1.10 Cantilever: Theory and Experiment......................
The cross-sectional area of the beam can be thought of as made up of a large number
of small elemental rectangles each of area”A. Therefore, the total moment of the forces
acting on the cross-section of the beam is the sum of the moments over all the elemental
rectangles that constitute the cross-section. This is the bending momentBMof the beam.
)BM=
ÿY
R(”Ax^2 )Or,BM=Y
R
(^1) ÿ
”Ax^2
2
(1.27)
The term(
q
”Ax^2 )=Igis a characteristic of the geometric shape of the cross-section
of the beam. Therefore,Igis called the geometric moment of inertia of the beam cross-
section.
(For a beam of circular cross-section,Ig=(fir^4 )/ 4 , whereris the beam radius. For
a beam of rectangular cross-section,Ig=(bd^3 )/ 12 , wherebis the breadth anddis the
thickness of the beam.)
Hence
BM=
YIg
R
(1.28)
At equilibrium, the moment of the external bending forces is balanced by the beam’s
bending moment (i.e., the moment of the internal restoring couple).
1.10 Cantilever: Theory and Experiment
A cantilever is a beam fixed horizontally at one end and loaded at the free end. Consider
a cantilever of lengthlloaded with a weightW=mgat the free end. (Heremis the
mass suspended at the free end of cantilever andg= 9.8 m/s^2. The mass of the cantilever
is assumed to be negligible here so that we do not need to consider it here.) Due to the
loading, the neutral filament AB of the cantilever forms an arc ABÕof radiusRand the
free end is depressed by a vertical distanceBBÕ=yas shown in Figure1.25. Consider two
points P and Q on the bent neutral filament ABÕat distancesxandx+dx, respectively
from the fixed end A. As can be seen from Figure1.25, PQ subtends an angled◊at the
centre (the point O) of the arc.
)PQ=(x+dx)≠x=dx=R◊Draw two tangents PC and QD to ABÕat points P and Q respectively such that the
vertical distance CD =dy. The angle between PC and QD isd◊since PC and and QD
are respectively perpendicular to OP and OQ. If the depressionyis small compared to
the radius of curvature, we can make the approximation,PC¥QD¥(l≠x).
)CD=dy=(l≠x)d◊Sincedx=R◊,
dy
dx
=
(l≠x)d◊
Rd◊PH8151 29 LICET