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Question : A shaft rotating at 200 r.p.m. drives another shaft at 300 r...

A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m. and transmits 6 kW through a belt. The belt is 100 mm wide and 10 mm thick. The distance between the shafts is 4m.The smaller pulley is 0.5 m in diameter. Calculate the stress in the belt, if it is 1. an open belt drive, and 2. a cross belt drive. Take \mu=0.3.

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Given :   N_{1}=200 r . p . m . ; N_{2}=300 r . p . m . ; P=6 kW =6 \times 10^{3} W ; b=100 mm ; t=10 mm ; x=4 m ; d_{2}=0.5 m ; \mu=0.3

Let    \sigma = Stress in the belt.

 

1. Stress in the belt for an open belt drive

First of all, let us find out the diameter of larger pulley \left(d_{1}\right). We know that

\frac{N_{2}}{N_{1}}=\frac{d_{1}}{d_{2}} \text { or } d_{1}=\frac{N_{2} \cdot d_{2}}{N_{1}}=\frac{300 \times 0.5}{200}=0.75 m

and velocity of the belt,       v=\frac{\pi d_{2} \cdot N_{2}}{60}=\frac{\pi \times 0.5 \times 300}{60}=7.855 m / s

Now let us find the angle of contact on the smaller pulley. We know that, for an open belt drive,

\sin \alpha=\frac{r_{1}-r_{2}}{x}=\frac{d_{1}-d_{2}}{2 x}=\frac{0.75-0.5}{2 \times 4}=0.03125 \text { or } \alpha=1.8^{\circ}

\therefore Angle of contact,     \theta=180^{\circ}-2 \alpha=180-2 \times 1.8=176.4^{\circ}

=176.4 \times \pi / 180=3.08 rad

Let     T_{1} = Tension in the tight side of the belt, and

T_{2} = Tension in the slack side of the belt.

We know that

2.3 \log \left(\frac{T_{1}}{T_{2}}\right)=\mu . \theta=0.3 \times 3.08=0.924

\therefore        \log \left(\frac{T_{1}}{T_{2}}\right)=\frac{0.924}{2.3}=0.4017 \text { or } \frac{T_{1}}{T_{2}}=2.52       …(i)                                                                               …(Taking antilog of 0.4017)

We also know that power transmitted (P),

6 \times 10^{3}=\left(T_{1}-T_{2}\right) v=\left(T_{1}-T_{2}\right) 7.855

 

\therefore        T_{1}-T_{2}=6 \times 10^{3} / 7.855=764 N …(ii)

From equations (i) and (ii),

T_{1}=1267 N , \text { and } T_{2}=503 N

We know that maximum tension in the belt \left(T_{1}\right),

1267=\sigma . b . t=\sigma \times 100 \times 10=1000 \sigma

 

\therefore           \sigma=1267 / 1000=1.267 N / mm ^{2}=1.267 MPa \ldots\left[\because 1 MPa =1 MN / m ^{2}=1 N / mm ^{2}\right]

 

Stress in the belt for a cross belt drive

We know that for a cross belt drive,

\sin \alpha=\frac{r_{1}+r_{2}}{x}=\frac{d_{1}+d_{2}}{2 x}=\frac{0.75+0.5}{2 \times 4}=0.1562 \text { or } \alpha=9^{\circ}

\therefore    Angle of contact,          \theta=180^{\circ}+2 \alpha=180+2 \times 9=198^{\circ}

=198 \times \pi / 180=3.456 rad

We know that

2.3 \log \left(\frac{T_{1}}{T_{2}}\right)=\mu . \theta=0.3 \times 3.456=1.0368

 

\log \left(\frac{T_{1}}{T_{2}}\right)=\frac{1.0368}{2.3}=0.4508 \text { or } \frac{T_{1}}{T_{2}}=2.82      …(iii)                                                    …(Taking antilog of 0.4508)

T_{1}=1184 N \text { and } T_{2}=420 N

We know that maximum tension in the belt \left(T_{1}\right),

1184=\sigma . b . t=\sigma \times 100 \times 10=1000 \sigma

\therefore       \sigma=1184 / 1000=1.184 N / mm ^{2}=1.184 MPa