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Stress & Strain

Introduction..... Symbols..... Stress..... Strain..... Young's Modulus..... Poissen's Ratio.....
Shear Stress..... Complementary Shear Stress..... Shear Strain..... Modulus of Rigidiy.....

Introduction

Any elastic material which is subject an applied force is deformed.   There are clearly defined relationships between the applied forces and the resulting deformations.  This page indentifies these relationships.



Symbols / Units
L = length (m)
F = Force (N)
σ1 = principal stress normal to σ2 & σ3(N/m2)
σ2 = principal stress normal to σ1 & σ3(N/m2)
σ2 = principal stress normal to σ1 & σ2(N/m2)
G = Modulus of Rigidity (N/m2)
E = Young's moudulus of elasticity (N/m2)
υ = Poissan's ratio
ε1 = strain in direction of σ1
ε2 = strain in direction of σ2
ε3 = strain in direction of σ3
δ = deflection(m)
τ = shear stress (N/m2)
φ = shear strain (radians)


Direct Stress

The average stress on a bar under a tensile load is defined as the force /the cross section area at right angles to the direction of the force.

σavg = F /A

The actual stress at a point a tensile load established by considering a small area Δ A with ΔF representing the normal force transmitted by the area. The stress is Δ Force /Δ Area as ΔA ->0 area at right angles to the direction of the force.

σ = limΔA ->0 ΔF /ΔA = dF /dA

Tensile stress is generally considered positive and compressive stress is negative/



Direct Strain

The average normal strain experienced by a bar under tensile load is simply defined as the average change in length /original length

εavg = δavg /L

The actual strain at a point as a result of a tensile load is established by considering a small length change Δ δ of small part of the original length ΔL . The strain is Δ δ as ΔL ->0 .

ε = limΔL ->0 Δ δ /Δ L = dδ /dL

Strain due to tensile stresses are generally considered positive and those resulting from compressive stress are negative/



Young's Modulus

Hook's law for an elastic material is that the strain is directly proportional to the stress.  The constant of proportionality is called Young's Modulus (E)

σ = E.ε

Note: the strain and strain used in engineering calculations are most often the average values.



Poissen's Ratio

The sketch below shows a very simplified view of an microscopic element of material subject to direct stress under three loading conditions.  There are assumed to be no shear stresses.

When an element is stressed in the x direction there is a proportional strain in the x direction.   The length increases due to tensile length.  In addition the section reduces in a proportional way.  The ratio of lateral strain to axial strain has been demonstrated to be proportional and the proportionality constant is called "Poissens ratio" υ.  This constant is different for different materials

υ = lateral strain /axial strain

The table below identifies the stress strain relationships for each of the three loading systems as illustrated.





Shear Stress

If an applied load consists of two equal and opposite forces which are not in the same line then the material being loaded will tend to shear as shown.

If area of the surface under shear load is A and the shearing force is F then the average shear stress is

τ av = F /A

Any section x-x between the tangential forces will experience the same shear stress.  The shear stress is tangential to the area on which it acts.  Dowels, keys, and rivets are generally designed for withstanding shear loads.  A shaft in torsion is mainly transmitted torque by shear loading.

The actual shear stress at a point due to a shear load is established by considering a small area Δ A with ΔF representing the tangential force transmitted by the area. The shar stress is Δ Force /Δ Area as ΔA ->0 area at right angles to the direction of the force.

τ = limΔA ->0 ΔF /ΔA = dF /dA



Complementary Shear Stress

Consider a small element under a shear stress condition.  For the element to be in equilibrium the horizontal forces and the vertical forces balanced and the sum of the moments on the area should also be zero.

The element assumed to be 1 unit thick and the couple resulting from the shear stress τ = τ .x.y. For equilibrium there must be a couple from a complementary shear stress τ ' this resulting couple will be τ '. y.x.

τ .x.y = τ ' y.x.... Therefore ....τ = τ '

For every shear stress the is an equal complementary shear stress on planes at right angles. The direction of the shear stresses on an element is either both towards or both away from the corners.

An important consequence of this is that near any free surface with no external applied forces, the shear stress on any cross section must be parallel to the boundary.  A round section subject to pure lateral shear is subject to shear stress as shown below.

The figure below shows a stress element and illustrates the method of identifying normal and shear stresses.  The directions are all positive



Shear Strain

A shear stress produces a shear strain as is shown below.   The shear strain is defined δ x / y which for the small deflections -> φ.  The shear strain is dimensionless and is measured in radians



Modulus of Rigidity

For elastic materials the shear strain is directly proportional to the shear strain within the limits of elasticity.. The ratio is called the modulus of rigidity (G).

Modulus of rigidity = G = τ / φ ( N/m2 )

This constant is different for different materials

Notes to be added



Links to Failure Criteria
  1. Engineering strain - Efunda Extremely clear information from reputable source
  2. Wiki Books - Solid mechanics ..An excellent online text


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Last Updated 09/09/2011