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Youngs Modulus reduced

Youngs Modulus
In mechanics, we first have learnt how to find the trajectory of a point particle using Newton’s
laws of motion. Later when we wanted to apply it to extended bodies, we realized there are the
set of bodies, we call rigid bodies, where we can use Newton’s laws with Euler’s modifications.
These were the bodies for which we can assume that the distance between any two points does not
change on application of a force. The assumption is justified as the individual atoms that makes
these rigid bodies, usually are arranged in a crystalline pattern inside and the forces that holds
into that pattern is very large compared to the externally applied forces. So we use rigid body
statics/dynamics to describe their state of motion. Examples of such rigid bodies are like metal,
wood, ceramic, glass, etc.
However, when we make a bow out of wood, it does bend. The springs are made out of metals
and they do stretch or contract. These examples tells us that the rigid body approximation is to
be questioned for studying certain systems. This is what we shall investigate in this experiment.
We shall consider the deformation of a rigid body, in this case a bar made out of iron.
Let us try to understand, using a simple picture, why the solid bodies deforms. We consider a
two dimensional plane. The plane is made up of regularly placed atoms. Forces between the atoms
are holding them together. Strength of this force depends on the material. Figure (1a) show a part
of this two dimensional plane, which has a few of this regularly placed atoms. The nearby atoms
are at a distance a. At this configurations, the atoms stay in equilibrium (that is they do not move
in the first approximation) when there is no external force on the body. If we apply an external
force horizontally, the atoms get an extra pull along horizontal direction. Hence, the interatomic
distance will increase from a to some a + a (see Figure (1b). But as it increase the pull of the
atoms to each other also increases. For a given force F , there will be a certain value of a ( for a
given material) at which the internal force because of the intra-atomic pull balances the external
force. The net result is an extension of the body. If you remove the force, the material will come
back to it’s original position. Now, the intra atomic force remain attractive for a certain value of
a + a. As the external force increases, some of the atomic bonds breaks and the material will not
come back to its original position when the force is removed. If we apply a large enough force, then
we can take it apart, that is breaking. The point where the material starts to give up its bonds
for external force varies from material to material. If the material is able to hold its integrity for a
large enough force, we call it an elastic material, if it can not then it is plastic. We shall consider
only elastic materials for further discussion. Metals, wood, glass are examples of elastic materials.
Let us consider that we have applied a longitudinal force to a wire and measuring its deformation.
The initial length of the wire was L and the cross section was A. On application of a longitudinal
force, let the material creates an internal force of magnitude F and it stretches by an amount L.
Let us assume we have done an experiment where we have applied di↵erent amount of forces and
measured di↵erent values of L. If we now plot F/A Vs L/L, the curve will look as Figure (1c) 1 .
The actual curve has more feature and looks more complecated
Figure 1: Left (a), cartoon demonstrating the stretching of two dimensional lattice. Right (b),
stress vs strain curve for a typical solid.
The quantity L/L is called strain and the quantity F/A is called stress. Clearly, till the point P,
the stress is proportional to strain. In this region, if we remove the force the material comes back
to its original length. This point is called the Yield point of the material. Afterwards, the linear
relationship do not hold and at point Q, the material breaks ( and so our curve ends).
The linear relation between the stress and the strain until the yield point is known as the Hook’s
law. Ratio of the linear stress to strain in this region is called Young’s modulus,
Young’s modulus is a property of the material. In this experiment we shall measure this property.
It is important to know the Young’s modulus, the yield and the breaking point of materials which
are used in constructions. For example, when you are making a bow, you will not like the material
to su↵er a permanent extension when a large force is applied. If you are making a steel bridge, you
will like it to extend as little as possible when a force is applied on it ( like a car moving through
it and pulling it downwards because of gravity), etc.
In the next section we shall discuss the main principal that we are using to measure the Young’s
modulus of iron. We shall then describe the set up and measurement procedures.
Theoretical Understanding
Figure (2a) shows the model of a rectangular beam of length L, breath a and width b. Figure (2b)
shows the schematic diagram of the experimental set up, where the beam is supported by the
knife edges and at the middle a weight is hung. This weight has cause the beam to bend and the
maximum depression from the original position is at the middle of the beam. In this experiment
we shall measure this depression against di↵erent weight to calculate the Young’s modulus of the
material of the beam.
Figure (2b) also shows the forces acting on the beam as a whole. The weight at the middle
at the beam gives a downward force of W . Since the beam is static this force is balanced by the
Figure 2: Left (a) shows the dimensions of the metal beam. Right (b) shows di↵erent forces acting
on the beam.
normal reactions of magnitude W
2 at the two knife edges. Moreover, at the centre of the bend beam,
the tangent to the beam is horizontal. This part do not su↵er any bending. If we take half the
beam from the centre to the edge and consider the beam upside down, it behaves like a cantilever
supported at one end and loaded at the other end. Near the support the cantilever do not have any
bending. This is shown in the Figure (3a). Length of this cantilever is L/2, the other dimensions
are same as that of the beam. We can calculate the depression of a cantilever of length L0 = L2 and
loaded at one end with a weight W 0 = W
2 . The beam in Figure 3a will have the same depression
at the middle.
Figure 3: Left (a) shows the forces acting on the cantilever. The length measured along the
cantilever is x at the place A and the corresponding depression is y. At the end of the cantilever
where the length is L0 , the depression is maximum and is . Right (b) shows di↵erent moments
acting on the AC section of the cantilever.
In Figure (3a), the cantilever has a depression of at the point C where it is loaded with a
weight W 0 . We consider a point A in the cantilever, Figure (3b) is an enlarged version of the
section AP to see di↵erent forces and torques acting on its sections. The red vertical arrows show
the direction of the weight acting on the beam at P and the normal reaction created at the end A.
These two forces are equal as the beam is in a steady state. However, as the two forces acts at the
opposite ends, they introduces a moment in the beam. Since the beam is not in rotation, this must
be balanced by an equal and opposite moment. This is what the elasticity provides. As the beam
bends, the middle layer of the beam ( shown by the dotted line) almost do not expand or contract.
This layer is called the neutral layer. However, for the bending to happen, the layers above this has
to expand. Inside these layers then a restoring force will be build up. At the steady state the force
that expands these layers is equal to the restoring force. Similarly, for the layers below the neutral
layers will contract and a restoring force will be build up there too. The forces responsible for the
expansion in the upper and contraction in the lower layers are shown by blue arrows. These forces
provides a equal and opposite moment required to balance the moment produced by the loading
weight and the normal reaction. Since the restoring force is proportional to Young’s modulus of the
material, we can now calculate the depression in the cantilever in terms of the Young’s modulus
of the beam and other parameters.
Figure 4: A infinitesimal section at a given x of the cantilever is shown with the neutral layer. The
geometry and the extension of di↵erent layers are highlighted.
Let us consider an infinitesimal section of the beam near to the point A. It is shown as the
section A0 A00 A01 A001 in Figure (4). AA1 in the figure marks the neutral layer. This infinitesimal
section bends as like the part of an arc. Let us assume that the two ends of the beam subtends
and angle of and the radius of the arc to the neutral layer is R. Clearly, from the geometry
AA1 = R .
We consider a layer above the neutral layer in the beam at a distance z away from the neutral
layer. This is marked by PQ in the Figure (4). As discussed above, this layer will expand. If we
draw a line parallel to the line A0 A00 through A1 , it cuts the PQ layer at the point R (see figure).
Clearly from the geometry, the extension of this layer compared to the neutral layer is RQ. We get
= PQ
= (R + z)
= z .
Hence, the strain developed in that layer would be given by the ratio of the original length AA1 to
the extension RQ, that is, strain s is
s= .
Using Hook’s law we can now calculate the stress produced in the material due to this expansion.
If we consider a region of z to z + dz near to the layer PQ, then the strain in that layer will be the
ratio of the force dFr produced in the material to the area adz. Hence we get
=Y ,
when Y is the Young’s modulus of the material. The moment of this force about the neutral layer
will be
Ya 2
dM = zdFr =
z dz.
To calculate the total moment developed about the neutral axis in the cantilever near to the point
A, we need to integrate above for all possible values of z, hence,
M =
z 2 dz
It is clear that the entire cantilever do not form like the part of an arc. At di↵erent parts of the
beam, it will have di↵erent curvature and hence di↵erent radius of curvature. Hence, we must write
M(x) =
Y ab3 1
12 R(x)
where x is length measured along the beam, it has a value 0 at the point of suspension of the beam
and at the loading end x = L0 . If we consider the depression at di↵erent point in the beam to be
y, then y = 0 at O and y = at C. Also we see, dx
= 0 at O. The radius of curvature R(x) of the
beam can be given as
d2 y
= 2,
Y ab3 d2 y
M(x) =
12 dx2
M(x) gives the internal bending moment of the beam at the position x and as we have discussed
before, this should be same as the external bending moment provided by the weight and the normal
reaction. Clearly from the Figure (3a) the bending moment at A because of the weight W 0 at C
would be (neglecting the weight of the rod) W 0 (L0 x), hence,
W 0 (L0
x) =
Y ab3 d2 y
12 dx2
We have already stated the boundary conditions for this di↵erential equation. Solving this we may
get the value of : at x = L, y = ,
4W 0 L03
Y ab3
Hence, in our original problem of the beam loaded at the middle, the depression , in terms of W
and L, will be
W L3
4Y ab3
In the experiment, we can measure the dimensions of the beam (that is L, a, b) and the depression
for di↵erent applied weight W . Hence we can invert the above equation to estimate the Young’s
modulus from the experiment as
W L3
Y =
4 ab3
This is the working formula for this experiment.
In reality there is an additional depression of the metal bar because of the shearing, but it is quite small and we
may neglect that here. You may recalculate the depression by including the weight of the metal bar. You will see,
the method that we adopt subsequently to estimate Young’s modulus will work even if you include the weight of the
metal bar.
• Objective of this experiment is to estimate the Young’s modulus of iron by measuring the
dimension of the beam and the depression at the middle when subjected to loading.
Experimental Setup
Figure 5: Picture of the actual experimental set up with its di↵erent components.
Picture of the actual experimental set up for the Young’s modulus is given in Figure (5). A
quick look at the eqn (14) tells us that we need to measure L and b relatively accurately. Length of
the metal bar L is of the order of one meter. We use a meter scale with one millimeter division to
measure L. This provides a relative accuracy of 0.2%, sufficient for the purpose. To measure the
width of the metal bar a, which is in the order of centimeters, we use a slide calipers. Thickness of
the metal bar b has to be measured with relatively higher accuracy as it comes with a cubic power in
the equation. For this purpose we use a screw gauge. Loading of the metal bar at the centre is done
with discrete disk-shaped masses of 500 grams each. These masses have a tolerance of 5 grams. We
use up to 3 kilograms of mass in steps of 500 grams for this experiment3 . The depression expected
for this amount of load is in orders of millimeters. We use a traveling microscope arrangement to
measure this. The measurements are done with the following instruments:
• Meter scale to measure L
• Slide caliper to measure a
• Screw Gauge to measure b
• Travelling Microscope to measure .
This manual assumes the students know how to use all these instruments, help may be acquired
from the teaching assistants.
3 kg of mass provides a mechanical stress well below the yield point for iron.
Experiment procedure
Here we adopt a methods to estimate the Young’s modulus based on eqn (14). We change the
mass for the loading from 0 gm to 3 kilograms in steps of 500 grams. For each amount of loading
we measure the depression. We plot the depression against the loading and from the slope of this
curve we estimate the Young’s modulus. The detailed procedure is given below.
Measurement procedure and precautions
• Place the two knife edges at a distance of one meters. Place the metal bar on top of the knife
edges with the load-hanger inserted in it. Measure the length of the metal bar in between
the two knife edge. This can be done by using the cotton thread provided. This is the length
L of the metal bar that you will be using for calculations.
• Place the load-hanger at the centre of the metal bar. Fix the load-hanger such that the metal
pin at the top remains vertical.
• Arrange the traveling microscope such that you can view the tip of the metal pin at the centre
of the cross-wire. The cross-wire should be focused properly. Also adjust the focusing knob
of the microscope to make the image of the metal pin sharp.
• You will be using the vertical movement of the microscope for this experiment. Make sure
that the vertical axis of the microscope is actually vertical.
• The traveling microscope can be moved in the vertical axis by means of two di↵erent adjustments, the corse adjustment and the fine adjustment. When the load is changed, the metal
pin as seen through the centre of the eye piece will move away from the centre of the crosswire. Moving the fine adjustment screw of the microscope leads to an apparent movement
of the image of the metal pin when seen through the microscope’s eyepiece. We use fine
adjustment screw to keep the metal pin at the centre of the cross wire and note down the
position of the metal pin from the microscope reading. Set up the traveling microscope such
that the position of the metal pin at no load and maximum load ( 3 kilogram) are covered
only by doing the fine adjustment.
• Figure out which direction the find adjustment knob is to be moved to move the metal pin’s
image in a given direction. This is important to minimize the back lash error while taking
the measurement.
• Find out the vernier constant/least count of the vertical scale of the traveling microscope.
Measure the width and the thickness of the metal bar using a screw gauge and a slide caliper
respectively. You may record these data as shown in Table (). Five measurements are to be
taken to measure the values of a and b and mean of these are to be considered for calculation.
• Keep the load hanger without any load. Use the fine adjustment knob such that the image
of the metal pin comes at the centre of the cross wire. Note down the vertical position of the
traveling microscope.
• Put a mass of 500 gm in the load-hanger. Use the fine adjustment screw such that the image
of the metal pin comes at the centre of the cross wire. Note down the vertical position of the
traveling microscope. We shall call this measurement as m. While putting the load the beam
will vibrate. Try to minimize the vibration by providing damping using your hand.
• Repeat the above with total masss increasing to 3 kilogram by steps of 500 grams.
• Decrease the total mass in steps of 500 grams successively and note down the vertical position
of the traveling microscope for each value of the mass. Note that you have to move the fine
adjustment screw in the opposite direction compared to when increasing the mass. If we
call the microscope readings as m, then the relation of the depression and m would be
m = m0 + , where m0 is the initial depression. Table () gives an idea how the recorded data
would look like.
Length of the metal bar
thickness (b)
least count = mm
main circular
± ] cm
width (a)
vernier const. = mm
main vernier
(L) = [
Table 1: Table for measurement of the dimensions of the metal bar.
Vertical Position (m)
least count/vernier const. = mm
main circular/
main circular/
total scale
(mm) (mm)
Table 2: Table for measurements of the position of the metal pin with respect to di↵erent load for
a constant length of the metal bar.
Figure 6: Figure showing the plot of measured depression values against di↵erent mass for increasing
and decreasing the mass. The best fit straight lines are also shown. Slope for the best fit line is
shown at the top of the figures.
To estimate the depression per unit mass, plot the value of the measured vertical positions of
the metal pin against the mass loading. This is shown in the Figure (6) for both increasing and
decreasing order of mass. The data points are fitted with a straight line by doing least square
regression. The best fit straight lines and the corresponding slopes are also shown in the figure. We
estimate the depression per unit mass by averaging over these two slopes. We calculate the value
of the Young’s modulus by
Y =
depression per unit mass
where g is the gravitational acceleration g ⇠ 9.81 m sec
SI units. For the sample data used here,
and all the quantities are measured in
Y = 234 G Pa,
where G Pa stands for Giga Pascal = 109 N m
Estimation of maximum possible error
To calculate the maximum possible error we use the eqn. (14)
Ymax = Y ⇥ 3
where for the quantity X, X corresponds to the accuracy of the measurement of X and Xmin
corresponds to the minimum measured value of X. Here we have neglected the error that may arise
because of incorrect tabulated values of the mass of the metal disks we have used for loading. You
may record the values required to estimate Ymax in a table like Table (). here Lmax would be
twice the smallest division of the meter scale, mmax will be twice the least count/vernier constant
of the traveling microscope. The values of amax and bmax would be given by the maximum
di↵erence between the five measured values of a and b.
Table 3: Quantities required to calculate the maximum possible error in the measured Young’s
With the sample data we used here we get the maximum possible error to be 30 G Pa.
The measured value of the Young’s modulus with error is
Y = [230 ± 30] G Pa ,
which has two significant digits.
Discussion and thoughts
We have already discussed about the precautions to be taken to perform this experiments already.
Here I will leave you with some observations and questions.
• How would you include the weight of the metal bar in calculation ? How will the result change
if you include it ?
• When we derived the equation for the depression as a function of the length along the cand2 y
tilever, we took the radius of curvature R = 1/ dx
2 . Try to derive this relation from the
• What happens if you do not put the load hanger at the middle of the metal bar but away
from the middle of the bar ?
• In Figure (6) the the value of m for the mass of 1500 gram is little bit higher than the expected
value of m from the best fit curve for both the increasing and decreasing cases. What do you
think can be the reason for this ? Based on this observation, can you suggest an improved
stratigy for the experiment ?
• The maximum error in this experiment is about 13%. What would you like to change in the
design/procedure/stratigy of the experiment such that the maximum error reduces to 5% ?
• If we use a cylindrical rod to do the experiment instead of a rectangular bar, how would the
expression for the Young’s modulus change ?