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PHYSICS 23 LABORATORY
E6-SHEAR MODULUS
NAME:
Partner’s Name:
Laboratory Instructor:
Laboratory Section:
Recitation Instructor:
Recitation Section:
OBJECTIVES: To determine the shear modulus of a rod.
Key Words:
Reading:
Introduction:
In this experiment we wish to get a feeling for what happens to a “rigid” body when we
apply forces to it. These forces (and torques) sum to zero so that the body remains in
static equilibrium after the forces are applied. This is a problem of fundamental concern
to many engineers. Will the bridge collapse when a heavy truck is driven over it? Will
the engine break loose from the wing during takeoff? Will the roof of the arena fall
under heavy wind and water loads?
As good scientists and engineers, we shall begin by trying very simple
approximations and testing them against our measurements in the laboratory to see if they
are sufficient. We know that rigid bodies are not really rigid. They distort under applied
forces. Our simple assumption will be that they distort like springs. That is, if we pull on
a rod, the change in length is proportional to how hard we pull. In other words, we
expect a behavior like F = -kx for a spring (remember x in this equation is the change in
length of the spring).
Suppose we pull on a rod of length L0 with a force F as shown. We would like
our results to be good for all rods of length L0 whether 1 inch in diameter or ten feet.
Define the stress, S, by
F
.
A
A is the cross sectional area of the rod. In this particular case, we have tensile stress
S=
because we are stretching the rod. Note the units of stress are newtons / m 2 or dynes /
cm 2 . The rod responds to our applied force by stretching an amount L. We wish to
have our results not dependent upon the original length of the rod, L0 so we define the
strain by the ratio of increase in length to the original length:
L
Tensile strain = L0
Now we make our assumption that the bar behaves like a spring, namely,
Stress = Y * Strain
Or
F
A
 Y LL0
L0
L
F
F
L0 + L
Y is the constant of proportionality called Young’s modulus or the stretch modulus. Note
that it does not depend upon the length of the rod or its cross sectional area. It is a
property of the material of which the rod is made. Since the strain is a pure number (just
the ratio of two lengths), Young’s modulus has the same dimensions as stress.
Let’s see how big an effect this is: Just how non-rigid is an aluminum rod. Being
naturally lazy, let’s take a rod one meter long and 1 cm in diameter. Thus A = πd 2 /4 =
π/4 cm 2 . Tables of Young’s modulus give for aluminum.
YAl = 0.70 X 10 12 dynes/cm 2 .
Suppose we suspend the rod vertically and hang a 10 kg weight on it and want to measure
the change in length of the rod. The gravitational force on a 10kg weight is F = mg = 10
X 10 3 gm X 980 cm/sec 2 = 9.8 X 10 6 dynes. Hence
L
L0

F
AY

9.8*106
12

4 *0.7*10
 178
. *105 meters
Or the rod stretches
L= 1.78 * 10 5 * 100 cm = 1.75 X 10 3 cm.
Previous experiments should have convinced you that you do not want to measure a
0.002 cm change in length in a meter rod using the equipment available in this
laboratory!
Note that eq. (l) can be written
F
YA
L0
L
So if we put k = YA / L0 and write the changes in length, L, as x, we have Hook’s law
F = kx. Since F here is the force we apply, this becomes F=-kx if F is the force the rod
exerts on us.
Since it appears difficult to measure L with our experiment, let’s consider the rotational
equivalent of the above experiment.
F

The rod shown is clamped on the left hand side so that the left hand end cannot rotate.
The right hand end is supported by bearings so it can rotate but not translate. None of
this is shown in the drawing. Note that when we apply F and hence a torque, = rF, the
rod twists through an angle. That little arrow on the end of the rod is supposed to be a


pointer so we can measure the angle . For rotation, recall that   F and   x, so
Hooke’s law, F = -kx, becomes  = - k’  for rotation. Once again, we would like to
assume that our rod behaves like a spring obeying this Hooke’s law for rotational
“stretching.” This analysis is more complicated than for the linear stretching case, but
once again the basic assumption is that shearing stress is proportional to shearing strain,
and the constant of proportionality, E, is called the shear modulus:
shearing stress = E * (shearing strain)
The shearing strain is just the angular rotation per meter of rod:
shearing strain 

L0
One finds eventually that
2 L0
E
 r 4
In our experiment, we apply F, not at the radius of the rod, r, but at a larger radius, R, so
 = FR.
The shearing stress is proportional to the torque  so that eq. (2) can be written
 = - k’  , (where  is the torque the rod exerts against us)
since the shearing strain is proportional to  . Note that this is just the rotational analog
of F =-kx.
Apparatus: Torsion lathe and specimens; Vernier caliper; weight pan and weights
Procedure:
1. Measure the dimensions of the specimen and place it in the torsion lathe,
setting the Vernier to zero with the weight pan in place.
2. Add loads to the pan up to the maximum load which will produce a twist
of no more than about 60 degrees (you need only use four masses ranging
up to 500 gms). Record the angle of twist both for the loading,  load,
and the unloading process,  unload. To help minimize friction effects use
1
the average value of  avg =
(  load +  unload).
2
3. Take the necessary measurements to enable you to convert load to torque.
Remember torque =  = RF = Rmg where R is the radius of the large
wheel.
4. Repeat steps 1-3 with a second specimen.
Measurements:
Analysis:
1. Make a table of torque in Newton-meters vs angle in rad. Plot a graph of
torque in Newton-meters vs angle in radians. From the slope of this graph,
determine E and compare this result with accepted values. Hint: Use eq. (3)
to find the relation between the slope of  vs.  and the constant E. Compute
the percent deviation between your measured value and the handbook value
for the same material.
2. Repeat for the second specimen.
Conclusions:
Material
Shear Modulus
Aluminum
Brass
Copper
Iron
Lead
Magnesium
Molybdenum
Nickel
Steel
Tungsten
1010 N m
2.37
3.53
4.24
7.0
0.54
1.67
14.7
7.06-7.55
7.79
14.81
2
From CRC Handbook of Chemistry and Physics