Question

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(1) (25 points) Load-elongation dataset were obtained from a tensile test of high-strength steel as shown in the table. The test specimen had a diameter of 0.505 in. and a gage length of 2.00 in. The Poisson's ratio of high-strength steel is found to be 0.33 At fracture, the elongation between the gage marks was 0.12 in and the minimum diameter was 0.42 in 1) Plot the engineering stress-strain curve using excel (or Elongation (in.) 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture Load (lb) 1,000 2,000 6,000 10,000 12,000 12.900 13.400 13,600 13,800 14,000 14.400 15.200 16.800 18.400 20,000 22,400 22,600 engineering paper) and determine: (a) the proportional limit (b) modulus of elasticity (c) yield stress using 0.2% offset method (d) ultimate stress (e) modulus of resilience (f) modulus of toughness (g) percent elongation in length and percent reduction in area (h) diameter of the test specimen under load when the axial stress reached 50 ksi? 2) If the specimen was loaded and stretched by 0.04 in./in. strain, what is the permanent strain after removing the load? 3) What is the new proportional limit and the new modulus of resilience?

Answer 1

Given data Diameter of test specimen, d -0.505 in Gage length, 1-2 in Poisson's ratio of high-strength steel, μ-033 Elongation between gage marks at fracture, A Minimum diameter, dmin 0.42 in 0.12 in 1) Write the expression to calculate the cross sectional area of the test specimen, 4 Substitute the value, π(0.5052 -0.20037 in2 Consider the following table

| Load |Stress(σ.L-1b/m2)| Strain(,- Elongatiorn Load., i. Strain * ΔΙ Modulus of elasticity Stress, σ Ibjin 1000 2000 6000 10000 12000 12900 13400 13600 13800 14000 14400 15200 16800 18400 20000 22400 22600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 0.12 4990.7671 9981.5342 59889.21 49907.67 59889.21 64380.9 66876.3 67874.43 68872.6 69870.74 71867.05 75859.66 83844.9 91830.12 99815.34 111793.2 112791.34 0.0001 0.0003 0.00095 0.00165 0.00195 0.00215 0.00235 0.0027 0.00315 0.0045 0.0051 0.0065 0.0115 0.0168 0.02535 0.0554 0.006 4.99x10 3.33x10 3.152 x10 3.024 x10 3.07×107 3x10 2.85x10 2.5 x10 2.1x10 1.5x10 1.4x10 1.21x10 0.73x107 0.54 x107 0.39 x107 0.202 x107 1.88x10 Table 1 From Hooke's law, proportionality limit is possible when stress is directly proportional strain and hence their ration is constant which is known as Proportionality constant. It is clear from the table that proportionality constant s about (3x10 lb/in1) . Hence, Proportional limit lies from (20001b) to (12900 1b)

Modulus of elasticity is also known as the proportionality constant or the Young's modulus. It is clear from the table that Modulus of Elasticity for the substance is 3x10 lb/in2 The end point of the proportionality limit gives the yield strength. It is clear from the table that the yield strength of the material is-g-64380 lb/in2 Hence, the yield stress using 0.2% offset method is | 64380 lb/in2 Write the expression to calculate the ultimate tensile strength, Max load Substitute the value, 22600 t0.20037 113x10 lb/in2 Hence, the ultimate tensile strength is 1.13x103 1b/in' Write the expression to calculate the modulus of resilience, 2E Here, (E-3x10' lb/in1) is the modulus of elasticity Substitute the values

(64380) 2x3x10 - 69.079 1b/in2 69.08 lb/in2 Hence, the modulus of resilience is | 69.08 lb/in2 Write the expression to calculate the modulus of toughness. Substitute the values 3x10 2(1+0.33) -1.12x10' lbfin Hence, the modulus of toughness is |1.12x10 lb/in2 Write the expression to calculate the Percentage elongation in length, Al Substitute the value, 0.42 21%

Hence, the Percentage elongation in length is 121% Write the expression to calculate the Percentage reduction in area, d-d nx 100 Substitute the values 0.5052-0.422 0.505 2x100 30.83% Hence, the Percentage reduction in area is 130.83%) Write the expression to calculate the strain in the specimen when stress have reached 50 ksi Substitute the values 1000 1b/in 1ksi 50 ksi x 3x10 lb/in1 - 1.67x10-3 in/in ε Write the expression to calculate the diametral strain, Substitute the values

0.33一1.67×10-3 E 5.06x10-3 in/in Write the expression to calculate the diameter of test specimen under specified load d-d Substitute the values, 5.06x10- 0.505-d" d' 0.50474 in 0.505 Hence, the diameter of test specimen under specified load is 0.50474 in 2) Write the expression to calculate the Permanent strain, Here, ( 0.04) is the strain when material is stretched Substitute the value, ep In(1+e) n(1+0.04) 0.0392 Hence, the Permanent strain after removing the load is 0.0392 3) Write the expression to calculate the permanent stress,

OpOu (1+e Here, (e- 0.04) is the strain when material is stretched Substitute the value, op 1.3x10)x(1+0.04) -1.1752x10 lb/in2 Write the expression to calculate the new Proportional limit, Substitute the values 1.1752x105 0.0392 2.997x10 lb/in Write the expression to calculate the modulus of resilience. 2E Here. E is the modulus of elasticity Substitute the values (64380) 2x2.997 x10 - 691.488 1b/in 691.491b/in2 Hence, the new proportional limit is, 2.997x10 lb/inand the new modulus of resilience is 691.49 1b/in2