PLAIN AND CIVIL: EXAMPLE 3.5.2. TORQUE CAPACITY OF TWO-ROW BOLT COUPLING

Wednesday 20 March 2019

EXAMPLE 3.5.2. TORQUE CAPACITY OF TWO-ROW BOLT COUPLING

A flanged bolt coupling consists of eight 10mm-diameter steel bolts on a 400mm diameter, and another set of six 10mm-diameter steel bolts on a concentric 300mm diameter bolt circle. What torque can be applied without exceeding a shear stress of 60 MPa in the bolts?

The relationship between the outer circle and the inner circle of bolts is:

$\frac{P_{1}}{R_{1}}=\frac{P_{2}}{R_{2}}$

Both radii are given, but between $P_{1}$ and $P_{2}$ , identify which bolt circle should the maximum shear stress be correlated. Since it is the maximum, this would signify using the outer radius of bolts.

We can solve for $P_{1}$ (for one bolt in the outer circle) which is located at the outer bolt circle:

$P_{1}=A\tau =\frac{\pi }{4}\left ( 10mm \right )^{2}\left ( 60\frac{N}{mm^{2}} \right )=1500\pi N$

With the value of $P_{1}$ determined, $P_{2}$ is computed from the previous relationship equation:

$P_{2}=\frac{P_{1}}{R_{1}}R_{2}$

$P_{2}=\frac{1500\pi N}{200mm}150mm$

$P_{2}=1125\pi N$

After these two parameters are known, solve for the summation of T.

$T=P_{1}R_{1}n_{1}+P_{2}R_{2}n_{2}$

$T=1500\pi N\left ( 200mm \right )\left ( 8 \right )+1125\pi N\left ( 150mm \right )\left ( 6 \right )$

${\color{Red} \mathbf{T=3412500\pi N\cdot mm\simeq 10.72kN\cdot m}}$

RELATED TOPIC:

3.0. TORSION

PLAIN AND CIVIL

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Wednesday 20 March 2019

EXAMPLE 3.5.2. TORQUE CAPACITY OF TWO-ROW BOLT COUPLING

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