Autorickshaw differential & foot brake

I calculated the design parameters for the differential and foot brake, then modeled the differential parts and assembly in SOLIDWORKS.

January 2025

Calculations

Design requirements

Parameter	Value
Engine power	6.5 kW at 4000 rpm
Differential ratio	4.2:1
Maximum speed	50 km/h
Vehicle mass / payload	400 kg / 500 kg
Rear wheel radius / track	0.25 m / 1.2 m
Stopping time	5 s
Incline	10°

Bevel gear geometry

I selected a 20-tooth pinion and an 84-tooth gear, with a 2.5 mm module and 25 mm face width.

\[\begin{aligned}i&=\frac{z_G}{z_P}=\frac{84}{20}=4.2\\d_P&=mz_P=50\ \mathrm{mm}\\d_G&=mz_G=210\ \mathrm{mm}\end{aligned}\]

\[\gamma=\tan^{-1}\left(\frac{20}{84}\right)=13.39^\circ,\qquad\Gamma=76.61^\circ\]

\[\begin{aligned}A_0&=\frac{d_P}{2\sin\gamma}=107.94\ \mathrm{mm}\\b&=\min(0.3A_0,10m)=25\ \mathrm{mm}\end{aligned}\]

Using 4000 rpm at the pinion:

\[\begin{aligned}v&=\frac{\pi d_Pn_P}{60\,000}=10.472\ \mathrm{m\,s^{-1}}\\W_t&=\frac{P}{v}=\frac{6500}{10.472}=620.70\ \mathrm{N}\end{aligned}\]

Gear design factors

I used carburized and case-hardened steel, with a bending design factor of 3.

\[\begin{aligned}Z_{NT}&=3.48N_L^{-0.0602}\\Y_{NT}&=1.683N_L^{-0.0323}\end{aligned}\]

\[\begin{aligned}Y_Z&=0.5-0.25\log_{10}(1-R)=1.075\\Z_Z&=\sqrt{Y_Z}=1.037\end{aligned}\]

Factor	Pinion	Gear
Bending life Y_NT	0.862	0.903
Pitting life Z_NT	1.00	1.09
Bending geometry Y_J	0.270	0.223
Contact geometry Z_I	0.094	0.094
Temperature K_θ	1	1
Lengthwise curvature Y_β	1	1

\[\begin{aligned}Y_x&=0.4867+0.008339m=0.50755\\Z_x&=0.00492b+0.4375=0.5605\end{aligned}\]

\[\begin{aligned}K_{H\beta}&=1+5.6\times10^{-6}b^2=1.0035\\Z_{xc}&=2\end{aligned}\]

For transmission accuracy number \(Q_v=6\):

\[\begin{aligned}B&=0.25(12-Q_v)^{2/3}=0.82548\\A&=50+56(1-B)=59.773\end{aligned}\]

\[K_v=\left(\frac{A+\sqrt{200v}}{A}\right)^B=1.59887\]

Gear bending and contact stress

Bending

I used \(K_A=1\) and an allowable bending stress number of 205 MPa.

\[\begin{aligned}\sigma_F&=\frac{W_tK_AK_vY_xK_{H\beta}}{bmY_\beta Y_J}\\\sigma_{FP}&=\frac{\sigma_{F\lim}Y_{NT}}{S_FK_\theta Y_Z}\end{aligned}\]

Result	Gear	Pinion
Bending stress	36.27 MPa	29.95 MPa
Permissible stress at S_F = 3	57.40 MPa	54.79 MPa
Bending factor	4.75	5.49

\(W_t\) is in N; \(b\) and \(m\) are in mm.

Contact stress model

\[\begin{aligned}\sigma_H&=Z_E\sqrt{\frac{W_tK_AK_vZ_xZ_{xc}K_{H\beta}}{bd_PZ_I}}\\\sigma_{HP}&=\frac{\sigma_{H\lim}Z_{NT}Z_W}{S_HK_\theta Z_Z}\end{aligned}\]

I used a contact stress number of 1380 MPa and \(S_H=\sqrt3\). Here, \(Z_E\) is the elastic coefficient and \(Z_W\) is the hardness-ratio factor.

Drum brake geometry

I used a 125 mm drum radius, 30 mm shoe width and friction coefficient of 0.32.

\[\begin{aligned}r&=0.125\ \mathrm{m},\quad b_s=0.030\ \mathrm{m}\\\theta_1&=0^\circ,\quad\theta_2=126^\circ,\quad\theta_a=90^\circ\end{aligned}\]

\[\begin{aligned}s&=r\theta_2=0.274889\ \mathrm{m}\\A_s&=b_ss=8.24668\times10^{-3}\ \mathrm{m^2}\end{aligned}\]

\[a=\sqrt{112^2+50^2}=122.654\ \mathrm{mm},\qquad c=212\ \mathrm{mm}\]

\(a\): distance from drum centre to shoe pivot; \(c\): actuation moment arm.

Brake force and torque

Vehicle braking requirement

For a 900 kg loaded vehicle stopping from 50 km/h in five seconds:

\[\begin{aligned}u&=13.8889\ \mathrm{m\,s^{-1}}\\|a_v|&=\frac{u}{5}=2.7778\ \mathrm{m\,s^{-2}}\\F_v&=m|a_v|=2500\ \mathrm{N}\end{aligned}\]

\[F_{v,\,10^\circ}=m(|a_v|+g\sin10^\circ)=4033.14\ \mathrm{N}\]

Assuming equal braking at the two rear wheels and neglecting drag and rolling resistance:

\[\begin{aligned}T_w&=\frac{F_vR_w}{2}=312.50\ \mathrm{N\,m}\\T_{w,\,10^\circ}&=504.14\ \mathrm{N\,m}\end{aligned}\]

Shoe forces

I used a sinusoidal pressure distribution, \(p(\theta)=P_a\sin\theta/\sin\theta_a\).

\[M_N=\frac{P_ab_sra}{\sin\theta_a}\left[\frac{\theta_2-\theta_1}{2}-\frac{\sin2\theta_2-\sin2\theta_1}{4}\right]\]

\[M_f=\frac{\mu P_ab_sr}{\sin\theta_a}\left[r(\cos\theta_1-\cos\theta_2)-\frac{a}{2}(\sin^2\theta_2-\sin^2\theta_1)\right]\]

\[\begin{aligned}F_{\mathrm{act}}&=\frac{M_N\mp M_f}{c}\\T_s&=\frac{\mu P_ab_sr^2}{\sin\theta_a}(\cos\theta_1-\cos\theta_2)\end{aligned}\]

The minus sign applies to the leading shoe and the plus sign to the trailing shoe. \(P_a\) is peak lining pressure; angular integration uses radians.

Shoe pressure example

For an assumed \(P_a=134.60\) kPa:

Quantity	Value
Normal moment M_N	82.79 N m
Friction moment M_f	25.57 N m
Leading-shoe actuation force	269.90 N
Shoe torque T_s	32.06 N m

Differential models and drawing