Solution to Problem 204 Stress-strain Diagram

Area, A = 0.25π(0.505²) = 0.0638π in₂
Length, L = 2 in
Strain = Elongation/Length
Stress = Load/Area
 

Load (lb)	Elongation (in.)	Strain (in/in)	Stress (psi)
0	0	0	0
2 310	0.0022	0.0011	11 532.92
4 640	0.0044	0.0022	23 165.70
6 950	0.0066	0.0033	34 698.62
9 290	0.0088	0.0044	46 381.32
11 600	0.011	0.0055	57 914.24
12 600	0.015	0.0075	62 906.85
14 000	0.02	0.01	69 896.49
14 400	0.025	0.0125	71 893.54
14 500	0.06	0.03	72 392.80
14 600	0.08	0.04	72 892.06
14 800	0.1	0.05	73 890.58
14 600	0.12	0.06	72 892.06
13 600	Fracture		67 899.45

 

 

From stress-strain diagram:

1. Proportional Limit = 57,914.24 psi
2. Modulus of Elasticity:
   E = 57914.24/0.0055 = 10,529,861.82 psi
   E = 10,529.86 ksi
3. Yield Point = 69,896.49 psi
4. Yield Strength at 0.2% Offset:
   Strain of Elastic Limit = ε at PL + 0.002
   Strain of Elastic Limit = 0.0055 + 0.002
   Strain of Elastic Limit = 0.0075 in/in
    

   The offset line will pass through Q (See figure below):
    

   

    

   Slope of 0.2% offset = E = 10,529,861.82 psi
    

   Test for location:
   slope = rise / run
   10,529,861.82 = (6989.64 + 4992.61) / run
   run = 0.00113793 0.0025, therefore, the required point is just before YP.
    

   Slope of EL to YP
   σ₁ / ε₁ = 6989.64/0.0025
   σ₁ / ε₁ = 2 795 856
   ε₁ = σ₁ / 2 795 856
    

   For the required point:
   E = (4992.61 + σ₁) / ε₁

   10 529 861.82 = (4992.61 + σ₁) / (σ₁ / 2 795 856)
   3.7662 σ₁ = 4992.61 + σ₁
   σ₁ = 1804.84 psi
    

   Yield Strength at 0.2% Offset
         = EL + σ₁
         = 62906.85 + 1804.84
         = 64 711.69 psi

5. Ultimate Strength = 73 890.58 psi
6. Rupture Strength = 67 899.45 psi