SME
Aerospace Structures
Thin-walled structures
Use this for “A 400 mm by 200 mm box with 2 mm walls (G = 27 GPa) carries 10 kN·m of torque. Find the shear flow, the shear stress and the twist per meter.”
Diagram
Wall
drawn
thicker
than
to
scale
(
b
t
=
100).
q
=
T
2Aₘ
is
the
same
all
round
one
closed
cell.
Shear stress τ = 31.25 MPa
This is an example. Type your own values to start a new problem.
q (N/mm)
=
T (kN·m)
2
×
Aₘ (mm²)
τ (MPa)
=
q (N/mm)
t (mm)
b
Width
example
h
Height
example
S
Mid-line perimeter
worked out (not typed)
G
Shear modulus
example
θ′
Twist rate
calculated
Formulas
Aₘ
=
b
×
h
80,000 = 400 × 200
S
=
2
×
(
b
+
h
)
1,200 = 2 × (400 + 200)
q
=
T
2 × Aₘ
62.5
=
10,000,000
2 × 80,000
τ
=
q
t
31.25
=
62.5
2
θ′
=
(
180
π
)
×
1,000
×
T
×
S
4 ×
Aₘ
2
×
t
×
G
0.4974
=
(
180
π
)
×
1,000
×
10,000,000
×
1,200
4 × 80,
000
2
× 2 × 27,000
Formulas are worked in N·mm, MPa. The step-by-step shows the conversions.
Assumptions
•
Bredt–Batho theory: one closed cell, its wall the same thickness t all round.
•
Width, height and perimeter run along the wall’s mid-line; Aₘ is the area inside it.
•
Worked in N–mm–MPa; θ′ is the twist over 1 m (1000 mm) of span, in degrees.
Step-by-step
WE KNOW
T = 10 kN·m, b = 400 mm, h = 200 mm, t = 2 mm, G = 27 GPa
FIND
enclosed area (Aₘ), mid-line perimeter (S), shear flow (q), shear stress (τ), twist rate (θ′)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N·mm, MPa).
T = 10 kN·m = 10,000,000 N·mm (1 kN·m = 10⁶ N·mm)
G = 27 GPa = 27,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find enclosed area (Aₘ)
Use Aₘ = b × h
The
mid-line
box
is
b
wide
and
h
high;
multiply
them.
Aₘ = 400 × 200
Aₘ = 80,000 mm²
Step 3 · Find mid-line perimeter (S)
Use S = 2 × (b + h)
Go once round the mid-line: two widths and two heights.
S
=
2
×
(400
+
200)
S
=
2
×
600
S
=
1,200
mm
Step 4 · Find shear flow (q)
Use q = T ÷ (2 × Aₘ)
Round
the
cell
the
shear
flow
adds
up
to
the
torque
2Aₘ
q
,
so
divide
T
by
twice
Aₘ.
q
=
10,000,000
2 × 80,000
q
=
10
7
160,000
q
=
62.5
N/mm
Step 5 · Find shear stress (τ)
Use τ = q ÷ t
Shear flow is force per length of wall; spread it through the wall’s thickness.
τ
=
62.5
2
τ
=
31.25
MPa
Step 6 · Find twist rate (θ′)
Use θ′ = (180 ÷ π) × 1,000 × T × S ÷ (4 × Aₘ² × t × G)
Bredt’s
twist
per
millimeter
is
T
(
S
÷
t
)
4Aₘ²
G
for
an
even
wall;
times
1000
mm
gives
it
per
meter,
and
180
π
turns
radians
into
degrees.
θ′
=
(
180
π
)
×
1,000
×
10,000,000
×
1,200
4 × 80,
000
2
× 2 × 27,000
θ′
=
180
π
×
1,000 ×
10
7
×
1,200
4 × (6.4 ×
10
9
) × 2 × 27,000
θ′
=
180
π
×
1,000 ×
10
7
×
1,200
(2.56 ×
10
10
) × 2 × 27,000
θ′
=
180
π
×
1,000 ×
10
7
×
1,200
(5.12 ×
10
10
) × 27,000
θ′
=
180
π
×
1,000 ×
10
7
×
1,200
1.3824 ×
10
15
θ′
=
57.3
×
1,000 ×
10
7
×
1,200
1.3824 ×
10
15
θ′
=
57,300 ×
10
7
×
1,200
1.3824 ×
10
15
θ′
=
(5.73
×
10
11
)
×
1,200
1.3824 ×
10
15
θ′
=
6.876 ×
10
14
1.3824 ×
10
15
θ′ = 0.4974 °/m
Check
Put the numbers back in. Both sides should match.
80,000 = 400 × 200 ✓
1,200 = 2 × (400 + 200) ✓
62.5
=
10,000,000
2 × 80,000
✓
31.25
=
62.5
2
✓
0.4974
=
(
180
π
)
×
1,000
×
10,000,000
×
1,200
4 × 80,
000
2
× 2 × 27,000
✓
Topic check
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