Question

A metallic cube whose each is 10 cm is subjected to a shearing force of 100kgf. The top face is displaced through 0.25 cm with respect to bottom. Calculate the shearing stress, strain and shear modulus.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{shearing}\mathrm{force},\mathrm{F}=100\mathrm{kgf}=100\times 9.8\mathrm{N} \\ \mathrm{side}\mathrm{of}\mathrm{cube},\mathrm{l}=10\mathrm{cm}=0.1\mathrm{m} \\ \mathrm{area}\mathrm{of}\mathrm{cube},\mathrm{A}=0.1\times 0.1=0.01{\mathrm{m}}^2 \\ ∆\mathrm{l}=0.25\mathrm{cm} \\ \mathrm{Shearing}\mathrm{stress},\mathrm{T}=\frac{\mathrm{F}}{\mathrm{A}} \\ \mathrm{T}=\frac{100\times 9.8}{0.01}=9.8\times {10}^4\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Strain}=\frac{∆\mathrm{l}}{\mathrm{l}} \\ \mathrm{Strain}=\frac{.25}{10}=0.025 \\ \mathrm{Shear}\mathrm{modulus}=\frac{\mathrm{Fl}}{\mathrm{A}∆\mathrm{l}} \\ \mathrm{Shear}\mathrm{modulus}=\frac{100\times 9.8}{0.01\times 0.025}=3.92\times {10}^6\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$