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Mechanics of Materials-ENGI 23413

LABORATORY REPORT

Title of experiment: Shear Strength

GROUP-D

GroupMembersName

1.Vipanjot Kaur-991311175

2.Simrat Kaur-991312062

3.Taljinder Singh-991306036

4.Gurinder Singh-991308470

PERFORMING DATE: 4 October, 2013

SUBMISION DATE: 11 October, 2013

SUBMITTED TO: James Forbes

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CONTENTS :-

b s t r a c t / E x e c u t i v e S u m m a r y 1

O b j e c t i v e s 2T h e o r y 3

M a t e r i a l s a n d E q u i p m e n t 4

P r o c e d u r e 5

D a t a / O b s e r v a t i o n 6

S a m p l e c a l c u l a t i o n s 7

R e s u l t s 8

D i s c u s s i o n 9

Conclusion&Recommendation 1 0

R e f e r e n c e s 1 1

p p e n d i x 1 2

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The units of shear stress in metric system are

ABSTRACT/EXECUTIVE SUMMARY:-

The purpose of this experiment is to become acquainted with the significance of the shear

strength by performing the shear strength test on the various samples/metals such steel,

aluminium and brass. The diameter measured of samples and the values of different loads

observed from the analog scale and tracer needle helped us to find out the area which further

continued to get the shear strength

Moreover, the observed values of the shear strength were compared to the theoretical values of

shear strength of particular sample/metals .And even the observed and theoretical ultimate

tensile strength values were compared considerably and their relation (ratio) was found

simultaneously.

OBJECTIVE:-

The objective of this lab was to understand the significance of shear strength by performing theshear strength experiments on different materials such as steel, brass and aluminum.

THEORY:-

Shear Stress:-

Shear stress is defined as the forces that act parallel to a given area and in the opposite

direction tending to produce a sliding of one portion past another.It is one of the three basic type

of stress, where compression and tension are other two.

or Pascal (Pa) and in Imperial system Pounds per square

inch which is represented by psi.

Shear Strength and its applications:-

Shear strength is equal to the maximum resisting force divided by the area under the shear:

= F/AShear strength is essential consideration in the designing and performance of many machine parts and

elements. Rivets, bolts and pins are often loaded properly, if not entirely, in the shear. Metalworking

operations such as shear cutting, blanking and punching require forces that are dependent on shear

strength of the material.

.

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fixture:-

Vernier

It have holes inside to putRound Stock double shear

- Caliper:-It

Shear test results are generally less precise than tensile tests because of a small amount of

bending stress and friction is usually present. Shear stress is most common on materials joined

together with rivets, bolts, screws and pins.

Shear tests are often made on flat stock as single or double shear tests. Round stock is most

commonly tested in double shear. For double shear as shown below in picture, the total area of

both cross-sections of the pin must be used in calculating the shear strength.

To calculate the percentage of error,

EQUIPMENT REQUIRED:

VEGA Hydraulic

require the ability

the sample and it

fixture consisting of three plates which are parallel to each other.

is the device used the measure the diameters of

used such as steel, aluminium and brass.

the metal pieces

-

1. Shear:-

Hydraulic shears are used in high volume metal fabricating operations that

to accurately and quickly cut and score sheets of metal.

2. is the

3.

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1 sample each of round stock:

Aluminum

Brass

Cold Rolled Steel

4. -

STAINLESS

NARDfHID

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8.

1.

2.

PROCEDURE:

Each samples diameter was measured and recorded on the observation table.

The sample was inserted into the test fixture so that it would pass through all 3

Plates of double shear fixture.

3. Do not center the specimen in the fixture but let one end project only a shortdistance, so that a second test can be performed on the same specimen.

4. The loaded fixture was then placed in the Vega Shear and the safety door wasclosed.

5. The Vega Shear pressure gauge/analog scale was then zeroed out.

6. A load was applied gradually by hand pumping the hydraulic lever until the

sample sheared.

7. The maximum load was recorded.

After recording the value the shear was loaded a little more so that the shearingslug passed into the relieved area of the lower part of the fixture. Hydraulic

pressure was then released and the fixture removed so the slugs could beremoved.

9. The analog scale and tracer needle were re zeroed and the procedure wasrepeated on the same sample.

10.The procedure again was repeated for the other 2 samples.

-

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distance,

1.

1. Record results in a tabulated form similar to below: lo o o

8. Re-zero

he hydraulic pressure, remove Ihe

7. After failure,

Mechamcs

^f f i t * ~

the

fixture from the machine,

of Materials Lab 2 2

EXPERIMENTAL PROCEDURE:

Measure and record the diameter of each sample

2. Insert a test sample through the hole in all three plates of the double shear fixture

Do No t centre the specimen in the fixture but let one end project only a short

so that a second test can be performed on the same specimen.

3. Place the loaded fixture under the hydraulic press and close the safety guard

4. Check that the analog scale and tracer needle are zeroed

5. Gradually apply load by pumping the hydraulic lever until the sample shears

6. Record the maximum load

continue to press the punch into the holder, so that the shearing

slug passes into the relieved area in the lower part of the holder. Then release

slide the punch

sideways out of the holder and eject the sheared slugs.

dial and make a second test on the same sample

9. Repeat the procedure for the other 2 samples

RESULTS:

X*-L

MATERIAL Maximum Load Average Load | Shear Strengjh

60x5.0

0- \ > O M t v[(A l u m'n k ,

2. Compare results to accepted values

3. Show their relation to tensile strengths

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Steel

Brass

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55"

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2.

Aluminum accepted values:

Tensile

RESULTS:-

Our results for Steel Shear = 66 ksi

Steel accepted values:- Shear Strength from 60 ksi

Ultimate Tensile Strength

Our results for Brass Shear = 45.6 ksi

Brass accepted values:- Shear Strength 43 ksi

Ultimate

3. Our results for Aluminum Shear= 31.9 ksi

Ultimate Tensile Strength 55ksi.

DISCUSSIONS:

Therefore, according to our calculations analysis the ultimate shear strength of the 66 ksi and

theoretical value is 60 ksi. We found the percentage error of 9 percent .The ultimate tensile

strength of the steel has the theoretical value of 63.8 ksi. So, the ratio of the ultimate tensile

strength and ultimate shear strength comes out to be 0.96.

Moreover, according to our calculations analysis the ultimate shear strength of the 45.6 ksi and

theoretical value is 43 ksi. We found the percentage error of 4.4 percent .The ultimate tensile

strength of the steel has the theoretical value of 50 ksi. So, the ratio of the ultimate tensile

strength and ultimate shear strength comes out to be 1.09.

Furthermore, according to our calculations analysis the ultimate shear strength of the 31.9 ksi

and theoretical value is 30 ksi. We found the percentage error of 5.9 percent .The ultimate

tensile strength of the steel has the theoretical value of 55 ksi. So, the ratio of the ultimate

tensile strength and ultimate shear strength comes out to be 1.72.

Hence, we observed that the values of the shear strength of aluminium and brass are almost

same as experimental and theoretical. Whereas, in case of steel there is a lot difference

between theoretical and experimental value. Even, percentage error in case of brass is least andin case of steel is more as 9%.

Results, show that shear strength of different materials is different .The ultimate shear strength

value for steel is higher as compared to brass and aluminium. This tells us that steel is very good

and strong metal, whereas, aluminium and brass are less strong. So, in order to cut steel the

large shearing force is required while brass and aluminium can be cut down easily.

Strength 50 ksi.

Strength

1.

63.8 ksi.

- Shear 30 ksi

-

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CONCLUSIONS AND RECOMMENDATIONS

We have noticed that ultimate shear strength is important aspect of material. Our result for the ultimate

shear strength of steel is 6 ksi more than the accepted value ; whereas our results for the brass andaluminum were just slightly higher than the maximum theoretical values. For the brass and aluminum

samples we did not have any information on the material type such as any alloys, heat treatment, cold

working, annealing or tempering. Any of these factors could have affected our shear strength results.

The steel sample we only knew was cold rolled and cold rolled low carbon steel (hard) had a value of 60

ksi shear strength, which is very close to our result of 66 ksi.

While the relation between the theoretical tensile strength and the ultimate shear strength was found in

all the cases it was almost 1 conveniently.

When calculating for tensile strength the force is a pulling force which is divided by the cross-sectional

area. Compared to tensile and compressive stress and strain, the shear forces act over an area which is

in linewith the forces. When calculating the shear strength the force is divided by twice the area since

the shear was at points on the samples. This would explain why the tensile strength is almost double of

the shear strength.

In conclusion our experiment was successful in showing the significance of shear strength. With more

technical information about each sample we may have been able to narrow the range of values for the

shear and ultimate strengths.

The followingrecommendationsare made in order to reduce the errors:-

The analog reader should be replaced by the digital reader to get the accurate readings.

We must know about the physical and mechanical properties of the samples used.

:-

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Bibliography:-

Lab handout

http://en.wikipedia.org/wiki/Ultimate_tensile_strength

http://www.diybanter.com/metalworking/11770-ultimate-shear-strength.

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