A motor turns a 250 mm pitch-diameter pulley at 1,450 rpm. The belt runs with 1,400 N on the tight side and 310 N on the slack side. The driven shaft is 45 mm in diameter and carries its hub on a parallel key 14 mm wide and 70 mm long. Walk the power path: the belt speed, the power the belt actually transmits, the torque that puts on the shaft, and the shear stress the key carries.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
d_p = 250 mm — Pulley pitch diameter
n = 1,450 rpm — Shaft speed
F₁ = 1,400 N — Tight-side tension
F₂ = 310 N — Slack-side tension
d = 45 mm — Shaft diameter
w = 14 mm — Key width
L = 70 mm — Key length
Determine
(a)the belt speed
(b)the power the belt transmits
(c)the shaft torque
(d)the shear stress in the key
Step 1 of 4(a) · solve for Belt speed
The rim is where rotation becomes translation: π times a quarter metre, 1,450 times a minute, is 19.0 m/s of belt. That figure is also the first design check in its own right — V-belts start complaining past about 30 m/s, and this one has margin.
Rearranged for v
v=πdn
Your values, in your units
v=π⋅(250mm)⋅(1,450rpm)
Answer
v=18.98m/s
Carried onward at full precision, not this rounded figure.
Only the DIFFERENCE in tension does work — the slack side's 310 N rides along to keep the belt gripping, contributing friction capacity and nothing else. The effective pull of 1,090 N at 19.0 m/s carries 20.7 kW.
Rearranged for P
P=(F1−F2)v
18.98 m/scarried from step 1
Your values, in your units
P=((1,400N)−(310N))⋅(18.9805m/s)
Answer
P=20.689kW
Carried onward at full precision, not this rounded figure.
P/ω gives 136.25 N·m — exactly, and not by luck. Chase the algebra and the speed cancels: T = (F₁−F₂)·d_p/2, the effective pull times the pulley radius. The reveal computes it the long way; the cancellation is the sanity check that the long way is right.
Rearranged for T
T=ωP
20.689 kWcarried from step 2
Your values, in your units
T=(1,450rpm)(20,688.7W)
Converted to base units
T=(1,450rpm)(20.6887kW)
Answer
T=136.25N⋅m
Carried onward at full precision, not this rounded figure.
Step 4 of 4(d) · solve for Shear stress in the key
The key sees the torque as a force at the shaft surface — 2T/d — spread over its shear plane. 136.25 N·m on a 45 mm shaft through a 14 × 70 key is 6.18 MPa: an order of magnitude under any steel key's allowable, which is what you want from the component whose whole job is to fail first, someday, cheaply.
Rearranged for τ
τ=dwL2T
136.25 N·mcarried from step 3
Your values, in your units
τ=(45mm)⋅(14mm)⋅(70mm)2⋅(136.25N⋅m)
Answer
τ=6.1791MPa
Carried onward at full precision, not this rounded figure.
Therefore the belt runs at 19.0 m/s carrying 20.7 kW, the shaft sees 136.25 N·m — the effective pull times the pulley radius, exactly — and the key works at 6.18 MPa, loafing by design.
Why this order
The chain is one quantity changing clothes: motion at the rim, force difference in the belt, power through the drive, torque in the shaft, stress in the key. The step worth slowing down on is (c), where the arithmetic comes out suspiciously round. It should: P = (F₁−F₂)·v and v = ω·r, so T = P/ω = (F₁−F₂)·r. The belt speed and shaft speed were only ever scaffolding — torque is effective pull times radius, and computing it through power and back is just the same fact wearing kilowatts in the middle. When a chain hands you an exact number, look for the cancellation; it is usually a law trying to be noticed.
The design lesson sits in step (d)'s margin. Keys are sized generously in shear not because torque estimates are poor but because the key is the drive's designated sacrifice: a five-dollar part in a milled slot, meant to shear before the shaft twists or the hub splits when something downstream jams. An overstrong key is not conservatism — it promotes the failure to whichever part is now weakest, and that part is never the cheap one.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
