0 y=

⁄l

0

W

YIg

(l≠x)^2 dx=

W

YIg

⁄l

0

(l^2 +x^2 ≠ 2 lx)dx=

W

YIg

C

l^2 x+

x^3

3

≠lx^2

Dl

0

Hence,

y=

Wl^3

3 YIg

(1.33)

IfW=mgandIg=(bd^3 )/ 12 ,

y=

(mg)l^3

3 Y

(^1) (bd (^3) )
12
(^2) =
4 Mgl^3
bd^3 Y

30