Sheartest

shear test, schmid's law

1. SHEAR TEST BATCH-7

2. DIRECT SHEAR STRESS Shearing stress is one that acts parallel to a plane, unlike the tensile and compressive forces that act normal to it. Shear stresses can be produced in structural bodies by various types of loading, such as direct shear, flexural, torsional . If the resultants of parallel but opposite forces act through the centroid of the sections that are spaced infinitesimal distances apart, the shearing stresses over the sections are uniform.

3. These stress conditions are called direct shear stress. A shear force P acting on a surface area A causes an average direct shear stress given by = P/A Since the concept of infinitesimal distance is only theoretical, such conditions are never completely realized in practical situations. However direct shear situations exist approximately in rivets or welded plates. Connections among truss members, beams and columns are often subjected approximately to direct shear stresses.

4. DIRECT SHEAR TEST Direct shear conditions are approximately simulated in laboratory tests by shearing forces distributed over very small lengths so that the bending moments created as a result are very small and the only possible failure mechanism remaining is through shearing of the specimens. In the direct shear test, the shearing stress is considered to be uniformly distributed over the cross-section. A round or flat stock specimen is clamped in a fixture and a shearing force is applied through a suitable shear tool.

5. The shear force is applied by a shear cutter. Two different cases of shearing may arise : (a) (b) (a) Single shear test and (b) Double shear test

6. (a) shows the case of single shear where shearing occurs across a single surface . Therefore, the shear stress in single shear test is given by = P/A (b) shows the case of double shear where shearing occurs across two surfaces. Therefore, the shear stress for double shear test is given by = P/(2A) where P is the total force applied on the specimen through the shear cutter and A is the cross-sectional area of the specimen.

7. The failure surfaces for the two tests are also somewhat different. Since the single shear is accompanied by bending moment across the shearing surface ,as the specimen is a cantilever beam, the failure surface is also bent, i.e., it is inclined to the original surface. But since the bending moment across the shearing surface for double shear is negligible the specimen is a simply supported beam, the failure surface is almost plane, i.e., similar to the original surface.

8. PRINCIPLE OF TEST Shear strength is determined by inserting a cylindrical specimen through round holes in three hardened steel blocks, the centre of which is pulled or pushed between the other two so as to shear the specimen on two planes. The test consists of subjecting a suitable length of cylindrical specimen to double shear loading using a suitable test rig in a testing machine under a compressive load or tensile pull and recording the maximum load P to fracture.

9. TEST PROCEDURE Mean diameter of the specimen must be determined. Specimen must be fixed in the shear tool. Set the UTM for the selected load range. The shearing tool is placed on movable cross-head of the testing machine. The specimen is loaded gradually until the specimen is sheared off. The ultimate load is recorded. Ultimate shear strength of the metal rod = P/2A

10. APPLICATION The main application of the shear test is to determine the shear strength of various materials and hence they can be used accordingly for suitable applications.