Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness $\mathrm{d}_1$ and $\mathrm{d}_2$ such that $\mathrm{d}_2=2 \mathrm{~d}_1$ and $l \gt \mathrm{d}_2$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force…

Question

Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness $\mathrm{d}_1$ and $\mathrm{d}_2$ such that $\mathrm{d}_2=2 \mathrm{~d}_1$ and $l \gt \mathrm{d}_2$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $	heta_2=2 	heta_1$. If the shear moduli of material 1 is $4 	imes 10^9 \mathrm{~N} / \mathrm{m}^2$, then shear moduli of material 2 is $\mathrm{x} 	imes 10^9 \mathrm{~N} / \mathrm{m}^2$, where value of $x$ is ________.

Aicharya · Accepted Answer

Deformation angle $\begin{aligned} & 2 	heta_1=	heta_2 \ & \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2} \end{aligned}$ $\begin{aligned} & \Rightarrow 2\left(\frac{\mathrm{~F}}{\ell \mathrm{~d}_1 \eta_1}ight)=\frac{\mathrm{F}}{\ell \mathrm{d}_2 \eta_2} \ & \Rightarrow \eta_2=\frac{\eta_1}{4}=1 	imes 10^9 \Rightarrow x=1\end{aligned}$

Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness…

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