Question

Problem 1: A round bar with a diameter of 32 mm, and 4012 mm in length was under tensile test to determine the mechanical properties of the material. The following table shows the load versus strain data collected from the testing machine and a strain gauge applied on the bar in the loading direction. The measured percentage elongations were 19.8 % and 14.6 % on gauge lengths of 100 mm and 200 mm respectively. Assume: the elongations include the necking; the strain outside the necking region was uniform. The griping lengths at both ends of the bar were neglected. Determine:

1. The 0.2% proof stress of the material (Unit: MPa)
2. The Young's modulus of the material (Unit: GPa)
3. The length of the bar when a tensile load of 502 kN is applied (Unit: mm)
4. The length of the bar at failure when the two broken pieces are fit together. (Unit: mm)

Load (kN)	Strain
0.0	0.00000
253.3	0.00163
371.0	0.00239
448.6	0.00289
506.5	0.00327
552.8	0.00360
591.3	0.00395
624.3	0.00443
653.1	0.00517
678.8	0.00637
701.9	0.00830
727.1	0.01207
749.6	0.01799
764.4	0.02401
775.4	0.03007
784.2	0.03614
791.6	0.04221
797.9	0.04828
803.4	0.05433
808.3	0.06037

Answer 1

Stress strain tabulation, Stress-strain curve and enlarged linear portion of curve is shown below.

A) 0.2% proof stress = Stress correponding to 0.002 strain ( Refer red lines in above graph)

  0.2% proof stress = 820 MPa

B) Young's modulus = Slope of curve in elastic region ( Refer green lines)

Young's modulus = (590-380)/(0.003-0.002)

Young's modulus = 210000 MPa

Young's modulus = 210 GPa

C) The length of the bar when a tensile load of 502 kN

change in length = PL/AE

change in length = (502 * 1000 * 4012) / (804.25 * 210000)

change in length = 11.925 mm

Therefore,

Length of bar = 4012 + 11.925

Length of bar = 4023.925 mm

D) The length of the bar at failure

- change in length in 100 mm gauge length = 19.8 * 100 / 100 = 19.8 mm

- change in length in 200 mm gauge length = 14.6 * 200 / 100 = 29.2 mm

- Remaining length = 4012 - 200 = 3812 mm

- change in length = PL/AE = (808300 * 3812) / (804.25*210000) = 230.13 mm

Therefore,

Length of bar at failure = 4012 + 29.2 + 230.13 = 4271.33 mm