Given the strain energy potential psi E lambda2 tr E2 mu t
Solution
solution:
1) here we are provided with strain energy density potential function or strain energy per unit volume function in term of strain tensor as
here psi=f
strain tensor=E
stress tensor=S
where constant=
lamda=m
mu=n
so strain potential function as
f=m/2*E^2+nE^2
where relation between stress and strain tensor as
s=df/dE
hence on differentiating above eqution woth respect to strain tensor, we get
df/dE=(d/dE)(m/2*Etranslational^2)+(d/dE)(nElateral^2)
m and n are constant so we get
S=mEtrans+2*n*Elateral
in this way above equation gives value of stress tensor and obeys hooks law which is
f=1/2*S*E
