1. Measure the diameter of the specimen at three different sections. Calculate the original
diameter by taking average of three readings. The minimum overall length of the specimen
shall be 20 times diameter plus 200 mm.
2. Mark the points over the grip length, with the punch such that the distance between two
consecutive points is half the gauge length.
3. Select a suitable loading range depending on the diameter of specimen. Start the UTM and
adjust the dead weight of movable heads and then set the load pointer to zero.
4. Fix the specimen bar between the grips of top and middle cross heads of loading frame.
5. Attach the extensometer on the bar at the central portion of the bar. The distance between
upper and lower pivots of extensometer shall be equal to gauge length.
6. Switch on the machine and open the control valve so that the load is increased gradually and
at the required rate.
7. Record the load at suitable interval from the digital display unit or the load dial.
8. Corresponding to loads, note the readings of extensometer.
9. For initial few observations, load and extension are in pace with each other. Record the yield
point load by observing the hesitation of load pointer. The extension readings are faster at
this moment.
10. Remove the extensometer; hence measure extension by divider or suitable scale
11. Record the maximum load. Observe the decrease in load and neck formation on the
specimen.
12. Record the load at fracture and put off the machine.
13. Remove the specimen. Observe the cup and cone formation at the fracture point. Rejoin the
two pieces, measure the final gauge length and the reduced diameter.
OBSERVATIONS:
Observations before test:
1. Diameter of bara)
d1 =………..mm, b) d2 =……….. mm, c) d3 =………… mm
2. Average diameter = d = (d1+d2+d3) / 3 =……………..mm
3. Gauge length = L0 = 5 d = ………………… = …………….mm
4. Least count of extensometer = L. C. = ……………….mm
Observations during test:
1. Range of loading = ……………………..
2. Load at elastic limit = ………………………kN
3. Load at upper yield point = ………………………kN
4. Load at lower yield point = ………………………kN
5. Ultimate load = ………………………kN
6. Breaking load =……………………….kN
Observations after test:
Final gauge length = L = …………............ mm
Reduced diameter = d1 = ..........................mm
CALCULATIONS:
Yield stress = Yield load / cross sectional area
= …………………/………………. = …………………N/mm2
Ultimate stress = Ultimate load / cross sectional area
= ……………../ …………….. = …………………. N/mm2
Breaking stress = Breaking load / cross sectional area
= ………………/… ………….. = ……….……… N/mm2
Actual breaking stress = Breaking load / reduced cross sectional area
= ………….... /…. ………… = ……….… N/mm2
= …………….. %
= …………….%
(Within Elastic limit)
Stress = σ = …………..N/mm2
Corresponding strain = ε = ……………
Modulus of Elasticity, E = σ/ ε = ................./……....…. = …………….. N/mm2
RESULT:
1. Yield stress = ………………. N/mm2
2. Ultimate tensile stress = ……………….. N/mm2
3. Nominal Breaking stress = …………..…… N/mm2
4. Actual breaking stress = ………………. .N/mm2
5. % Elongation = ………………..%
6. % reduction in area = ………………..%
7. Modulus of Elasticity, E = ………………… N/mm2 (By graph)
8. Modulus of Elasticity, E =…………………. N/mm2 (By calculation)