Question

4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of 2000 rev/min and transmits 30 hp to a 50-tooth gear. If the face width is 1 in, estimate the bending stress.

Answer 1

Answer:

Answer is 15.07.

Refer below for the explanation.

Explanation:

As per the values mentioned in the question,

Pressure = 20°

Pitch = 5 teeth/in

Pinion = 2000

Power = 30 hp

Teeth gear= 50

Teeth pinion= 20

Face width = 1 in

Diameter,

Diameter=N/p

=20/5=4

Y = 0.321

Formula to find velocity,

V=^dnp

=^×4×2000/12

=2094.40

For cut:

Kv=2000×V/2000

Kv=2000×2094.40/2000

Kv=63025.

The bending stress formula,

W^t=t/d/2

=472.69

=Kvwtp/Fy

=2.0472×472.69×5

=15.07

Answer 2

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle ($\Phi$) = 20°

Pinion speed ($n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ($N_G$) = 50

Teeth on pinion ($N_p$) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

=$\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ($n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load$(W^t)$, in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi