Youngs-modulus.pdf

7/27/2019 Youngs-modulus.pdf

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TO DETERMINE YOUNGS MODULUS OF ELASTICITY OF THE MATERIAL OF A

BAR BY THE METHOD OF FLEXURE

Aim: To determine the Youngs modulus of elasticity of the given material.

Apparatus: sprit level, meter scale, screw gauge, travelling microscope and slide caliper

Theory: If a light bar of breadth b and depth d is placed horizontally on two Knife-edges

separated by a distanceL, and a load of mass m, applied at the mid-point of the bar, produces a

depression l of the bar, then Youngs modulus Yof the material of the bar is given by

(1)where g is the acceleration due to gravity. This is the working formula of the experiment, and is

valid so long as the slope of the bar at any point with respect to the unstrained position is much

less than unity. Here Yis determined by measuring the quantities b, d,L and the mean depression

l corresponding to a load m. Ifb, d, L and l are measured in cm, m in gm, g is expressed in

cm/sec2, and then Yis obtained in dyne/cm

2.

Description: A rectangular bar AB of uniform cross-section and about one meter long is taken

(Fig. 1). The bar is placed horizontally and symmetrically on two steel knife-edges N1 and N2.

The knife-edges in turn are mounted on two steady iron stands. At the mid-point of the bar is

placed a light frame F containing a vertical pointer P and a scale pan or hanger S. Weights are

placed on the hangers to depress the bar. A microscope, capable of sliding along a vertical scale,

is conveniently placed in front of the pointer.

Photograph of Youngs modulus by method of Flexure setup


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Procedure:

(i) Measure the length of the given bar with a meter scale and mark its mid-point. Drawmarks on the scale corresponding to some length L1 L1

/of the bar (say 70 cm).

(ii) Mount on the bar, the frame F carrying a knife-edge. Now place the bar with its leastdimension vertical, on the knife-edges N1 and N2 such that the L1 L1

/marks coincide with

the knife-edges. Mount a spirit level on the bar and adjust the leveling screws until the

bar is horizontal.

(iii) Bring the knife-edge of the frame F on the central transverse mark of the bar. Place themicroscope and view the pointer P. Adjust the leveling screws of the microscope until the

vertical scale is perfectly vertical and the axis of the microscope is horizontal. Focus the

eye-piece on the cross-wires by keeping one of the cross-wires horizontal. Focus the tip

of the pointer and adjust the vertical position of the microscope until the image of the tip

of the pointer touches that of the horizontal cross-wire. As far as possible, avoid parallax.

Figure 1: Schematic diagram of Flexure method arrangement

(iv) Determine the vernier constant of the microscope. With zero loads on the hanger, recordthe position of the microscope on the vertical scale.

(v) Place a load of 50 gm on the hanger. This will produce a depression of the bar. Alter thevertical position of the microscope until the image of pointer touches that of the

horizontal cross-wire. Note again the vertical scale reading of the microscope. The

difference of the two microscope readings gives the depression of the bar for the load of

50 gm.

(vi) Increasing gradually the load in steps of 50 gm and at each observation, and at each steprecord the vertical scale reading of the microscope. Now decrease the load to zero in the

same steps as used for increasing the load, and record the corresponding vertical scale-

P

S

AB

F


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readings of the microscope. Determine the mean of these two readings, and calculate the

depression by subtracting the zero-load reading.

(vii) Remove the bar without disturbing the position of the stands, and measure accurately thedistance between the knife-edges (i.e. L1 L1

/) by placing vertically the marked face of a

meter scale across the knife-edges.

(viii) Determine the vernier constant of the slide callipers and measure with it the breadth b ofthe bar at three different places. Calculate the mean breadth of the bar. Note the zero

error, if any, of the slide callipers and find the correct value ofb.

(ix) Determine the least count of the screw gauge and measure depth dof the bar at a numberof places along the length of the bar. Find the mean value. Note the zero error, if any of

the screw gauge and obtain the correct value ofd.

(x) Draw a graph with the load m in gm along the X-axis and the corresponding depression lin cm along the Y-axis and determine the value ofY.

Experimental Results:

Table-1

Vernier Constant (v.c.) of the microscope

.. Divisions (say m) of the vernier scale = divisions (say n) of the main scale.

Value of l smallest mainscale division (l1)

Value of 1 vernier division

( ) (cm)Vernier constant

v.c. = ( (cm). ..

Table-2 : Load-depression data for chosen length

Distance between the knife-edges L = cm

No.

of

obs.

Load

in

(kg)

Microscope reading for

Increasing load (cm)

Microscope reading for

decreasing load (cm)

Mean

reading

(cm)

Depression

l (cm)

Mainscale

Vernier Total Mainscale

Vernier Total

1

23

..

..

0

0.51.0

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

..

.(a)

..(b).(c)

0

(b)-(a(c)-(a)


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Table-3

Vernier Constant (v.c.) of the slide calipers

.. Divisions of the vernier scale = divisions of the main scale.

Value ofl smallest mainscale division (l1)

Value of 1 vernier division

(

) (cm)

Vernier constant

v.c. = (

(cm)

. ..

Table-4: Measurement of breadth (b) of the bar by slide calipers

No.

of

obs.

Readings (cm) of

the

Total

reading b

(cm)

Mean b

(cm)

Zero error

(cm)

Correct b

(cm)

Main

scale

Vernier

Table-5: Least count (L.C.) of the screw gauge

Pitch of the screwp (cm) No. of divisions n on the

circular scale

Least count =p/n (cm)

.

Table6: Measurement of depth (d) of the bar by the screw gauge

No.

ofobs.

Readings (cm) of

the

Total

readingd(cm)

Mean d

(cm)

Zero error

(cm)

Correct d

(cm)

Main

scale

Vernier

1

2

3

Discussions:

Even though entire scale or bar deforms by applying stress to given bar or scale, length L in

correspond to the portion between two knife edges but not the total length of thegiven bar or meter scale. Why? Refer to Appendix and find the answer.


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Computation of proportional error: We have . The quantities L, b, d, and aremeasured in this experiment. The maximum proportional error in Y due to errors in the

measurement ofL, b, d, and is given by

3 3 Precautions: (i) In the expression forY, both the lengthL between the knife-edges and the depth

d of the bar occur in powers of three. But as d is much smaller than L, much care should be taken

to measure to minimize the proportional error in Y.

(ii) Care should be taken to make the beam horizontal and to lead the bar at its mid-point.

(iii) Try to avoid parallax and back-lash errors during measurements.


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Appendix: Derivation of expression for Youngs modulus

Let us consider a beam initially unstressed as shown in fig 1(a). Now the beam is subjected to a

constant bending moment (i.e. Zero Shearing Force') along its length as would be obtained by

applying equal couples at each end. The beam will bend to the radius R as shown in Fig 1(b).

As a result of this bending, the top fibers of the beam will be subjected to tension and the bottom

to compression. It is reasonable to suppose, therefore, that somewhere between the two there are

points at which the stress is zero. The locus of all such points is known as neutral axis. The

radius of curvature R is then measured to this axis.

In order to compute the value of bending stresses developed in a loaded beam, let us consider the

two cross-sections of a beam HE and GF, originally parallel as shown in fig 1(a).when the beam

is to bend it is assumed that these sections remain parallel i.e. H'E' and G'F', the final position

of the sections, are still straight lines, they then subtend some angle .

Consider now fiber AB in the material, at a distance y from the N.A, when the beam bends this

will stretch to A'B'

Therefore, strain in fibre AB = . But from the figures 1(a) and 1(b), AB

= CD and CD = C'D'

Therefore, strain= . However, .Therefore

(A1)Consider any arbitrary cross section of beam as shown in Fig.2. Strain on a fibre at a distance y,

from the N.A, is given by the expression


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Fig.2

If the shaded strip is of area dA then the force on the strip is Bending moment (or the moment of resistance to bending) about the neutral axis = F y =

The total moment for the whole cross-section is therefore equal to = Now the term is the property of the material and is called geometrical moment of inertia(Ig) or second moment of inertia.

Therefore, bending moment (A2)Bending of the rod with a load at the centre is shown in Fig. 3. Consider right hand sided half

portion of the rod described with solid lines. It is equivalent to a cantilever with a fixed end at O.

Fig.3

P

W/2

xO

W/2

L/2W

l


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Considering the section PB of the cantilever OB, at a distance x away from its fixed end O,

Moment of bending couple due to load W/2 is equal to

(A3)

The rod being in equilibrium, this must be balanced by bending moment (or the moment of the

resistance due to bending)

From (A2) and (A3) (A4)

For small curvatures, (try to derive yourself)

By substituting for R and integrating, (A5)

Since at x=0, dl/dx = 0, implies C = 0, substituting C = 0 in (A5) and on further integrating

between the limits x= 0, and x = L/2 gives (A6)For a rectangular cross section with breadth b and depth d, (how? clue: For arectangular cross section bar )By substituting forIg in A6, or

References:

1) Mechanics by D.S.Mathur- S.Chand & Company (Pvt.) Ltd, New Delhi2) NPTEL course notes on strength of materials http://www.nptel.iitm.ac.in/3) Introduction to Structural Mechanics by P.S.Smith- Palgrave McMillan (2007)4) Solid Mechanics by Ray Hulse, Keith Sherwin and Jack Cain, Palgrave McMillan (2004)

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