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.
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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.
s
s

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.