GATE Papers >> Mechanical >> 2016 >> Question No 45

Question No. 45 Mechanical | GATE 2016

A slider crank mechanism with crank radius 200 mm and connecting rod length 800 mm is shown. The crank is rotating at 600 rpm in the counterclockwise direction. In the configuration shown, the crank makes an angle of $90^\circ$ with the sliding direction of the slider, and a force of 5 kN is acting on the slider. Neglecting the inertia forces, the turning moment on the crank (in kN-m) is __________

 


Answer : 0.9 : 1.1


Solution of Question No 45 of GATE 2016 Mechanical Paper

$ \because\mathrm T=\mathrm F.\mathrm r\left(\mathrm{sinθ}+\frac{\sin2\mathrm\theta}{2\sqrt{\mathrm n^2-\sin^2\mathrm\theta}}\right) $

Here, $ \mathrm\theta=90^\circ $

$ \begin{array}{l}2\mathrm\theta=180^\circ\\\therefore\mathrm{sinθ}=\sin90^\circ=1\\\sin2\mathrm\theta=\sin180^\circ=0\\\therefore\mathrm T=\mathrm F.\mathrm r\left(1+0\right)=\mathrm F.\mathrm r=5\times0.2=1\;\mathrm{kN}.\mathrm m\end{array} $

Alternate:

 

$ \begin{array}{l}\mathrm s\mathrm i\mathrm n\mathrm\phi=\frac{200}{800}\\\mathrm\phi=14.47^\circ\\\therefore\mathrm\theta=180^\circ-104.47^\circ\\\;\;\;\;\;=75.57^\circ\end{array} $

For piston, (neglecting inertia forces)

$ \begin{array}{l}\mathrm T\;\mathrm c\mathrm o\mathrm s\mathrm\phi=5000\\\mathrm T=\frac{5000}{\cos14.47^\circ}=5163.803\mathrm N\end{array} $

$ \therefore $ Turning moment on crank

$ \begin{array}{l}=\left(\mathrm T\;\mathrm{sinθ}\right)\times\left(0.2\;\mathrm m\right)=5163.803\times\sin75.53^\circ\times0.2\;\mathrm{Nm}\\=1000\;\mathrm{Nm}=1\;\mathrm{kN}-\mathrm M\end{array} $

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