SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Shaft size for a power and speed
Use this for “Size a solid shaft to carry 30 kW at 1200 rpm with 60 MPa allowed.”
Diagram
J
=
πd⁴
32
=
54,240
mm⁴;
τ
=
Tr
J
grows
from
0
at
the
centre
to
τₘₐₓ
=
16T
πd³
=
60
MPa.
T
=
P
ω
,
ω
=
2π
n
60
=
125.7
rad/s.
The shaft’s diameter and length are not to one scale.
Angular speed ω = 125.7 rad/s
This is an example. Type your own values to start a new problem.
ω (rad/s)
=
2π
×
n (rpm)
60
T (N·m)
=
P (kW)
×
1,000
ω (rad/s)
d (mm)
=
3
16
×
T (N·m)
×
1,000
π
×
τallow (MPa)
Solving τallow = 16T ÷ πd³ for d gives the cube root; P × 1,000 turns kW into W, so T is in N·m, and T × 1,000 makes it N·mm.
Formulas
ω
=
2π
×
n
60
125.6637
=
2π
×
1,200
60
T
=
P
×
1,000
ω
238.7324
=
30
×
1,000
125.6637
d
=
∛(16
×
T
×
1,000
π × τallow
)
27.2632
=
∛(16
×
238.7324
×
1,000
π × 60
)
Assumptions
•
A straight round shaft of one material, linear elastic: stresses stay under the yield strength.
•
Steady power at a steady speed; the shaft is sized on torsion shear alone, no bending.
•
Formula units are N, mm and MPa; the power is in kW and the torque in N·m.
Step-by-step
WE KNOW
P = 30 kW, n = 1,200 rpm, τallow = 60 MPa
FIND
angular speed (ω), torque (T), diameter needed (d)
Step 1 · Find angular speed (ω)
Use ω = 2π × n ÷ 60
Each turn is 2π radians, and a minute is 60 s.
ω
=
2π
×
1,200
60
ω
=
125.6637
rad/s
Step 2 · Find torque (T)
Use T = P × 1,000 ÷ ω
Power
is
torque
times
angular
speed;
×
1,000
turns
kW
into
W,
so
T
comes
out
in
N·m.
T
=
30
×
1,000
125.6637
T
=
30,000
125.6637
T
=
238.7324
N·m
Step 3 · Find diameter needed (d)
Use d = ∛(16 × T × 1,000 ÷ (π × τallow))
Set
τₘₐₓ
=
16
T
πd³
equal
to
the
allowed
stress
and
solve
for
d
;
×
1,000
turns
N·m
into
N·mm.
d
=
∛(16
×
238.7324
×
1,000
π × 60
)
d
=
∛(3,819.7184
×
1,000
60π
)
d
=
∛(
3,819,718.4
60π
)
d
=
∛20,260
d
=
27.2632
mm
Check
Put the numbers back in. Both sides should match.
125.6637
=
2π
×
1,200
60
✓
238.7324
=
30
×
1,000
125.6637
✓
27.2632
=
∛(16
×
238.7324
×
1,000
π × 60
)
✓
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