# Stress, Strain and Young's Modulus

## Stress is force per area - strain is deformation of a solid due to stress

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### Stress

Stress is "*force per unit area*" - the ratio of applied force *F* and cross section *-* defined as "force per area".

*tensile stress*- stress that tends to stretch or lengthen the material - acts normal to the stressed area*compressive stress*- stress that tends to compress or shorten the material - acts normal to the stressed area*shearing stress*- stress that tends to shear the material - acts in plane to the stressed area at right-angles to compressive or tensile stress

#### Tensile or Compressive Stress - Normal Stress

Tensile or compressive stress normal to the plane is usually denoted "**normal stress**" or "**direct stress**" and can be expressed as

σ = F_{n}/ A (1)

where

σ = normal stress ((Pa) N/m^{2}, psi)

F_{n}= normal component force (N, lb_{f }(alt. kips))

A = area (m^{2}, in^{2})

*a kip*is a non-SI unit of force - it equals*1,000 pounds-force**1 kip = 4448.2216 Newtons (N) = 4.4482216 kilonewtons (kN)*

#### Example - Tensile Force acting on a Rod

A force of *10 kN* is acting on a circular rod with diameter *10 mm*. The stress in the rod can be calculated as

*σ = (10 10 ^{3} N)*

*/ (π ((10 10*

^{-3}m) / 2)^{2})* = 127388535 (N/m ^{2}) *

* = 127 (MPa)*

#### Example - Force acting on a Douglas Fir Square Post

A compressive load of *30000 lb* is acting on short square *6 x 6 in* post of Douglas fir. The dressed size of the post is *5.5 x 5.5 in* and the compressive stress can be calculated as

*σ = (30000 lb)* */ ((5.5 in) (5.5 in) ^{})*

* = 991 (lb/in ^{2}, psi) *

#### Shear Stress

Stress parallel to the plane is usually denoted "**shear stress**" and can be expressed as

τ = F_{p}/ A (2)

where

τ = shear stress ((Pa) N/m^{2}, psi)

F_{p}= parallel component force (N, lb_{f})

A = area (m^{2}, in^{2})

### Strain

Strain is defined as "deformation of a solid due to stress" and can be expressed as

ε = dl / l_{o}

= σ / E (3)

where

dl = change of length (m, in)

l_{o}= initial length (m, in)

ε = unit less measure of engineering strain

E = Young's modulus (Modulus of Elasticity)(N/mpsi))^{2}(Pa), lb/in^{2}(

- Young's modulus can be used to predict the elongation or compression of an object.

#### Example - Stress and Change of Length

The rod in the example above is *2 m* long and made of steel with Modulus of Elasticity *200 GPa*. The change of length can be calculated by transforming (3) as

* dl = σ l_{o }/ E*

* = (127 10 ^{6} Pa) (2 m) / (200 10^{9} Pa) *

* = 0.00127 (m)*

* = 1.27 (mm)*

### Young's Modulus - Modulus of Elasticity (or Tensile Modulus) - Hooke's Law

Most metals deforms proportional with imposed load over a range of loads. Stress is proportional to load and strain is proportional to deformation as expressed with Hooke's law

E = stress / strain

=σ/ε

= (F_{n}/ A) / (dl / l_{o})(4)

where

E = Young's modulus (N/m^{2}) (lb/in^{2}, psi)

Modulus of Elasticity, or Young's Modulus, is commonly used for metals and metal alloys and expressed in terms *10 ^{6} lb_{f}/in^{2}, N/m^{2} or Pa*. Tensile modulus is often used for plastics and is expressed in terms

*10*.

^{5}lb_{f}/in^{2}or GPa### Shear Modulus

*S = stress / strain *

* = τ / γ *

* = (F _{p} / A) / (s / d) (5)*

*where *

*S = shear modulus (N/m ^{2}) (lb/in^{2}, psi)*

* γ = unit less measure of shear strain *

*F _{p} = force parallel to the faces which they act*

*A = area (m ^{2}, in^{2})*

*s = displacement of the faces (m, in)*

*d = distance between the faces displaced (m, in)*

### Elastic Moduli

Material | Young's Modulus | Shear Modulus | Bulk Modulus | |||

10^{10} N/m^{2} | 10^{6} lb/in^{2} | 10^{10} N/m^{2} | 10^{6} lb/in^{2} | 10^{10} N/m^{2} | 10^{6} lb/in^{2} | |

Aluminum | 7.0 | 10 | 2.4 | 3.4 | 7.0 | 10 |

Brass | 9.1 | 13 | 3.6 | 5.1 | 6.1 | 8.5 |

Copper | 11 | 16 | 4.2 | 6.0 | 14 | 20 |

Glass | 5.5 | 7.8 | 2.3 | 3.3 | 3.7 | 5.2 |

Iron | 9.1 | 13 | 7.0 | 10 | 10 | 14 |

Lead | 1.6 | 2.3 | 0.56 | 0.8 | 0.77 | 1.1 |

Steel | 20 | 29 | 8.4 | 12 | 16 | 23 |

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