Solution to Problem 203 Stress-strain Diagram

Problem 203
The following data were recorded during the tensile test of a 14-mm-diameter mild steel rod. The gage length was 50 mm.

Load (N)	Elongation (mm)	Load (N)	Elongation (mm)
0	0	46 200	1.25
6 310	0.010	52 400	2.50
12 600	0.020	58 500	4.50
18 800	0.030	68 000	7.50
25 100	0.040	59 000	12.5
31 300	0.050	67 800	15.5
37 900	0.060	65 000	20.0
40 100	0.163	65 500	Fracture
41 600	0.433

Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limits; (b) modulus of elasticity; (c) yield point; (d) ultimate strength; and (e) rupture strength.

Solution 203

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Area, A = 0.25π(14²) = 49π mm²
Length, L = 50 mm
Strain = Elongation/Length
Stress = Load/Area

Load (N)	Elongation (mm)	Strain (mm/mm)	Stress (MPa)
0	0	0	0
6 310	0.010	0.0002	40.99
12 600	0.020	0.0004	81.85
18 800	0.030	0.0006	122.13
25 100	0.040	0.0008	163.05
31 300	0.050	0.001	203.33
37 900	0.060	0.0012	246.20
40 100	0.163	0.0033	260.49
41 600	0.433	0.0087	270.24
46 200	1.250	0.025	300.12
52 400	2.500	0.05	340.40
58 500	4.500	0.09	380.02
68 000	7.500	0.15	441.74
59 000	12.500	0.25	383.27
67 800	15.500	0.31	440.44
65 000	20.000	0.4	422.25
61 500	Failure		399.51

From stress-strain diagram:

Proportional Limit = 246.20 MPa
Modulus of Elasticity
E = slope of stress-strain diagram within proportional limi
E = 246.20/0.0012 = 205 166.67 MPa = 205.2 GPa
Yield Point = 270.24 MPa
Ultimate Strength = 441.74 MPa
Rupture Strength = 399.51 MPa

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