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DETERMINATION OF YOUNGS MODULUS OF ALUMINIUM BY
DEFLECTION METHOD
Aim:
To determine the Youngs modulus of Aluminium specimen by
conductingdeflection on simply supported beam.
Apparatus Required:
Beam test set up, simply supported beam, weights, dial gauge 1
No. one hook
measuring tape.
M. Steel beam wts-500gm-2 Nos.
Formula:
The equation for Youngs Modulus from the deflection of a flat
rectangular
section beam of the materials simply supported and loaded at the
centre is given
by
EYoungs modulus in N/mm2
Fa
2(3L - 4a)
= -------------------48 EI
bd3
IMoment of inertia of beam in mm4 I= ------ mm4
12
= Deflection of beam in mm Yc (measured in dial gauge)
W measured mass of weight in Kg
g 9.81 acceleration due to gravity m/sec2.
L Distance between supports point (1000mm for
guidance).aDistance from support end to loading point ( 450mm for
guidance)
b = breadth = 25mm
d = depth = 6mm
1
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Procedure:
Both side Hinged condition.
The longitudinal and the cross sectional dimensions of the M.
steel beam are
noted. The positions of the support are known, so as to obtain
the given span length.
The beam is laid over the Hinged support in order to form simply
supported
beam.
The deflection meter (dial gauge) is fitted with the place near
to loading
point. Initial readings are note down when dial gauge tip
touches the beam,
consider that is zero.
Selection of points is close for dial gauge and loading
point.
The loading point is placed in center of the beam measured from
the
support.
The length of the beam (i.e.) distance in between two supports
are taken as
.
The load is gradually applied and corresponding deflection is
noted.
For given load, the youngs modulus E is found.
Result:
Youngs modulus of given Aluminium material, E= 70103
N/mm
2
Sl.
No.
Weight
applied
(Kg)
DeflectionYoungs modulus
E 103 N/mm2Loading (Y)Div Mm
1 0.5 80 0.80 68.97
2 1.0 163 1.63 67.7
DialHinged C b
d
bd3 25 63
I = ----- = --------- = 450mm4
12 12
2
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Roller a
hinged C b
case d = 6(dial gauge) 25
IF LOAD APPLIED TO CENTER
FL3
= --------------192EI
With the load applied the deflection at distance a from the
fixed
support (where a is less than or equal toL/2) is:
Fa2(3L - 4a)
= -------------------
48 EI
SAMPLE CALCULATION(ONLY FOR GUIDANCE):
Fa2(3L - 4a) = -------------------
48 EI
19.81(450)2((31000) (4450)
E= ------------------------------------------------------
481.63450
E= 67.70 103N/ mm2
RESULT:
Experimental youngs modulus value should be less than or equal
to
theoretical .
Eexp =67.70 103N/ mm2
E the =70103N/ mm2
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