SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Hollow shaft
Use this for “A steel tube shaft, 60 mm outside and 40 mm inside, 1.5 m long, carries 2 kN·m. Find the largest shear stress and the twist.”
Diagram
J
=
π(
d
4
−
dᵢ
4
)
32
=
1.021 ×
10
6
mm⁴;
τ
=
Tr
J
grows
from
39.2
at
the
bore
to
τₘₐₓ
=
58.76
MPa.
φ
=
TL
GJ
=
0.0382
rad
=
2.19°:
the
end
view
shows
it
to
scale,
the
shaft’s
side
×10
so
it
shows.
The shaft’s diameter and length are not to one scale.
Polar moment of area J = 1,021,000 mm⁴
Shear modulus G = 77 GPa
Twist in degrees φ° = 2.186°
This is an example. Type your own values to start a new problem.
J (mm⁴)
=
π
×
(
(
dₒ (mm)
)
4
−
(
dᵢ (mm)
)
4
)
32
τₘₐₓ (MPa)
=
T (N·m)
×
1,000
×
(
dₒ (mm)
÷
2)
J (mm⁴)
φ (rad)
=
T (N·m)
×
L (mm)
G (GPa)
×
J (mm⁴)
T is typed in N·m: × 1,000 makes it N·mm, to go with mm and MPa.
φ°
Twist in degrees
calculated · °
Formulas
J
=
π
×
dₒ
4
−
dᵢ
4
32
1,021,017.6124
=
π
×
60
4
−
40
4
32
τₘₐₓ
=
T
×
1,000
×
dₒ ÷ 2
J
58.7649
=
2,000
×
1,000
×
60 ÷ 2
1,021,017.6124
φ
=
T
×
L
G
×
J
0.03816
=
2,000
×
1,500
77 × 1,021,017.6124
φ
°
=
φ
×
180
π
2.1864
=
0.03816
×
180
π
Assumptions
•
A straight round shaft of one material, linear elastic: stresses stay under the yield strength.
•
Plane sections stay plane, and the torque is the same all along the length.
•
Steel unless said: G = 77 GPa. Formula units are N, mm and MPa.
•
The bore is centered; shear runs from τₘₐₓ × dᵢ ÷ dₒ at the bore to τₘₐₓ outside.
Step-by-step
WE KNOW
T = 2,000 N·m, dₒ = 60 mm, dᵢ = 40 mm, L = 1,500 mm, G = 77 GPa
FIND
polar moment of area (J), largest shear stress (τₘₐₓ), angle of twist (φ), twist in degrees (φ°)
Step 1 · Find polar moment of area (J)
Use J = π × (dₒ⁴ − dᵢ⁴) ÷ 32
The polar moment of a ring: the whole disc’s less the bore’s.
J
=
π
×
60
4
−
40
4
32
J
=
π
×
12,960,000 − 2,560,000
32
J
=
π
×
10,400,000
32
J
=
10,400,000π
32
J
=
1,021,018
mm⁴
Step 2 · Find largest shear stress (τₘₐₓ)
Use τₘₐₓ = T × 1,000 × (dₒ ÷ 2) ÷ J
Shear
grows
with
the
radius,
so
it
is
largest
at
the
surface,
c
=
d
2
;
×
1,000
turns
N·m
into
N·mm,
giving
MPa.
τₘₐₓ
=
2,000
×
1,000
×
60 ÷ 2
1,021,018
τₘₐₓ
=
2,000
×
1,000
×
30
1,021,018
τₘₐₓ
=
2,000,000
×
30
1,021,018
τₘₐₓ
=
60,000,000
1,021,018
τₘₐₓ = 58.7649 MPa
Step 3 · Find angle of twist (φ)
Use φ = T × L ÷ (G × J)
The twist grows with torque and length and falls with the stiffness GJ; the × 1,000 for N·mm and the × 1,000 for GPa cancel.
φ
=
2,000
×
1,500
77 × 1,021,018
φ
=
2,000
×
1,500
78,618,386
φ
=
3,000,000
78,618,386
φ
=
0.03816
rad
Step 4 · Find twist in degrees (φ°)
Use φ° = φ × 180 ÷ π
A
radian
is
180
π
degrees.
φ
°
=
0.03816
×
180
π
φ
°
=
6.8688
π
φ
°
=
2.1864°
Check
Put the numbers back in. Both sides should match.
1,021,018
=
π
×
60
4
−
40
4
32
✓
58.7649
=
2,000
×
1,000
×
60 ÷ 2
1,021,018
✓
0.03816
=
2,000
×
1,500
77 × 1,021,018
✓
2.1864
=
0.03816
×
180
π
✓
Topic check
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