Calculate the change in volume:
If the bar is 1 m long with rectangular cross section of 300 mm deep and 400 mm wide, compute the change in volume of the solid because of a longitudinal compressive force of 720 kN now if the elastic constants E and υ for the material are called as 120 kN/mm2 and 0.2 respectively.
Solution
Area of cross section of the member = 300 × 400 = 120000 mm²
Longitudinal strain ε = P/AE = - 720 × 1000/120000 × 120 × 103 = - 0.00005
(Note that all the values have to be converted to consistent units; here, it is N for forces and mm for length.)
∴ Total change in length δ = 1000 × (- 0.00005) = - 0.05 mm.
Lateral strain εl = -υε = - 0.2 × (- 0.00005) = 0.00001
Change in depth = 0.00001 × 300 = 0.003 mm
Change in width = 0.00001 × 400 = 0.004 mm
∴ Change in volume of the solid,
= (1000 - 0.05) (300 + 0.003) (400 + 0.004) - (1000 × 400 × 300)
= 999.95 × 300.003 × 400.004 - (1000 × 400 × 300)
= - 3600.108 mm3
Let us consider an alternate approximate method also.
Change in volume, dV = (V + dV) - V
= (l + Δl) (b + Δb) (d + Δd) - l . b . d
where Δl, Δb, and Δd are changes in length, breadth and depth of the solid.
i.e. dV = l (1 + ε1) × b (1 + ε2) × d (1 + ε3) - l . b . d
where ε1, ε2 and ε3 are the strains in the three mutually perpendicular directions.
∴ dV = l bd × (1 + ε1) (1 + ε2) (1 + ε3) - l bd
= l bd × (1 + ε1 + ε2 + ε3 + ε1 ε2 + ε2 ε3 + ε3 ε1 + ε1 ε2 ε3) - l bd
= l bd × (ε1 + ε2 + ε3 + ε1 ε2 + ε2 ε3 + ε3 ε1 + ε1 ε2 ε3)
Neglecting the second order products,
dV = V × (ε1 + ε2 + ε3)
Now let us calculate the change in volume of the given solid using Eq.
Change in volume, dV = V × (ε1 + ε2 + ε3)
= 1000 × 300 × 400 (- 0.00005 + 0.00001 + 0.00001)
= - 3600 mm3
By there is a small error, the approximation is quite satisfactory (As an exercise you might calculate the percentage error in the value). If you are extremely particular about accuracy, you use the subsequent formulation:
dV = V × (ε1 + ε2 + ε3 + ε1 ε2 + ε2 ε3 + ε3 ε1 + ε1 ε2 ε3)