---
title: "Open flat belt power and cross belt length"
subject: "Machine Design I"
chapter: 8
key: "semester_6/machine_design_i:open-flat-belt-power-and-length"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/open-flat-belt-power-and-length
site: BCT Engineering
---

# Open flat belt power and cross belt length

Machine Design I · Chapter 8: Design of belts

- Asked in: year not given, 8 marks
- Video: [Length of open belt drive](https://www.youtube.com/watch?v=lMJjGYECc3U) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/open-flat-belt-power-and-length
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-8

## Question

An open flat belt drive connects two pulleys of diameters 400 mm (driver, 600 rpm) and 800 mm with centre distance 2.5 m. The belt is 150 mm wide and 8 mm thick, with density 1000 kg/m^3, permissible stress 2 MPa and coefficient of friction 0.3. (a) Find the length of the belt, the angle of contact on the smaller pulley, the centrifugal tension, the tensions in the tight and slack sides, and the power transmitted. (b) Find the length of belt and the angle of contact if the same pulleys are connected by a cross belt.

## Short answer

Belt length L=2C+π(D+d)/2+(D−d)²/(4C)=6.90 m, contact angle θ=180°−2sin⁻¹[(D−d)/(2C)]=170.8°. Centrifugal tension Tc=mv²=189.5 N. With T₁=σbt=2400 N and e^(μθ)=2.446, T₂=1093 N. Power=(T₁−T₂)v=16.4 kW. Cross belt: L=7.03 m, θ=207.8°.

## Answer

Data: $d = 0.4$ m, $D = 0.8$ m, $C = 2.5$ m, $n = 600$ rpm, $b = 0.15$ m, $t = 0.008$ m, $\rho = 1000$ kg/m$^3$, $\sigma = 2$ MPa, $\mu = 0.3$.

#### (a) Open belt

**Length**

$$
L = 2C + \frac{\pi}{2}(D + d) + \frac{(D - d)^2}{4C} = 5 + 1.885 + 0.016 = 6.901\ \text{m}
$$

**Angle of contact on the small pulley**

$$
\sin\alpha = \frac{D - d}{2C} = \frac{0.4}{5} = 0.08, \qquad \alpha = 4.589^\circ
$$

$$
\theta = 180^\circ - 2\alpha = 170.82^\circ = 2.981\ \text{rad}
$$

**Speed and masses**

$$
v = \frac{\pi d n}{60} = \frac{\pi \times 0.4 \times 600}{60} = 12.57\ \text{m/s}
$$

$$
m = \rho b t = 1000 \times 0.15 \times 0.008 = 1.2\ \text{kg/m}
$$

**Tensions**

$$
\begin{aligned} T_{max} &= \sigma b t = 2 \times 150 \times 8 = 2400\ \text{N} \\ T_c &= m v^2 = 1.2 \times 12.57^2 = 189.5\ \text{N} \\ T_1 - T_c &= 2400 - 189.5 = 2210.5\ \text{N} \end{aligned}
$$

The tight side tension is the maximum permissible value, $T_1 = T_{max} = 2400$ N. Using $\dfrac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta}$:

$$
e^{\mu\theta} = e^{0.3 \times 2.981} = e^{0.894} = 2.446
$$

$$
T_2 - T_c = \frac{2210.5}{2.446} = 903.7\ \text{N} \quad\Rightarrow\quad T_2 = 903.7 + 189.5 = 1093.2\ \text{N}
$$

**Power**

$$
P = (T_1 - T_2)\,v = (2400 - 1093.2) \times 12.57 = 16\,420\ \text{W} = 16.4\ \text{kW}
$$

#### (b) Cross belt

$$
\sin\alpha = \frac{D + d}{2C} = \frac{1.2}{5} = 0.24, \qquad \alpha = 13.89^\circ
$$

$$
\theta = 180^\circ + 2\alpha = 207.8^\circ
$$

$$
L = 2C + \frac{\pi}{2}(D + d) + \frac{(D + d)^2}{4C} = 5 + 1.885 + 0.144 = 7.029\ \text{m}
$$

The angle of contact is the same on both pulleys, and larger than for the open belt, so a cross belt can transmit more power, but the belt rubs where it crosses and wears faster.

**Answer: (a) L = 6.90 m; theta = 170.8 degrees; Tc = 189.5 N; T1 = 2400 N, T2 = 1093 N; P = 16.4 kW. (b) L = 7.03 m; theta = 207.8 degrees.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives the length of an open belt?
   A. L = 2C + (pi/2)(D + d) + (D - d)^2/(4C)
   B. L = 2C + (pi/2)(D + d) + (D + d)^2/(4C)
   C. L = C + pi (D + d)
   D. L = 2C + pi (D - d)
   Answer: A. For an open drive the correction term uses (D - d)^2; the cross belt uses (D + d)^2. Here L = 6.90 m. (Key point: Open belt length)

2. How is the angle of contact on the small pulley found for the open belt?
   A. theta = 180 + 2 alpha with sin(alpha) = (D + d)/(2C)
   B. theta = 180 - 2 alpha with sin(alpha) = (D - d)/(2C)
   C. theta = 180 degrees exactly
   D. theta = 180 - alpha with sin(alpha) = (D - d)/C
   Answer: B. sin(alpha) = 0.4/5 = 0.08 so alpha = 4.59 degrees and theta = 170.8 degrees (2.981 rad). (Key point: Contact angle)

3. How are T2 and power found once Tc = 189.5 N is known and T1 = Tmax = 2400 N?
   A. T2 = T1 / e^(mu theta); P = T1 v
   B. (T1 - Tc)/(T2 - Tc) = e^(mu theta) gives T2 = 1093 N; P = (T1 - T2) v = 16.4 kW
   C. T2 = T1 - Tc; P = T2 v
   D. T2 = Tc e^(mu theta); P = (T1 + T2) v
   Answer: B. The tight side is the maximum allowable tension; the ratio with centrifugal tension gives T2, and power uses the tension difference times speed. (Key point: Tensions and power)

4. How does a cross belt compare with an open belt on the same pulleys?
   A. Smaller contact angle (207.8 vs 170.8 reversed), same wear
   B. Larger contact angle on both pulleys (207.8 degrees), so more power, but the belt wears where it crosses
   C. Same contact angle but shorter belt
   D. Larger contact angle but needs a smaller centre distance
   Answer: B. For the cross belt theta = 180 + 2 alpha with sin(alpha) = (D + d)/(2C), giving 207.8 degrees on both pulleys; rubbing at the crossing increases wear. (Key point: Cross belt)

## Related in Machine Design I

- [Tension ratio of a flat belt drive](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Abelt-tension-ratio-derivation) — Machine Design I, chapter 8
- [Roller chain drive: length, speed and factor of safety](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aroller-chain-drive-selection) — Machine Design I, chapter 8
- [V-belt compared with flat belt](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Avbelt-versus-flat-belt) — Machine Design I, chapter 8
- [Number of V-belts for a motor drive](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Av-belt-number-of-belts) — Machine Design I, chapter 8
- [Weighted decision matrix for drive selection](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aweighted-decision-matrix-drive) — Machine Design I, chapter 1

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