How is the tangential load found?

F_t = P / v = 10000 / 5.655 = 1768 N

What does the deformation factor C = 0.111 e / (1/Ep + 1/Eg) depend on, and what is its value?

Tooth error and moduli of elasticity; 166.5 N/mm

Tooth width and speed; 11.1 N/mm

What is the limiting wear load?

F_w = 120 x 60 x 1.5 x 1.1 = 11880 N, with Q = 2 z2 / (z1 + z2)

F_w = 400 x 60 x pi x 6 x 0.1084

Which conclusion is correct?

Wear governs because F_w = 11.9 kN is only 1.5 times F_d = 7.9 kN; both exceed F_d so design is safe

Bending governs because F_s is largest

The design fails since F_d is greater than F_t

The design is unsafe because F_w < F_s

Practice · 8 marks

Dynamic and wear load check, spur pair

Short answer

Dynamic load Fd = 7.92 kN, beam strength Fs = 49.0 kN, wear load Fw = 11.9 kN. Both Fs and Fw exceed Fd, so the design is safe; wear governs.

A steel spur pinion of 20 teeth and module 6 mm, face width 60 mm, meshes with a steel gear of 60 teeth (20 degree full depth) and transmits 10 kW at 900 rpm of the pinion. The tooth error is 0.015 mm and E = 200 GPa for both gears. Take endurance strength of the teeth as 400 MPa, y = 0.154 - 0.912/z, the deformation factor C = 0.111 e/(1/Ep + 1/Eg) in N/mm (e in mm, E in N/mm^2), and the load-stress factor K = 1.1 N/mm^2. Find the dynamic load by Buckingham's equation, the beam strength, the limiting wear load, and state whether the design is safe.

Data: $z_1 = 20$ , $z_2 = 60$ , $m = 6$ mm, $b = 60$ mm, $P = 10$ kW, $n_1 = 900$ rpm, $e = 0.015$ mm.

Velocity and transmitted load

d_1 = mz_1 = 120\ \text{mm}, \qquad v = \frac{\pi d_1 n_1}{60\,000} = \frac{\pi \times 120 \times 900}{60\,000} = 5.655\ \text{m/s}

F_t = \frac{P}{v} = \frac{10\,000}{5.655} = 1768\ \text{N}

Dynamic load (Buckingham)

Deformation factor:

C = \frac{0.111\, e}{\frac{1}{E_p} + \frac{1}{E_g}} = \frac{0.111 \times 0.015}{2/200\,000} = 166.5\ \text{N/mm}

F_d = F_t + \frac{21 v\,(bC + F_t)}{21 v + \sqrt{bC + F_t}}

\begin{aligned} bC + F_t &= 60 \times 166.5 + 1768 = 11\,758\ \text{N} \\ \sqrt{11\,758} &= 108.4, \qquad 21v = 118.8 \\ F_d &= 1768 + \frac{118.8 \times 11\,758}{118.8 + 108.4} = 1768 + 6147 = 7915\ \text{N} \end{aligned}

Beam strength

y = 0.154 - \frac{0.912}{20} = 0.1084

F_s = \sigma_e\, b\, \pi m\, y = 400 \times 60 \times \pi \times 6 \times 0.1084 = 49\,040\ \text{N}

(The weaker is the pinion, as both are steel and the pinion has smaller $y$ .)

Wear load

Q = \frac{2 z_2}{z_1 + z_2} = \frac{2 \times 60}{80} = 1.5

F_w = d_1\, b\, Q\, K = 120 \times 60 \times 1.5 \times 1.1 = 11\,880\ \text{N}

Check

Quantity	Value (N)	Ratio to $F_d$
Dynamic load $F_d$	7915	1.0
Beam strength $F_s$	49 040	6.2
Wear load $F_w$	11 880	1.5

Both $F_s > F_d$ and $F_w > F_d$ , so the design is safe against bending and wear. Wear is the controlling criterion (factor 1.5 only).

Answer: Fd = 7.92 kN; Fs = 49.0 kN; Fw = 11.9 kN; the design is safe (wear governs).

✓ Answer checked by Claude Opus 5.5 (high effort)

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