Question

Relation between volumetric strain and linear strain

Answer 1

Answer:

Linear strain is the ratio of the change in length of the body due to the deformation to its original length in the direction of the force while Volumetric strain is the ratio of the change in volume of the body to the deformation to its original volume.

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Find:

• Relation between volumetric$(E_v)$ and longitudinal (linear) strain $(E_x,E_y\&E_z)$

Solution:

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(3.5,6.1){\sf c}\put(7.7,6.3){\sf a}\put(11.3,7.45){\sf b}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture} We\:know\:that, \\ \\ E_x =\dfrac{\Delta a}{a} \\ \\ E_y =\dfrac{\Delta b}{b} \\ \\ E_z =\dfrac{\Delta c}{c}$ $</p><p>\setlength{\unitlength}{0.74 cm}\begin{picture1}\thicklines\put(3.5,-3.9){\sf {c + \Delta c}}\put(7.7,-3.7){\sf{ a + \Delta a}}\put(11.3,-2.55){\sf {b + \Delta b}}\put(6,-4){\line(1,-10){5}}\put(6,-1){\line(1,-10){5}}\put(11,-1){\line(0,-11){3}}\put(6,-4){\line(0,-9){3}}\put(4,-2.7){\line(1,-10){5}}\put(4,0.3){\line(1,-10){5}}\put(9,0.3){\line(0,-11){3}}\put(4,-2.7){\line(0,-11){3}}\put(6,-4){\line(1,-10){5}}\put(6,-1){\line(1,-10){5}}\put(11,-1){\line(0,-11){3}}\put(6,-4){\line(0,-11){3}}\put(4,-2.7){\line(1,-10){5}}\put(4,0.3){\line(1,-10){5}}\put(9,0.3){\line(0,-11){3}}\put(4,-2.7){\line(0,-9){3}}\put(6,6){\line(-3,-8){2}}\put(6,-1){\line(-3,-8){2}}\put(11,-1){\line(-3,-8){2}}\put(11,-4){\line(-3,-8){2}}\end{picture1}$ $We\:know\:that, \\ \\ E_v =\dfrac{\Delta v}{v} \\ \\ E_v =\dfrac{v_f - v_i}{v_i} \\ \\ => v_f = ( a + \Delta a )( b + \Delta b )( c + \Delta c ) \:\:\:and\: v_i = abc \\ \\ \\ \implies E_v =\dfrac{v_f - v_i}{v_i} \\ \\ \implies E_v =\dfrac{ ( a + \Delta a )( b + \Delta b )( c + \Delta c ) - abc}{abc} \\ \\ \implies E_v = \dfrac{ abc\!\!\!\!\!/ \left[\left(1+\dfrac{\Delta a}{a}\right) \left(1+\dfrac{\Delta b}{b}\right) \left(1+\dfrac{\Delta c}{c}\right) - 1\right]}{ abc\!\!\!\!\!/} \\ \\ \implies E_v = 1\!\!/ + \dfrac{\Delta a}{a} + \dfrac{\Delta b}{b} + \dfrac{\Delta c}{c} + \dfrac{\Delta a*\Delta b}{a*b} + \dfrac{\Delta a*\Delta c}{a*c} + \dfrac{\Delta b*\Delta c}{b*c} + \dfrac{\Delta a*\Delta b*\Delta c}{a*b*c} - 1\!\!/ \\ \\ \because ^{\Delta a}/_a ,\: ^{\Delta b}/_b \:and\: ^{\Delta c}/_c\:are\:very\:small\\ \therefore their\: multiplications\:are\: negligible,\:i.e., \\ \implies E_v = \dfrac{\Delta a}{a} + \dfrac{\Delta b}{b} + \dfrac{\Delta c}{c} \\ \\ \implies E_v = E_x + E_y + E_z \\ So,\\ \boxed{Volumetric\:Strain = \Sigma Longitudinal\:Strain}$

Formula Used:

• $E_x =\dfrac{\Delta a}{a}, E_y =\dfrac{\Delta b}{b} \& E_z =\dfrac{\Delta c}{c}$
• $v_f = ( a + \Delta a )( b + \Delta b )( c + \Delta c ) \:\:\:and\: v_i = abc$

Hope it helps!