FACULTY OF CHEMICAL ENGINEERING

THERMODYNAMICS AND MATERIAL ENGINEERING LABORATORY

Group Member : Laboratory Stamp

Date of Submission: Date of Experiment:

Lecturer’ Name: Section:

EXPERIMENT 1 : DANDANG MARCET (Marcet Boiler)

1.0 Objectives To study the relationship between pressure and temperature for saturated steam 2.0 Apparatus Marcet boiler 3.0 Theory Thermodynamic properties of a pure substance can be determined by two of the following characteristics: T, P, H, S, V etc. Data characteristics can be obtained either from experiments or equations, or using certain. Between two thermodynamic properties that are easily measured T and P. Relations between these properties can be expressed in the form of graphs or equations. Clausius-Clapeyron equation one has published the following: where Vf = volume of water Vg = volume of steam Hf = Enthalpi water Hg = Enthalpi steam Hfg = (Hf-Hg) = Heat dormant of vaporization Dormant for the heat value of water can be obtained from the equation.

4.0 Method 1. Fill water into the boiler by opening the filler plug water. Make sure the valve

overflow is open, and let the water out of there when the boiler full. Close the filler plug back.

2. Heat the water in the boiler has been completed while the valve overflow is opened.

3. When the temperature around 97C to 100C show cloase valve overflow. 4. Note the increase in temperature for each pressure of 0.5 bar until the

pressure reaches 5 bars. 5. Switches off heating and record the fall of temperature decrease for each 0.5 bar

pressure. 6. Record results in Table 1.

5.0 Results

i. Draw P versus T graphs for the process of heating and cooling process.

ii. Find the slope graph

iii. Calculate the value TVg / Hfg using Table 2 and compare with the slope of the

graph you draw.

6.0 Discussion

Discuss your observations during the tests conducted. Discuss the graph that plotted and indicate the relationship between temperature and pressure steam. 7.0 Conclusion State the conclusions that can be obtained. 8.0 References Cengel, Y.A. and Boles, M.A. Thermodynamics: An Engineering Approach, 7th ed. McGraw Hill, New York Wylen G.V. , R. Sonntag, and Borgnakke C., Fundamentals of Thermodynamics, 6th ed., John Wiley, New York.

Table 1: Results of tests

Gauge

Pressure

bar

Absolute

Pressure

bar

Steam Temperature

Measured

Slope

dT/dP

Calculated

Slope TVg/Hfg Increasing

Pressure

Decreasing

Pressure

Mean

EXPERIMENT 2: UNIT PEMAMPAT SATU-PERINGKAT (IMBANGAN TENAGA)

(Single-stage compressor (energy balance))

1.0 Objectives

i. To evaluate how quickly a single stage compressor heats up

ii. To analyze the steady-state heat loss of a single stage compressor

2.0 Apparatus

Single stage compressor

3.0 Theory

In a standard single stage vane compressor (Figure 1), a fixed volume of air is trapped

between two sliding vanes driven by an offset rotor. This volume of air is pushed to the outlet

where the area (and hence volume) is reduced. Its pressure is thus higher and it is squeezed out

of the outlet port.

Figure 1: Schematic of a Single-Stage Compressor Rig

4.0 Method

1) Turn the ‘power’ switch on.

2) Open the green inlet valve fully (i.e. turn the valve ‘anti-clockwise’ to open it).

3) Record initial readings for all parameters given in Table 1 at the beginning of the

experiment. Use a tachometer to record the motor speed, N.

4) Turn the ‘motor’ switch on.

5) Using the red outlet valve, set the compressor outlet pressure, p2 to 1 bar gauge (i.e. turn

the valve ‘clockwise’ to increase pressure).

6) Adjust the spring balance to ensure that the indicator on the motor is between the two

black lines prior (i.e. before) recording any motor ‘force’ values.

7) Record the temperature of the outlet air, T2 at one minute intervals (e.g. use a stop watch)

until two consecutives readings are the same (i.e. € < 0.5%).

To stop the experiment:-

1) Using the red valve, reduce the compressor outlet pressure, p2 to 0 bar gauge (i.e. turn the

valve ‘anti-clockwise’ to decrease pressure).

2) Turn the ‘motor’ switch off.

3) Turn the ‘power’ switch off.

5.0 Analysis

Specific volume through flow meter, (m3 kg

-1), with R = 0.287 kJ kg

-1 K

-1.

*Don’t forget to convert pressures and temperatures unit into kPa (absolute) and oK,

respectively.

Volumetric flow rate, (m3s

-1)

Mass flow rate, (kg s-1

)

Energy added to air by compressor, = , (W)

Power out of motor, = (W) Heat emitted from compressor, = -

6.0 Results

Data:

Length of torque arm L = 0.2 m

Compressor swept volume V Vcomp = 267 cm3 rev

-1

Specific heat capacity of air cp = 1005 J kg-1

K-1

Table 1: Parameters consisting of important parameters for the single stage compressor unit

Parameters

Initial readings

Air temperature into apparatus, To (oC)

Air temperature into compressor, T1 (oC)

Air temperature out of compressor, T2 (oC)

Atmospheric pressure (from barometer), Po (bar)

Vacuum into compressor (gauge), p1 (bar)

Air pressure out of compressor (gauge) p2 (bar)

Flow meter, Vin (l min-1

)

Motor speed, (rev min-1

), N (rev min-1

)

Motor spring balance, F (kg)

Electric power in, (Watt)

*Record all the above data at the beginning of the experiment

Table 2: Outlet temperature data over time (i.e. at 1 minute time interval)

Time, t (min) Temperature, T2 (oC)

7.0 Discussion

Plot the outlet temperature, T2 data over time (i.e. data from Table 2). Additionally, calculate and

plot energy added to air by compressor, and heat emitted from the compressor, against time,

respectively.

Discuss the followings:

a. Why is the heating up curve this shape?

b. What is happening to the energy that is being input to the system as it heats up?

c. How does the amount of energy lost as heat compare to

i) The work input to the compressor?

ii) The heat added to the air?

8.0 Conclusion

Draw conclusion about the behavior (i.e. dynamic profile) of the single stage compressor as it

heats up. Also, speculate the results (i.e. heating curve) if the compressor outlet pressure, p2 is

reduced to 0.5 bar.

9.0 References

Cengel, Y.A. and Boles, M.A. Thermodynamics: An Engineering Approach, 7th ed. McGraw Hill, New York

Wylen G.V., Sonntag, R.and Borgnakke, C., Fundamentals of Thermodynamics, 6th

ed., John

Wiley, New York.

EXPERIMENT 3 : HUKUM PERTAMA DAN KEDUA TERMODINAMIK

(Application Of First And Second Law Of Thermodynamics )

1.0 Objective

To apply the first law of thermodynamics and the concept of the second thermodynamic

law in a process

2.0 Apparatus

Hot Plate, Balance and Thermometer

3.0 Theory Energy balance or the first law of thermodynamics is a basic calculation of heat transfer. The energy balance for the system is

Ein - Eout = Esystem

U system = 0

Usystem = U hot water + U cold water = 0

mc (T2-T1)hot water + mc (T2-T1)cold water = 0

where m = mass, c = heat capacity and T = temperature.

Clausius statement saying it is not possible to heat flow from the reservoir to the cold heat

reservoir without doing any work. In other words, heat can be transferred from the hot reservoir

to the cold reservoir easily without doing work.

4.0 Method

1) Weigh two empty containers(different size). 2) Fill the large container with a little cold water and weigh the mass of cold water.

3) Fill the smaller container with hot water (80-90 C) and weigh the mass of hot water.

4) Place the hot water container inside the cold water container. Note the reading of both water temperature for every 1 minute until they reach the same temperature (thermal equilibrium).

Figure 1: Schematic Diagram of the Experiment

5.0 Results

Data: Mass of hot water used :__________

Mass of cold water used :__________

Time (min) Hot water (C) Cold water (C)

Cool water

Hot water

6.0 Analysis

i. Draw water temperature against time to get the equation for the graph (curve fitting)

T for hot water and cold water. ii. Get the change of temperature ( Calculate the energy change for both.

7.0 Discussion Discuss your observations during the tests conducted. Compare changes in energy for hot water and cold water. Try associate with the second law of thermodynamics. 8.0 Conclusion State the conclusions that can be obtained. 9.0 References Cengel, Y.A. and Boles, M.A. Thermodynamics: An Engineering Approach, 7th ed. McGraw Hill, New York. Wylen G.V. , R. Sonntag, and Borgnakke C., Fundamentals of Thermodynamics, 6th ed., John Wiley, New York.

EXPERIMENT 4: SISTEM PENYEJUKAN MAMPATAN WAP (Vapour Compression Refrigeration)

1.0 Objective

To determine Coefficient of Performance (COP) for vapor compression cooling system 2.0 Apparatus Refrigeration Cycle Demonstration Unit, R633, PA Hilton is using R-141b refrigerant fluid 3.0 Theory Cooling system receives heat from the cold space and transfer heat to space. This process can only be done if the work is supplied to the system as depicted in Figure 1.

Figure 1: Schematic Diagram of a Refrigeration System

Coeeficient of Performance (COP) can be calculated based on the following equation:

COP = Cooling Effect /Work, net in

4.0 Method Identify the following instruments: Evaporator, Condenser, Compressors, expansion valves, wattmeter, water flows in and out, and the flow of fluid R-141b as shown in Figure 2 (Diagram 1). Make sure the thermometer each position. Do not blow out the Vent Valve.

Hot Tank

Cool Tank

Q2

Q1

W

Evaporator

1

2

3 4

Compressors Injap

Throttle

Q2

Q1

Condenser

Diagram 1

Diagram 2

Diagram 3

Figure 2: The Experimental Set-Up

Launch operations and take readings 1. Open the valve as shown in Diagram 2. 2. Urn on the water pipe. Set the water flow to the evaporator at the rate of flow

25 g/s and flow rate of water flow rate to Condenser as maximum. 3. Then switch on the device. 4. Allow cooling system to stable (approximately 10 min). Record readings

As shown in Table 1 5. Repeat the experiment by changing the flow rate of water flow rate to Condenser

In between 35 to 40 g/s. Repeat 4 times.

a. Results Record your data as in Table 1.

Terminating Operations When experiment is done, make sure the device is closed in accordance to these procedure. 1. Switch off the main switch. 2. After one minute, Turn off the water supply pipes. 3. After one minute, close the valves to the "Shut Down Position" as the shown in

Figure 2(Diagram 3). 7.0 Discussion i. Calculate the rate of heat transfer that occurs in the condenser and the

evaporator for the five tests run. Discuss the results. ii. What are the effects of changing the condenser flow rate of T1-T6 to compressor

power. Draw graph. Discuss the results obtained. 8.0 Conclusion State the conclusions that can be obtained. 9.0 References Cengel, Y.A. and Boles, M.A. Thermodynamics: An Engineering Approach, 7th ed. McGraw Hill, New York. Wylen G.V. , R. Sonntag, and Borgnakke C., Fundamentals of Thermodynamics, 6th ed., John Wiley, New York.

Table 1: Data for Vapour Compression Refrigeration

Test No. 1 2 3 4 5

Absolute condenser

Condenser Gauge

pressure

Absolute evaporator

evaporator Gauge

pressure

Compressor Power input, W

Temp T1

Temp T2

Temp T3

Temp T4

Temp T5

Temp T6

Temp T7

Heat Transfer evaporator

Heat Transfer condenser

EXPERIMENT 5: MOMEN LENTUR (BENDING MOMENT)

1.0 ABSTRACT

This is an introductory experiment of bending moment. The objectives of the experiment

include to comprehend the action of moment of resistance in a beam and also to measure the

bending moment at a normal section of a loaded beam and to check its agreement with theory.

An experimental beam is used to carry out the bending moment test. The beam is divided

into two parts and is connected by a joint, thus forming a “frictionless” hinge. A moment of

resistance about the hinge is provided by an under slung spring balance, which acts at a lever arm

of 150mm. There are three load hangers placed on the beam to induce load onto the beam.

In this experiment, the experimental beam will be induced with external load, which the

load will be put on the load hanger. The beam will bend after the load is put on. Spring balanced

is done each time after the load is altered. Then, the reading of the spring balance is recorded

after that. The more load applied onto the beam, the more the beam will bend, especially where

the load is altered at the joint of the beam.

From all the data taken during the experiment, the bending moment for each loading

arrangement at the section is calculated. The bending moment diagram is drowned to compare

the measured value with it. The measured bending moment in N.mm will be 150 times the

change in the spring balance reading. The similarity and difference between the experimental

value and theory will be identified and discussion is made accordingly.

Lastly, conclusion is made based on the bending moment diagram, the importance of the

relationship between load and bending moment in engineering application especially in the field

of construction and manufacturing. The values from experiment may differ from theory as errors

may happen in the process of experiment.

2.0 OBJECTIVE

To study the property of material (bending moment) when subject to different load

3.0 INTRODUCTION

3.1 Background

Beams are design to carry loads perpendicular to their longitudinal axis. Internal shear

forces and bending moments develop along the span of a beam. An important consideration in

beam design is the amount of load it can carry. This depends on factor such as support and

loading condition, geometry (beam shape and dimension), and beam material. In designing a

beam for maximum efficiency, it is critical to determine the internal shear forces and bending

moment distribution along the beam. This is accomplished by constructing shear and bending

moment diagram (Figure 1).

3.2 Theory

When a beam is loaded by forces or couples, internal stresses and strains are created. To

determine these stresses and strains, we first must find the internal forces and internal couples

that act on cross sections of the beam. As an illustration, consider a cantilever beam acted upon

by a vertical force P at its free end (Figure 1a). Now imagine that we cut through the beam at a

cross section mn located at distance x from the free end and isolate the left hand part of the beam

as a free body (Figure 1b). The free body is held equilibrium by the force p and by the stresses

that act over the cross section; all we know is that the resultant f these stresses must be such as to

maintain equilibrium of the free body selected.

P

m

n

x

(a)

P

x

(b)

V

M

(c)

Figure 1: Bending Moment of a Beam

It is convenient to reduce the resultant to a shear force v acting parallel to the across

section and a bending couple of moment M. Because the load P is transverse to the axis of the

beam, no axial force exists at the cross section. Both the shear force and bending couple act in

the plane of the beam, which means that that the moment vector for the couple is perpendicular

to the plane of the figure. The moment of the bending couple is called bending moment M.

Because shear forces and bending moments, like axial forces in bars and twisting couples in

shafts, are known collectively as stress resultants.

The shear force and bending moment are assumed to be positive when they act on the

left-hand part of the beam in the directions shown in Fig 1-2b. If you consider the right-hand part

of the beam (Fig 1-2c), then the direction of the same stress resultant is reversed.

Shearing Force

Qx

+ve +ve

RA RB

Figure 2: Vertical Equilibrium of Moment

For vertical equilibrium of part (A) there must be a shearing force Qx acting as shown in

Figure 2 and equal to -RA. For vertical equilibrium of part (B) the shearing force Qx , evidently

acts as shown and has the value of RA-P. To clarify thr sign convention must be used. The

normal one are shown. Which leads to the –ve values above.

Bending Moment Qx P l

MX MX (A) (B)

RA a b RB

Figure 3: Equilibrium of Moment

For equilibrium of moments take an axis as shown (Figure 3) in the section XX to

eliminate Qx. Then using the sign convention given for part (A) of the beam,

(A) (B)

X

X

+ve +ve

Mx = RA x a

And for part (B) of the beam,

Mx = RB x b – P (b-l)

It can easily be proved that these have the same value by substituting for RB and Pl as follows,

MA = -b (P-RB) + Pl = -b x RA + RA (a + b) = RA x a

4.0 APPARATUS

The experimental beam is in to parts, the smaller being (A) and the larger (B). At the

section normal to the beam axis where they joint a pair of ball bearing pinned in (B) fits in half

housings fixed in (A) thus forming a “frictionless” hinge. A moment of resistance about the

hinge is provided by an under slung spring balance which acts at a lever arm of 150mm

Two and bearings on stands support the beam, and several stirrup shaped load hangers

can be threaded onto the two part of the beam. To introduce a load hanger at the joint between

part (A) and (B) the beam comes apart so that the hanger can be threaded on.

It is generally sufficient to line up parts (A) and (B) by eye, and to re-align them by

adjusting the screws on the spring or spring balance each time a load is altered.

5.0 PROCEDURE

Part 1

The beam was set up so that the bearing pin in part (B) is 300mm from the left hand

support and 600mm to the right hand support (Figure 4). Position one load hanger on the middle

of the smaller part (A) of the beam, one on the larger part (B), and the third one at the joint just

over the bearing pin (in the groove provided). After that, align the two parts of the beam using

the adjustment on the spring balance and the initial reading noted. Next, 10N (1kg) weight hang

on part (B), and re-align the beam and note the new reading on the spring balance. The

difference between the two readings is the effect of applying the 10N (1kg) weight on the beam.

The distance recorded from this weight to the right hand support. The procedure repeated by

using the hanger over the bearing pin, and then the hanger on part (A). Finally the whole

procedure repeated by using a 20N (2kg) load.

Part 1.

Bending moment at C (Mc) for loading shown

Figure 4

Table 1: Data for Part 1

Load (N) Balance Reading (N)/ Net force (N) for load at

P1 P2 P3

0 / / /

10 / / /

20 / / /

Bending moment (N.mm) and theoretical value

10 / / /

20 / / /

Part 2

Position the three load hangers as desired, the beam aligned and noted the initial reading

of the spring balance. Some weights (masses) hang on all three hangers and then the beam re-

aligned and noted the new balance reading. As time permits repeat with a different positions, and

with different loadings.

900mm

600mm

300mm

100mm

A B

C

P1 P2 P3

RA RB

Part 2.

a) Bending moment, Mc for loading shown, Figure 4a (superposition)

Figure 4a

Table 2: Data for Part 2

Loading (N) Balance

reading (N)

Net force (N) Bending

moment(N.mm)

Bending moment

(Theoretical

value) (N.mm)

0

P2=5

P1=P3=10, P2=5

b) Bending moment, Mc for various loading shown, Figure 4b

Figure 4b

900mm

500mm

300mm

100mm

A B

C

P1 P2 P3

RA RB

A B

C

P1 P2 P3

RA RB

Table 3: Data for Part 2b

Loading (N) Balance

reading (N)

Net force (N) Bending moment

(N.mm)

Bending moment

(Theoretical

value) (N.mm)

0

P1=5, P3 =12

P1=5, P2 =10,

P3=2

6.0 RESULT

For each loading arrangement calculate the bending moment at the section, draw the

bending moment diagram, and compare the measured value with it. The measured bending

moment in N mm will be 150 times the change in the spring balance reading. If masses in

kilogrammes were used remember to multiply them by 9.81 to get the load in Newtons.

7.0 DISCUSSION

When the load was doubled in Part 1 of experiment, did the bending moment double?

Compare the average figure which experiment the comparison between the experiment and

theoretical bending moments (use either a ratio or percentage error).

8.0 CONCLUSION

Did the experiment result verify with theory?

EXPERIMENT 6: KILASAN PALANG (TORSION OF BAR)

1.0 ABSTRACT

This experiment was done to investigate the relationship between torque, T and the angle

of twist, ø of a circular cross – section specimen and to acknowledge the students about to

determine the Modulus of Rigidity, G of materials .A base frame which has a clamp at one end

and a ball bearing in a housing (plummer block) at the other. A short shaft in the bearing has a

three jaw chuck facing the clamp and a torsion head at the outward side. A hanger cord is wound

round the torsion head with an effective diameter of 75 mm. Specimens in the forms of lengths

of rod are gripped by the fixed clamp and the rotating chuck 450 mm away. An arc shaped scale

of degrees is mounted on a base which can be moved along the length of the specimen. A pointer

on a spring steel strip registers the rotation of the specimen when a load is applied to the hanger

cord .The data collected is then being calculated and the angle of twist, ø versus torque, T

diagram is plotted. The Modulus of Rigidity, G is then being calculated and would be compared

to the theory .The conclusions are made based on the calculated data and the plotted diagram.

The value of the experiment might differ from the theory as errors might occur in the process of

the experiment.

2.0 INTRODUCTION

Consider a circular shaft that is attached to a fixed support at one end. When a torque, T

is applied to the other end, the shaft with free end rotating through an angle called the angle of

twist, ø.

Consider the small square element formed by two adjacent circles and two adjacent

straight lines traced on the surface of a cylinder of radius ρ before any load is applied. As the

shaft is subjected to a torsion load, the element deforms into a rhombus.

3.0 THEORY

Shearing strain, γ is measured by the change in the angles formed by the sides of that

element. It follows that

Lγ = ρø

Then,

γ = ρø / L

where γ = shearing strain

ρ = length from the center to any point in the circle of the shaft

ø = angle of twist

L = length of shaft

It also follows that the shearing strain is maximum on the surface of the shaft, where

ρ = c. Therefore

γ max = cø / L

where γ max = maximum shearing strain

c = radius of the shaft

ø = angle of twist

L = length of shaft

Now a relationship between the angle of twist, ø of a circular shaft and the torque, T

exerted on the shaft will be derived. Considering first the case of a shaft of length, L and of

uniform cross-section of radius, c subjected to a torque, T at its free end. The angle of twist and

the maximum shearing strain are related as

γ max = cø / L

but in the elastic range, the yield stress is not exceeded anywhere in the shaft. Hooke’s Law

applies, and γ max = τ max / G.

γ max = τ max / G

= Tc / JG

Then, solving for ø, it shows that

ø = TL / JG

where ø = angle of twist

T = torque exerted on the shaft

L = length of the shaft

J = polar moment of inertia, [π (c24 – c1

4)] / 2

G = Modulus of Rigidity

4.0 EXPERIMENT

MATERIAL

1. Steel ,aluminium alloy and nylon rod

2. Torsion bars testing machine

3. Ruler

4. Digital Caliper

Procedure:

1. The aluminum alloy rod is clamped in position and the load hanger is placed on the cord.

2. The rotation scale and pointer was set 400 mm from the fixed clamp and the pointer is set

to zero.

3. A load of 40 N by 5 N increments is added and the twist of the specimen for each

increment is recorded in a table.

4. After the load is removed, the rotation scale and pointer is changed to 200 mm from the

clamp and the above procedure is repeated.

5. The diameter of the rod is measured and recorded.

6. Now, the specimen is changed for the steel rod and the procedure is repeated for the 400

mm length only.

7. Finally, the nylon rod is clamped in position and twisted over 400 mm and 200 mm is

measured when a load up to 40 N by increments of 5N is applied to the torsion head. The

load is removed and whether full elastic recovery had occurred is noted.

5.0 DISCUSSION

a. To what extent does the experiment verify the torque/twist formula based on

graph plotted (angle of twist against torque).

b. Determine the modulus of rigidity (G) of material.

c. Handbook values for modulus of rigidity (G)n are 26200 N/mm² for aluminium

alloy and about 79000 N/mm² for steel and 75000 N/mm² for nylon (Teflon).

6.0 CONCLUSION

Give conclusion from the objective of the experiment that you have carried out.

Table 1

Material : Aluminium Alloy

Diameter:

Hanger

Load (N)

Torque (Nmm) Torque (Nmm)

Over 200mm

Twist of rod

Over 400 mm Over 400 mm (°) Over 200 mm (°)

5

10

15

20

25

30

35

40

Table 2

Material : Steel

Diameter:

Hanger Load

(N)

Torque (Nmm) Twist over 400 mm

( ° )

5

10

15

20

25

30

35

40

Table 3

Material : Nylon

Diameter:

Hanger

Load (N)

Torque (Nmm)

Over 400 mm

Torque (Nmm)

Over 200mm

Twist of rod

Over 400 mm (°) Over 200 mm (°)

5

4

3

2

1

0

EXPERIMENT 7: UJIAN MAMPATAN (COMPRESSION TEST)

1.0 ABSTRACT

This experiment was done to determine the properties of materials in compression and to

compare the properties in compression and in tension. By using a 10 ton Universal Testing

Machine, Model ET-2201-UTM, the modulus of elasticity, E of the materials and the ultimate

strength in compression can be determined. The purpose of this experiment was to measure and

study the differences between compression and tension. From this experiment, we made a

relation between load and compression of a specimen which will be used to determine modulus

of elasticity, E. The data collected is then being calculated and the stress, σ versus strain, ε

diagram is plotted. The Modulus of Elasticity, E in compression is then being calculated and

compared to tension. The conclusions are made based on the calculated data and the plotted

diagram. The value of the experiment might differ from the theory as errors might occur in the

process of the experiment.

2.0 INTRODUCTION

Like tensile test, compression test was done to investigate the mechanical properties of

specimen in compression such as modulus of elasticity, yield stress, ultimate stress, and strain.

If a specimen made of ductile material were loaded in compression instead of tension, the

stress-strain curve obtained would be essentially the same through its initial straight line portion

and through the beginning of the portion corresponding to yield and strain hardening.

Particularly noteworthy is the fact that for given steel, the yield strength is the same in both

tension and compression. For larger values of the strain, the tension and compression stress-

strain curves diverge, and it should be noted that necking cannot occur in compression. For most

brittle materials, the ultimate strength in compression is much larger than the ultimate strength in

tension. An example of brittle material with different properties in tension and compression is

provided by concrete, whose stress-strain diagram is shown in Figure 1.

On the tension side of the diagram, we first observe a linear elastic range in which the

strain is proportional to the stress. After the yield point has been reached, the strain increases

faster than the stress until rupture occurs. The behaviour of the material in compression is

different. First the linear elastic range is significantly larger. Second, rupture does not occur as

the stress reaches its maximum value. Instead, the stress decreases in magnitude while the strain

keeps increasing until rupture occurs. Note that the modulus of elasticity, which is represented by

the slope of the stress-strain curve in its linear portion, is the same in tension and compression.

This is true of most brittle materials.

Figure 1: Stress-strain Diagram

3.0 THEORY

Based on the collected data, stress and strain of the materials can be calculated and stress-strain

diagram can be plotted. Modulus of elasticity, E can be determined then.

Stress, σ = P/A Strain, ε = δ/L

where P = load applied (N) where δ = deformation (m)

A = area of the specimen (m²) L = original length (m)

Modulus of elasticity, E = σ/ε

4.0 PROCEDURE

1. The knob on the upper right hand side of the pump was turned to release the master

cylinder as far as it will go and the load frame being moved down to its lower position

and tighten back the knob like before.

2. Measuring the diameter of the specimen provided in 3 parts before the experiment began.

3. The specimen shall be placed on the compression plate (yellow).

4. Slowly and continuously leaded by pumping pump until the upper steel compression

plate touched the surface of the specimen.

σ U, compression

Linear elastic range

Rupture, compression

Rupture, tension σ U, tension

5. After the specimen was properly placed on the plate the compression measurement dial

gauge was set up.

6. Loaded pumping slowly and constantly with a low raising stroke of the handle.

7. The dial gauge and sample were observed. Read the force from Pressure gauge behind

the pump every 10 mm and make of it with the corresponding compression.

8. The pressure reading should be taken until the pressure reaches 200 kg/cm² (± 20).

5.0 RESULT

a. Calculate cross-sectional area for each specimen tested.

b. Calculate stress for each load taken.

c. Calculate strain for each load taken.

d. Plot the stress vs strain diagram. Determine the modulus of elasticity ( E ).

6.0 DISCUSSION

Discussion about the graphs and the result obtained. Compare the modulus of Elasticity ( E) in

compression and in tension. Give comments.

7.0 CONCLUSION

Briefly explain your conclusion from the experiments.

TABLE 1: EXPERIMENT DATA

Material:

Diameter (mm):

Length (mm):

Elongation, δ (mm) Load, P

Strain, ε

Stress, σ (MPa)

TABLE 2: EXPERIMENT DATA

Material:

Diameter (mm):

Length (mm):

Elongation, δ

(mm)

Load, P

Strain, ε

Stress, σ

(MPa)