IB Structures 2022-23
Handout 1
Thin-walled structures
Filled Version
Text and pictures in grey are omitted from the version in lectures
DO NOT DISTRIBUTE DIGITALLY OR IN HARD COPY
Cambridge University Engineering Department
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IB Structures 2022-23
1.1 Introduction
In this section we will consider the stresses and strains generated in thin-walled sections due to
various loads. In general, the wall-thickness of the section is assumed to be very much less than
the other dimensions of the structure, and this allows us to make a number of assumptions about
the nature of the stresses:
1. Through-thickness stresses are zero.
2. The stress state is uniform through the section.
no stress on this face
stress here = stress here
These are usually reasonable for e.g. a circular cylinder with radius/thickness ≥ 20.
We will also examine the effect of stiffeners. Real thin-walled structures, such as the boxgirder bridge or the aircraft fuselage below, have additional stiffeners:
• to prevent local buckling of the walls;
• to carry locally concentrated loads (for instance heavy local loads on the bridge shown
below);
• as a fail-safe device (see Examples Paper 2/1, q1).
Thin walled flange
Thinwalled
web
stiffeners
Part of an aircraft fuselage
Section of a box-girder bridge
The stiffeners shown are schematic. The actual cross-section is likely to be:
or
or
Handout 1. Thin-walled structures
3
1.2 Stresses
Much of this will be revision from 1A, except the section on torsion. It will be particularly
helpful to review your 1A notes on Bending and Shearing Stresses in Beams.
1.2.1 Circular cylinder due to internal pressure
σt — through-thickness
σl — longitudinal
σh - hoop
Despite our assumption, the through-thickness stress cannot be constant across the wall. If the
pressure internally is p, and externally is 0, then
σt = −p
σt = 0
on inner face
on outer face
However, we shall assume σt = 0 everywhere (|σt | ≤ p ≪ σh for thin-walled).
Hoop stress
p
σh
l
Equilibrium l
2 p r l = σh l 2t
pr
σh =
(n.b. ≫ p)
t
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Longitudinal stress
The longitudinal stress will vary, depending on the support conditions.
Example 1: Closed end cylinder
σl
closed end,
no forces
p
Equilibrium ↔
p π r2 = σl 2π rt
pr
σl =
2t
Example 2: Pipeline with bellows expansion joint. Bellows allow expansion of the pipelines,
due to e.g. temperature change.
σl
Equilibrium ↔
0 longitudinal force = 0
at bellows joint
p
p
σl = 0
Effect of stiffeners
Circumferential stiffeners must reduce the average hoop stress:
Handout 1. Thin-walled structures
5
Consider an average hoop stress σav
σav A = p 2rl;
A > 2lt
⇒
σav <
pr
t
where A is the wall area including stiffeners. But a shorter free body that excluded stiffeners
would give
pr
σav = .
t
Clearly the hoop stress must vary along the section. However, locally, away from stiffeners, the
hoop stress may well reach its familiar value — σh = prt is a useful conservative estimate.
Longitudinal stiffeners (if connected to the end of the cylinder) will carry the same load as the
skin, and hence the longitudinal stress will be reduced.
stiffeners carry full share of load
σl A = p π r2
where A is the wall area including stiffeners — σl will not vary around the section.
Try Questions 1 and 2, Examples Sheet 2/1
1.2.2 Case study introduction — lifting Storebælt approach spans
The Storebælt link is an 18 km combined bridge and tunnel that forms part of a road and rail link
connecting the mainland of Denmark to the island of Zealand. The link opened in 1998. The
most spectacular part of the link is a suspension bridge making up part of the East Bridge, which
has a main span of 1624 m. This study, however, will concentrate on the approach spans, a series
of steel box girder bridges, each 193 m in length. More details of the project will be found on
the world-wide web at http://www.storebaelt.dk
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Great Belt
Approach spans
(note vertical scale is distorted)
The cross-section used for the box-girder bridges is shown below
The approach spans were lifted into place in one piece. We will calculate the stresses that
this generated in the bridge — but first some theory.
1.2.3 Stresses due to axial forces, bending moments and shear forces
We will consider general, non-circular cylinders that have a vertical axis of symmetry, under the
action of vertical loads:
Handout 1. Thin-walled structures
7
Why the axis of symmetry? It ensures that vertical loads and bending moments (symmetric
applied loads):
Cause vertical deflections, curvatures in a vertical plane (symmetric deformations)
Do not cause sideways deflections, twist (antisymmetric deformations)
There will be more about symmetry and antisymmetry in Handout 3.
Axial force
σl =
P
A
=
force
area
Resultant force acts through centroid
Bending moments
σl
M
Neutral axis
(no bending stress)
is at the centoidal level
σl
y
centroid
σl =
My
I
from 1A, structures data book p.5
M
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Example: Longitudinal stresses in a circular section due to applied moments
Calculate the distribution of longitudinal stresses in a circular section due to an applied moment.
M
σl
t
y
θ r
M
σl =
My
I
y = r cos θ
I=
Z
y2 dA = π r3t
(From the mechanics data book, or by integration)
σl =
M cos θ
π r 2t
Handout 1. Thin-walled structures
9
Shear Stress
A Reminder About Complementary Shear Stress If a shearing stress τ is applied
to a block of material, what is the shearing stress τ ′ ?
τ'
τ
b
a
τ
depth = c
τ'
Moment equilibrium: τ ′ (bc)(a) = τ (ac)(b) ⇒ τ ′ = τ
A state of simple shear requires equal shear stress on all four faces of an arbitrary
small block. Stresses τ ′ are complementary to τ (and vice-versa).
Example (from first principles): Cylindrical cantilever subject to end load
Calculate from first principles the stresses for this simple case to show how the shear stress
formula was derived.
bending moment = Sz
z
S
No longitudinal forces
at the free end
φ
τ
τ
Only longitudinal forces shown
From earlier example,
σ = |{z}
Sz
moment
r cos θ
I
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Shear forces acting along the cut edge must balance the end stresses.
2 τ zt =
Z φ
−φ
σl t r d θ
shear stress × area = total force due to longitudinal stresses
Z
Sz φ 2
=
r t cos θ d θ
I −φ
Szr2t
2 sin φ
=
I
shear stress × area
z
S 2
= 2r t sin φ
I
S
= As ȳ — as data book
I
Total shear force/unit length =
where As ȳ = 2r2t sin φ is the first moment of area of the cut out section. You can check this in the
Mechanics Data Book.
How to derive the formula for shear stress
Shear Force
on section
Change in bending
moment along beam,
dM/dx = S
Shear Stress
in section
Longitudinal shear stress
Change in longitudinal
stress along cut-out
section of beam
Handout 1. Thin-walled structures
11
Shear stress due to applied shear force
SAs ȳ
I
(1A, and structures data book p6)
shear force/unit length =
cut out this section
S
y
τ
centroidal level
centroidal level
As
ȳ
As ȳ
cross-sectional area of the ‘cut out’ section, shown shaded.
distance from the centroid of the whole section to the centroid of the ‘cut out’.
first moment of area of the ‘cut out’ section.
S
applied shear force.
I
second moment of area of whole section.
The shear stress can then be found by dividing the shear force by the area of the longitudinal
cut, shown shaded below.
t
t
ength
unit l
shear force
in this case
1 × 2t
The maximum shear stress will be found with the minimum area, so the cut should be perpendicular to the wall
τ=
Effect of stiffeners
Longitudinal stiffeners will increase the cross-sectional area of the structure, and hence reduce
the stress due to axial loads
P
σl =
|{z}
A
|{z}
reduced
increased
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The stiffeners will also increase the second moment of area for the section, and hence also reduce
the stresses due to bending.
My
σl =
|{z}
I
|{z}
reduced
increased
The shear stresses due to shear-loading will not be changed significantly. The increase in I will
generally be matched by an increase in As ȳ. The stiffeners may slightly redistribute the stresses.
increased
z}|{
S As ȳ
shear force/unit length =
I
|{z}
negligible overall effect
increased
To avoid the full complications of calculating exact section properties including stiffeners, the
stiffeners are often ‘smeared’ out by making the thin walls a little thicker.
Equal areas
Section including stiffener
Thicker, ‘smeared’ section
It is necessary to have a good understanding of the problem to know when to use the smeared,
and when to use the actual thickness. The case study should help to clarify with examples.
Handout 1. Thin-walled structures
13
1.2.4 Case study part 1 — Stresses during the lifting of a Storebælt approach span
The exact cross-section used for the spans was shown earlier. Below is a slightly simplified crosssection that we will use, which omits the central diaphragm. The stiffeners have been ‘smeared’.
The actual, and the smeared thickness, is shown for each part of the cross-section.
All dimensions in mm.
t =13 (21), area = 25100×21 = 527100
25100
Values in brackets are
smeared thicknesses
2523
6700
4177
97
y
26
datum
11000
t = 10 (14)
area = 9726×14 = 136200
t = 20 (24)
area = 11000×24 = 264000
Section properties
Total Area
A = 527100 + 2 × 136200 + 264000 all mm2
= 1.064 × 106 mm2 = 1.064 m2
Mass/Unit length
ρ A = 7800 kg/m3 × 1.064 m2 = 8300 kg/m
Centroid
Aȳ = ∑ As ȳs
for each section
= 527100 × 6700 + 2 × 136200 × 6700/2
mm3
= 4.444 × 109 mm3
∴ ȳ = 4177 mm
2nd moment of area Consider a section at a time (around the centroid!), ignoring the 2nd moment of area of the flanges about their own centroid, (the bd 3 /12 terms) as negligible.
Top flange
Itop = 527100 × 25232 = 3.355 × 1012 mm4
Bottom flange
Ibottom = 264000 × 41772 = 4.606 × 1012 mm4
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Webs — are more difficult. There are various shortcuts that we Rcould take (talk to your supervisor), but we’ll do the calculation from first principles, Iwebs = y2 dA, to show that it is quite
straightforward.
Consider a thin slice at a distance y from the centroid:
14 mm × (9726/6700)
= 20.3 mm
dA = (20.3×2) dy
dy
14 mm
dy
y
Iwebs =
Z
y2 dA =
Z 2523
−4177
y2 × 40.6 dy
y3
= 40.6
3
2523
−4177
12
4
= 1.206 × 10 mm
Total
I = (3.355 + 4.606 + 1.206) × 1012 mm4
= 9.167 m4
Stresses due to lifting
Floating Crane
Fixed Crane
Existing span
New span
Barge
193 m
Loading due to self-weight = 8300 kg/m × 9.81 N/kg = 81420 N/m
Free-body diagram of the bridge:
Handout 1. Thin-walled structures
7.857
´ 10
6
15
81420 N/m
N
7.857
´ 10
6
N
Maximum bending moment, at centre = 7.857 × 106 × 96.5 − 81420 × 96.5 × 48.25
= 379.1 × 106 Nm
(= wl 2 /8)
Longitudinal Stresses
σmax =
Mmax ymax
I
Maximum tensile stress in bottom flange — ymax = 4.177 m
379.1 × 106 × 4.177
9.167
= 172.7 × 106 N/m2
σmax =
Shear Stresses
shear force/unit length = q =
SAsȳ
I
To maximize q we should maximize S and As ȳ
Smax = 7.857 × 106 N
at ends of section
To maximize As ȳ we should ‘cut’ the section at the centroidal level:
area = 136200×(2523/6700) = 51288
area = 527100
25100
2523
Cut at centroidal level
(As ȳ)max = 527100
|
{z× 2523} + 51288
| {z × 2} ×
top flange
3
= 1.459 m
area of two webs
2523/2
| {z }
distance to their centroid
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qmax =
7.857 × 106 × 1.459
= 1.251 × 106 N/m
9.167
To find the maximum shear stress, we must cut through the wall thickness. The minimumlength cut will be perpendicular to the actual wall, hence we should use the actual, and not the
smeared thickness, t = 10 mm = 0.01 m.
minimum-length cut, perpendicular to the wall
×
τmax =
qmax
1.251 × 106
= 62.55 × 106 N/m2
=
length of cut
2 ×0.01
|{z}
two webs
The complementary shear is perpendicular to the cut, and hence ‘flows’ around the section.
Try Questions 3 and 4, Examples Sheet 2/1
Handout 1. Thin-walled structures
17
1.2.5 Torsion
Uniform thin-walled circular cylinders
The shear stress will be constant because of the axi-symmetry
t
r
T
T
small segment
τ
T
τ
δθ
Calculate the torque about the centre, δ T , in equilibrium with the small segment shown
δ T = τ tr
δ θ} × |{z}
r
| {z
area
} lever arm
| {z
force
Integrate around the circle to find T
T=
Z T
dT =
0
Z 2π
0
= τ r 2 2π t
τ=
T
2π r 2 t
τ r2t d θ
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General thin-walled cross-section
Because of the lack of symmetry, we can no longer assume that the shear stress is constant around
the section.
What is τ?
T
T
Shear Flow
For thin-walled structures, the concept of a shear flow, q, is useful. The shear flow
is the shear force per unit wall length of the structure. It is the shear stress times the
wall thickness:
q = τ ×t
For example, the shear flow in a thin-walled circular cylinder is q = T /2π r2 .
Just as there must always be a complementary shear stress, there must also always
be a complementary shear flow.
τ'
τ
τ
τ'
t
τ ′ = τ ⇒ τ ′ t = τ t ⇒ q′ = q
Consider a section of thin-walled structure, of varying thickness, in a state of pure shear (no
normal stresses)
τ2
t2
τ2
shear stress
varies around
face
τ1
τ1
l
t1
Handout 1. Thin-walled structures
Equilibrium ↔
19
τ2 t2 l = τ1 t1 l
but
q1 = τ 1 t 1 ,
q2 = τ 2 t 2 ,
∴ q1 = q2
Shear flow due to applied torque is constant around a thin-walled section
What applied torque, T , is in equilibrium with this constant shear flow? Consider the torque
in equilibrium with a small element of the wall:
O
r
δAe
τ
δl
δ T = τ |{z}
t δ l × |{z}
r
area
| {z
} lever arm
force
but τ t = q = constant
and δ lr = 2δ Ae — the shaded area enclosed to mid-thickness
∴ δ T = 2qδ Ae
Integrate around the section
T=
H
I
dT =
I
2q dAe
q is constant, dAe = Ae ,
T = 2q Ae
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Thus, for an arbitrary thin-walled closed section subject to an applied torque,
or, as q = τ × t
q=
T
2Ae
τ=
T
2Aet
Important: These formulae only apply to closed sections:
These two
sections
have very
different
torsional
response
Impossible
to have
any shear
flow at
break
Effect of stiffeners
The stiffeners will not affect Ae , the area enclosed by the section, and hence the shear flow q will
be unaffected. Locally, at a point as shown below, the shear stress τ = q/tskin will be unaffected
by the stiffeners.
tskin
1.2.6 Case study part 2 — stresses due to twisting of a Storebælt approach
span
As a worst case, consider what would happen if one end of the span was lifted by only one corner.
7.857 ×
106
7.857 × 106 N
N
98.61 × 106 Nm
≡
q=
12.55 m
T
2Ae
T = 98.61 × 106 Nm
25.1 + 11
× 6.7 = 120.9 m2
Ae =
2
Handout 1. Thin-walled structures
21
∴ q = 407.8 × 103 N/m
To find the maximum shear stress, cut at the thinnest section (τ = q/t). In the web, t = 10 mm =
0.01 m,
τ = 40.78 × 106 N/m2
n.b. this is due to torque alone — a complete answer requires us to also superimpose stresses
due to the shear force. In the webs, the stresses will reinforce in one web, and tend to cancel in
the other web.
1.3 Strain in 3-dimensions
We assume in the definitions in this section that the strains are small, of the order of 10−3 . Note
that strains are dimensionless.
1.3.1 Normal strain
You are already familiar with one-dimensional strain as being extension/original length. In 3
dimensions, there are three such strains, in three orthogonal directions, called the normal strains.
A cube, subject to normal strains in directions perpendicular to its faces, will become some
general brick shape. The angle between every face will remain at 90◦ .
normal
strains
z
y
x
The normal strains in the x, y, z directions are called, respectively, εxx , εyy , and εzz .
1.3.2 Shear strain
In addition to normal strains there are also shear strains, e.g. in two dimensions:
y
y-face
(perpendicular to y-axis)
x-face
x
shear
strain
y n.b. γ measured in radians
— it’s a ratio of lengths
γxy
x
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For small strains, the shear strain γxy is the change in angle between faces that were originally
perpendicular, in this example the x and y faces. Note that the same shear strain could have
drawn in a number of ways, by superposing a rotation (a rigid body motion which causes no
strain). Obviously, γyx = γxy .
y
y
y
γxy
γxy/2
γxy/2
γxy
x
x
x
Equal shear strains
In three dimensions, there will be three shear strains, γxy , γyz , γzx . A cube, subjected only to
shear strains, will become a parallelepiped, and the lengths of its sides will be unchanged.
shear
strains
1.4 Stress-strain relationships
Real material behaviour is complex, but the key features can often be captured in a simple model.
We will use a linear-elastic, homogeneous, isotropic, time-independent model. What do these
terms mean?
1.4.1 Material model
Linear-elastic
σ
elastic
linear
ε
elastic — deformation disappears when load is removed.
linear — stress-strain curve is a straight line.
Homogeneous
Doesn’t vary with position.
Isotropic
Doesn’t vary with direction (not true for e.g. fibre-reinforced materials.)
Handout 1. Thin-walled structures
Time-independent
23
No creep.
1.4.2 Normal strain due to uniaxial normal stress
y
σxx
σxx
x
E = Young’s Modulus, ν = Poisson’s Ratio
εxx =
σxx
E
εyy = εzz = −νεxx
σ xx
= −ν
E
Typical values for E: Steel, 210 × 109 N/m2 ; Aluminium alloy, 70 × 109 N/m2 ; Rubber, 0.1 ×
109 N/m2 .
Typical values for ν : Steel, 0.3; Aluminium alloy, 0.33; Cork, ≈ 0.
Example: Anticlastic curvature
Bend an eraser between your fingers — note the anticlastic curvature:
shorter
longer
due to Poisson’s ratio effects
— causes curvature across beam
shorter
longer
1.4.3 Normal strain due to temperature change
An unrestrained body subject to a temperature rise will undergo a uniform expansion, without
changing shape. Thus the body undergoes a uniform normal strain in all directions, and no shear
strain. The normal strain is (approximately) linear with change in temperature ∆T .
εxx = εyy = εzz = α ∆T
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α is the coefficient of linear expansion.
Typical values for α : Steel 11 × 10−6 K−1 ; Aluminium Alloy, 20 × 10−6 K−1 .
1.4.4 Normal strain due to 3D normal stress
The strain due to normal stress in three perpendicular directions can be found by superposition:
1
σxx
|E{z }
εxx =
−
due to σxx
1
ν σyy
| E{z }
−
due to σyy
1
ν σzz
| E{z }
due to σzz
If a temperature change is also superposed, then we get the expression on page 1 of the data
book:
1
εxx = (σxx − νσyy − νσzz ) + α ∆T
E
with similar expressions in the other directions.
In an isotropic material, there is never any shear strain due to normal stress and temperature
change alone
1.4.5 Shear strain
A shear stress on the x-face acting in the y direction (and its complementary stress on the y-face
in the x direction) will cause a shear strain in the x-y plane, but no shear strains on the y-z or z-x
planes.
y
τ
τ
x
γxy =
γ
τ
τ
1
τxy
G
G is the shear modulus. G is related to E and ν by the formula G = E/2(1 + ν ) (proved in
Examples sheet 2/2).
In an isotropic material, a shear stress will cause no normal strain in any direction,
εxx = εyy = εzz = 0
For an isotropic material:
• Shear stress only causes shear strain in its own plane;
• Normal stress only causes normal strain
Handout 1. Thin-walled structures
25
Example 1: Change of dimension of a closed pressurised cylinder
t
σh
σt
r
σl
σl
σt
l
Stresses
σt ≈ 0,
σl =
pr
,
2t
σh =
pr
t
Strains
Longitudinal
1
(σl − νσh − νσt )
E
pr
(1 − 2ν )
=
2Et
εl =
Hoop
1
(σh − νσl − νσt )
E
pr
(2 − ν )
=
2Et
εh =
Through-wall
1
(σt − νσh − νσl )
E
pr
=−
(3ν ) — becomes thinner
2Et
εt =
Note: zero stress in a direction doesn’t imply zero strain.
Change in dimension
Length
∆l
= εl ,
∆l = εl l
l
Circumference
∆C
= εh ,
C
∆C = εhC
Radius — Note that this is nothing to do with the strain in the through-wall direction, εt !
∆r 2π ∆r ∆C
=
=
= εh ,
r
2π r
C
∆r = εh r
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Change in volume
Define a Volumetric Strain as
∆V
.
V
Volume enclosed by cylinder, V = π r2 l
∂V
= π r2 ,
∂l
∂V
= 2π rl
∂r
For small variations around the current position,
∆V =
∂V
∂V
∆l +
∆r = π r2 ∆l + 2π rl ∆r
∂l
∂r
∆l 2 ∆r
∆V
=
+
V
l
r
pr
(5 − 4ν )
= ε l + 2ε h =
2Et
Volumetric strain =
∆V =
pr
pπ r 3 l
(5 − 4ν ) × π r2l =
(5 − 4ν )
2Et
2Et
Volume of metal in wall, Vm = 2π rtl
∂ Vm
∂ Vm
= 2π rt,
= 2π tl,
∂l
∂r
For small variations around the current position,
∂ Vm
= 2π rl
∂t
∆Vm = 2π rt ∆l + 2π tl ∆r + 2π rl ∆t
Volumetric strain =
∆Vm ∆l ∆r ∆t
=
+
+
Vm
l
r
t
= εl + εh + εt =
pr
(3 − 6ν )
2Et
pπ r 2 l
pr
∆Vm =
(3 − 6ν ) × 2π rtl =
(3 − 6ν )
2Et
E
Handout 1. Thin-walled structures
27
Example 2: Thermal Stress in a Restrained Cylinder
A cylinder is initially unstressed, and just fits between two rigid walls. The temperature of the
cylinder then increases by 100◦ C. Calculate the stresses in the cylinder
t
σh
σt
r
σl
σl
σt
l
In the longitudinal direction this is a statically indeterminate problem. There are two unknowns,
and only one equation of equilibrium, and so is impossible to calculate the stresses from equilibrium alone. (There will be much more about statically indeterminate problems later.)
Free-body diagram
σl
F — wall force — unknown
The key to statically indeterminate problems is to use compatibility. In this case, we know that
there is zero longitudinal strain — the cylinder cannot change in length.
εl =
1
(σl − νσh − νσt ) + α ∆T = 0
E
σh = 0 — no internal pressure
∴ σl = −E α ∆T
e.g. for Al-alloy, E = 70 × 109 N/m2 ,
α = 22 × 10−6
⇒ σl = −154 × 106 N/m2
• Could cause a weak alloy to yield.
• Would cause a thin cylinder (r/t ≈ 1000) to buckle.
The skin of Concorde (Al-alloy) heated up by approximately 100◦ C during flight — this would
cause problems if the outer skin was rigidly connected to the interior of the aircraft!
Try Questions 5, 6 and 7, Examples Sheet 2/1
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1.5 Torsional rigidity
We have already considered equilibrium relationships for torsion — the shear stress due to an
applied torque for a thin-walled section. In this section, we will find the stiffness relationship
between applied torque and resultant twist of a section. For general sections this is difficult, as
the cross-sections will warp — plane sections do not remain plane. However, we will start with
a simple thin-walled circular cylinder, where symmetry shows that warping will not occur.
1.5.1 Uniform thin-walled circular section
Consider a thin-walled cylinder of thickness t, radius r, length l, whose base is fixed, while a
torque T is applied to the top surface. What is the resultant rotation θ ? Or, better (as it doesn’t
depend on l), what is the resultant twist/unit length φ ?
T
t
r
l
θ
line
before
twist
line
after
twist
γ
Equilibrium
τ=
Compatibility
T
2π r 2 t
γl = θ r
Define the twist/unit length φ = θ /l
γ=
Material law
θ
r = φr
l
τ = Gγ
Substitute for τ and γ to give the required stiffness relationship between T and φ
T
= Gφr
2π r 2 t
T = G 2π r 3 t φ
Handout 1. Thin-walled structures
29
Thus the torsional rigidity, or torsional stiffness, T /φ is given by
T
= G 2π r 3 t
φ
and the shear stress (constant in the section) is
τ = Grφ
1.5.2 General thin-walled section
T
l
T — Applied torque
θ — Rotation of one end
relative to the other
T, θ
τ
In this case, the compatibility between twist and shear strain (which was straightforward for
the circular section) is tricky, because of warping. For non-circular cross-sections, plane sections
do not remain plane when twisted — they warp.
T
T
We shall use Virtual Work to find the compatibility relationship — this is possible because we
have already worked out the equilibrium relationship q = τ t = T /2Ae.
Virtual Work
Equilibrium set: torque T in equilibrium with shear stresses τ (s).
Compatible set: rotation θ (unknown), compatible with shear strain γ .
External Work
External Work = T θ (torque × rotation)
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IB Structures 2022-23
Internal Work Consider a small element, subject to a shear stress τ , that undergoes a small
shear strain γ
small deformation
τ
τ
c
τ
τ
b
τ
γ
τ
a
only this
stress does
work
τ
τ
Work done in the small element
δ Wi = |{z}
τ bc ×
force
= τγ ×
γa
|{z}
displacement
δV
|{z}
small volume
Now consider a strip of the section as the volume δ V
l
t(s)
δs
s
δ V = lt(s)δ s
δ Wi = τ (s)γ (s)lt(s)δ s
γ , t and τ may all vary around the section
Total internal work
I
Wi =
section
τ (s)γ (s)lt(s) ds
random start point
on section
Handout 1. Thin-walled structures
31
τ (s) and t(s) may vary around the section, but q = τ (s)t(s) is constant (recall that the shear flow
is uniform)
I
Wi = ql γ (s) ds
Virtual Work The internal work must equal the external work
T θ = ql
I
γ (s) ds
We can use this expression to find the correct compatibility relationship, because we already
know the equilibrium relationship T = 2Ae q.
∴ 2Ae θ = l
I
γ (s) ds
This is the compatibility relationship for the section, which we can write as a relationship between φ = θ /l and the shear deformation γ (s).
Compatibility
H
γ (s) ds
φ=
2Ae
The equilibrium and stress-strain relationship are straightforward
Equilibrium
T
q = τ (s)t(s) =
2Ae
Material law
q = τ (s)t(s) = Gγ (s)t(s)
Torsional stiffness Substitute for q in the material law to give
T
= Gγ (s)t(s)
2Ae
T
γ (s) =
2Ae Gt(s)
Substitute into the compatibility law to give a stiffness relationship between T and φ
I
T
1
ds
φ=
2Ae 2Ae Gt(s)
Only t varies around the section, so the torsional stiffness is given by:
G 4A2e
T
= H ds
φ
t(s)
Torsional stiffness is often denoted as GJ, where J is the torsion constant
T = (GJ)φ
cf. for bending, M = EI κ .
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IB Structures 2022-23
Example: Relative stiffness of closed sections with same wall length and thickness
4
7
8/π
1
4
Ae
20.4
16
7
Relative
stiffness
∝ Ae2
1
0.62
0.12
1.5.3 Case study part 3 — torsional stiffness of a Storebælt approach span.
As well as being necessary to calculate deflections due to e.g. off-centre loads, it is essential
to know the torsional stiffness of a bridge to be able to calculate its aerodynamic response —
Tacoma Narrows was too flexible!
All dimensions in mm.
t =13 (21)
Values in brackets are
smeared thicknesses
25100
6700
97
26
11000
t = 10 (14)
t = 20 (24)
Local stiffeners will not greatly affect the torsional rigidity — the flexible sections between
stiffeners will dominate the calculation. Hence we will use actual, not smeared, thicknesses.
Ae = 120.9 m2
I
25100
ds
9726
11000
=
+2 ×
+
= 4426
t
| 13
| 20
{z } | {z10 }
{z }
top flange
two webs
bottom flange
G = 81 × 109 N/m2 for steel
Handout 1. Thin-walled structures
33
T
G 4A2 81 × 109 × 4 × 120.92
= H ds e =
φ
4426
t(s)
= 1.07 × 1012
Torsion constant J =
Nm
radians/m
1.07 × 1012
= 13.2 m4
81 × 109
1.5.4 Axi-symmetric shafts
For the special case of shafts with circular symmetry, the results for thin sections can be used to
make calculations for thick, or solid sections.
T, θ
Thin shell cut
from solid
τ(r)
δT
T
R2
r δr
R1
Torsional rigidity
For the thin shell
δ T = 2 π r 3 δ r Gφ
Integrate over the whole shaft
T = 2 π Gφ
=
Z R2
R1
r3 dr
2 π Gφ 4
(R2 − R41 )
4
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IB Structures 2022-23
For circular shafts, the torsional rigidity is given by
T
π
= G (R42 − R41 )
φ
2
For Rcircular shafts only, the torsion constant J is given by the polar second moment of area,
J = r2 dA
Shear stress due to torque
For the thin shell
τ = Grφ
For circular shafts, the shear stress is given by
τ
T
= Gφ =
r
J
cf. for bending,
σ
y
= Eκ =
M
I
Try Questions 8 and 9, Examples Sheet 2/1