Question

Problem 1: A tension test was performed on a steel specimen having an original diameter of 0.50 in. and gage length of 2.00 in. The final diameter is de .38 in. The data is listed in the table. 1) Plot the stress-strain diagram on engineering paper, so you can calculate the area easily; 2) Determine approximately the modulus of elasticity; 3) Find the yield stress by 0.2% offset method; 4) Calculate approximately the modulus of resilience; 5) Determine approximately the modulus of toughness 6) Calculate the percent of elongation and reduction of area. Note: the percent of elongation = (LL)/L) x 100% the percent of reduction of area =((Ag-A)/A.) x 100% Load (kip) Elongation (in.) 1.50 4.60 8.00 11.00 12.00 16.60 20.00 21.50 23.00 20.50 18.50 17.20 0.0005 0.0015 0.0025 0.0035 0.0050 0.0080 0.0200 0.0400 0.1000 0.2800 0.4000 0.4600

Answer 1

1.

Load	Elongation	Area	Length	Stress= load/area	Strain = elongation / length*100
kips	in	in^2	in	ksi	%
0	0	0.19635	2	0.00	0
1.5	0.0005	0.19635	2	7.64	0.025
4.6	0.0015	0.19635	2	23.43	0.075
8	0.0025	0.19635	2	40.74	0.125
11	0.0035	0.19635	2	56.02	0.175
12	0.005	0.19635	2	61.12	0.25
16.6	0.008	0.19635	2	84.54	0.4
20	0.02	0.19635	2	101.86	1
21.5	0.04	0.19635	2	109.50	2
23	0.1	0.19635	2	117.14	5
20.5	0.28	0.19635	2	104.41	14
18.5	0.4	0.19635	2	94.22	20
17.2	0.46	0.19635	2	87.60	23

Stress vs strain curve 140.00 120.00 100.00 80.00 Stress ksi 60.00 40.00 20.00 0.00 10 15 20 25 Strain, %

2.

2.

Modulus of elasticity is slope of strss strain curve at origin.

E = (56.02-0)/(0.175-0)*100

E = 32011.428 ksi

3. draw a line parallel to initial slope of curve and strat that line from 0.2% strain, for your understanding i have increase the stresss strain curve and now you can see how line is plotted .

corresponding to stress where this line is cutting the stress strain curve is known as yield stress corresponding to 0.2% offset

140.00 Stress vs strain curve 120.00 Stress ksi 20.00 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 28 3 32 3.4 3.6 3.8 4 42 4.4 4.6 43 5 52 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8 82 8.4 8.6 8.8 9 92 9.4 9.6 9.5 10 10.2 10.4 10.5 10.5 11 11.2 11.4 11.5 11.8 12 12.2 12.4 12.6 12.5 13 13.2 13.4 13.6 13.8 14 14.2 14.4 14.6 14.8 15 15.2 15.4 15.5 15.8 16 16.2 16.4 16.6 16.8 17 17.2 17.4 17.6 17.8 18 18.2 18.4 18.6 18.8 19 19.2 19.4 19.6 19.8 20 20.2 20.4 20.5 20.8 21 21.2 21.4 21.6 21.8 22 22.2 22.4 22.5 22.8 23 23.2 23.4 23.6 23.8 24 24.2 Strain,

Yield stress = 93 ksi (approximate)

4.

Slope is lineraly upto 56.02 ksi hence modulus of resilence is area of strain stress curve upto elastic point i.e. upto linear curve of stress strain curve.

Modulus of resilence = 0.5*56.02*0.175/100 = 0.049 ksi

5. 6.

2 Too 0.5 modulus of toughness = area under stress curve = $x146.37 X0.4 + +(14637 +202.80)x( Too (5-0.42 +(202180 +151166 )(23-5) 100 l = 40.225 ksi Qib lo elongatim = 0.46 x 100 =23% olo reduction of area= $x052 -7, X0138? x 100 2 Txo.52 = 42.24%

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