# Stress, Strain and Young's Modulus

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

### Stress

Stress is the ratio of *applied force F to a cross section area* *-* defined as "*force per unit 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 (lb_{f}/in^{2}))

F_{n}= normal force acting perpendicular to the area (N, lb_{f})

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

*a kip*is an imperial unit of force - it equals*1000 lb*_{f}(pounds-force)*1 kip = 4448.2216 Newtons (N) = 4.4482216 kilo Newtons (kN)*

A normal force acts perpendicular to area and is developed whenever external loads tends to push or pull the two segments of a body.

#### 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 a plane is usually denoted as "**shear stress**" and can be expressed as

τ = F_{p}/ A (2)

where

τ = shear stress (Pa (N/m^{2}), psi (lb_{f}/in^{2}))

F_{p}= shear force in the plane of the area (N, lb_{f})

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

A shear force lies in the plane of an area and is developed when external loads tend to cause the two segments of a body to slide over one another.

### Strain (Deformation)

Strain is defined as "deformation of a solid due to stress".

- Normal strain - elongation or contraction of a line segment
- Shear strain - change in angle between two line segments originally perpendicular

Normal strain 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 = strain - unit-less

E = Young's modulus (Modulus of Elasticity)(Pa , (N/m))^{2}), psi (lb_{f}/in^{2}

- Young's modulus can be used to predict the elongation or compression of an object when exposed to a force

Note that strain is a dimensionless unit since it is the ratio of two lengths. But it also common practice to state it as the ratio of two length units - like *m/m* or *in/in*.

- Poisson's ratio is the ratio of relative contraction strain

#### 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 (200 10 ^{9} N/m^{2})*. The change of length can be calculated by transforming

*(3)*to

* dl = σ l_{o }/ E*

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

* = 0.00127 m*

* = 1.27 mm*

#### Strain Energy

Stressing an object stores energy in it. For an axial load the energy stored can be expressed as

*U = 1/2 F _{n} dl *

*where *

*U = deformation energy (J (N m), ft lb)*

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

Most metals deforms proportional to 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 of Elasticity - or Modulus of Rigidity

*G = stress / strain *

* = τ / γ *

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

*where *

*G = Shear Modulus of Elasticity - or Modulus of Rigidity (N/m ^{2}) (lb/in^{2}, psi)*

*τ * = shear stress ((Pa) N/m

^{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)*

### Bulk Modulus Elasticity

The Bulk Modulus Elasticity - or Volume Modulus - is a measure of the substance's resistance to uniform compression. Bulk Modulus of Elasticity is the ratio of stress to change in volume of a material subjected to axial loading.

### Elastic Moduli

Elastic moduli for some common materials:

Material | Young's Modulus- E - | Shear Modulus- G - | Bulk Modulus- K - |
---|---|---|---|

(GPa)(10^{6} psi) | (GPa)(10^{6} psi) | (GPa)(10^{6} psi) | |

Aluminum | 70 | 24 | 70 |

Brass | 91 | 36 | 61 |

Copper | 110 | 42 | 140 |

Glass | 55 | 23 | 37 |

Iron | 91 | 70 | 100 |

Lead | 16 | 5.6 | 7.7 |

Steel | 200 | 84 | 160 |

*1 GPa = 10*^{9}Pa (N/m^{2})*10*^{6}psi =10*1 Mpsi =*^{3}ksi

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