AIM :- To perform the shear test on UTM.

APPARATUS USED :- A UTM, Specimen, shearing attachment, vernier caliper etc.

THEORY :- A type of force which causes or tends to cause two contiguous parts of the body to slide relative to each other in a direction parallel to their plane of contact is called the shear force. The stress required to produce fracture in the plane of cross-section, acted on by the shear force is called shear strength.

PROCEDURE :-

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. OBSERVATION :-

· Applied compressive force (F) =——– kgf.

· Diameter of specimen =——— mm.

CALCULATION :-

The shear strength shall be calculated from the following formulae : τ_s = (F/2) / (πd²/4) = 2F / πd²

where ‘d’ is the actual diameter of the specimen

PRECAUTIONS :-

· The specimen should be all place equal dia.

· Measure the diameter of specimen carefully.

· The specimen should be properly grip between the test rig.

· Take reading more carefully.

· After shearing specimen stop to m/c.

CONCLUSION :-

VIVA-QUESTIONS :-

· Does the shear failure in wood occur along the 45° shear plane ?

· What is buldging? Why does it occur?

· What is single & double shear ?

· What is find in shear test?

· What is unit of shear strength ?

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2. Lab Manual | Perform compression test on UTM
3. Lab Manual | study of Impact Testing m/c and Perform charpy impact test

AIM:- To Perform compression test on UTM.

APPARATUS :- A UTM or A compression testing m/c, cylindrical or cube shaped specimen of cast iron, Alumunium or mild steel, vernier caliper, liner scale, dial gauge (or compressometer).

THEORY :- Several m/c and structure components such as columns and struts are subjected to compressive load in applications. These components are made of high compressive strength materials. Not all the materials are strong in compression. Several materials, which are good in tension, are poor in compression. Contrary to this, many materials poor in tension but very strong in compression. Cast iron is one such example. That is why determine of ultimate compressive strength is essential before using a material. This strength is determined by conduct of a compression test.

Compression test is just opposite in nature to tensile test. Nature of deformation and fracture is quite different from that in tensile test. Compressive load tends to squeeze the specimen. Brittle materials are generally weak in tension but strong in compression. Hence this test is normally performed on cast iron, cement concrete etc. But ductile materials like aluminium and mild steel which are strong in tension, are also tested in compression.

TEST SET-UP, SPECIFICATION OF M/C AND SPECIMEN DETAILS :

A compression test can be performed on UTM by keeping the test-piece on base block (see in fig.) and moving down the central grip to apply load. It can also be performed on a compression testing machine. A compression testing machine shown in fig. it has two compression plates/heads. The upper head moveable while the lower head is stationary. One of the two heads is equipped with a hemispherical bearing to obtain. Uniform distribution of load over the test- piece ends. A load gauge is fitted for recording the applied load.

SPECIMEN :- In cylindrical specimen, it is essential to keep h/d < 2 to avoid lateral instability due to bucking action. Specimen size = h < 2d.

PROCEDURE :-

1. Dimension of test piece is measured at three different places along its height/length to determine the average cross-section area.

2. Ends of the specimen should be plane . for that the ends are tested on a bearing plate.

3. The specimen is placed centrally between the two compression plates, such that the centre of moving head is vertically above the centre of specimen.

4. Load is applied on the specimen by moving the movable head.

5. The load and corresponding contraction are measured at different intervals. The load interval may be as 500 kg.

6. Load is applied until the specimen fails.

OBSERVATION :-

1. Initial length or height of specimen h =————– mm.

2. Initial diameter of specimen do =——————– mm.

S.No.

Applied load (P) in Newton

Recorded change in length (mm)

CALCULATION :-

· Original cross-section area Ao =——

· Final cross-section area Af =——–

· Stress =——-

· Strain =——-

For compression test, we can

· Draw stress-strain (a-s) curve in compression,

· Determine Young’s modulus in compression,

· Determine ultimate (max.) compressive strength, and

· Determine percentage reduction in length ( or height) to the specimen. PRECAUTIONS :-

· The specimen should be prepared in proper dimensions.

· The specimen should be properly to get between the compression plates.

· Take reading carefully.

· After failed specimen stop to m/c.

RESULT :- The compressive strength of given specimen =——————- Nmm 2

CONCLUSION :-

VIVA-QUESTIONS :-

· Compression tests are generally performed on brittles materials-why ?

· Which will have a higher strength : a small specimen or a full size member made of the same material ?

· What is column action ? how does the h/d ratio of specimen affect the test result ?

· How do ductile and brittle materials in their behaviour in compression test ?

· What are bi-modulus materials ? Give examples.

Related posts:

1. Lab Manual | To perform the shear test on UTM
2. Lab Manual | study the UTM and perform the tensile test
3. Lab Manual | study of Impact Testing m/c and Perform charpy impact test