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AIM

To determine the shear strength (ultimate shear stress) of the mild steel specimen supplied using double shear method.

EQUIPMENT

UTM, Shear attachment to the UTM, shear dies and Venire Calipers.

THEORY AND PRINCIPLE

Shear strength of the material is the ultimate shear stress ( ) attained by the specimen, which under double shear given by,

Where,

F = Maximum load at which the specimen breaks, and

A = cross-sectional area of the specimen.

The load range to which the machine is to be set for the test is selected bases on the expected maximum load F to be applied on the specimen. This is calculated from the yield stress fy and the factor of safety   , as follows:

Permissible shear stress t for mild steel is,

And therefore,

________________________(2)

PROCEDURE

Measure the diameter of the specimen. Calculate the maximum load expected to be applied on the specimen using equation (2) and select the load range to be used. Set the UTM for the selected load range. Set the correct set or dies to assemble the shear attachment with the right set of dies in it. Insert the specimen in to the dies so that it projects equally on either side. Place the entire bear assembly with the specimen in it centrally over the baring plate on the lower table. Bring the lower cross- head close to the top surface of the assembly. Float the lower table and set the load pointer to zero. Apply the load gradually until the specimen breaks. Note the ultimate load applied on the specimen.

OBSERVATION

The observation taken are tabulated in Table

Calculations

(i) Range

Using equation (2) and taking fy = 250 N/mm2 and =3 and, expected maximum load to be applied on the specimen is,

F’ = 3×0.45×250/2A

(ii) Shear strength

Using equation (1) the ultimate shear stress is= F/2=___________

RESULT

Ultimate shear stress of the material is fount to be N/mn^2

Questions- Do you know?

1. Distinguish between single shear and double shear?

2. Distinguish between average shear stress and maximum shear stress?

3. Why Modulus of rigidity is not determined from shear test?

4. Why structural component is designed mainly by considering double shear strength?