Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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 Step 1

Given data:

Solid shaft diameter (d) = 60 mm = 0.06 m

Determine,

Shear stress at points A and B

$\mathrm{1 }. {\mathrm{\tau }}_A$

$\mathrm{2 }. {\mathrm{\tau }}_B$

Step 2

We know that, from torsion equation

$\frac{\mathrm{T }}{\mathrm{J }}= \frac{\mathrm{\tau }}{R}$

$\mathrm{\tau }= \frac{\mathrm{T }}{\mathrm{J }}\times R$

Where,

$\mathrm{\tau }=$shear stress, N/m ²

$\mathrm{T }=$Torque, N.m

$\mathrm{R }=$Radius, m

$\mathrm{J }=$Polar moment of inertia, m ⁴

^{$\mathrm{J }= \frac{\mathrm{\pi }}{\mathrm{2 }}\times {\mathrm{d }}^{\mathrm{4 }}{\mathrm{m }}^4$}

^{$\mathrm{J }= \frac{\mathrm{\pi }}{\mathrm{2 }}\times {( \mathrm{0 }. \mathrm{06 }\mathrm{m }) }^{\mathrm{4 }}= \mathrm{2 }. \mathrm{03 }\times {\mathrm{10 }}^{- \mathrm{5 }}{\mathrm{m }}^4$}

Step 3

From the diagram

$$\begin{gathered} {\mathrm{\tau }}_{\mathrm{A }\mathrm{C }}= \mathrm{400 }\mathrm{N }. \mathrm{m } \\ {\mathrm{\tau }}_{\mathrm{AC }}= \mathrm{200 }\mathrm{N }. \mathrm{m }( \mathrm{600 }\mathrm{N }. \mathrm{m }- \mathrm{400 }\mathrm{N }. \mathrm{m }) \end{gathered}$$

${\mathrm{\tau }}_{\mathrm{A }}= \frac{{\mathrm{T }}_{\mathrm{A }\mathrm{C }}}{\mathrm{J }}\times \mathrm{R }\mathrm{N }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{A }}= \frac{\mathrm{400 }\mathrm{N }. \mathrm{m }}{\mathrm{2 }. \mathrm{03 }\times {\mathrm{10 }}^{- \mathrm{5 }}{\mathrm{m }}^{\mathrm{4 }}}\times \mathrm{0 }. \mathrm{03 }\mathrm{m }\mathrm{N }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{A }}= \mathrm{591133 }\mathrm{N }/ {\mathrm{m }}^{\mathrm{2 }}= \mathrm{591 }. \mathrm{133 }\mathrm{kN }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{B }}= \frac{{\mathrm{T }}_{\mathrm{B }\mathrm{C }}}{\mathrm{J }}\times \mathrm{R }\mathrm{N }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{B }}= \frac{\mathrm{200 }\mathrm{N }. \mathrm{m }}{\mathrm{2 }. \mathrm{03 }\times {\mathrm{10 }}^{- \mathrm{5 }}{\mathrm{m }}^{\mathrm{4 }}}\times \mathrm{0 }. \mathrm{03 }\mathrm{m }\mathrm{N }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{B }}= \mathrm{295566 }. \mathrm{5 }\mathrm{N }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{B }}= \mathrm{295 }. \mathrm{566 }\mathrm{kN }/ {\mathrm{m }}^2$

The Shear stresses at Point A and B are

${\mathrm{\tau }}_{\mathrm{A }}= \mathrm{591 }. \mathrm{133 }\mathrm{kN }/ {\mathrm{m }}^2$

${\mathrm{\tau }}_{\mathrm{B }}= \mathrm{295 }. \mathrm{566 }\mathrm{kN }/ {\mathrm{m }}^2$

Step 4

The shear stress on volume elements are