SME
Machine Design
Shafts and bearings
Use this for “A 40 mm shaft carries a 400 N·m bending moment and a 600 N·m torque. With Sᵧ = 420 MPa, find σ, τ, σ′ and the factor of safety, or the diameter for n = 2.”
Diagram
J
=
πd⁴
32
=
251,300
mm⁴;
τ
=
Tr
J
grows
from
0
at
the
centre
to
τₘₐₓ
=
16T
πd³
=
47.75
MPa.
At
the
surface
σ
=
32M
πd³
=
63.66
MPa
and
τ
=
16T
πd³
;
σ′
=
√
σ
2
+ 3τ²
=
104.4
MPa.
The shaft’s diameter and length are not to one scale.
Von Mises stress σ′ = 104.4 MPa
Factor of safety n = 4.024
Yield strength Sᵧ = 420 MPa
This is an example. Type your own values to start a new problem.
σ (MPa)
=
32
×
M (N·m)
π
×
(
d (mm)
)
3
τ (MPa)
=
16
×
T (N·m)
π
×
(
d (mm)
)
3
n
=
Sᵧ (MPa)
σ′ (MPa)
σ′ = √(σ² + 3τ²) joins the first two lines to the third. M and T are typed in N·m: the formulas’ × 1,000 makes them N·mm, to go with mm and MPa.
Formulas
σ
=
32
×
M
×
1,000
π ×
d
3
63.662
=
32
×
400
×
1,000
π ×
40
3
τ
=
16
×
T
×
1,000
π ×
d
3
47.7465
=
16
×
600
×
1,000
π ×
40
3
σ
′
=
√
σ
2
+ 3 ×
τ
2
104.3649
=
√
63.662
2
+ 3 ×
47.7465
2
σ
′
=
16
×
√
4 ×
M
2
+ 3 ×
T
2
×
1,000
π ×
d
3
104.3649
=
16
×
√
4 ×
400
2
+ 3 ×
600
2
×
1,000
π ×
40
3
n
=
Sᵧ
σ
′
4.0243
=
420
104.3649
Assumptions
•
A solid round shaft under steady loads (no fatigue); the stresses stay under the yield strength.
•
M and T act at the same section; the worst point is on the surface, where σ is largest.
•
Shear from the transverse force is left out (it is zero at the surface).
•
Formula units are N, mm and MPa; M and T are typed in N·m.
Step-by-step
WE KNOW
d = 40 mm, M = 400 N·m, T = 600 N·m, Sᵧ = 420 MPa
FIND
bending stress (σ), torsion shear stress (τ), von Mises stress (σ′), factor of safety (n)
Step 1 · Find bending stress (σ)
Use σ = 32 × M × 1,000 ÷ (π × d³)
Bending
stress
at
the
surface
is
Mc
I
,
and
for
a
solid
round
section
that
is
32
M
πd³
;
×
1,000
turns
N·m
into
N·mm.
σ
=
32
×
400
×
1,000
π ×
40
3
σ
=
32
×
400
×
1,000
64,000 × π
σ
=
32
×
400
×
1,000
201,100
σ
=
12,800
×
1,000
201,100
σ
=
12,800,000
201,100
σ
=
63.662
MPa
Step 2 · Find torsion shear stress (τ)
Use τ = 16 × T × 1,000 ÷ (π × d³)
Torsion
shear
at
the
surface
is
Tc
J
,
and
for
a
solid
round
section
that
is
16
T
πd³
;
×
1,000
turns
N·m
into
N·mm.
τ
=
16
×
600
×
1,000
π ×
40
3
τ
=
16
×
600
×
1,000
64,000 × π
τ
=
16
×
600
×
1,000
201,100
τ
=
9,600
×
1,000
201,100
τ
=
9,600,000
201,100
τ
=
47.7465
MPa
Step 3 · Find von Mises stress (σ′)
Use σ′ = √(σ² + 3 × τ²)
Von Mises (distortion energy) joins the normal and shear stress into one stress to compare with Sᵧ.
σ
′
=
√
63.662
2
+ 3 ×
47.7465
2
σ
′
=
√
4,052.9 + 3 × 2,279.7
σ
′
=
√
4,052.9 + 6,839.1
σ
′
=
√
10
,892
σ
′
=
104.3649
MPa
Step 4 · Find factor of safety (n)
Use n = Sᵧ ÷ σ′
How many times the working stress the material takes before it yields.
n
=
420
104.3649
n
=
4.0243
Check
Put the numbers back in. Both sides should match.
63.662
=
32
×
400
×
1,000
π ×
40
3
✓
47.7465
=
16
×
600
×
1,000
π ×
40
3
✓
104.3649
=
√
63.662
2
+ 3 ×
47.7465
2
✓
104.3649
=
16
×
√
4 ×
400
2
+ 3 ×
600
2
×
1,000
π ×
40
3
✓
4.0243
=
420
104.3649
✓
Topic check
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