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The data shown in the accompanying table are From a tensile test of high-strength steel. The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.5-3). At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area. TENSILE-TEST DATA FOR PROB. L.5-7 Laid (lb) Elongation (in,) 1000 0.0002 2000 0.0006 6000 0.0019 10,000 0.0033 12,000 0.0039 12,900 0.0041 13,400 0.0047 13,600 0.0054 13,800 0.0063 14,000 0.0090 14,400 0.0102 15,200 0.0130 16,800 0.0230 18,400 O.0336 20,000 O.05O7 22,400 0.1108 22,600 Fracture

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Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

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BuyFindarrow_forward

Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 1, Problem 1.5.7P
Textbook Problem
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The data shown in the accompanying table are From a tensile test of high-strength steel. The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.5-3). At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area.

TENSILE-TEST DATA FOR PROB. L.5-7

Laid (lb)   Elongation (in,)

1000  0.0002

2000  0.0006

6000  0.0019

10,000  0.0033

12,000   0.0039

12,900   0.0041

13,400   0.0047

13,600   0.0054

13,800   0.0063

14,000   0.0090

14,400  0.0102

15,200   0.0130

16,800   0.0230

18,400   O.0336

20,000   O.05O7

22,400   0.1108

22,600   Fracture

To determine

The proportional limit.

The modulus of elasticity.

The yield stress at 0.1% offset.

The ultimate stress.

The percent elongation in area.

The percent reduction in area.

Explanation of Solution

Given information:

The diameter of the specimen is 0.505in , the length is 2.00in , the elongation between the gage marks is 0.12in , the minimum diameter is 0.42in

Write the expression for the initial area of the specimen.

  Ai=πdi24 .....(I)

Here, the diameter of the specimen is di and the initial area is Ai .

Write the expression for final area of the specimen.

  Af=πdf24 .....(II)

Here, the final diameter of the specimen is df and the final area is Af .

Write the expression for the stress in the specimen.

  σ=PAi.....(III)

Here, the load on the specimen is P and the stress in specimen is σ .

Write the expression for the strain in terms of elongation.

  ε=ΔLLi ….. (IV)

Here, the change in lengths is ΔL and the strain is ε .

Write the expression for the modulus of elasticity.

  E=linear stresslinear strain .....(V)

Here, the modulus of elasticity is E .

Write the expression for the percentage elongation.

  εL=(LfLiLi)100 .....(VI)

Here, percentage elongation is εL .

Write the expression for the percentage reduction

  εA=(AiAf)Ai×100.....(VII)

Here, percentage reduction in area is εA .

Calculation:

Substitute 0.505in. for d in Equation (I).

  Ai=π×( 0.505in)24=π×0.255025in24=0.200in2

Substitute 0.42in. for df in Equation (II).

  Af=π( 0.42in)24=π×0.1764in24=0.138in2

Substitute 0.200in.2 for Ai in Equation (III).

  σ=P0.200σ=5P

Substitute 2in. for Li , in Equation (IV).

  ε=ΔL2ε=0.5ΔL

    Load (lb) Elongation ΔL Stress σ(lb/in.2) Strain ε
    1000 0.0002 4992.6 0.0001
    2000 0.0006 9985.2 0.0003
    6000 0.0019 29955.6 0.00095
    10000 0.0033 49926.0 0.00165
    12000 0.0039 59911.20 0.00195
    12900 0.0043 64404.54 0.00215
    13400 0.0047 66900.84 0

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