Axial Deformation

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In the linear portion of the stress-strain diagram, the tress is proportional to strain and is given by
 

$ \sigma = E \varepsilon $

 

since $ \sigma = P / A $ and $ \varepsilon = \delta / L $, then $ \dfrac{P}{A} = E \dfrac{\delta}{L} $
 

$ \delta = \dfrac{PL}{AE} = \dfrac{\sigma L}{E} $

To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the stress must not exceed the proportional limit.
 

If however, the cross-sectional area is not uniform, the axial deformation can be determined by considering a differential length and applying integration.
 

Non-uniform cross-section

 

$ \delta = \displaystyle \dfrac{P}{E} \int_0^L \frac{dx}{L} $

where A = ty, and y and t if variable, must be expressed in terms of x.
 

For a rod of unit mass ρ suspended vertically from one end, the total elongation due to its own weight is
 

$ \delta = \dfrac{\rho g L^2}{2E} = \dfrac{MgL}{2AE} $

where ρ is in kg/m3, L is the length of the rod in mm, M is the total mass of the rod in kg, A is the cross-sectional area of the rod in mm2, and g = 9.81 m/s2.
 

Stiffness, k
Stiffness is the ratio of the steady force acting on an elastic body to the resulting displacement. It has the unit of N/mm.
 

$ k = \dfrac{P}{\delta} $

 

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