SME
Advanced Solid Mechanics (Solid Mechanics II)
Plates and shells
Use this for “A 10 mm steel plate, 400 mm across and clamped round its edge, takes 100 kPa. How far does its center sag, and what is the largest stress?”
Diagram
Clamped edge: the plate leaves the wall level; the stress is largest at the edge.
wₘₐₓ
=
f
×
pa⁴
64D
,
with
D
=
Et³
12(1 −
ν
2
)
.
The sag is drawn bigger than life.
This is an example. Type your own values to start a new problem.
D (N·mm)
=
E (GPa)
×
(
t (mm)
)
3
12(1
−
ν
2
)
wₘₐₓ (mm)
=
×
p (kPa)
×
(
a (mm)
)
4
64
×
D (N·mm)
edge
Edge support
example
σₘₐₓ
Largest bending stress
calculated
Formulas
D
=
E
×
t
3
12 × (1 −
ν
2
)
1.8315 ×
10
7
=
200,000 ×
10
3
12 × (1 −
0.3
2
)
f
=
1
1 = 1
wₘₐₓ
=
f
×
p
×
a
4
64 ×
D
0.1365
=
1
×
0.1
×
200
4
64 × (1.8315 ×
10
7
)
σₘₐₓ
=
3
×
p
×
a
2
4 ×
t
2
30
=
3
×
0.1
×
200
2
4 ×
10
2
Formulas are worked in MPa. The step-by-step shows the conversions.
Assumptions
•
A thin circular plate, t well under the radius, with small deflections under one even pressure p.
•
The edge factor f is 1 when clamped and (5 + ν) ÷ (1 + ν) when simply supported.
•
σₘₐₓ is 3pa² ÷ (4t²) at a clamped edge, or 3(3 + ν)pa² ÷ (8t²) at a simply supported center.
•
The steps work in N, mm and MPa (1 MPa = 1 N/mm²).
Step-by-step
WE KNOW
E = 200 GPa, ν = 0.3, t = 10 mm, p = 100 kPa, a = 200 mm, edge: clamped
FIND
flexural rigidity (D), edge factor (f), center deflection (wₘₐₓ), largest bending stress (σₘₐₓ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to MPa).
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
p = 100 kPa = 0.1 MPa (1 MPa = 1000 kPa)
Step 2 · Find flexural rigidity (D)
Use D = E × t³ ÷ (12 × (1 − ν²))
The
plate’s
bending
stiffness:
like
EI
for
one
unit
of
width,
t
3
12,
raised
by
1
1 −
ν
2
because
the
sides
hold
each
other.
D
=
200,000 ×
10
3
12 × (1 −
0.3
2
)
D
=
200,000 ×
10
3
12 × (1 − 0.09)
D
=
200,000 ×
10
3
12 × 0.91
D
=
200,000 ×
10
3
10.92
D
=
200,000
×
1,000
10.92
D
=
200,000,000
10.92
D
=
1.8315 ×
10
7
N·mm
Step 3 · Find edge factor (f)
Use f = 1
Held
level
at
the
edge:
the
clamped
case
is
the
base
one,
so
f
=
1.
clamped chosen: clamped
f
=
1
Step 4 · Find center deflection (wₘₐₓ)
Use wₘₐₓ = f × p × a⁴ ÷ (64 × D)
Integrate
the
plate
equation
D
∇⁴w
=
p
with
the
edge
held:
pa
⁴
64
D
when
clamped,
f
times
that
otherwise.
wₘₐₓ
=
1
×
0.1
×
200
4
64 × (1.8315 ×
10
7
)
wₘₐₓ
=
0.1
×
200
4
1.172 ×
10
9
wₘₐₓ
=
0.1
×
1.6 ×
10
9
1.172 ×
10
9
wₘₐₓ
=
1.6 ×
10
8
1.172 ×
10
9
wₘₐₓ = 0.1365 mm
Step 5 · Find largest bending stress (σₘₐₓ)
Use σₘₐₓ = 3 × p × a² ÷ (4 × t²)
Clamped:
the
bending
moment
is
largest
at
the
edge,
pa
²
8
per
unit
width,
and
σ
=
6M
t
2
.
edge factor 1: clamped
σₘₐₓ
=
3
×
0.1
×
200
2
4 ×
10
2
σₘₐₓ
=
3
×
0.1
×
200
2
4 × 100
σₘₐₓ
=
3
×
0.1
×
200
2
400
σₘₐₓ
=
3
×
0.1
×
40,000
400
σₘₐₓ
=
0.3
×
40,000
400
σₘₐₓ
=
12,000
400
σₘₐₓ = 30 MPa
Check
Put the numbers back in. Both sides should match.
1.8315 ×
10
7
=
200,000 ×
10
3
12 × (1 −
0.3
2
)
✓
1 = 1 ✓
0.1365
=
1
×
0.1
×
200
4
64 × (1.8315 ×
10
7
)
✓
30
=
3
×
0.1
×
200
2
4 ×
10
2
✓
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Thick-walled cylinder (Lamé)
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