Lecture 4 - 2003
Pure Twist
pure twist around center of rotation D => neither axial (σ) nor bending forces (Mx, My) act on
section; as previously, D is fixed, but (for now) arbitrary point.
as before:
a) equilibrium of wall element: b) compatibility (shear strain) d
d 
q +  σ ⋅t = 0
ds
dx 
d
d
u+ v=γ=0
ds
dx
small deflections
c) tangential displacement (δv) in terms of η, ζ and φ (geometry)
δv
δx
=
δη
δx
⋅cos( α ) +
δζ
δx
⋅sin( α ) + hp⋅
δφ
δx
N.B. hp =>hD from definition of problem
further assumptions:
1) preservation of cross section shape => ζ = ζ(x); η = η(x) φ = φ(x)
d
d 
2) shear though finite is small ~ 0 =>
u = − v
ds
dx 
δu
axial stress
3) Hooke's law holds => σ = E⋅
δx
----------------------------- from equilibrium -------------------------------⌠
 σ dA = Nx
⌡
⌠
 τ⋅h dA =
p

⌡
⌠
 σ⋅y dA = -M.z
⌡
⌠
 σ⋅z dA = M y
⌡
⌠
 q⋅ h ds = T
p
p

⌡
pure twist
⌠
 σ dA = Nx = 0
⌡
⌠
⌠
 τ⋅cos( α ) dA =  q⋅cos( α ) ds = Vy
⌡
⌡
⌠
 σ⋅y dA = −M z = 0
⌡
⌠
⌠
 τ⋅sin( α ) dA =  q⋅sin( α ) ds = Vz
⌡
⌡
⌠
 σ⋅z dA = M y = 0
⌡
pure twist also => only φ is finite i.e. other displacements (and derivatives) ζ = η = 0 =>
δv
δx
=
δη
δx
⋅cos( α ) +
δζ
δx
⋅sin( α ) + hp⋅
using negligible shear assumption
δφ
δx
becomes
δv
δx
= hD⋅
δφ
δx
δφ
d
d 
d
and integration along s =>
u = − v => u = −hD⋅
δx
ds
dx  ds
1
notes_13_pure_twist.mcd
u=−
δφ ⌠
⋅  hD ds + u0( x)
δx 
⌡
previously u = −η'⋅Y − ζ' ⋅ Z + u0( x) which showed u linear with y and z => plane sections plane.
 ⌠

here - only if h is constant so it can come outside hD  1 ds ⌡
D


is u (longitudinal
displacement) linear. u is defined as warping displacement (function).
stress analysis can be made analogous for torsion and bending IF the integrand hD⋅ds thought to
be a coordinate. calculation of stresses will involve statical moments, moments of inertia and
products of inertia which will be designated "sectorial" new coordinate = Ω
Ω wrt arbitrary origin and ω wrt normalized sectorial coordinate (as before like wrt center of area)
dΩ = hD⋅ds = dω
the warping function then becomes:
u=−
δφ
δx
⋅Ω + u0( x) = −φ'⋅ Ω + u0( x)
b) warping stresses
as before: axial strain = du/dx => u' = −φ''⋅ Ω + u'0( x) and
σ = E⋅ u' = −E⋅ φ''⋅Ω + E⋅ u'0( x)
⌠
 σ dA = 0
⌡
E⋅ u'0( x) = E⋅φ''
determines u'0( x)
⌠
 Ω dA
⌡
⌠


⌡
(−E⋅ φ''⋅Ω + E⋅ u'0(x)) dA = 0
and stress becomes:
A


⌠

 Ω dA


⌡
that is: σ = −E⋅ φ''⋅  Ω −
= −E⋅ φ''⋅ω
A


σ = −E⋅ φ''⋅Ω + E⋅φ''⋅
where
=>
⌠
 Ω dA
⌡
A
= −E⋅ φ''⋅ω
⌠
 Ω dA
⌡
ω=Ω−
A
this defines the normalized coordinate in the same sense as y and Y etc.
..............................................................
as an aside:
u' = −φ''⋅ ω
=>
u = −φ'⋅ ω + constant
dω = hD⋅ds
in this sense, ω is defined as the unit warping function
displacement per unit change in rotation
dependent only on s within a constant
................................................................
2
notes_13_pure_twist.mcd
shear flow follows from integration of
d
d 
q = − σ ⋅t
ds
dx 
d
d 
q +  σ ⋅t = 0 along s as above and leads to :
ds
dx 
⌠
d
q( s, x) = −
σ⋅t ds + q1( x)
 dx
⌡
=>
using the expression for axial stress σ = E⋅ u' = −E⋅ φ''⋅ω
⌠
q( s, x) = q1( x) − 

⌡
s
0
s
 ⌠ s

 d σ  ⋅t ds = q ( x) − ⌠

E⋅
(
)
+
E⋅φ'''
−
φ'''⋅ω⋅t
ds
=
q
x

ω⋅t
ds

1
1
 ⌡0
⌡
dx 
0


where q1( x) is f(z) and represents the shear flow at the start of the region. it is 0 at a stress free
boundary which is convenient for an open section: q1( x) = 0
as before if we designate the integrals which are the static moments of the cross section
area: e.g. Qy and Qz:
⌠
Qω =  ω dA =
⌡
s
⌠
 ω⋅t ds
⌡
= "sectorial statical moment of the cut-off portion of the cross section"
0
therefore:
q( s, x) = E⋅φ'''⋅Qω
designate torsional moment wrt D by Tω
⌠
⌠
now, since dΩ = hD⋅ds = dω =>  q⋅ hD ds =  q dω

⌡
⌡
parts:
⌠
⌠
 u dv = (u⋅v) ( b) − ( u v) ( 0) −  v du
⌡
⌡
⌠
 τ⋅h dA =
D

⌡
⌠
 q⋅ h ds = T
D
ω

⌡
and using integration by parts
u=q
v=ω
du = dq
dv = dω
integration along s and as dq = δq/δs*ds
⌠
⌠
⌠
δq
 q dω = q⋅ω(s = b) − q⋅ω(s = 0) −  ω dq = q⋅ω(s = b) − q⋅ω(s = 0) −  ω⋅ ds

δs
⌡
⌡
⌡
q⋅ω(s = b) = 0
and q⋅ω(s = 0) = 0
now using equilibrium:
as q(s=b) and q(s=0) = 0 (stress free ends)
d
d 
q +  σ ⋅t = 0
ds
dx 
3
notes_13_pure_twist.mcd
⌠
⌠
⌠
δq
d
 q dω = 0 −  ω⋅ ds =  ω⋅ σ⋅t ds


dx
δs
⌡
⌡
⌡
d
σ = −E⋅ φ'''⋅ω =>
dx
substituting σ = −E⋅ φ''⋅ω from above
⌠
⌠
⌠
d
 q dω =  ω⋅ σ⋅t ds = −E⋅ φ'''⋅ ω⋅ ω⋅t ds = −E⋅ φ'''⋅Iωω

dx
⌡
⌡
⌡
⌠
⌠
Iωω =  ω⋅ ω dA =  ω⋅ ω⋅t ds
⌡
⌡
where
similar to Iz
⌠
⌠
Iz =  y ⋅ y dA =  y ⋅ y⋅t ds
⌡
⌡
N.B. sometimes this
is represented by Iyy
⌠
going back to the relationship for torsional moment, where we have derived relationships for  q⋅ hD ds

⌡
=>
⌠
⌠
Tω =  q⋅ hD ds =  q dω = −E⋅ φ'''⋅Iωω

⌡
⌡
−Tω
therefore: φ''' =
E⋅ Iωω
Tω
if we think of a distributed
torsional load (moment/unit
length) mD;
mD
equilibrium over element dz =>


−Tω + mD⋅dx +  Tω +
 d T  ⋅dx = 0
 ω

dx  
Tω + dTω/dx*dx
=>
T'ω = −mD
and just as M'y = Vz the warping moment M'ω may be defined as M'ω = Tω
thus: φ'' =
−M ω
E⋅ Iωω
σ = −E⋅ φ''⋅ω =
Mω
Iωω
and the stresses are as follows:
⋅ω
and ... from q( s, x) = E⋅φ'''⋅Qω
4
q( s, x) = τ ⋅t =
−Tω
Iωω
⋅Qω
notes_13_pure_twist.mcd
c) Center of twist
calculation of the sectorial quantities assumes center of twist yD and zD are known. the
second and third equilibrium conditions above require:
⌠
 σ⋅y dA = 0
⌡
⌠
 σ⋅z dA = 0
⌡
for pure twist
⌠
⌠
using σ = −E⋅ φ''⋅ω this requires  ω⋅y dA = 0 and  ω⋅z dA = 0 as E ≠ 0 and φ'' ≠ 0
⌡
⌡
now for some geometry: determine distance from tangent to wall from hD in terms of the coordinates of the
center, the angle (α) that the y axis would have to rotate to line up with the positive direction of the tangent and
the perpendicular distance from the origin of the centroidal coordinates hC
ds
a = y D⋅cos( β )
hC
hD
D β
hD = hC − a − b
y
tangent
a
β=α−
π
2
β
yD
β
b


cos( β ) = cos α −


sin( β ) = sin α −
α
π
2
= sin( α )
π
2
= −cos( α )
C
zD
z
b = zD⋅sin( β )
parallel to
tangent
hD = hC − a − b = hC − y D⋅cos( β ) − zD⋅sin( β ) = hC − y D⋅sin( α ) + zD⋅cos( α )
α
multiply by ds and apply geometry:
hD⋅ds = hC⋅ds − y D⋅sin( α ) ⋅ds + zD⋅cos( α ) ⋅ds
ds
dy
β
dz
ds⋅cos( β ) = dz = ds⋅sin( α )
ds⋅sin( β ) = −dy = ds⋅ ( −cos( α ))
y,η
dy = ds⋅cos( α )
z,ζ
hD⋅ds = hC⋅ds − y D⋅sin( α ) ⋅ds + zD⋅cos( α ) ⋅ds = hC⋅ds − y D⋅dz + zD⋅dy
sectorial coordinate ω = hds =>
hD⋅ds = dωD = hC⋅ds − y D⋅dz + zD⋅dy = dωC − y D⋅dz + zD⋅dy
5
notes_13_pure_twist.mcd
which is now integrated:
ωD = ωC − y D⋅z + zD⋅y
and introduced into the equilibrium equations above where ω is ωD :
⌠
⌠
 ω⋅y dA = 0 and  ω⋅z dA = 0 =>
⌡
⌡
⌠
 ω ⋅y dA = 0 =
D

⌡
⌠


⌡
(ωC − yD⋅z + zD⋅y)⋅y dA
⌠
 ω ⋅z dA = 0 =
D

⌡
and ....
⌠


⌡
(ωC − yD⋅z + zD⋅y)⋅z dA
now using second moment nomenclature (including treating ω as a coordinate) =>
⌠
⌠
⌠
 ω ⋅y dA − y  y ⋅ z dA + z ⋅ y ⋅ y dA = 0
D
D
C

⌡
⌡
⌡
becomes
recall that Iyωc is referred to C for ω
Izωc − y D⋅Iyz − zD⋅Iz = 0
and .......
⌠
⌠
⌠
 ω ⋅z dA − y  z⋅ z dA + z ⋅ y ⋅ z dA = 0
D
D
C

⌡
⌡
⌡
becomes
Iyωc − y D⋅Iy + zD⋅Iyz = 0
which provides two equations in two unknowns y D and zD
Given
Iyωc − y D⋅Iy + zD⋅Iyz = 0
yD →
Iyωc⋅Iz − Iyz⋅ Izωc
2
Iy ⋅Iz − Iyz
and for principal axes Iyz = 0
yD →
and ...
Iyz := 0
zD →
y D :=
Iyωc
Iy
 yD 
:= Find( y D , zD)
 zD 
Izωc − y D⋅Iyz + zD⋅Iz = 0
6
2
Iy ⋅Iz − Iyz
(Iyωc⋅Iz − Iyz⋅Izωc)
zD →
and ...
−Izωc⋅Iy + Iyz⋅Iyωc
 I ⋅I − I 2
 y z yz 
zD :=
(−Izωc⋅Iy + Iyz⋅Iyωc)
 I ⋅II − I 2
 y z yz 
−Izωc
Iz
notes_13_pure_twist.mcd