engineering mechanics labdmce.ac.in/newdmcewebsite/others/resource/humanities/engg... ·...

DATTA MEGHE COLLEGE OF ENGINEERING

AIROLI, NAVI MUMBAI

ENGINEERING MECHANICS LAB

Experiment No.

Name of Experiment

Beam Reaction

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO.-

BEAM REACTION

AIM :-To verify the condition of equilibrium by finding reactions at the

support of a beam.

APPARATUS :- A Calibrated beam supported on The compression balance, weights,

and hangers.

DESCRIPTION :- It consists of a wooden beam having equivalent grooves across its

length at 10cm intervals. The beam is supported by 2 balances. The

distance of the weight from the support can be suitably adjusted.

The experimental values of the reaction can be directly known from

the balance.

THEORY : - The condition of the equilibrium are:

1) The algebraic sum of the horizontal components of all forces

must be zero.

2) The algebraic sum of vertical components of all forces

must be zero.

3) The algebraic sum of moment of all forces about any point

must be zero.

PROCEDURE :-

1. Study the working of the apparatus.

2. Note down the zero errors if any.

3. Note down the self–weight of the beam.

4. Apply weighs on hangers from left & hold down the position

from the left hand support.

5. Take five such readings carefully at different points,

different reading with different position of load.

6. Calculate reading on the support by applying condition of

equilibrium.

OBSERVATIONS : -

1. Self weight of beam = _____________N.

2. Span of beam = ____________Cm.

3. Initial reading of left balance = _____________N.

4. Initial reading of right balance = ______________N.

5. C.G measured from L.H.S. = _____________ Cm.

6. Weight of the hanger = ______________ N.

OBSERVATION TABLE:-

sr. no. Weight applied (N) Distance from

L.H.S cm.

Observed reading Calculated reading

W1 W2 W3 X1 X2 X3 R1(N) R2(N) R1(N) R2(N)

1

2

3

4

5

CALCULATION :-

R2 = (W1X1 + W2X2 + W3X3) =

L

R1 =[ (W1+W2+W3) – R2] =

PRECAUTION :

1. The readings should be taken carefully.

2. The weighs should be added carefully.

CONCLUSION :

1. It is observed that the observed values

and calculated values are same.

2. It can be concluded that the experiment

Carried out is correct.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment Link Polygon

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO.____

LINK POLYGON

AIM :- To find the reactions of the support of the beam by graphical

method by constructing the link polygon.

APPARATUS :- Link polygon apparatus, Weights, Hangers, Scales, Strings, etc.

THEORY :- Link polygon means solving the forces along such directions that

their components cancel each other, finally having only two

forces in the system. For equilibrium, link polygon must be

closed.

PROCEDURE :-

1) Carefully study the working of the apparatus.

2) Note down the initial reading on the balances.

3) Apply the weights at the different position on the links and note

down the distances from left hand support.

4) Note down the corrected readings from the balances.

5) Five such readings are taken.

6) Calculate the reaction in each case by graphical representation

method by constructing link polygon.

PRECAUTION :- 1) All readings should be taken carefully.

2) The weights should be added gently.

3) The distances should be accurately measured.

OBSERVATIONS :-

1) Initial reading on left balances R1 = __________N.

2) Initial reading on right balances R2 = __________N.

3) Length of beam L = __________Cm.

OBSERVATION TABLE :-

Sr.

No.

Load (N) Distance from L.H.S. (cm) Corrected Reading

W1 W2 W3 W4 W5 X1 X2 X3 X4 X5 R1

(N.)

R2

(N.)

1

2

3

4

5

Calculated Reading (N) Value from graph (N)

R1 (N.) R2 (N.) R1 (N.) R2 (N.)

SAMPLE CALCULATION :-

FORMULA :-

R2 = W1X1 + W2X2 + W3X3 + W4X4 + W5X5

L

R1 = (W1 + W2 + W3 + W4 + W5) – R2

CONCLUSION :- The calculated values by Equilibrium equations and graphical

methods tally fairly with observed values at the support.

Hence the link polygon experiment is correct.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment

Jib Crane

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO ____

JIB CRANE

AIM :- To Verify Lami’s theorem by finding the forces in the

Jib-crane.

APPRATUS :- Jib-crane apparatus consists of a vertical part, the chain, Compression jib,

weights, scale spring etc.

THEOREM :- Lami’s Theorem:

If three forces are acting on a system & if it is

Equilibrium, then these forces can be represented by

three sides of triangle & each force is proportional to the

sine of the angle between the remaining two forces.

PROCEDURE :-

1) Carefully study the working of apparatus.

2) Note down the error, if any on the spring balance.

3) Apply weights at the apex points using a hanger frame.

4) Note-down the readings on the spring balance.

5) Note down the length of vertical part, tie & jib as well as

horizontal balance.

6) Draw the space diagram of jib crane using suitable scale.

7) Measure the angles at the apex point from the diagram.

8) Using Lamin’s theorem find the forces in the jib & tie.

9) Draw the triangle of forces to suitable scale.

10) Find the forces in the jib & tie from the triangle of

forces by graphical method.

PRECAUTIONS :-

1) The initial readings should be taken carefully.

2) Measures the angles accurately.

OBSERVATIONS:

1) Initial reading of tie = N.

2) Initial reading of jib balance = N.

OBSERVATION TABLE:-

Sr.

No.

Load ( N )

W

Length in Cm. Observed Reading Corrected

Reading

JIB TIE POST JIB( N ) TIE( N ) JIB( N ) TIE

( N )

1

2

3

4

5

Sr. No. α β θ Value From Graph ( N)

JIB TIE

1

2

3

4

5

SAMPLE CALCULATION :- When W = _____ N.

W = T = J ;

Sin β sin α Sin θ

T = Tension in Tie = N.

J = Compression in Jib = N

CONCLUSION :- The experimental value of forces in the jib & tie are nearly

same as calculated from lami’s theorem & forces triangle.

Thus, Lami’s theorem is verified.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment

Friction

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO.-

FRICTION

AIM :- To find the co-efficient of friction between_________ and ________

surface using inclined plane.

APPARATUS :- Inclined plane, apparatus with wooden surface with arrangement

to try and measure angle of inclination at any disc reed portion,

trolley with wooden base, weight pan, spirit level1.

THEORY :- Co-efficient of friction :-

It is the ratio of limiting friction to normal reaction.

Angle of response :-

It is the angle made by the inclined plane with horizontal, when

the body just starts moving down the plance.

Cone of friction :-

It is the cone generated by a line making an angle equal to

angle of friction with normal surface of contact having its value in

the surface.

PROCEDURE :- 1) Working of apparatus is carefully studied.

2) Plane is made horizontal with the help of spirit level and reading

on pan is noted.

3) Weight of trolley and pan are noted down.

4) Plane is inclined at an angle ‘θ’.

5) Same weights are kept in trolley; keep it loosely on any individual

point.

6) String is taken over pulley and pan is attached at other end.

7) Weights are put on the pan till trolley just starts moving.

8) Three such different readings are taken with different angle of

inclination.

9) String is detached from trolley and angle of inclination of plane is

slowly increased till just pulley moves down the plane.

10) Angle of repose is noted down.

11) Co-efficient of friction is calculated using Formula

µ= (P – Wsin θ)

W cos θ

PRECAUTION :- 1) Weights should be gently added.

2) String should be parallel to inclined plane and accordingly the

pulley should be adjusted.

OBSERVATIONS :-

1) Self weight of the trolley = -----------‘N’

2) Self weight of the pan = _______‘N’


Sr. No.

Angle of

Inclination

θ

W = weight

in trolley +

weight in (N)

it

P= weight

of pan +

weight in

(N) it

µ= (P – Wsin θ) =

W cos θ

Average µ

1. 2. 3. 4. 5.

SAMPLE CALCULATIONS :-

µ= (P – Wsin θ) =

W cos θ

CONCLUSION :- Average value of co-efficient of friction between________ and

________surface is ___________


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment Simple Pendulum

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO.____

SIMPLE PENDULUM

AIM :- To determine time period of oscillation ‘T’ and acceleration due

to gravity ‘g’ by using simple pendulum.

APPARATUS :- Pendulum bob, stand, split cork, stop watch, thread, meter scale.

PROCEDURE :- 1) Measure the radius of the given sample pendulum.

2) Tie a long thread to the hook of the pendulum bob. The tread is

passed through a split cork fixed in the clamp. Thus, the

pendulum is suspended.

3) Measure the length of the string from the point of suspension up to

the surface of the bob, using a scale. Let it be ‘|’ cm.

4) Displace the pendulum bob, from its position of rest through a

small angle and release it. The bob will start oscillating.

5) Start the stopwatch when the bob comes to the middle position and

stop it when the bob comes to the same position in the same

direction 20th times. This gives the time for 20 oscillations.

6) Repeat the procedure and take 2 more readings for same length.

Find the mean time, from that calculate periodic time.

7) Find mean L / T2. Calculate ‘g’ from these observations and also

find slope of graph (‘ℓ’) against T2.

OBSERVATIONS :-

r = radius of pendulum = m.

s = distance from center of pendulum to

length of end of thread = m.

ℓ = length of thread = m.

L = ℓ + s = m.

g = acceleration due to gravity = m/sec2

(for calculation purpose)

(Assume


Sr.

No

.

Length

of

thread

‘ℓ’

(m)

Length

of

pendulu

m L = (ℓ

+ s)

(m)

Time for 20

oscillations

(sec)

Mean t

=

t1+t2+t3

3

(sec)

Periodic

time

from

Expt T

= t.

20

(sec)

T2 L

T2

‘g’ from

Expt. = 4

2 (L)

T2

(m/sec2)

Periodic

time ‘T’ by

formula

(g=9.81

assume) T

= 2 √L

9.81

(sec)

RESULT :- i) Acceleration due to gravity by experiment =

ii) Acceleration due to gravity by graph =

iii) Periodic time ‘T’ from experiment =

iv) Periodic time ‘T’ from formula =

CONCLUSION :- The acceleration due to gravity by experimental & by graph are

same, so the experiment is correct.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment

Fly-Wheel

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO.-

FLY WHEEL

AIM : To determine experimentally, the moment of inertia of a

flywheel and shaft & compare the result with calculated

values.

APPARATUS : Cast-Iron flywheel mounted on a steel shaft & running in

ball bearings, light cord, hanger, small weights, tape and

stopwatch.

PROCEDURE : 1) Loop the chord around the pin on the shaft & attach the

free end to the hanger, the latter resting on the floor of

the flywheel.

2) Place a small weight on the hanger & rotate the flywheel

by hand through a convenient number of revolutions,

taking care to avoid double coiling of the cord on the

shaft. 3) Measure the height of the base of the hanger from the

floor and place a reference mark on the rim of the

flywheel, thus enabling the no. of subsequent revolutions

to be counted.

4) Release the wheel allowing the weight to descend to the

floor level & with the help of stop-watch, count the total

time for which the flywheel is in motion from start to

stop position. Also count the total no. of revolutions

made.

5) Determine the moment of inertial of flywheel by

Substituting in the formula & compare the values

Obtained by calculations.

THEORY : Let

W = Weight of hanger and load

h = vertical distance between weight & floor

R = radius of the shaft

N = total no. of revolutions of flywheel till the end

n = no. of revolutions of flywheel till the touch of hanger

to ground level

F = Equivalent frictional force at shaft reading radius

t = total time of motion of flywheel & shaft

ω = angular velocity of flywheel & shaft

v = linear velocity of hanger & load at attachment

Ө = 2ПN

ωt/2 = 2ПN

ω = 4ПN/t

P.E lost by ‘ω’ = (Total friction work in “N” rev. + K.E of ‘ω’ at detachment)

Therefore Wh = F(2ПRN) + W(v)2 /2g

But, v = Rω

Therefore, F = [ Wh - W(v)2 /2g ]/ (2ПRN)

F = [ Wh - W(Rω)2 /2g ]/ (2ПRN)

Work done against friction in

(N-n) revolutions = K.E of flywheel & shaft at detachment

Therefore Fx2ПRN(N-n) = Iω2

Therefore I = 2Fx2ПR(N-n)/ ω2

Mean I = 4FПR(N-n)/ ω2

Where F = [ Wh - W(Rω)2 /2g ]/ (2ПRN)

OBSERVATION TABLE :

Sr.

No.

h =

height

of pan

from

G.L

W =

Wt. of

hanger

+ Wt.

in it

‘N’

Rev.

‘n’

Rev.

‘t’

=

total

Tim

e

in

sec.

ω =

(4ПN/t)

F =

[Wh -W(Rω)2/2g ] ______________ (2ПRN)

I=

4FПR(N-n)

_________

ω2

Observations:

Wt. of Pan =

Diameter of Flywheel ( D1 ) =

Width of flywheel ( b1 ) =

Diameter of shaft ( D4) =

Length of shaft ( L) =

Density of steel =

W1 = Weight of Flywheel without shaft = П [ D12 - D22 ]xLxDensity/4 =

W2 = Weight of shaft = П [D22 ]xLxDensity/4 =

Calculations:

I1 = Mass Moment of Inertial of flywheel excluding shaft = W1x[ D12 - D22 ] /8g=

I2 = Mass Moment of Inertial of shaft = [W2xD42 ] /8g=

I = I1+ I2 = Total Mass Moment of Inertia of flywheel & shaft=

CONCLUSION: From the above experiment, it is observed that the Moment of

Inertia of flywheel & shaft is same as that of theoretical values.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment

Coplanar,

Concurrent Forces System

Date of Performance

Date of Submission

Grade given

EXPT. NO.__

POLYGON LAW OF COPLANAR FORCES

(CONCURRENT)

AIM:- To verify the conditions of equilibrium of a coplanar concurrent

force system and to find the resultant of the force system when

other forces are known.

APPARATUS:- Polygon of Force Apparatus, Weights, Strings, Drawing Paper

and Mirror Piece.

THEORY:- The Conditions of equilibrium of coplanar concurrent force

system is given by the equations as

∑FX = 0

∑FY = 0

Where ∑FX and ∑FY are the sum of Horizontal and

vertical components of the given forces along the ‘X’ and ‘Y’ axis

respectively.

PROCEDURE:-

1)Fix a Drawing sheet on the Board. Make a knot of 5 strings and

pass 4 of the 5 strings over the pulleys mounted on the board.Load

them with weights. To the other end of the fifth string , tie a pan

and add some known weight to it till the system attains equilibrium

position in a particular direction.

2)Since all the five strings are meeting at the knot, we can say that

Tensile forces in the strings are concurrent and they are in

equilibrium. With the help of pencil, make the directions of the

forces and centre point of the force system on the paper.

3)Remove the drawing paper and make ‘X’ and ‘Y’ axis at the point

of concurrency. Calculate the angles made by the forces with

respect to X axis. Resolve all the forces along X and Y axis

and calculate ∑FX and ∑FY , which should be equal to zero

for a perfect case of equilibrium. Repeat the procedure by

changing the weights in all the 5 directions.

4) A force polygon is drawn with all the five concurrent

forces. Ideally, a closed polygon is to be obtained, if there is no

error.

5) With the help of above information, calculate fifth force by

knowing the four forces, which is nothing but a Resultant to the

four forces in magnitude and direction.

OBSERVATION TABLE :

SET

NO.

FORCE

MAGNITUDE

(N.)

ANGLE

MADE

BY THE

FORCE

WITH

X-Axis

‘Ө’

F X

=

(FCOSӨ)

∑ F x

F Y

=

(FSIN Ө)

∑ F Y

I

F1

F2

F3

F4

F5

Ө1=

Ө2=

Ө3=

Ө4=

Ө5=

II

F1

F2

F3

F4

F5

Ө1=

Ө2=

Ө3=

Ө4=

Ө5=

Resultant of Four Forces:

Resultant of the four forces i.e F1, F2, F3 and F4 is given by the equation as

R= F5 = √Rx²+ Ry² , where

Rx = (F1COSӨ1 + F 1COSӨ2 + F 1COSӨ3 + F 1COSӨ4 )

Ry = (F 1sinӨ1+ F 1sinӨ2 + F 1sinӨ3 + F 1sinӨ4)

F 5 = √Rx²+ Ry² =

SAMPLE CALCULATION:-

F1 =

F2 =

F3 =

F4 =

F5 =

Ө1=

Ө2=

Ө3=

Ө4=

Ө5=

∑Fx =

∑FY =

RESULT :- The conditions of equilibrium of coplanar concurrent force system

is verified, as the error is very low and negligible.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment

COPLANAR NON-CONCURRENT, NON-

PARALLEL FORCE SYSTEM

Date of Performance

Date of Submission

Grade given

EXPT.NO.-

COPLANAR NON-CONCURRENT, NON-PARALLEL FORCE SYSTEM

AIM:- To verify the conditions of equilibrium of a Coplanar

Non-Concurrent, Non- Parallel force system and to find the

Error, if any .

APPARATUS:- Polygon of Force Apparatus, Drawing Sheet, Rectangular

Card Board, Drawing Pins, Torchlight and Weights .

THEORY:- The Conditions of Equilibrium of a Coplanar Non-

Concurrent, Non- Parrallel force system is given by the

Equations

∑Fx =0; ∑Fy=0; ∑M=0,

Where ∑Fx is the sum of Horizontal Components of the given force

system,

∑Fy is the sum of Vertical Components of the given force system

and

∑M is the sum of Moment of all the given forces about any

convenient point.

PROCEDURE:-

Fix a drawing sheet on the board. Four strings are attached to the four corners of a

Rectangular Cardboard and the fifth is attached to the centre of it. The other ends of the

strings are connected to the four corners of the Cardboard taken through the smooth pulleys

and known weights are attached to them.Tie a pan to the free end of the fifth string

(Vertical) and add weights to it till the cardboard comes to equilibrium at a convenient

position.

Note the direction of all the four forces acting on the cardboard.Choose the direction of the

string attached to the pan as Y-axis and draw the line perpendicular to it as the X-axis

passing through the center point as ‘O’. Extend the lines of action of the forces to meet the

X-axis and resolve all the forces acting at the position in ‘X’ and ‘Y’ directions. Note down

the distances of the forces from the point ‘O’.

Calculate the sum of all the horizontal components as ∑Fx and sum of all the

vertical components as ∑Fy. Mathematically, ∑Fx and ∑Fy must be zero for the

equilibrium.There could be minor error, if any.

Similarly, calculate the Moment of all the horizontal and vertical components

of the forces about point ‘O’.This should be equal to ∑M=0. There could be some minor

error, if any. Draw the force Polygon. Ideally, it should be a closed figure, without any error.

OBSERVATION TABLE

SET

NO.

NAME

OF

FORCE

FORCE

(N.)

ANGLE

WITH

X-AXIS

‘Ө’

DISTANCE

OF

FORCE

FROM

POINT’O’

IN

X-

DIRECTIO

N (CM.)

FX

=

(FCOSӨ)

∑Fx

FY

=

(FSIN Ө)

∑FY

MO

=

(Moment

Abount

Point ‘O’)

∑MO

CLOCK-

WISE MOMEN

T

(N-Cm.)

ANTI-

CLOCK-WISE

MOMEN

T

(N-Cm.)

I

F1

F2

F3

F4

F5

Ө1=

Ө2=

Ө3=

Ө4=

Ө5=

II

F1

F2

F3

F4

F5

Ө1=

Ө2=

Ө3=

Ө4=

Ө5=

SAMPLE CALCULATIONS:- F1 = ; Ө1=

F2 = ; Ө2=

F3 = ; Ө3=

F4 = ; Ө4=

F5 = ; Ө5=

∑Fx = ; ∑FY = ; ∑ MO=

RESULT:- The conditions of equilibrium for Non-Concurrent, Non- Parallell forces is

verified and the error, if any is found to be negligible as per the calculations.


AIROLI, NAVI MUMBAI


Experiment No.

Name of Experiment

BELL-CRANK LEVER

Date of Performance

Date of Submission

Grade given

EXPERIMENT NO.__

BELL-CRANK LEVER

Aim : To determine balancing moment about hinged center of Bell Crank Lever.

Apparatus: Bell Crank Lever, Scale, Different weights set

Theory : To balance rotational motion of body, we have to apply an equal and opposite

moment of same magnitude at the same point. For equilibrium, total clockwise

moment should be equal to anti-clockwise moment.

Formula : 1. ∑M=0

2. (W)x(X) = (TxY)

From the above equation, T = [(W)x(X)]/ (Y) = __ N.

Procedure :

i. Set the Bell- crank lever using the pointer.

ii. Note down the initial reading of spring balance.

iii. Apply some load(weight) at a distance “X’ from “O”.

iv. By rotating the handle, bring it to initial position.

v. Note down the final reading of spring balance.

vi. The difference between final and initial values of spring balance, will give

experimental tension.

vii. Calculate the tension analytically also, by using the above formula.

viii. Compare the answers of experimental and analytical method.

ix. Both the values should be the same.

Result : The difference in values of experimental and analytical method is negligible or

equal to zero.

Conclusion : As the values of experimental and analytical method is equal, it can be concluded

that ,it has satisfied one of the equilibrium equation ∑M=0.

engineering mechanics labdmce.ac.in/newdmcewebsite/others/resource/humanities/engg... ·...

Documents