SME
Mechanics of Materials (Solid Mechanics I)
Torsion
Use this for “A solid steel shaft 50 mm across and 1.5 m long carries 2 kN·m. Find the largest shear stress and the twist.”
Diagram
J
=
πd⁴
32
=
613,600
mm⁴;
τ
=
Tr
J
grows
from
0
at
the
centre
to
τₘₐₓ
=
16T
πd³
=
81.49
MPa.
φ
=
TL
GJ
=
0.0635
rad
=
3.64°:
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 = 613,600 mm⁴
Shear modulus G = 77 GPa
Twist in degrees φ° = 3.638°
This is an example. Type your own values to start a new problem.
J (mm⁴)
=
π
×
(
d (mm)
)
4
32
τₘₐₓ (MPa)
=
T (N·m)
×
1,000
×
(
d (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
×
d
4
32
613,592.3152
=
π
×
50
4
32
τₘₐₓ
=
T
×
1,000
×
d
÷ 2
J
81.4873
=
2,000
×
1,000
×
50 ÷ 2
613,592.3152
φ
=
T
×
L
G
×
J
0.0635
=
2,000
×
1,500
77 × 613,592.3152
φ
°
=
φ
×
180
π
3.6381
=
0.0635
×
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.
Step-by-step
WE KNOW
T = 2,000 N·m, d = 50 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 × d⁴ ÷ 32
The polar moment of area of a solid round section.
J
=
π
×
50
4
32
J
=
π
×
6,250,000
32
J
=
6,250,000π
32
J
=
613,592.3
mm⁴
Step 2 · Find largest shear stress (τₘₐₓ)
Use τₘₐₓ = T × 1,000 × (d ÷ 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
×
50 ÷ 2
613,592.3
τₘₐₓ
=
2,000
×
1,000
×
25
613,592.3
τₘₐₓ
=
2,000,000
×
25
613,592.3
τₘₐₓ
=
50,000,000
613,592.3
τₘₐₓ = 81.4873 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 × 613,592.3
φ
=
2,000
×
1,500
47,250,000
φ
=
3,000,000
47,250,000
φ
=
0.063497
rad
Step 4 · Find twist in degrees (φ°)
Use φ° = φ × 180 ÷ π
A
radian
is
180
π
degrees.
φ
°
=
0.063497
×
180
π
φ
°
=
11.43
π
φ
°
=
3.6381°
Check
Put the numbers back in. Both sides should match.
613,592.3
=
π
×
50
4
32
✓
81.4873
=
2,000
×
1,000
×
50 ÷ 2
613,592.3
✓
0.063497
=
2,000
×
1,500
77 × 613,592.3
✓
3.6381
=
0.063497
×
180
π
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.