Question 6.8: Show that dε in Equations (6.21) and (6.22) reduces to dε1 i......
Show that d \bar{\varepsilon } in Equations (6.21) and (6.22) reduces to d{\varepsilon }_{1} in a 1-direction tension test.
d \bar{\varepsilon }=(√2/3 )\left[(d\varepsilon_{2}-d\varepsilon_{3})^2 + (d\varepsilon_{3}-d\varepsilon_{1})^2 +(d\varepsilon_{1}-d\varepsilon_{2})^2 \right]^{1/2} (6.21)
d \bar{\varepsilon }=(2/3 )^{1/3} (d\varepsilon_{1}^{ \ \ 2} + d\varepsilon_{2}^{ \ \ 2} + d\varepsilon_{3}^{ \ \ 2})^{1/2}, (6.22)
In a 1-direction tension test d{\varepsilon }_{2} =d{\varepsilon }_{3}=-(1/2) d{\varepsilon }_{1} . Substituting into Equation (6.21),
d \bar{\varepsilon }=(√2/3 )\left\{(0)^2 +\left[-(1/2)d\varepsilon_{1}-d\varepsilon_{1}\right]^2 +\left[d\varepsilon_{1}-(-1/2)d\varepsilon_{1}\right]^2 \right\}^{1/2}
=(√2/3) \left[(9/4)d\varepsilon_{1}^{ \ \ 2} +(9/4)d\varepsilon_{1}^{ \ \ 2 } \right]^{1/2} =d\varepsilon_{1}.
Substituting d{\varepsilon }_{2} =d{\varepsilon }_{3}=-(1/2) d{\varepsilon }_{1} into Equation (6.22),
d \bar{\varepsilon }=√(2/3 )\left[d\varepsilon_{1}^{ \ \ 2}+(-1/2d\varepsilon_{1})^2 + (-1/2d\varepsilon_{1})^2\right] ^{1/2}
=√(2/3) \left[d\varepsilon_{1}^{ \ \ 2}+d\varepsilon_{1}^{ \ \ 2}/4 + d\varepsilon_{1}^{ \ \ 2}/4\right]^{1/2} = d\varepsilon_{1}.