SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Bending, torsion and axial load on a shaft
Use this for “A 50 mm solid shaft carries 20 kN of tension, a 600 N·m bending moment and an 800 N·m torque. Find the principal stresses and τₘₐₓ at the top.”
Diagram
σavg
=
σₓ + σᵧ
2
=
59.1 + 0
2
σavg = 29.5 MPa
R
=
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
=
44
MPa.
σ₁
=
σavg
+
R
=
73.5
MPa;
σ₂
=
σavg
−
R
=
−14.4
MPa;
τₘₐₓ
=
R.
tan
2θₚ
=
2τₓᵧ
÷
(σₓ
−
σᵧ),
so
θₚ
=
23.9°:
the
turned
element
carries
σ₁
on
the
faces
turned
θₚ
and
no
shear.
τ
is
plotted
positive
down,
so
X
turns
to
σ₁
by
2θₚ
the
way
the
element
turns.
This is an example. Type your own values to start a new problem.
σ (MPa)
=
4
×
P (kN)
π
×
(
d (mm)
)
2
+
32
×
M (N·m)
π
×
(
d (mm)
)
3
τ (MPa)
=
16
×
T (N·m)
π
×
(
d (mm)
)
3
σ₁
Larger principal stress
calculated
σ₂
Smaller principal stress
calculated
τₘₐₓ
Largest in-plane shear
calculated
Formulas
σ
=
4
×
P
π ×
d
2
+
32
×
M
π ×
d
3
+59.0783
=
4
×
20,000
π ×
50
2
+
32
×
600,000
π ×
50
3
τ
=
16
×
T
π ×
d
3
32.5949
=
16
×
800,000
π ×
50
3
τₘₐₓ
=
√(
(
σ
2
)
2
+
τ
2
)
43.9885
=
√(
(
59.0783
2
)
2
+
32.5949
2
)
σ₁
=
σ
2
+
τₘₐₓ
+73.5277
=
59.0783
2
+
43.9885
σ₂
=
σ
2
−
τₘₐₓ
−14.4494
=
59.0783
2
−
43.9885
Formulas are worked in N, N·mm. The step-by-step shows the conversions.
Assumptions
•
The point is on the top surface, where bending tension and torsion shear both peak.
•
A solid round shaft with no stress raisers, elastic, so the axial and bending stresses add.
•
Formula units are N, mm and MPa: N·mm ÷ mm³ is MPa.
Step-by-step
WE KNOW
d = 50 mm, P = +20 kN, M = 600 N·m, T = 800 N·m
FIND
normal stress (σ), shear stress (τ), largest in-plane shear (τₘₐₓ), larger principal stress (σ₁), smaller principal stress (σ₂)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N, N·mm).
P = +20 kN = 20,000 N (1 kN = 1000 N)
M = 600 N·m = 600,000 N·mm (1 N·m = 1000 N·mm)
T = 800 N·m = 800,000 N·mm (1 N·m = 1000 N·mm)
Step 2 · Find normal stress (σ)
Use σ = 4 × P ÷ (π × d²) + 32 × M ÷ (π × d³)
The
axial
stress
P
A
and
the
bending
stress
Mc
I
at
the
top
add:
both
act
along
the
shaft.
σ
=
4
×
20,000
π ×
50
2
+
32
×
600,000
π ×
50
3
σ
=
4
×
20,000
2,500 × π
+
32
×
600,000
125,000 × π
σ
=
4
×
20,000
7,854
+
32
×
600,000
392,700
σ
=
80,000
7,854
+
19,200,000
392,700
σ
=
10.19
+
48.89
σ
=
+59.0783
MPa
Step 3 · Find shear stress (τ)
Use τ = 16 × T ÷ (π × d³)
Torsion
shear
at
the
surface:
Tc
J
with
c
=
d
2
and
J
=
πd⁴
32
.
τ
=
16
×
800,000
π ×
50
3
τ
=
16
×
800,000
125,000 × π
τ
=
16
×
800,000
392,700
τ
=
12,800,000
392,700
τ
=
32.5949
MPa
Step 4 · Find largest in-plane shear (τₘₐₓ)
Use τₘₐₓ = √((σ ÷ 2)² + τ²)
Mohr’s
circle
with
σᵧ
=
0:
its
radius
from
the
center
σ
2
out
to
the
point
(
σ
,
τ
).
τₘₐₓ
=
√(
(
59.0783
2
)
2
+
32.5949
2
)
τₘₐₓ
=
√
29.54
2
+
32.5949
2
τₘₐₓ
=
√
872.6116 + 1,062
τₘₐₓ
=
√
1
,934.6116
τₘₐₓ = 43.9885 MPa
Step 5 · Find larger principal stress (σ₁)
Use σ₁ = σ ÷ 2 + τₘₐₓ
The circle’s right end: its center plus the radius.
σ₁
=
59.0783
2
+
43.9885
σ₁
=
29.54
+
43.9885
σ₁
=
+73.5277
MPa
Step 6 · Find smaller principal stress (σ₂)
Use σ₂ = σ ÷ 2 − τₘₐₓ
The circle’s left end: its center less the radius.
σ₂
=
59.0783
2
−
43.9885
σ₂
=
29.54
−
43.9885
σ₂
=
−14.4494
MPa
Check
Put the numbers back in. Both sides should match.
+59.0783
=
4
×
20,000
π ×
50
2
+
32
×
600,000
π ×
50
3
✓
32.5949
=
16
×
800,000
π ×
50
3
✓
43.9885
=
√(
(
59.0783
2
)
2
+
32.5949
2
)
✓
+73.5277
=
59.0783
2
+
43.9885
✓
−14.4494
=
59.0783
2
−
43.9885
✓
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