How is W_t found for the 25-tooth, m = 4 pinion at 1500 rpm?

W_t = P / V with V = pi d n / 60 = 7.854 m/s, so 1910 N

Why is Y_N = 0.939 below 1?

Because N = 9 x 10^8 cycles exceeds 10^7, where Y_N falls with the number of cycles

What is the factor of safety and conclusion?

n = 221.6 x 0.939 / 86.8 = 2.4, safe in bending

n = 86.8 / 221.6 = 0.39, unsafe

Practice · 8 marks

Bending stress and safety factor of a spur pinion

Short answer

Calculate tangential load Wt = P/V = 15 kW / 7.854 m/s = 1910 N, apply AGMA bending equation with given factors to get sigma = 86.8 MPa. Allowable stress St = 221.6 MPa, life factor YN = 0.939, resulting factor of safety n = 2.40.

A spur pinion of module 4 mm with 25 teeth and face width 40 mm transmits 15 kW at 1500 rpm. Use the AGMA bending stress equation sigma = Wt Ko Kv Ks Km Kb/(b m J) with Ko = 1.25, Ks = 1, Km = 1.3, Kb = 1 and J = 0.34. The gears are of quality number Qv = 6, so Kv = ((A + sqrt(200 V))/A)^B with B = 0.25 (12 - Qv)^(2/3) and A = 50 + 56 (1 - B), V in m/s. The pinion is through-hardened Grade 1 steel of 250 HB with allowable bending stress St = 0.533 HB + 88.3 MPa. The life required is 10000 hours, and the life factor is YN = 1.3558 N^(-0.0178). Take KT = KR = 1. Find the bending stress and the factor of safety.

Pitch line velocity and tangential load

d = mz = 4 \times 25 = 100\ \text{mm}, \qquad V = \frac{\pi d n}{60} = \frac{\pi \times 0.1 \times 1500}{60} = 7.854\ \text{m/s}

W_t = \frac{P}{V} = \frac{15\,000}{7.854} = 1910\ \text{N}

Dynamic factor ( $Q_v = 6$ )

\begin{aligned} B &= 0.25\,(12 - 6)^{2/3} = 0.25 \times 3.302 = 0.8255 \\ A &= 50 + 56\,(1 - 0.8255) = 59.77 \\ K_v &= \left(\frac{A + \sqrt{200 V}}{A}\right)^B = \left(\frac{59.77 + \sqrt{1570.8}}{59.77}\right)^{0.8255} = (1.663)^{0.8255} = 1.522 \end{aligned}

Bending stress

\sigma = \frac{1910 \times 1.25 \times 1.522 \times 1 \times 1.3 \times 1}{40 \times 4 \times 0.34} = \frac{4724}{54.4} = 86.8\ \text{MPa}

Allowable stress and life factor

S_t = 0.533 \times 250 + 88.3 = 221.6\ \text{MPa}

Number of cycles:

N = 1500 \times 60 \times 10\,000 = 9\times10^{8}

Y_N = 1.3558\,(9\times10^8)^{-0.0178} = 0.939

Factor of safety

n = \frac{S_t\, Y_N}{K_T\, K_R\, \sigma} = \frac{221.6 \times 0.939}{1 \times 1 \times 86.8} = 2.40

The pinion is safe in bending (n > 1) with a reasonable margin. Surface (contact) strength should be checked separately, because pitting often governs the life of hardened gears.

Answer: sigma = 86.8 MPa; n = 2.4.

✓ Answer checked by Claude Opus 5.5 (high effort)

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