The double shear test is a commonly used method for determining the ultimate shear strength of mild steel specimens. This test involves subjecting the specimen to a load that causes two contiguous parts of the body to slide relative to each other in a direction parallel to their plane of contact. The shear force is the load that produces this sliding motion. The shear strength is defined as the maximum load that can be applied normal to a fastener’s axis before it fractures. In the case of double shear, the load is applied in one plane, which causes the fastener to be cut into three pieces, while in single shear, the fastener would be cut into two pieces.
To perform the double shear test, a Universal Testing Machine (UTM) is used. There are two types of UTM available, namely the screw type and the hydraulic type. Of the two, the hydraulic type is generally considered easier to operate. The UTM applies the load to the specimen, and the resulting data is used to determine the ultimate shear strength of the material. This information is useful in designing structures that require fasteners to withstand shear forces.
Apparatus
A Universal Testing Machine, commonly referred to as a UTM, is a piece of equipment used to perform mechanical tests on various materials. One of the essential components of a UTM is the Shear Tool Assembly, which allows for the application of forces and strains on a test specimen in a controlled manner.
The Specimen is the material being tested and is placed between the two jaws of the Shear Tool Assembly, which can apply either tensile or compressive forces to the specimen. The machine can also test the specimen’s shear strength by applying forces parallel to the material’s surface.
To measure the dimensions of the specimen accurately, Venire Calipers are used. These calipers are precision instruments that allow the user to measure the distance between two points with high accuracy, making them ideal for measuring the thickness or diameter of a test specimen. This information is essential in calculating the mechanical properties of the material being tested.
Theory and Principle
The shear strength of a material is defined as the maximum shear stress (Tmax) that a specimen can withstand under double shear, which is given by the equation Tmax = F/2A, where F is the maximum load at which the specimen breaks and A is the cross-sectional area of the specimen.
To conduct a shear strength test, the load range of the testing machine is selected based on the expected maximum load F, which is calculated from the yield stress fy and the factor of safety (F.S) using Equation 4: F = 0.9(F.S)fyA.
The permissible shear stress for mild steel is given by Equation 2: T = 0.45fy. Therefore, using Equation 3, we can determine that Tmax = (F.S)0.45fy.
Procedure
The task at hand is to determine the diameter of a given rod using a Vernier caliper. To achieve this, the diameter of the specimen will be measured at three different sections. The next step is to calculate the maximum load that is expected to be applied on the specimen by using equation (2) and then selecting the appropriate load range to be used.
After selecting the load range, the Universal Testing Machine (UTM) will be set accordingly. The correct set or disc will then be assembled with the shear attachment and the specimen will be inserted into the disc so that it projects equally on either side, as shown in Figure 3.
Next, the entire shear assembly with the specimen will be placed centrally over the baring plate on the lower table. The lower cross-head will be brought close to the top surface of the assembly and the lower table will be floated. The load pointer will then be set to zero.
Gradually, the load will be applied to the specimen until it breaks, as shown in Figure 4. The ultimate load applied on the specimen will be noted. Finally, the shear strength of the steel specimen will be computed based on the gathered data.
Observations
The observation taken are tabulated in Table.
Calculations
Using equation (1) the ultimate shear stress is= F/2A=___________
Result
Ultimate shear stress of the material found to be N/mn^2