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GOVERNMENT POLYTECHNIC

MUZAFFARPUR

Lab Manual

Civil Engineering Department

CIVIL 5TH SEM

THEORY OF STRUCTURE LAB Subject code-1615506

Prepared by:-

DEEPAK KUMAR (Civil Engg. Dept.)

Reference to: - Prof. Dr. C. S. Singh (Civil Engg. Dept.)

Prof. Dr. B. B. Chaudhur (Civil Engg. Dept.)

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CONTENTS

LIST OF PRACTICALS

1. To Verify Strain in an externally loaded beam with the help of a strain gauge indicator and to verify

Theoretically.

2. To study behavior of different types of Columns:

(i)

Both ends fixed (ii) One end fixed and other Pinned (iii) Both ends pinned (iv) One end

fixed

And other free.

3. To find Euler’s buckling load for different types of Columns :

(i) Both ends fixed (ii) One end fixed and other pinned. (ii) Both ends pinned (iv) One end fixed

And other free.

4. To Study two hinged arch for the horizontal displacement of the roller end for a given system of loading and to compare the same with those obtained analytically.

5. Determination of Shear force and loading.

6. Compression test on metal.

7. Determination of deflection of beam.

8. Determination of Moment of Inertia of fly wheel.

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Theory of structure Lab manual

EXPERIMENT NO. 1

Aim: - To verify strain in an externally loaded beam with the help of a strain gauge

indicator and to verify theoretically.

Apparatus: - Strain gauge Indicator, weights, hanger, scale, Vernier caliper.

Formula: - f = M y

I

Theory : - When a beam is loaded with some external loading, moment & shear force

are set up at each strain. The bending moment at a section tends to deflect the

beam & internal stresses tend to resist its bending. This internal resistance is

known as bending stresses.

Following are the assumpsions in theory of simple bending.

1. The material of beam is perfectly homogeneous and isotropic (i.e. have same

elastic properties in all directions.)

2.The beam material is stressed to its elastic limits and thus follows Hook’s

law.

3. The transverse section which are plane before bending remains plane after

bending also.

4. The value of young’s modulus of elasticity ‘E’ is same in tension and

compression. The bending stress at any section can be obtained by beam equation.

f = (M/I) y

Where, M= moment at considered section.

f = extreme fiber stresses at considered section.

I = Moment of inertia at that section.

y= Extreme fiber distance from neutral axis.

fmax = maximum stress at the farthest fiber i.e. at Ymax from neutral

axis.

Digital strain indicator is used to measure the strain in static condition. It incorporates

basic bridge balancing network, internal dummy arms, an amplifier and a digital display

to indicate strain value.

In resistance type strain gauge when wire is stretched elastically its length and diameter

gets altered. This results in an overall change of resistance due to change in both the

dimensions. The method is to measure change in resistance, which occurs as a result of

change in the applied load.

Strain can be calculated analytically at the section by using Hook’s law. Distrain

indicator is used to measure the extreme fiber at particular section. It basically

incorporates basic bridge balancing network, internal dummy arms, amplifier & digital

display to indicate strain value.

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Two -Arm Bridge requires two strain gauge and will display the strain value two times of

actual. Four -Arm Bridge requires four strain gauge and will display the strain value

four times of actual.

Procedure: - i) Mount the beam with hanger, at the desired position and strain gauges, over it supports properly and connect the strain gauges to the digital indicator as per the circuit diagram. ii) Connect the digital indicator to 230(+/- 10%) colts 50 Hz single phase A.C. power

supply and switch ‘ON’ the apparatus.

iii) Select the two/four arm bridge as required and balance the bridge to display a ‘000’

reading.

iv) Push the ‘GF READ’ switch and adjust the gauge factor to that of the strain gauge

used (generally 2.00)

v) Apply load on the hanger increasingly and note the corresponding strain value.

Observation: - 1) Width of the beam model, = B (cm) =

2) Depth of the beam model, = D (cm)

3) Span of the beam, = L (cm)

4) Moment of inertia of beam,= I

5)

Ymax = D/2

Modulus of elasticity of beam material, E

Observation Table:-

S.No Load applied on Moment at the f max= (M/I) Ymax Theoretical Observed the hanger P mid span section Strain strain on the ( kg) ( kg cm ) = PL/4 Ø = f max display

E

1

2

3

4

5

Sample Calculation: - For reading No. …….

Load applied on the hanger P (kg)

Moment at the mid span section (kg cm) = PL/4

f max= (M/I) Ymax Theoretical Strain Ø = f max

E

Observed strain on the display

Result : - From observation table, it is seen that, the theoretical and observed value of

strain is same.

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EXPERIMENT NO. 2

Aim: - To study behavior of different types of columns:

(i) Both end fixed (ii) One end fixed and other pinned (iii) Both end pinned

(iv) One end fixed and other free.

Apparatus: - Column Buckling Apparatus, Weights, Hanger, Dial Gauge, Scale, Vernier

caliper. Diagram:-

Theory :-

Column – a bar or a member of a structure inclined at 900 to the horizontal

and carrying an axial compressive load is called a column.

If compressive load is applied on a column, the member may fail either by crushing or by

buckling depending on its material, cross section and length. If member is

considerably long in comparison to its lateral dimensions it will fail by

buckling. If a member shows signs of buckling the member leads to failure

with small increase in load. The load at which the member just buckles is

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called as crushing load. The buckling load, as given by Euler, can be found by

using following expression.

Procedure:

End condition – a loaded column and struts can have only one of the

following four end condition -:

(a) Both end hinged or pin jointed – In this case the end of the column cannot have any lateral displacement

but can take slope when the column buckle on loading as shown in

figure.. (b) Both end fixed –

In this case both ends are rigidly fixed. The end cannot have any lateral

displacement and also cannot take slope as shown in figure.. (c) On end fixed and other hinged-

In this case one end of the column and struts is hinged and the other end

is fixed. The fixed end can neither more laterally nor it take any slope but

the hinged end can take slope when the column is loaded as shown in

figure.. (d) One end fixed and other free-

In this case one end is secured both in position and direction and the

other end is free to take any position and slope as shown in figure..

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EXPERIMENT NO. 3

Aim: - To find Euler’s buckling load for different types of columns:

(i) Both end fixed (ii) One end fixed and other pinned (iii) Both end pinned

(iv) One end fixed and other free.

Apparatus: - Column Buckling Apparatus, Weights, Hanger, Dial Gauge, Scale, Vernier

caliper. Diagram:-

Theory :-If compressive load is applied on a column, the member may fail either by

crushing or by buckling depending on its material, cross section and length. If

member is considerably long in comparison to its lateral dimensions it will fail

by buckling. If a member shows signs of buckling the member leads to failure

with small increase in load. The load at which the member just buckles is

called as crushing load. The buckling load, as given by Euler, can be found by

using following expression.

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P = π² EI

le²

Where,

E = Modulus of Elasticity

= 2 x 105 N/mm

2 for steel

I = Least moment of inertia of column section

Le = Effective length of column

Depending on support conditions, four cases may arise. The effective length for each of

which are given as:

1. Both ends are fixed le = L/ 2

2. One end is fixed and other is pinned le = L/√ 2

3. Both ends are pinned le = L

4. One end is fixed and other is free le = 2L

Procedure: - i) Pin a graph paper on the wooden board behind the column.

ii) Apply the load at the top of columns increasing gradually. At certain stage of loading

the columns shows abnormal deflections and gives the buckling load. iii) Not the buckling load for each of the four columns.

iv) Trace the deflected shapes of the columns over the paper. Mark the points of change

of curvature of the curves and measure the effective or equivalent length for each case

separately.

v) Calculate the theoretical effective lengths and thus buckling loads by the expressions

given above and compare them with the observed values.

Observation: -

1) Width of strip (mm) = b

2) Thickness of strip (mm) = t

3) Length of strip (mm) = L

4) Least moment of inertia

I = bt³

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Observation Table:-

Sr. End Euler’s Effective Length (mm)

No condition Buckling load

Theoretical Observed Theoretical Observed 1 Both ends

fixed

2 One end

fixed and

other

pinned

3 Both ends

pinned

4 One end

fixed and

other free.

Sample Calculation: - End condition: Both ends fixed

Euler’s buckling load. = P = π² EI

le²

Effective Length (mm) =.

Result :-The theoretical and experimental Euler’s buckling load for each case is found

nearly same.

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EXPERIMENT NO. 4

Aim: - To study two hinged arch for the horizontal displacement of the roller end for

a given system of loading and to compare the same with those obtained

analytically.

Apparatus: - Two Hinged Arch Apparatus, Weight’s, Hanger, Dial Gauge,

Scale, Vernier caliper.

Formula: - H = 5WL(a – 2a³ + a4)

8r

Where,

W= Weight applied at end support.

L= Span of two hinged arch.

r= rise of two hinged arch.

a = dial gauge reading. Diagram:-

Theory :-The two hinged arch is a statically indeterminate structure of the first degree.

The horizontal thrust is the redundant reaction and is obtained y the use of

strain energy methods. Two hinged arch is made determinate by treating it as a

simply supported curved beam and horizontal thrust as a redundant reaction.

The arch spreads out under external load. Horizontal thrust is the redundant

reaction is obtained by the use of strain energy method.

Procedure: - i) Fix the dial gauge to measure the movement of the roller end of the model and keep the

lever out of contact.

ii) Place a load of 0.5kg on the central hanger of the arch to remove any slackness and

taking this as the initial position, set the reading on the dial gauge to zero.

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iii) Now add 1 kg weights to the hanger and tabulated the horizontal movement of the

roller end with increase in the load in steps of 1 kg. Take the reading up to 5 kg load.

Dial gauge reading should be noted at the time of unloading also. iv) Plot a graph between the load and displacement (Theoretical and Experimental)

compare. Theoretical values should be computed by using horizontal displacement

formula. v) Now move the lever in contact with 200gm hanger on ratio 4/1 position with a 1kg

load on the first hanger. Set the initial reading of the dial gauge to zero.

(e) Place additional 5 kg load on the first hanger without shock and observe the dial

gauge reading. (f) Restore the dial gauge reading to zero by adding loads to the lever hanger, say the

load is w kg. The experimental values of the influence line ordinate at the first hanger position

shall be 4w

5.

Repeat the steps 5 to 8 for all other hanger loading positions and tabulate. Plot the

influence line ordinates.

x) Compare the experimental values with those obtained theoretically by using

equation. (5).

Precaution : - Apply the loads without jerk. : - Perform the experiment away from vibration and other disturbances.

Observation Table:-

Table: - 1 Horizontal displacement

Sr.No. Central load ( kg ) 0.0 1.0 2.0 3.0 4.0 5.0 6.0

Observed horizontal

Displacement ( mm )

Calculated horizontal

Displacement Eq. (4)

Sample Calculation: - Central load (kg) =……….. Observed horizontal Displacement (mm). =

Calculated horizontal Displacement = H = 5WL (a – 2a³ + a4)

8r

=………

Result :-The observed and horizontal displacement is nearly same.

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EXPERIMENT NO. 5

AIM :- Determination of shear force and loading.

APPARATUS USED :- Apparatus of simply supported beam.

THEORY :-

BEAM :- It is a structural member on which the load act perpendicular to axis. It is that

whenever a horizontal beam is loaded with vertical loads, sometimes it bends due to

the action of the loads. The amounts by which a beam bends, depends upon the amount

and types of loads, length of beam, elasticity of the beam and the type of beam. In

general beams are classified as under:

1. Cantilever beam :- It is a beam whose one end is fixed to a rigid support

and the other end is free to move.

2. Simply supported beam:- A beam supported or resting freely on the walls

or columns at its both ends is known as simply supported beam.

3. Rigidly fixed or built-in beam:- A beam whose both the ends are rigidly

fixed or built in walls is called a fixed beam.

4. Continuous beam :-A beam support on more than two supports is known as a

continuous beam. It may be noted that a continuous beam may not be

overhanging beam.

TYPES OF LOADING :

1. Concentrated or point load :-A load acting at a point on a beam is known

as concentrated or a point load.

2. Uniformly distributed load :-A load, which is spread over a beam in such a

manner that each unit length is loaded to a same extent.

3. Uniformly varying load :-A load, which is spread over a beam, in such a

manner that its extent varies uniformly on each unit length.

SHEAR FORCE :- The shear force at the cross-section of a beam may be defined as

the unbalanced vertical forces to the right or left of the section.

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IMPORTANT POINTS :-

If loading is uniformly distributed load then shear force diagram will be a curve of first

degree and

B.M. diagram will be a curve of second degree.

1. If the loading is point load then its corresponding S.F. diagram would be a curve of

zero degree and the B.M. diagram would be a curve of first degree.

2. If the loading is uniformly varying load its S.F. diagram would be curve of second

degree and BMD will be of third degree.

3. Bending moment is maximum where shear force is zero.

4. In case of simply supported beam the first step is to calculate the reactions at the

support, then we proceed in usual manner.

5. In case of cantilever beam there is no need of finding reaction and start from the

free end of the beam.

6. Point of flexural is the where BM changes its sign.

7. B.M. at the support is zero for simply supported beam.

Example :-A simply supported beam 4m. long is subjected to two point loads of 2KN &

4KN each at a distance of 1.5m and 3m from the left end draw the S.F & B.M diagram

for the beam.

RESULT :-

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EXPERIMENT NO. 6

AIM:-Compression test on metal

APPARATUS :-A UTM or A compression testing m/c, cylindrical or cube shaped

specimen of cast iron, Aluminium or mild steel, vernier caliper, liner scale, dial gauge (or

compressometer).

THEORY :-Several m/c and structure components such as columns and struts are

subjected to compressive load in applications. These components are made of high

compressive strength materials. Not all the materials are strong in compression. Several

materials, which are good in tension, are poor in compression. Contrary to this, many

materials poor in tension but very strong in compression. Cast iron is one such example.

That is why determine of ultimate compressive strength is essential before using a

material. This strength is determined by conduct of a compression test.

Compression test is just opposite in nature to tensile test. Nature of deformation and

fracture is quite different from that in tensile test. Compressive load tends to squeeze the

specimen. Brittle materials are generally weak in tension but strong in compression.

Hence this test is normally performed on cast iron, cement concrete etc. But ductile

materials like aluminium and mild steel which are strong in tension, are also tested in

compression.

TEST SET-UP, SPECIFICATION OF M/C AND SPECIMEN DETAILS :

A compression test can be performed on UTM by keeping the test-piece on base block

(see in fig.) and moving down the central grip to apply load. It can also be performed on a

compression testing machine. A compression testing machine shown in fig. it has two

compression plates/heads. The upper head moveable while the lower head is stationary.

One of the two heads is equipped with a hemispherical bearing to obtain. Uniform

distribution of load over the test-piece ends. A load gauge is fitted for recording the

applied load.

SPECIMEN :-In cylindrical specimen, it is essential to keep h/d ≤ 2 to avoid lateral

instability due to bucking action. Specimen size = h ≤ 2d.

PROCEDURE :-

1. Dimension of test piece is measured at three different places along its height/length

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to determine the average cross-section area.

2. Ends of the specimen should be plane .for that the ends are tested on a bearing plate.

3. The specimen is placed centrally between the two compression plates, such that the

centre of moving head is vertically above the centre of specimen.

4. Load is applied on the specimen by moving the movable head.

5. The load and corresponding contraction are measured at different intervals. The load

interval may be as 500kg.

6. Load is applied until the specimen fails.

OBSERVATION :-

1. Initial length or height of specimen h =------mm.

2. Initial diameter of specimen do =-------------mm.

S.No. Applied load (P) in Newton Recorded change in length (mm)

CALCULATION :-

Original cross-section area Ao =-----

Final cross-section area Af =--------

Stress =-------

Strain =-------

For compression test, we can

Draw stress-strain (σ-ε) curve in compression,

Determine Young’s modulus in compression,

Determine ultimate (max.) compressive strength, and

Determine percentage reduction in length ( or height) to the specimen.

PRECAUTIONS :-

The specimen should be prepared in proper dimensions.

The specimen should be properly to get between the compression plates.

Take reading carefully.

After failed specimen stop tom/c.

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RESULT :-The compressive strength of given specimen = -------Nmm2 CONCLUSION :-

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EXPERIMENT NO. 7

AIM: Determination of deflection of beams (Effect of beam length and width)

1. OBJECTIVE

The objective of this laboratory experiment is to find the relationship between the deflection (y)

at the centre of a simply supported beam and the span, width.

2. MATERIALS - APPARATUS

Steel Beams, Deflection measuring device, 500g weight

3. INTRODUCTORY INFORMATION

The deflection of a beam, y, will depend on many factors such as: -

The applied load F (F=m•g).

The span L.

The width of the beam b, and its thickness h.

Other factors such as position, method of loading, the material of which the beam is made will

also influence the deflection.

If we wish to find the relationship between y and one of the possible variables it is necessary to

keep all the other possible variables constant throughout the experiment.

3.1 Length calculation

In this experiment the same beam is used throughout and the centrally applied point load is kept

constant.

Thus keeping all possible variables other than the deflection y and the span L constant we may

investigate the relationship between y and L.

Let yLn where n is to be found

Then y = k•Ln where k is a constant

Taking logarithms:

log y = n log L + log k which is in the straight line form (y = mx + C).

Thus plotting logy against log L will give a straight-line graph of slope “n” and “k” may be

determined.

3.2 Width calculation

In this experiment beams of the same material but of different width are used. The span and

loading are kept the same for each beam. Hence keeping all possible variables other than width

and deflection constant the relationship between y and b is determined.

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Let ybn where n is to be found.

Then y = k•bn where k is a constant.

Taking logarithms,

log y = n log b + log k which is in the straight line from : (y = mx + C).

Thus plotting logy against log b will give a straight line of slope “n” and “K” may be

determined.

4.1 PROCEDURE (Length calculation)

a) Mark the centre of the beam on each side of this point mark off distances off 500, 600,

1000 mm.

b) With a span of 500 mm measure the height of the central point on the deflection -

measuring device. Apply a central load of 500g and measure the new height.

c) Repeat 2 for spans of 600, 1000.

d) Enter your results in the table below and complete the table

e) Plot the graph of log y against log L with log y on the “y” axis and log L on the “x” axis.

f) Draw the mean straight line of the graph and measure its slope to determine n.

4.1.1 Results

A/A Width b

(mm)

Length L

(mm)

Deflection y

(mm)

Log L Log y

1

30

2

3

4

5

6

SLOPE = n = yLn

4.2 PROCEDURE (Width calculation)

a) Mark the beams with the same span so that they will be supported near their ends and

also mark the mid- point of the span.

b) Take the beam of largest width, measure the width with the vernier Calipers.

c) Support the beam at the two marked supporting points and measure the height of the mid-

point with the deflection measuring device.

d) Apply the 500 g load as the mid -point and once again measure the height at the centre.

e) Repeat 2, 3, and 4 for each beam.

f) Enter your results in the table below and complete the table.

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g) Plot the graph of logy against log b with logy on the “y” axis and log b on the “x” axis.

h) Draw the mean straight line of the graph and measure its slope to determine n.

4.2.1 Results

A/A Length

L

(mm)

Width b

(mm)

Deflection y

(mm)

Log b Log y

1

800

2

3

4

5

6

SLOPE = n = ybn

5. QUESTIONS

Plot the graph of log y against log L with log y on the “y” axis and log L on the “x” axis.

Determine slope n. How does your result compare with the generally accepted

relationship?

Plot the graph of logy against log b with logy on the “y” axis and log b on the “x” axis.

Determine slope n. How does your result compare with the generally accepted

relationship?

Calculate the corresponding deflections y, during length calculation (b has constant

value), according to the formula shown below.

Calculate the corresponding deflections y, during width calculation (b has variables

values), according to the formula shown below.

Compare the observed and calculated values of deflections y.

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h=0.004 m

=F*L

3

y48*E*I

L = length of beam (m)

y = deflection of beam (m)

F = force (N)

E = Young's Modulus (N/m2)

I = moment of inertia of beam (m4)

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Experiment: Deflection of beams (Macaulay’s Method) 1. OBJECTIVE

To determine experimentally the deflection at two points on a simply-supported beam carrying

point loads and to check the results by Macaulay’s method.

2. APPARATUS

Beam deflection apparatus, steel beam, two dial test-indicators and stands, micrometer, rule, two

hangers, weights.

3. PROCEDURE (Experimental)

Assemble the apparatus as shown in fig. 1 with the beam simply supported at its ends A and B.

Place load hangers at point C and D distant a and b

W1 W2

Y1 Y2

A α C D B

R1 b R2

l

Figure 1

Respectively from end A. Select two points X and Y approximately in positions shown in the

figure and set up the dial gauges to bear at these points on the upper surface of the beam. Zero

the dial gauges with the hangers in position.

Apply suitable loads W1 and W2 at C and D respectively and note the deflections at X and Y as

indicated by the dial gauges. Record the values of W1 and W2 and the corresponding deflections

at X and Y. Sketch the arrangement and indicate on the sketch the distances a, b, and l. Also the

distances of points X and Y from end A.

Measure the cross-sectional dimensions of the beam, using a micrometer.

Calculate the deflections at X and Y, using Macaulay’s method and compare the values with the

observed results.

4. THEORY

Consider the simply-supported beam loaded as shown in fig.2.

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W1 W2

X

A α C D B

RA b RB

x l

Figure 2

For values of x between b and l

bxWaxWxRM lAXX 2 (i)

For values of x between a and b

axWxRM lAXX (ii)

For values of x between o and a

xRM AXX (iii)

Eqn. (i) gives the bending moment at any section of the beam provided bracketed terms are

discarded when they become negative. For this reason, the bracketed terms are known as the

“Macaulay Ghost Terms”.

Since Mdx

ydEI

2

2

EI )()( 212

2

bxWaxWxRdx

ydA (iv)

In Macauley’s method, the bracketed terms are intergraded as a whole. This is justified since

1

)(

x

xx

dxax 1

2

)()(

x

x

axdax

EI AbxW

axWx

Rdx

dyA 22

2

)(2

2)(

22 (v)

EIy BAXbxW

axWx

RA 333

)(6

2)(

6

1

6 (vi)

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By integrating the bracketed quantities as a whole, the constants A and B have the same values

for all values of x.

This may be shown to be the case as follows:

Put x = a in eqn. (v) and omit the term in (x-b) since it is then negative.

Then, AaaWIa

Rdx

dyEI A 2

2

)(22

Aa

RA 2

2

For values of x between o and a

EI xRdx

ydA

2

2

Integrating

EI 1

2

2A

xR

dx

dyA

Putting x = a

EI 1

2

2A

aR

dx

dyA

Since the two equations concern the slope dy/dx at the same point that the constants A and A1

must be equal. Similarly by putting x = b it may be shown that the constant is again A.

The actual values of the constants A and B are obtained from the boundary conditions, that is, in

eqn. (vi):

y = o when x = o and

y = o when x = 1

In the particular case considered, B = o.

5. PROCEDURE (Calculations)

a) Set up an expression for the bending moment for any section in the extreme right-hand

panel of the beam, measuring x from the left-hand end. Put in square brackets, the

‘ghost’

b) Integrate to obtain the slope equation and again to obtain the deflection equation and

again to obtain the deflection equation, adding the constants A and B respectively at each

stage. Integrate the ‘ghost’ terms as a whole.

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c) Calculate the constants A and B from the condition that the deflection y is zero at the two

values of x corresponding with the supports. Omit negative ‘ghost’ terms.

d) To determine slope or deflection at a particular point on the beam substitute the

corresponding value of x in the appropriate expression and omit any ‘ghost’ term which

may become negative.

5.1 Results

Width of beam, b (m)

Thickness of beam, d (m)

Span, l (m)

Load W1 (g)

Load W2 (g)

Distance a (m)

Distance b (m)

Deflection at Y1 (mm)

Deflection at Y2 (mm)

Young’s Modulus, E = 210 GPa (assumed)

5.2 Calculations

Second moment of area of beam cross-section I= 43

12m

bd

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Reaction RA =…………………… Reaction RB =……………………..

Flexural rigidity EI =

By means of Macaulay’s method calculate the deflection at the points X and Y using the

appropriate values of x and tabulate the results, as follows:

Point Observed

Deflection

Calculated

Deflection

1

2

6. CONCLUSION

Compare the observed and calculated values of deflection at the two points and comment on

probable causes of discrepancy.

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EXPERIMENT NO. 8

Aim:

Determination of moment of inertia of a flywheel.

Apparatus:

Fly wheel, weight hanger, slotted weights, stop watch, metre scale.

Theory:

The flywheel consists of a heavy circular disc/massive wheel fitted with a strong axle

projecting on either side. The axle is mounted on ball bearings on two fixed supports.

There is a small peg on the axle. One end of a cord is loosely looped around the peg

and its other end carries the weight-hanger.

Let "m" be the mass of the weight hanger and hanging rings (weight assembly).When

the mass "m" descends through a height "h", the loss in potential energy is

The resulting gain of kinetic energy in the rotating flywheel assembly (flywheel and

axle) is

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Where

I -moment of inertia of the flywheel assembly

ω-angular velocity at the instant the weight assembly touches the ground.

The gain of kinetic energy in the descending weight assembly is,

Where v is the velocity at the instant the weight assembly touches the ground.

The work done in overcoming the friction of the bearings supporting the flywheel

assembly is

Where

n -number of times the cord is wrapped around the axle

Wf - work done to overcome the frictional torque in rotating the flywheel assembly

completely once

Therefore from the law of conservation of energy we get

On substituting the values we get

Now the kinetic energy of the flywheel assembly is expended in rotating N times

against the same frictional torque. Therefore

and

If r is the radius of the axle, then velocity v of the weight assembly is related to r by

the equation

Substituting the values of v and Wf we get:

Now solving the above equation for I

Where, I = Moment of inertia of the flywheel assembly

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N = Number of rotation of the flywheel before it stopped

m = mass of the rings

n = Number of windings of the string on the axle

g = Acceleration due to gravity of the environment.

h = Height of the weight assembly from the ground.

r = Radius of the axle.

Now we begin to count the number of rotations, N until the flywheel stops and also note

the duration of time t for N rotation. Therefore we can calculate the average angular

velocity in radians per second.

Since we are assuming that the torsional friction Wf is constant over time and angular

velocity is simply twice the average angular velocity

Applications:

Flywheels can be used to store energy and used to produce very high electric power

pulses for experiments, where drawing the power from the public electric network

would produce unacceptable spikes. A small motor can accelerate the flywheel between

the pulses.

The phenomenon of precession has to be considered when using flywheels in moving

vehicles. However in one modern application, a momentum wheel is a type of flywheel

useful in satellite pointing operations, in which the flywheels are used to point the

satellite's instruments in the correct directions without the use of thrusters rockets.

Flywheels are used in punching machines and riveting machines. For internal

combustion engine applications, the flywheel is a heavy wheel mounted on the

crankshaft. The main function of a flywheel is to maintain a near constant angular

velocity of the crankshaft.

Thanks.

government polytechnic muzaffarpurgpmuz.bih.nic.in/docs/tos.pdf · buckling depending on its...

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