Mos Manual PDF Version

DEPARTMENT OF CIVIL ENGINEERING DATE:

PAGE NO.

PRACTICAL NO.1

EQULIBRIUM OF COPLANAR,

CONCURRENT FORCES


PAGE NO.

FIGURE:

OBSERVATION TABLE:

FORCE

VALUE OF Pi

ANGLE αi

COMPONENTS

FX= Pi COS θi FY=Pi SIN θi

P1

P2

P3

P4

P5

∑FX= ∑FY=

R = error

=

CALCULATIONS: resultant force is obtained analytically as well as graphically.


PAGE NO.

EQUILIBRIUM OF COPLANAR, CONCURRENT

FORCES

OBJECT: To verify the polygon law of forces for a system in equilibrium, both

analytically and graphically.

APPARATUS: A force table with five attached pulleys, weights.

THEORY: Concurrent forces in equilibrium can be represented by a closed polygon

the respective sides being parallel and proportional to their respective forces. If the

forces are not in equilibrium then the closing side of polygon shows the resultant of the

system of forces.

POLYGON LAW OF FORCES (STATEMENT):

PROCEDURE: Five strings are radiating from a single point. These strings pass over

smooth pulleys freely slide able along the circumference of a horizontal force table

with graduations for measuring angles. Attach five known weights to five strings

arranged at particular angles. Note down the readings when the central ring is at centre

of force table. As this system is in equilibrium, the resultant force is obtained

analytically as well as graphically which should be negligible in magnitude. Draw

space diagram by drawing the angle as measured on force table and show the

respective forces. Using Bow‟s notation draw force diagram, which will be the

polygon. For the system is in equilibrium. The sides of polygon should be five but due

to error polygon remains open & hence the closing side in reverse order will represent

the resultant.


PAGE NO.

CONCLUSIONS:

ASSIGNMENT:

1. State the law and comment on the verification of law?

2. What are the reasons for these errors?

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.2

EQULIBRIUM OF COPLANAR, NON

CONCURRENT & NON PARALLEL FORCES


PAGE NO.

FIGURE:

OBSERVATION TABLE:

SR

NO.

P1 P2 P3 P4 θ1 θ2 θ3 θ4

1.

2.

3.

4.

5.

CALCULATION TABLE FOR OBSERVATION NO.-

SR

NO.

FORCE(Pi)

ANGLE (θi)

FX= Pi COS θi

FY=Pi SIN θi Moment @ A

FX x Y FY x X

1.

2.

3.

4.

= =

R (Analytical) = & α =

R (Graphical) = & α =


PAGE NO.

EQUILIBRIUM OF COPLANAR, NON CONCURRENT

& NON PARALLEL FORCES

OBJECT: To determine the weight of a plate by equilibrium of coplanar, non

concurrent, non parallel Forces (Graphically and analytically).

THEORY:

Force system is having five forces P1, P2, P3, P4, & self weight (W) of the plate. As

force system is in equilibrium, resultant (R) of four forces must be equal, opposite &

collinear with the weight of the plate, If four forces P1 to P4 represented by sides of

polygon the closing side of polygon will give the magnitude & direction of weight of

the plate.

APPARATUS:

Board, sheet of paper, pulleys, strings, mild steel plate of unknown weight, known

weights.

PROCEDURE:

1) Attach the sheet of paper on a vertical board.

2) Adjust position of pulleys so that the strings passed through them are attached to

four corners A, B, C, & D of the plate remains in same vertical plane.

3) Apply different weights at the end of the string to keep the plate in equilibrium at

the middle portion of the attached sheet.Note down the magnitude of weights P1,

P2, P3, & P4 attached to string passing through corners A, B, C, D and the centre of

the plate.


PAGE NO.

4) Mark the projections of strings and plate perpendicular to the paper.

5) Remove the sheet, mark angle θ1, θ2, θ3, θ4 respectively with horizontal.

6) Find out the value of W analytically as well as graphically.

7) Change the position of lower pulleys and repeat procedure with different weight

and changed position of plate.

8) Draw funicular polygon for any one observation.

CONCLUSION:

ASSIGNMENT:

1. Comment about graphical value vs. analytical value (calculated)

2. Check Varignon‟s theorem & comment on its verification.

3. Enlist causes due to which error may occur.

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.3

BELL CRANK LEVER


PAGE NO.

FIGURE:

OBSERVATION TABLE

SR.NO. W

(Kg)

D cm P1

(Kg)

P2

(Kg )

P= P2-P1

(Kg)

P.L

Kg- cm

W.D

Kg -cm

%

Error

OBSERVATION: Length of short arm L = cm

CALCULATIONS:


PAGE NO.

BELL CRANK LEVER

OBJECT: To verify the principle of moment.

THEORY: Principle of moment: If the given system of force is in equilibrium, algebraic

sum of moments about any point is equal to zero.

APPARATUS: Bell crank lever including spring balance, weights.

PROCEDURE:

1) Measure length L of vertical arm of lever.

2) Adjust the screw of the balance and make long arm horizontal

3) Note down the reading P1 of the balance

4) Suspend known weight W at convenient point along the long arm and note its

distance from fulcrum (D).

5) Adjust screw of the balance to make long arm again horizontal.

6) Note down the reading P2 of the balance and compute P= P2-P1

7) Similarly change the value of W and D and repeat the same procedure.

8) Verify the principle of moment (P x L) = (W x D) for each case.


PAGE NO.

CONCLUSION:

ASSIGNMENT:

1)State and explain principle of Moment.

2)Comment on the verification of the principle of moment?

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.4

SIMPLY SUPPORTED BEAM


PAGE NO.

OBSERVATION TABLE:

SR NO.

APPLIED LOAD (Kg) OBSERVED

REACTIONS

DISTANCE OF LOAD

FROM SUPPORT (cm)

W1 W2 W3 R1 R2 X1 X2 X3

1.

2.

3.

4.

CALCULATION TABLE FOR FORCES:

SR NO. R1+ R2 W1+W2+W3+W4 % ERROR IN FORCES

1.

2.

3.

4.

CALCULATION TABLE FOR MOMENTS:

SR.NO SUM OF MOMENTS

@ SUPPORT 2

MOMENT DUE TO

REACTION 1

% ERROR IN MOMENT

1.

2.

3.

4.


PAGE NO.

SIMPLY SUPPORTED BEAM

OBJECT: To determine the reactions at supports and verify the condition of equilibrium

for a beam simply supported at ends.

THEORY: Beam is a structural element carrying transverse loads.

Analytical condition of equilibrium

1) Algebraic sum of horizontal forces must be equal to zero. i.e. ∑FX=0

2) Algebraic sum of vertical forces must be equal to zero.i.e ∑FY=0

3) Algebraic sum of moments of forces about any point is zero.

Graphical conditions of equilibrium

1) Force polygon should close which indicates that there is no motion of translation.

2) Funicular polygon should close which indicates that there is no rotating motion.

PROCEDURE:

1) Note down the self weight of the beam, weight of hook and weight of hangers.

2) Put the known three loads at various point along the length of the beam, so that

weights are W1, W2, and W3.

3) Measure the horizontal distances of these loads from one of the support.(X1, X2, X3)

4) Observe the reactions in the compression spring balances provided at each support.


PAGE NO.

FIGURE

OBSERVATIONS:

1) Self weight of the beam (Ws) =

2) Weight of the hook=

3) Length of the beam=_

CONCLUSION:

ASSIGNMENT:

1)Explain different types of supports with sketches

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.5

COEFFICIENT OF STATIC FRICTION


PAGE NO.

FIGURE

OBSERVATION TABLE:

SR.NO TYPE OF

THE

MOTION

WOOD TO

WOOD

WOOD TO BRASS WOOD TO

ALUMINIUM

UP THE

PLANE

P W µ P W µ P W µ

1.

2.

3.

4.

DOWN THE

PLANE

5.

6.

AVERAGE µ= AVERAGE µ= AVERAGE µ=


PAGE NO.

COEFFICIENT OF STATIC FRICTION

OBJECT: To determine the co-efficient of static friction between wood to Wood, wood

to Aluminum and wood to brass surfaces.

APPARATUS:

Inclined plane, Boxes with Wood, Aluminum & Brass bottom Surface and Weights.

THEORY:

When a body is lined over another body and is gently pushed, it doesn‟t move.

The frictional force prevents the motion. This shows that the force of friction has

increased itself, so as to become equal and opposite of the applied force. Thus the force

of friction has a remarkable property of adjusting its magnitude so as to become

exactly equal to applied force which tends to produce motion.

The force of friction cannot increase beyond certain unit. If applied force

exceeds this limit, the force of friction cannot balance it and block begins to move in

the direction of the applied force. This maximum value of frictional force which comes

into play when the body just begins to slide over the surface of the other body is known

as limiting friction and the ratio of the limiting friction to the normal reaction between

the two bodies is known as Coefficient of static Friction and is denoted by µ.

Where,

F= Frictional force acting upward along the plane when body tends to move down.

R= Normal reaction acting at right angle to the plane


PAGE NO.

CALCULATION:

Case-1: Motion up the plane

Where,

θ= Angle of inclination of plane with horizontal

W= Weight of box + Weight in box sliding over surface

P= Weight in pan + Weight of Pan

Case-2: Motion down the plane

Case-3: Motion down the plane at angle of repose

µ= tanα= tan θ=


PAGE NO.

θ= Angle of friction, which is also equal to angle of repose (α), i.e. angle of

inclined plane.

PROCEDURE:

1.) Fix the angle of inclined plane and clean the surface.

2.) Put a box on the clean surface whose co-efficient of static friction is required to be

found, at the lower end of inclined plane. Connect the box with a string which will

pass over the frictionless pulley fixed at higher end. Connect pan with it.

3.) Put a limiting value of weight which causes a uniform upward sliding of the surface

box.

4.) Find out the co-efficient of friction by the formula. Draw free body diagram and

prove the formulae (A).

5.) Set the angle of inclined plane as higher value, out fixed value of weight pan and

vary the weight in the box placed on surface so that minimum value of weight in the

box may cause uniform motion of the box down the plane.

6.) Draw the free body diagram for second case (motion down the plane) prove the

formula (B).

7.) Remove pan and weight in the pan, put the box on the inclined plane and find the

minimum angle of inclination when particular box slide with uniform motion. This

angle is called angle of repose (α) & it is numerically equal to angle of friction (θ).

Draw free body diagram and prove the expression (C)

CONCLUSION:


PAGE NO.

µ= WOOD TO WOOD =

µ= WOOD TO BRASS =_

µ= WOOD TO ALUMINIUM =_

ASSIGNMENT:

1) Write law of static friction.

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.6

SYSTEM OF PULLEYS


PAGE NO.

Figure (1) Figure (2)

Figure (3) Figure (4)


PAGE NO.

SYSTEM OF PULLEYS

OBJECT: To Determine the mechanical advantage, velocity ratio, ideal effort,

frictional effort, ideal load, frictional load and efficiency of given system of pulleys.

THEORY: lifting machines: In all type of machines, one needs to have maximum

efficiency so that with minimum effort maximum output will be available. Efficiency

is defined as ratio of output to input.

Where,

W= Actual load lifted by machine, P= Actual effort applied on machine

X= distance moved by load, Y= Distance moved by effort

From the above equation efficiency can also be obtained as ratio of Mechanical

Advantage and Velocity Ratio.

In Ideal machines, there is no friction; therefore no work is wasted in overcoming the

frictional resistance. Let Pi and Wi represent ideal effort and ideal load. Therefore, (P-

Pi) represents effort required to overcome friction and (Wi-W) represents load

equivalent to reduce friction.

We, know that,


PAGE NO.

OBSERVATION TABLE:

S.NO. TOTAL

LOAD

(W)

TOTAL

EFFORT

(P)

MECHANICAL

ADVANTAGE

MA= W/P

IDEAL

EFFORT

PI=W/VR

FRICTIONAL

EFFORT

PF=P-(W/VR)

IDEAL

LOAD

WI= PVR

FRICTIONAL

LOAD

WF= PVR-W

EFFICIENCY

η = (W/P)VR x

100

OBSERVATION: Velocity ratio for,

1) Differential pulley:

V.R. = Distance moved by effort

Distance moved by load

πD D

= =

πd d

2) First system of pulley = V.R. = 2n

3) Second system of pulley = n

4) Third system of pulleys = V.R. = 2n-1 =


PAGE NO.

For ideal machine efficiency = 1 therefore W/ P= V.R. but for ideal machine effort

required= Pi

Hence,

Therefore,

Frictional effort= P-Pi

=

Also, for ideal machine load lifted= Wi therefore,

Wi = P V.R.

Frictional load = Wi- W

= (P V.R.) – W

The relationship between actual load lifted (W) and actual effort (P) applied is called

law of machine and it is given as P = m W+ C

Where, m = coefficient of friction represented by slope of graph between W & P

C= Constant which represents initial machine friction and is given by

intercept of graph on Y- axis.


PAGE NO.

System of Pulleys: The most common example of lifting machines is a pulley.

To get maximum efficiency, one needs to have maximum M.A. and minimum

V.R. To achieve such a condition, pulleys are arranged in different manners,

such as,

1) Differential Pulley: Pulleys of different diameter are coupled together. Load

is suspended from smaller diameter pulley and effort on large diameter

pulley as shown in fig(1).

2) First system of Pulley: Pulleys are arranged together in the manner shown in

fig (2). If load moves by distance ‟X‟ then effort moves by 2n

X distance,

where n= number of pulleys. Here in this system last pulley is fixed, it is

only for changing the direction of effort.

V.R = Distance moved by effort (2

n X)

Distance moved by load X

V.R. = 2n (Here n= 2)

3) Second system of pulleys: As shown in fig (3) it is also known as simple

block which consists of fixed (upper) block of pulley and movable (lower)

block of pulley to which load is attached. There are two types of

arrangement either equal number of pulleys in upper block. There is only

one string passing over all the pulleys and effort is connected to it. Sketch

the arrangement and free body diagrams for both the types. If X is distance

travelled by load, the effort by distance nX


PAGE NO.

Where, n= no of portions of the string supporting lower block.

4) Third system of pulleys: The load is attached to a movable platform and each

string which is passing over different pulley is attached to earlier pulley. The

effort is suspended from string passes through 1st

pulley. All the pulleys are

movable; no fixed pulley is required as in case of 1st

system of pulleys as shown

in fig (4).

V.R. = 2n-1.

PROCEDURE:

(1) Sketch all three types of system of Pulleys

(2) Study the motion of effort and load to be lifted.

(3) As per free body diagram of system of pulley derive the expression of Velocity

Ratio.


PAGE NO.

CONCLUSION:

ASSIGNMENT:

1) Explain law of machine.

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.7

WHEEL AND DIFFERENTIAL AXLES


PAGE NO.

FIGURE

OBSERVATIONS:

1. Diameter of effort wheel (D). =

2. Diameter of bigger axle (d1) =

3. Diameter of small axle (d2) =

4. Velocity ratio VR= = 2D =_

d1 -d2

5. Weight of load pulley= 130 gms


PAGE NO.

WHEEL AND DIFFERENTIAL AXLES

OBJECT: To determine the mechanical advantage, velocity ratio, ideal effort, frictional

effort, ideal load, frictional load, and efficiency of a given wheel and differential axles.

THEORY: The wheel and differential axles act on the differential principle. It is a

simple device to achieve very high velocity ratio by keeping very small difference in

diameter of two axles. The axle is made up of two cylinders B and C of different

diameters. The wheel A and cylinder B and C turn about common axis. One string is

wound round the wheel A and to the one end of which the effort P is applied. The

second string is wound round the two cylinders in such a way that as the cylinder turn,

it unwinds on the smaller cylinders and wind at the same time on the larger cylinder,

lifting the load W attached to the pulley. For one revolution of the wheel and axle, the

displacement of the effort P is equal to the length of the string that unwinds from the

wheel A=y= πD

The length of the string that unwinds from the smaller cylinder C = πd2

The length of the string that winds on the larger cylinder B= πd1

The load string shortens by = πd1- πd2

Therefore, the displacement of the load W

X = πd1- πd2

2


PAGE NO.

OBSERVATION TABLE

S.NO. TOTAL

LOAD

(W)

TOTAL

EFFORT

(P)

MECHANICAL

ADVANTAGE

MA= W/P

IDEAL

EFFORT

PI=W/VR

FRICTIONAL

EFFORT

PF=P-(W/VR)

IDEAL

LOAD

WI= PVR

FRICTIONAL

LOAD

WF= PVR-W

EFFICIENCY

η = (W/P)VR x

100

CALCULATION:

1. Give sample calculation for all quantities for any one reading of observation table.

1) Maximum mechanical advantage= 1/m=

2) Maximum efficiency= 1/ (m VR) =


PAGE NO.

Velocity Ratio= Y = πD

X

1 (πd1- πd2)

2

= 2D

d1 -d2

For a larger velocity ratio, d1 and d2 are made nearly equal.

EQUIPMENT: Wheel and differential axle, weights.

PROCEDURE:

1) Measure the dimensions of wheel axle, & note total applied load to the system.

2) Increase the effort gradually in such a way that a minimum effort the load just starts,

moving upward note the value of effort.

3) Change the applied load and find corresponding effort.

4) Take six sets of readings

5)Plot the graphs of load (W) on X- axis Vs i) actual effort (P) ii) ideal effort (Pi)

iii)frictional effort (P-W/VR) iv) Efficiency (η) on Y axis. Use same scale for all effort

curves and different scale for efficiency, it is desirable to draw all graph on same graph

paper.

6) Work out the law of machine P= m W+C where, P is actual effort required to raise

the load W and m is the slope of the graph, C is the initial effort required to start the

machine which is obtained as the intercept on Y- axis.


PAGE NO.

CONCLUSION:

Law of machine P= W + (write the value of m & C)

ASSIGNMENT:

1) Mention which graph passes through the origin and why?

2) State whether m/c is reversible or irreversible, why?

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.8

SINGLE PURCHASE CRAB


PAGE NO.

FIGURE

OBSERVATION:

1. Diameter of wheel (D) =_

2. Diameter of axle (d) =

3. No. of teeth on spur wheel (Ts)=

4. No. of teeth on pinion wheel(Tp)=_

5. Velocity ratio (VR)= D x Ts =

D x Tp

OBSERVATION TABLE

S.NO. TOTAL

LOAD

(W)

TOTAL

EFFORT

(P)

MECHANICAL

ADVANTAGE

MA= W/P

IDEAL

EFFORT

PI=W/VR

FRICTIONAL

EFFORT

PF=P-(W/VR)

IDEAL

LOAD

WI= PVR

FRICTIONAL

LOAD

WF= PVR-W

EFFICIENCY

η = (W/P)VR x

100


PAGE NO.

SINGLE PURCHASE CRAB

OBJECT: To determine the mechanical advantage, velocity ratio, ideal effort, frictional

effort, ideal load, frictional load and efficiency of a given single purchase crab.

THEORY: The single purchase crab is a simple lifting machine in which velocity ratio

is achieved using Gearing.

EQUIPMENTS: Single purchase crab, weights, caliper, meter ruler etc.

PROCEDURE:

1) Measure the required dimensions of machine. Note the total load applied and

corresponding effort required to raise the load with slow uniform motion.

2) Take six readings with uniform increase in load and corresponding effort required.

3) Plot the graphs with (i)actual effort (ii) ideal effort (iii) frictional effort and iv)

efficiency on Y- axis and load on X- axis. Use the same scale for all effort curves.

4) Work out the law of machine P= mW+C where, P is actual effort required to raise

the load W, “m” and “C” are constants of a given machine.


PAGE NO.

CALCULATION:

1) Give sample calculation for all quantities for any one reading of observation table.

2) For law of machine, P= mW+C

P= W+_

3) Maximum mechanical advantage= 1/m=

4) Maximum efficiency= 1/(m VR) =

CONCLUSION:

ASSIGNMENT:

1) Define self locking machine and state whether the machine is self locking or not?

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.9

COMPRESSION TEST ON METALS


PAGE NO.

Fig 9.1

Fig 9.2 Fig 9.3


PAGE NO.

COMPRESSION TEST ON METALS

OBJECT: To determine the compressive strength and to study the behavior of mild

steel and cast iron specimens when subjected to compressive load up to failure.

EQUIPMENT: Universal testing machine with an attachment to test specimen in

compression.

APPARATUS: Vernier caliper

MATERIAL: Mild steel and Cast iron

RELATED I.S. CODE IS 1378:1993: HARD METAL- COMPRESSION TEST

SALIENT FEATURES

There are several materials used in engineering practice that are primarily meant to

carry compressive loads. Steel is one such material used as compression member.

Relevant mechanical properties are determined by loading a specimen up to failure or

a predetermined load.

When a specimen is subjected to an axial load on the end surface producing crushing

action, an internal resistance is set up against shortening of its length. This resistance

is called compressive resistance. The intensity of the resistance is termed as a

compressive stress. Thus, compressive stress is a ratio of compressive load (P) to the

cross section area (A) resisting it. Fig 9.1 shows a typical stress- strain curves for

brittle and ductile materials.

Specimen for compression test has tendency to fail by buckling if the length of the

specimen is more as compared to the cross-sectional dimensions. Therefore, an


PAGE NO.

OBSERVATION TABLE:

TYPE OF

SPECIMEN

DIMENSIONS

BEFORE TEST AFTER TEST

DIA

(mm)

HEIGHT

(mm)

ASPECT

(mm)

HEIGHT

(mm)

TOP

DIA

MIDDLE

DIA

(mm)

BOTTOM

DIA

MILD 1.

STEEL

2.

CAST 1.

IRON

2.

CALCULATION:

TYPE OF

SPECIMEN

ASPECT

RATIO

MAXIMUM

LOAD (N)

MAXIMUM STRESS

(N/mm 2)

MILD STEEL

1.

2.

CAST IRON

1.

2.


PAGE NO.

attempt is made to obtain the uniform stress distribution of direct stress over the

cross-section normal to the direction of the load by limiting the length of the

specimen with respect to cross-sectional dimensions, holding the specimen properly

and applying truly axial load.

Ductile material viz. mild steel when subjected to this test, keeps on expanding in

lateral directions (known as bulging) as the load increases. This bulging enables it to

resist almost indefinitely large force without fracture. Therefore, for ductile materials,

compressive strength is the strength with respect to an arbitrary chosen deformation.

This ability of a material to undergo large plastic deformation under compression is

called malleability of ductile material. Fig 9.2 shows the behavior of ductile material

under compression load with different aspect ratios. Brittle materials such as C.I.

usually fails by shearing along the plane inclined (45 + Ø/2)0

with longitudinal axis,

where Ø is defined as angle of internal friction and is defined as maximum angle at

which the particle of a matter doesn‟t slip when placed on an incline. Fig 9.3 shows a

specimen with its angle of internal friction.

SPECIMEN:

For uniform stressing of the compression specimen, a circular section is to be

preferred over other shapes. An aspect ratio (defined as ratio height to diameter) of 2

is generally adopted to avoid buckling of the specimen. In this test, two 16 mm

diameter specimen of aspect ratio 1.25 and 3.75 are adopted to understand the effect,

of slenderness of a specimen when subjected to compressive load.


PAGE NO.

PRECAUTIONS

1) Specimen should be placed carefully in the centre so that applied load will be

exactly an axial load and no eccentricity for the load would occur for the proper

transfer of load.

2) End surfaces of the specimen should be perfectly smooth.

3) When the load is nearing its failure value, it is preferable to remain at a safe distance

from the machine.


PAGE NO.

PROCEDURE:

1) Measure the length and diameter of both M.S. Specimen with the help of vernier

caliper.

2) Place the specimen with the aspect ratio 1.25 properly on the lower cross-head and

bring the middle cross head to a level just touching the upper surface of the specimen.

3) Adjust the capacity of the machine as per the anticipated value to be applied.

4) Apply the compressive load gradually

5) After reaching to the specified value of load, stop the machine and unload the

specimen

6) Repeat steps 1 to 5 for another mild steel specimen with aspect ratio of 3.75.

7) Repeat step 1 to 3 for specimen of C.I. and measure only the maximum load at

fracture.

8) Test the specimen up to its fracture

9) Unload the specimen and repeat steps 7 to 8 for another cast iron specimen with

aspect ratio 3.75.

10) Examine and study the type of fracture/ deformation characteristics for all the

specimens.


PAGE NO.

CONCLUSION: Study the fracture/ deformation characteristics of the specimen with

respect to

1. Different aspect ratios

2. Different materials used.

ASSIGNMENT:

1) Identify different elements under compression and write a short note on it

GRADE

SIGNATURE:


PAGE NO.

PRACTICAL NO.10

COMPRESSION TEST ON TIMBER


PAGE NO.

Figure 10.1


PAGE NO.

COMPRESSION TEST ON TIMBER

OBJECT: To determine the compressive strength of the timber specimen and to study

its behavior when subjected to load up to fracture.

EQUIPMENT: Universal testing machine with an attachment to test specimen in

compression

APPARATUS: Vernier caliper

MATERIAL: Timber

RELATED I.S.CODES: I.S. 1708-1986: Methods of testing of small clear specimens

of timber.

SALIENT FEATURES

Wood exhibits, under compressive loading, a behavior peculiar to itself. It is a

homogeneous material, being composed of cells formed by organic growth which

align themselves to form a series of tubes or column in the direction of grain

alignment. As a result of this structure, the elastic limit is relatively low, there is no

definite yield point, and considerable set takes place before failure. These properties

vary with the orientation of the load with respect to the direction of the grain. For

loads normal to the grain, not only the elastic limit is important but also the strength at

rupture. Rupture often occurs because of collapse of the tubular fiber structure.

Important variables which affect compressive strength of timber are:

a) Shape and size of the specimen

b) Capping on top and bottom of the specimen

c) Speed of testing


PAGE NO.

Figure 10.2.a Figure 10.2.b


PAGE NO.

In this test, standard specimens are compressed between two platens under axial

load.

Figure 10.1 shows the line diagram of the machine. The mechanical properties such as

yield strength, ultimate compressive strength, modulus of toughness are determined

by studying the behavior of the specimen under load. Since in this experiment the

basic aim is to find and compare the strength of timber under different orientation,

therefore, importance is given only to find ultimate strength of the specimen. The

possible patterns of the failure of the specimen are shown in the figure. Fig 10.2 a and

10.2 b shows the failure of the specimen under adopted orientations.

SPECIMEN:

The size of the test specimen shall be 50 x 50 x 200 mm or 20 x 20 x 80 mm if it is to

be subjected to load parallel to the grain and its size shall be 50 x 50 x 150 or 20 x 20

x 100mm if it is subjected to load perpendicular to the grain. Figure 2.3 a and 2.3 b

shows the arrangement of the specimen when subjected to load. The specimen shall be

free from defects and shall not have slope of grain more than 1 in 20 parallel to its

longitudinal edges. The end plane of the specimen shall be perfectly at right angle to

the length of the specimen.

PROCEDURE:

1) Check the specimen for any defects and ensure that the ends are plane and fibers

are parallel to its longitudinal edges.

2) Measure the cross- section of the specimen and length of the specimen with

vernier calipers.

3) Place the specimen in the lower cross-head such that the load is applied parallel

to the grain.

4) Apply the load up to failure of the specimen and observe the type of failure.


PAGE NO.

Figure 10.3 a

Figure 10.3.b


PAGE NO.

5) For second case, place another specimen such that the load is applied

perpendicular to the grain.

6) Repeat step 4.

PRECAUTIONS:

1) Specimens should be placed carefully in the centre so that the applied load will be

exactly an axial load and no eccentricity for the load would occur for the proper

transfer of load.

2) When the load is nearing its failure value, it is preferable to remain at a safe

distance from the machine.

LIMITATIONS:

1) It is difficult to ensure that the load applied on the specimen is truly concentric and

axial.

2) There is always a tendency for bending stress to be set up which may result into

slipping of the specimen as the load increases.

3) Friction between the heads of the testing machine and the end surfaces of the

specimen due to lateral expansion of the specimen may alter results considerably in

comparison to those obtained in the absence of such lateral restraint.

4) The relatively larger cross-section area of the specimen is required to obtain a

proper degree of stability of the specimen.


PAGE NO.

OBSERVATION TABLE:

CROSS- SECTIONAL

AREA

(mm2)

LOAD CASE

LOAD APPLIED

PARALLEL TO GRAIN

LOAD APPLIED

PERPENDICULAR TO

THE GRAIN

LOAD AT FAILURE

(N)

ULTIMATE

STRENGTH (N/mm2)

CONCLUSION:

Comment on the nature of fracture and the strength of the specimens.

ASSIGNMENT:

1) Explain effect of moisture on the strength of timber.

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.11

TENSION TEST ON METALS


PAGE NO.

Fig 11.1

Fig 11.2

Fig 11.3


PAGE NO.

TENSION TEST ON METALS

OBJECT: To study the behavior of metallic materials under the tensile load with the

help of stress-strain curve on universal testing machine.

EQUIPMENT: Universal testing machine

APPARATUS: Vernier- Calliper.

MATERIALS: Mild steel and Cast Iron

CALIBRATION

The machine is calibrated with the help of pivoting ring. The pivot ring is kept on

lower cross head and compressive force is applied on it. The deflection of the dial of

pivot ring is compared with the standard values provided by manufacturer. If the

values so obtained do not fit in desired accuracy of 1%, the counterweight on the

pendulum is adjusted to bring the readings within permissible limits.

RELATED I.S. CODES

IS: 1608-1995: Mechanical testing of metals- Tensile testing

IS: 1816-1979: Methods of tensile test for light metals & their alloys

SALIENT FEATURES

This test is performed on Universal Testing machine. The machine is also used for

testing materials subjected to tension, compression, bending and shear; Fig 10.1 shows

the line diagram of the Universal testing machine. The test consists of straining a test

piece by force, generally to fracture for the purpose of determining one or more

mechanical properties viz Percentage elongation,


PAGE NO.

OBSERVATION TABLE

SR.NO LOAD

(N)

ELONGATION

(mm)

STRESS

(N/mm2)

STRAIN

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15


PAGE NO.

Percentage reduction in area, modulus of elasticity, Yield stress, Ultimate stress and

breaking stress.

Metals used in engineering structures are mostly ductile e.g. steel, aluminum,

brass, copper, but some metals like cast iron are brittle. When specimen is tested under

tensile load, both these types behave differently. Ductile metals undergo comparatively

large elongations, along the direction of loading accompanied necessarily, with

contraction in the traverse direction. At a certain stress level, the specimen ceases to

take any additional load but the elongation continues to a certain level, where after

additional load is required to increase the elongation. At fracture, ultimately, the

specimen breaks at or very near the narrowest section. The fracture is not along a plane

right angle it is along an inclined plane and has an appearance of a cup and cone shape.

Fig 11.1 shows a typical fracture of a mild steel specimen. Obviously therefore,

fracture is not due to insufficient tensile resistance but due to insufficient shear

resistance on inclined planes.

Thus the stages which a ductile material undergoes are yield-ultimate- breaking

stress. The value of braking if calculated with respect to its original area of the

specimen, comes to be lesser than ultimate stress and is known as nominal breaking

stress, whereas if breaking stress is calculated taking into consideration the

instantaneous cross-sectional area, its value comes to be greater than ultimate stress

and is known as actual breaking stress or true stress, since engineering design takes

into consideration the loads with respect to yield stress, true stress is generally not

plotted in the stress-strain curve. Fig 11.3 shows the behavior of mild steel when

subjected to tensile load. Brittle metals have inadequate tensile resistance, they do not

undergo any worthwhile elongation, they fracture is comparatively abrupt and the

fractured surface


PAGE NO.

DIMENSIONS OF THE SPECIMEN BEFORE TESTING

SR.NO PARAMETERS MILD STEEL CAST IRON

1 Gauge length (L0) mm

2 Diameter (d0) mm

3 Cross-sectional Area (A0) mm2

DIMENSIONS OF THE SPECIMEN AFTER TESTING

SR.NO PARAMETERS MILD STEEL CAST IRON

4 Final length (Lf) mm

5 Diameter (df) mm

6 Cross-sectional Area (Af) mm2

LOAD AT CRITICAL POINT OF OBSERVATIONS

7. Yield load (Py) N

8. Ultimate load (Pu) N

9. Breaking load (Pf) N


PAGE NO.

is a clear right section. In such metals, the maximum load is also the breaking load fig

11.3 shows the behavior of cast iron when subjected to tensile load. Typical fractures

of important materials under tension are shown in fig 11.2

SPECIMEN

The specimen is specially prepared as per prescribed standard guidelines, which are so

framed as to induce failure within the length earmarked for observation called the

gauge length. Test piece, the original gauge length of which is related to the original

cross sectional area (A0) by the equation L0 = 5.65 A0 are called proportional test piece.

The portion in which the specimen is gripped is also highly stressed, but the

dimensions are so adjusted that these stresses do not affect failure within the gauge

length. Fig 11.1 shows the details of fillet on a specimen.

PROPERTIES

Test is essentially meant to provide data which can be used for engineering design.

PROCEDURE

1. Measure the cross-sectional dimensions of the given specimen and mark gauge

length points on the specimen with the help of hammer & punch as per the instructions.

2. Adjust the capacity of U.T.M as per the anticipated value of load at failure of the test

specimen. 3. Fix the specimen in machine between the fixed cross head and upper

cross head also see that the specimen is symmetrically with respect to its longitudinal

axis through out the length.

4. Apply the load gradually so as to avoid sudden shocks.

5. Note the ultimate load and breaking load. Measure the distance between gauge

points and reduced diameter of failure.


PAGE NO.

6. Observe the type of failure.

7. Plot a graph of load verses deflection and find the value of modulus of elasticity.

8. Determine the stress-strain characteristics and comment on the type of material from

the nature of the fracture.

9. Draw a graph of load vs. Deflection for both the materials.

CONCLUSIONS

ASSIGNMENT:

Comment or compare the following items as obtained in the experiment with their

corresponding standard theoretical values:

1. Stress-Strain characteristics

2. The various stress limits

3. Yield and elongation

4. Area under the curve significance

5. Fracture surface

GRADE:

SIGNATURE:


PAGE NO.

Material Elastic Curve Yield Point Inelastic curve

Mild steel,

Structural steel

Linear Upper & Lower yield

point both distinctly

visible

Yield- ultimate-

breaking

Aluminum,

copper & their

alloys

Linear Only one yield point Yield- ultimate-

breaking

High yield

deformed bar

Linear Equivalent to proof

stress

Yield- ultimate-

breaking

Cast Iron

Concrete

Linear non-

Linear

Ultimate


PAGE NO.

CALCULATIONS

SR.NO PARTICULARS FORMULA MILD STEEL CAST IRON

1 Yield

stress(N/mm2)

Py/Ao

2 Ultimate

stress(N/mm2)

Pu/Ao

3 Nominal

breaking stress

(N/mm2)

Pf/Af

4 Actual breaking

stress(N/mm2)

Pf/Af

5 Percentage

reduction in area

%

Ao -Af/Ao x

100

6 Percentage

elongation %

Lf –Lo /L0 x

100

7 Modulus of

Elasticity

(N/mm2)

E


PAGE NO.

PRACTICAL NO.12

SHEAR TEST


PAGE NO.

Figure 12.1

Figure 12.2


PAGE NO.

SHEAR TEST

OBJECT: To determine shear strength of given mild steel specimen.

EQUIPMENT: Universal Testing Machine

APPARAUS: Vernier Caliper

SPECIAL TOOL: Shear Attachment

Material: Mild Steel

RELATED I.S. CODES

I.S. 5242-1979: Method of test for determining shear strength of metals.

SALIENT FEATURES

The test is carried on UTM with the help of a special tool known as shear attachment

and is shown in fig 12.1 this test consists of breaking the specimen under single or

double shear.

This test is essential, as many machine parts and structural members are

subjected to shear stress and such static properties in shear play an important role in

design. A member is said to be subjected to shear if the line of action of force is

parallel to the plane of the cross-sectional resisting it. If only one plane resists, it is

called as single shear and if two planes resist shear, it is called as double shear. A

shearing stress, acts parallel to a plane, whereas tensile and compressive stresses act

normal to the plane.

This specimen is tested in the form of a prismatic bar which is held rigidly at one end

(in single shear) or at both ends (in double shear) and load is applied through a


PAGE NO.

OBSERVATION TABLE

SR.NO. TYPE OF

STRESS

DIAMETER(mm) C/S AREA

(mm2)

LOAD

(N)

1. SINGLE SHEAR

2. DOUBLE

SHEAR

CALCULATIONS

STRESS (N/mm2)

CASE

AVERAGE SINGLE SHEAR DOUBLE

SHEAR


PAGE NO.

metallic piece having a hole through which the specimen passes. The load application

is very near the support to avoid the bending of the specimen, the line sketch is shown

in fig 12.2

The test consists of gradual application of load till the specimen fail in shear.

Maximum load divided by the area under shear is called as the shear strength of the

specimen

Shear Strength= shear force at failure

Cross-sectional area resisting shear.

Following are the important definitions relevant to this test:

1. Complementary shear:

This shear balances the externally applied shear to keep the element in equilibrium.

The value of this is equal to externally applied shear and acts in the opposite direction.

2. Pure Shear:

When an element is acted upon by the shear stress, it is said to be in pure shear. The

condition of pure shear is difficult to attain. The test carried over here is a theoretical

example of an element subjected to pure shear.

3. Punching Shear:

When a body bears on another the type of stress developed in the second one is called

punching shear.


PAGE NO.

SPECIMEN

The specimen is specially prepared and possesses smooth surface and the diameter is

just equal to the diameter of dies. It is important to note that there should not be any

PRECAUTIONS

1. It is advisable to remain away from the machine when test is being carried out.

2. Loads should be applied at a gradual rate so as pre the requirement of

application of static load.

3. Specimen should have diameter just equal to the inner diameter of the dies.

LIMITATIONS

1. It is not possible to calculate the modulus of rigidity owing to impossibility of

measuring strain during the test.

2. The accuracy of the result depends on the hardness and the sharpness of the

edges of the dies used.


PAGE NO.

gap left between the inside diameter of the dice and the diameter of the specimen, as it

induces bending, and the fracture occurs due to combined effect of bending and shear,

which is undesirable.

APPLICATIONS

The result of this test is useful in designing rivets, crank pins etc which are subjected to

shear loads. Also, determination of the shear strength is useful to design the members

subjected to combined effect of bending and shear such as machine shaft etc.

PROCEDURE:

1. Determine the mean diameter of the specimen with the help of vernier caliper.

2. Place the specimen in the assembly for single/ double shear.

3. Keep the assembly with the assembly with specimen on the lower cross-head and

bring the middle crosshead just above the shear attachment.

4. Apply the load gradually till the specimen fails in single/ double shear and note the

corresponding load from the dial gauge.

5. Study the type of fracture.


PAGE NO.

CONCLUSION

ASSIGNMENT:

1) State the types of shear.

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.13

BREINELL HARDNESS TEST


PAGE NO.

Figure 13.1

Figure 13.2


PAGE NO.

BRINELL HARDNESS TEST

OBJECT: To determine the Brinell hardness number of given metal specimens by

Brinell hardness test.

EQUIPMENT: Brinell hardness testing machine

APPARATUS: Microscope

MATERIAL: Steel, Brass, Aluminum, Copper, Cast Iron

RELATED I.S. CODES

IS 3034 -1965: Method for brinell Hardness test for copper & copper alloys

IS 1500-1983: Methods for Breinell Hardness test for steel

IS 1790-1961: Methods for Breinell Hardness test for light metals & their alloys

SALIENT FEATURES

Hardness is basically an important mechanical property of the metals and is defined

as the resistance given by metal specimen to indentation, scratching or abrasion on its

surface. Brinell hardness is a method of finding hardness of given specimen by

indentation and was introduced by J.A.Breinell in 1900. This method uses a steel ball

gradually. The hardness tester is shown in fig 13.1.

The hardness number is then calculated from the depth of the indentation (of

diameter d) produced by the load applied. Thus, it is an indirect method of finding

hardness. Breinell hardness is defined as the ratio of the load applied to the spherical

area of the indentation formed on the specimen surface and is equivalent to kgf/mm2.

Fig 13.2 shows the indentation on the metal surface.


PAGE NO.

OBSERVATION TABLE

SR.NO METAL BALL

DIAMETER

D (mm)

TEST LOAD

P (kg)

DIAMETER OF

INDENTATION

D (mm)

1. MILD STEEL

2. CAST IRON

3. BRASS

4. COPPER

5. ALUMINIUM

CALCULATIONS

SR.NO METAL BHN REMARK

1. MILD STEEL

2. CAST IRON

3. BRASS

4. COPPER

5. ALUMINIUM


PAGE NO.

BHN= TEST (kgf) /surface area of indentation (mm2)

Where,

h= depth of indentation

F=force applied in kgf

D= diameter of ball in mm (generally 1, 2, 2.5, 5 and 10 mm)

Load to be applied in kgf is shown as follows according to the provisions of Indian

standards.

PROCEDURE

1. Check the specimen as per I.S.Code specifications.

2. Select the appropriate ball indenter and the corresponding load to be applied on the

specimen selected.

3. Place the specimen on the anvil and bring the indenter in the contact with the surface

of the specimen selected.

4. Apply the load at a gradual rate till the test load „F‟ is attained. The time from the

initial application of force until the full test load is reached shall not be less than two

seconds nor greater than 8 seconds. The test load shall maintain for ten to fifteen

seconds for steel 30 ± 2 seconds for light metal and 15 to 20 seconds for grey cast

iron.

5. Unload the specimen by releasing the lever.


PAGE NO.

6. Measure the Diameter of the indentations at right angles with the help of the

microscope. The average of these readings gives the diameter of the indentation.

7. Calculate the BHN of the given specimen by the formula given in the theory.

PRECAUTIONS

1. The surface of the test specimen shall be smooth and even free from oxide scale

foreign matter and in particular, completely free from lubricants.

2. Preparation of the test specimen shall be carried out in such a way that any

alteration of the surface due to heat or cold working is minimized. Ambient

temperature for carrying out the test is 100

to 350 C.

3. The distance between the centre of any indentation and the edge of the test piece

shall be at least 2.5 times the mean diameter of the indentation in case of steel, cast

iron, copper and copper alloys and at least three times the mean diameter of the

indentation in case of light metals, lead, tin and their alloys.

4. The distance between the centre‟s of the two adjacent indentation shall be at least

four times the mean diameter of the indentation in the case of steel, cast iron,

copper and copper alloys and at least six times the mean diameter of the indentation

in case of light metals, lead, tin and their alloys.

LIMITATIONS

1. Specimens with lesser thickness viz. blades, if tested by this methods, would not

give accurate results as the usual indentation might be greater than the thickness of

the specimen and lesser thickness of the specimen shall be at least eight times the

depth of the indentation


PAGE NO.

2. It is recommended that the steel ball indenter should be used for materials with a

Brinell hardness not exceeding 350. For higher up to 630. Indentor of carbide metal

should be used.

APPLICATIONS

This property of the metal has a very large application in the industry, right from a

small needle to gigantic aircraft. The quality of the metal to be used for a specific

purpose is controlled by this test e.g. the drill shall have hardness greater than the

material to be worked on. Similarly, strength of jobs like forging, alloying of case-

hardening is determined with this test.

LOAD (kgf) MATERIAL REMARKS

30 D2 STEEL -

30 D2 GREY C.I -

10 D2 GREY C.I For HB less than 140

5D2, 10D2, 15D2 Light metal and alloys 5D2 for HB less than 55

5D2 Copper and its alloys For HB less than 40

10D2 Copper and its alloys For HB 35 to 200

30D2 Copper and its alloys For HB greater than 190

I.S limit for Breinell hardness test on steel is HB 450 and balls of diameter 2.5mm

and 1mm should not be used for conducting hardness test on cast metal


PAGE NO.

CONCLUSIONS

ASSIGNMENT:

1) Define hardness and state types of hardnes.

GRADE:

SIGNATURE:


PAGE NO.

PRACTICAL NO.14

IZOD IMPACT TEST


GIT – GANDHINAGAR INSTITUTE OF TECHNOLOGY PAGE NO.

Figure 14.1

Figure 14.2



IZOD IMPACT TEST

OBJECTIVE: To study the behavior of metals under impact loads.

EQUIPMENTS: Izod impact testing machine

MATERIALS: Mild steel, Cast Iron, Aluminum and Brass

REFERENCES

IS: 1598:1997: Methods for Izod impact test on metals.

SALIENT FEATURES

Components of many structures of machines have to resist dynamic loads which

are either impact loads or rapid flucating loads. Impact loads are applied suddenly and

this provides information about the behavior of materials under such loads. The

predominant consideration in impact resistance is the capacity of rupture. A material to

absorb energy which depends on toughness of material (energy required to rupture a

material). In this test, toughness is not measured through strain or deflection

measurements but it is ascertained by measuring energy to rupture a specimen. Fig

14.1 shows the line diagram of the machine.

In such tests, the load may be applied in flexure, tension, compression, or torsion and

impact may be delivered through dropping weight, swinging pendulum or rotating fly

wheel. In this test, the specified cantilever length of the specimen is projecting out

from vise with notch at the base as shown in fig14.2. and swinging pendulum strikes

on the edge of the specimen giving a flexural type of loading. Indian standards lay

down the following major requirements for the impact testing machines.



OBSERVATION TABLE

SR.

NO.

SPECIMEN GROSS

ENERGY

(Kg-m)

ENERGY

LOSSES

(Kg-m)

IMPACT

VALUE

(Kg-m)

MODE OF

FAILURE

1. MILD STEEL

2. COPPER

3. BRASS

4. CAST IRON

5. ALUMINIUM



PROCEDURE:

1. Check the specimen measure its dimensions and ascertain that the IS requirements

are satisfied.

2. Fix the specimen tightly in the vise and ascertain that the longitudinal axis of the

specimen lines in the plane of swinging of the hammer.

3. Plane of symmetry of notch should coincide with the top surface of the grips. The

hammer should strike on the same side of the notch.

4. Set the pointer to read the energy of the blow of the pendulum.

5. Release the pendulum and allow it to strike the test specimen.

6. Read the indicator and obtain impact value of the specimen by deducting the energy

losses from the reading.

7. Study the type of fracture and correlate it with fracture of specimens of different

material.

8. Repeat steps 4 to 7 without specimen the indicator reading shows the energy losses

due to friction. Note down this value.

9. The impact value of the material is obtained by subtracting reading obtained in step

8 from step 6.

LIMITATIONS

To determine the impact resistance, temperature should be 320

to 380

because it has a

very marked effect on impact resistance of notched bar.



REQUIREMENTS IZOD IMPACT TEST

Distance between the base of notch and

point of specimen hit by hammer

22 mm± 0.5 mm

Angle at tip hammer 750

Speed of hammer 3 to 4 m/s

Striking energy of hammer 16.56 ±0.34 kgm

Accuracy of graduation of scale ±0.14 kg

Specimen used in this test are notched type which causes high localized stress

concentration, artificially reduces ductility and tends to introduce a brittle type of

fracture.

SPECIMEN

Specimen with notch at suitable position is used for carrying out the impact test on

metals for steel specimen. Requirement is as follows:

SHAPE OF THE

SPECIMEN

TYPE TYPE OFNOTCH REFERENCE

Square or round Cantilever beam

Fixed at notch

V notch at base Fig no14.2.



PRECAUTIONS

1. It is not advisable to stand near or in front of machine when the pendulum is to be

released.

2. Notch should be exactly in the line of action of the pendulum.

3. When specimen is being fitted, care should be taken that pendulum does not get

released.

APPLICATIONS

The Impact test is basically designed to check the suitability of materials subjected to

dynamic loads e.g. vehicles travelling over bridges, hull of ship subjected to waves or

hammer falling on nail. The ductile material like steel has more impact value and

therefore has wide acceptability in structural and other mechanical applications.

CONCLUSION

ASSIGNMENT:

1)Students should observe the systems which are subjected to impact load and try to

enlist few of them.

GRADE:

SIGNATURE:

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