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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