Stress, Strain, and Material Properties

Normal Stress and Strain

Normal Stress: σ = P A \sigma = \frac{P}{A} (Force/Area). Can only be valid if stress is uniformly distributed.

Normal Strain: ε = δ L \varepsilon = \frac{\delta}{L} (total elongation/total length).

Notation: Tensile stress or strain is positive. Compressive stress or strain is negative.

Stress-Strain Diagrams

This is the stress strain diagram for steel:

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Proportional Limit (A): Where the proportionality between stress and strain ends.

Modulus of Elasticity (Slope of A-O): Measures a material’s stiffness; E = σ ε \frac{\sigma}{\varepsilon}

Yield Stress (B): This point represents where permanent deformation of steel occurs.

Ultimate Stress (D): The maximum stress steel can withstand before necking begins

Ductility and Elongation

Ductility: A measure of the amount of permanent strain a material undergoes before failure occurs. Quantified using percent elongation and percent reduction in area.

Percent Elongation = L 1 − L 0 L 0 ⋅ 100 \frac{L_{1} - L_{0}}{L_{0}} \cdot 100

Percent Reduction = A 0 − A 1 A 0 ⋅ 100 \frac{A_{0} - A_{1}}{A_{0}} \cdot 100

Mechanical Properties

Elasticity: The material returns to its original dimensions during unloading.

Plasticity: The material undergoes inelastic strains beyond strain at the elastic limit

Creep: Strains that develop when a load is sustained for long periods of time.

Hooke’s Law

The linear relationship between stress and strain for a material in simple tension or compression. Only relates longitudinal stresses and strains.

σ = E ε \sigma = E\varepsilon (psi, ksi, or pascals)

Poisson’s Ratio

The relationship of lateral strain and axial strain of a bar after loading (think of stretching a rubber band).

ν = − l a t e r a l s t r a i n a x i a l s t r a i n = − ε ′ ε \nu = - \frac{lateral\ strain}{axial\ strain} = - \frac{\varepsilon'}{\varepsilon} (dimensionless)

Example 1:
Problem 1.4-3: Find maximum normal stress when the thickness of the tube is ¾ in.

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  1. Use normal stress equation:

σ = P A \sigma = \frac{P}{A}

  1. Find inner diameter:

d = 3 i n d = 3in , t = 0.75 i n t = 0.75in

d i = d − 2 t d_{i} = d - 2t

d i = 3 i n − ( 2 ⋅ 0.75 i n = 1.5 i n d_{i} = 3in - (2 \cdot 0.75in = 1.5in

  1. Calculate cross sectional area:

A = π 4 ( d 2 − d i 2 ) A = \frac{\pi}{4}(d^{2} - d_{i}^{2})

A = π 4 ( ( 3 i n ) 2 − ( 1.5 i n ) 2 = 5.3 i n 2 A = \frac{\pi}{4}({(3in)}^{2} - ({1.5in)}^{2}\ = 5.3{in}^{2}

  1. Calculate normal stress:

σ = 3 k i p s 5.3 i n 2 = 0.566 k s i \sigma = \frac{3kips}{5.3{in}^{2}} = 0.566\ ksi

Example 2:
Problem 1.7-7: Find the change in diameter and magnitude of P for the monel metal bar when L=9in, d=0.225in, and the bar elongates 0.0195in.

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Change in diameter:

  1. Find axial strain:

ε = δ L = 0.0195 i n 9 i n = 0.002167 \varepsilon = \frac{\delta}{L} = \frac{0.0195\ in}{9\ in} = 0.002167

  1. Find lateral strain:

Poison’s ration for monel (Appendix I): ν = 0.32 \nu = 0.32

ν = − ε ′ ε ⇒ ε ′ = − ν ε = − 0.32 ⋅ 0.002167 = − 0.000693 \nu = - \frac{\varepsilon'}{\varepsilon} \Rightarrow \varepsilon' = - \nu\varepsilon = - 0.32 \cdot 0.002167 = - 0.000693

  1. Find change in diameter

ε ′ = Δ d d ⇒ Δ d = ε ′ d = − 0.000693 ⋅ 0.225 i n = − 0.000156 \varepsilon' = \frac{\Delta d}{d} \Rightarrow \Delta d = \varepsilon'd = - 0.000693 \cdot 0.225in = - 0.000156

Magnitude of P:

  1. Find normal stress

Modulus of elasticity for monel (Appendix I): E = 25,000 ksi

σ = E ε = 25 , 000 k s i ⋅ 0.002167 = 54.167 k s i \sigma = E\varepsilon = 25,000\ ksi\ \cdot \ 0.002167 = 54.167\ ksi

  1. Find cross-sectional area

A = π 4 d 2 = π 4 ( 0.225 i n ) 2 = 0.0397 i n 2 A = \frac{\pi}{4}d^{2} = \frac{\pi}{4}{(0.225in)}^{2} = 0.0397{in}^{2}

  1. Calculate P

σ = P A ⇒ P = σ A = 54.167 k s i ⋅ 0.0397 i n 2 = 2.15 k i p s \sigma = \frac{P}{A} \Rightarrow P = \sigma A = 54.167\ ksi\ \cdot 0.0397{in}^{2} = 2.15\ kips