Question

The drilling pipe on an oil rig is made from steel pipe which having thickness of 5-mm and an outside diameter of 90-mm. Calculate the maximum shear stress occur in the pipe if the pipe is turning at 650 rev/min while being powered using 12kW motor.

Answer 1

Answer:

$\tau_{max}= 3.28 \ MPa$

Explanation:

outside diameter = 90 mm

inside diameter = 90- 2× t=  90- 2× 5 = 80mm

where t is thickness of pipe.

power (P)  = 12 kW

Revolution (N)= 650 rev/min

we

Power = torque × angular velocity

P= T× ω

ω =  $\frac{2 \pi N}{60}$

$P=T \times \frac{2\pi N}{60}\\12 \times 10^3=T\times \frac{2\pi \times 650}{60}$

T=  176.3 Nm

for maximum shear stress

$\frac{\tau_{max}}{y_{max}}=\frac{T}{I_p}$

where ymax is maximum distance from neutral axis.

$y_{max}=\frac{d_0}{2}= \frac{90}{2}$= 45 mm

$I_p$= polar moment area

          = $\frac{\pi}{32} (d_o^4-d_i^4)=\frac{\pi}{32} (90^4-80^4)$

          = 2,420,008 mm⁴

$\dfrac{\tau_{max}}{45}=\dfrac{176.3 \times 10^3}{2,420,008}$

$\tau_{max}= 3.28 \ MPa$