Question

Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape of a rectangular parallelepiped of dimensions l1,l2, and l3. Show that the volume constancy requires that the following expression be satisfied: e1+e2+e3 = 0

Answer 1

Answer:  ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

                         l₀l₀l₀=l₁l₂l₃

                        $\frac{lo*lo*lo}{l1*l2*l3}=1.0$

                      taking natural log on both sides

                              $ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)$

Considering the logarithmic Laws of division and multiplication :

                                ln(AB) = ln(A)+ln(B)

                                ln(A/B) = ln(A)-ln(B)

                           $ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0$

Use the image attached to see the definition of true strain defined as

                         ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0