EXPERIMENT
OBJECTIVE:
To find out the Shear Strength of a given specimen using Universal Testing Machine.
APPARATUS REQUIRED:
- Universal testing machine
- Shear test attachment
- Given Specimen
- Vernier caliper etc.
THEORY:
When two equal and opposite forces having different lines of action tend to cause angular distortion in the body; the forces applied are known as shear forces.
The stress required to produce fracture in the plane of cross-section, acted on by the shear force is called shear strength.
PROCEDURE:
- Insert the specimen in position and grip one end of the attachment in the upper portion and one end in the lower position.
- Switch on the main switch of Universal Testing Machine.
- Bring the drag indicator in contact with the main indicator.
- Select the suitable range of loads and space the corresponding weight in the pendulum and balance it if necessary with the help of small balancing weights.
- Operate (push) the button for driving the motor to drive the pump.
- Gradually move the head control ever in left hand direction till the specimen shears.
- Note down the load at which the specimen shears.
- Stop the machine and remove the specimen.
- Repeat the experiment with other specimens.
OBSERVATIONS:
Record the following data:
- Diameter of the specimen, d = ….. mm
- Cross-sectional area of the specimen (in double shear) = 2 x [frac up=”Π” down=”4″] x d2
- Load taken by the specimen at the time of failure, W = …… N
- Strength of the specimen against shearing, τ =[frac up=”W” down=”2 x (Π/4) x d2“] = ….. [frac up=”N” down=”mm2“]
PRECAUTIONS:
- The measuring range should not be changed at any stage during the test.
- The inner diameter of the hole in the shear stress attachment should be slightly greater than the specimen.
- Measure the diameter of the specimen accurately.
- The method for determining the shear strength consists of subjecting a suitable Length of steel specimen in full cross-section to double shear, using a suitable test rig, in a testing m/c under a compressive load or tensile pull and recording the maximum load ‘F’ to fracture.
RESULT:
Shear strength of specimen = …..
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