Question 9.11: Repeat the above problem for the strain condition: ∈xx = +40...
Repeat the above problem for the strain condition:
∈_{x x}=+400 \mu, \quad ∈_{y y}=+200 \mu, \quad \gamma_{x y}=+375 \mu
Mohr’s circle for the strain is shown in Figure 9.38:
(a) Now, from the Mohr’s circle, we can say the principal strains are:
∈_1= OC + CE =(300+212.5) \mu=512.5 \mu
and \epsilon_2= OF = OC – CF =(300-212.5) \mu=87.5 \mu
Therefore,
\tan 2 \phi=\frac{187.5}{(400-300)} \Rightarrow \phi=30.96^{\circ} ⦪
where \phi is the angular orientation of the principal strain. The situation is shown in the Figure 9.39.
(b) Maximum in-plane shear strain is
\frac{\left(\gamma_{x y}\right)_{\max }}{2}=212.5 \mu \Rightarrow\left(\gamma_{x y}\right)_{\max }=425 \mu
which occurs at angle
\phi_{ s }=\frac{1}{2}\left(90^{\circ}-2 \phi\right)
or \phi_s=45^{\circ}-\phi=45^{\circ}-30.96^{\circ}=14.04^{\circ} ⦨
(c) Finally considering the plane-strain condition, we get our principal strains as:
∈_1=512.5 \mu, \quad ∈_2=87.5 \mu, \quad ∈_3=0
and the maximum shear strain \gamma_{\max } is given by:
\frac{\gamma_{\max }}{2}=\frac{1}{2}\left(\varepsilon_1-\varepsilon_3\right)=\frac{512.5}{2} \mu
or \gamma_{\max }=512.5 \mu
