Lecture 3 - 2003
Kollbruner Section 5.2 Characteristics of Thin Walled Sections and ..
Kollbruner Section 5.3 Bending without Twist
thin walled => (cross section shape arbitrary and thickness can vary)
axial stresses and shear stress along center of wall govern
normal (to curved cross section) stress neglected
position determined by curvelinear coordinate s along center line of cross section
St. Venant torsion not a player as K ~ t^3
y, η
use shear flow:
z => ζ
x => ξ
z, ζ
a) equilibrium of wall element:
x
σ
τ = τ xs = τ sx
ds
dx
q
s
x
t
q + dq/ds*ds
σ + dσ dx*dx
στ
 σ +  d σ  ⋅dx ⋅t⋅ds − σ⋅t⋅ds +




dx  
 q + d q⋅ ds  ⋅dx − q⋅ dx = 0

 ds 
using q = τ ⋅t
as it includes τ(s) and t(s)
d
d 
q +  σ ⋅t = 0
ds
dx 
1
notes_12_bending_wo_twist.mcd
s
ds
←
→
↓
↑
↔
b) compatibility (shear strain)
du/ds
↑
du/ds*ds=du
d
d
u+ v=γ
ds
dx
dx
dv/dx
v is displacement in s direction
u is displacement in x direction
↔
→
←↓
dv/dx*dx=dv
x
view looking ⊥ to surface
c) tangential displacement (δv) in terms of η, ζ and φ
v is displacement in s direction
η is displacement in y direction
ζ is displacement in z direction
and φ is rotation
all at point s on cross section.
α
dv
dη
dv
α
dζ
δ v is differential over distance dx
δη is component of δv in y
δζ is component of δv in z
direction
direction
and hp*δφ is component due to
differential rotation between x and
x+dx
z component
y component
hp*dφ
dφ
p
dφ
hp
.
dv
superposition =>
rotation component
dv = dη⋅ cos( α ) + dζ⋅ sin( α ) + hp⋅ dφ
η, ζ and φ depend on s and x while α and hp are independent of x (prismatic section)
rewrite as:
2
notes_12_bending_wo_twist.mcd
δv
δx
=
δη
δx
⋅cos( α ) +
δζ
δx
⋅sin( α ) + hp⋅
δφ
δx
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
equilibrium for the cross section:
⌠
 σ dA = Nx
⌡
⌠
 σ⋅y dA = −M z
⌡
Nx = axial force
Mz and My are bending moments wrt y and z respectively
note integral expressions
Tp is torsional moment wrt cross sectional point P
Qx and Qy shear forces
⌠
 σ⋅z dA = M y
⌡
⌠
 τ⋅h dA =
p

⌡
⌠
 q⋅ h ds = T
p
p

⌡
⌠
⌠
 τ⋅cos( α ) dA =  q⋅cos( α ) ds = Vy
⌡
⌡
⌠
⌠
 τ⋅sin( α ) dA =  q⋅sin( α ) ds = Vz
⌡
⌡
5.3 Bending without twist
⌠
 σ dA = 0
⌡
⌠
 q⋅ h ds = 0
p

⌡
3
possible only if lateral loads pass
through P
notes_12_bending_wo_twist.mcd
δv
=
δx
using
δη
δx
⋅cos( α ) +
δζ
δx
⋅sin( α ) + hp⋅
δφ
δx
from above,
δφ
d
d 
=0
u = − v with no twist =>
δx
ds
dx 
becomes: δu
δs
=−
dη
dx
⋅cos( α ) −
dζ
dx
⋅sin( α ) which can be integrated to become:
⌠
⌠
dζ
= ζ' (prime is control F7)
u = −η'⋅ cos( α ) ds − ζ'⋅ sin( α ) ds + u0( x) where
dx
⌡
⌡
and
δη
δx
is η', and comes outside the integral due to our prismatic assumption
dY (=dy) =ds*cos(α), dZ (=dz) =ds*sin(α),
dz
where Y and Z refer to a coordinate system
with an arbitrary origin, whereas x and y are
defined centroidal (refer to center of area) =>
α
⌠
⌠
u( x) = −η'⋅ 1 dY − ζ'⋅ 1 dZ + u0( x)
⌡
⌡
y,η
dy
ds
z,ζ
u = −η'⋅Y − ζ' ⋅ Z + u0( x)
which says longitudinal displacement u is distributed linearly across cross section (plane
sections remain plane)
u0 is the constant of integration which is f(x)
axial strain = du/dx => u' = −η''⋅Y − −ζ''⋅Z + u'0( z) and
σ = E⋅ u' = −E⋅ η''⋅Y − ζ''⋅Z + E⋅ u'0( z)
⌠
now with  σ dA = 0
⌡
⌠
⌠
⌠
⌠
 σ dA =  E⋅ u' da = − E⋅η''⋅Y dA −  E⋅ζ''⋅Z dA +
⌡
⌡
⌡
⌡
⌠
⌠
⌠
−E⋅ η''⋅ Y dA − E⋅ζ''⋅ Z dA + E⋅ u'0( z)⋅ 1 dA = 0
⌡
⌡
⌡
4
⌠
 E⋅ u' ( z) dA = 0
0

⌡
=>
=>
notes_12_bending_wo_twist.mcd
E⋅ u'0( z) = E⋅η''⋅
⌠
 Y dA
⌡
A
⌠
+ E⋅ζ''⋅ Z dA
⌡
σ = −E⋅ η''⋅Y − −E⋅ ζ''⋅Z + E⋅η''⋅
⌠
 Y dA
⌡
A
+ E⋅ζ''⋅
determines E⋅ u'0( z) in stress σ =>
⌠
 Z dA
⌡
A




⌠
⌠


 Y dA
 Z dA




⌡
⌡
σ = −E⋅ η''⋅  Y −
− −E⋅ ζ''⋅  Z −
A
A




⌠
 Y dA
⌡
A
y=Y−
but .............
is the definition of the y position of the centroid and
⌠
 Y dA
⌡
A
and
z=Z−
rearranged becomes
⌠
 Z dA
⌡
⌠
 Z dA
⌡
A
the z position =>
and σ = −E⋅ η''⋅y − E⋅ζ''⋅z
A
where y and z are the position of the point s in the centroidal coordinate system.
η'' and ζ'' are determined by the equilibrium conditions
⌠
 σ⋅y dA = −M z
⌡
⌠
⌠
 σ⋅y dA = 
⌡
⌡
⌠
 σ⋅z dA = M y
⌡
σ = −E⋅
η''⋅y − E⋅ζ''⋅z
⌠

 ⌠

y ⋅ y dA ⋅ −E⋅ζ''⋅  z⋅ y dA = −M z
⌡



⌡

(−E⋅ η''⋅y − E⋅ζ''⋅z) ⋅y dA = −E⋅ η''⋅  
5
notes_12_bending_wo_twist.mcd
⌠
⌠
 σ⋅z dA = 
⌡
⌡
⌠
⌠
y ⋅ z dA − E⋅ζ''  z⋅ z dA = M y
⌡
⌡
( −E⋅ η''⋅y − E⋅ζ''⋅z)⋅z dA = −E⋅ η'' 
⌠
⌠
⌠
noting that  z⋅ z dA = Iy ,  z⋅ y dA = Iyz and  y ⋅ y dA = Iz and solving the two
⌡
⌡
⌡
equations for two unknowns E⋅η'' and E⋅ζ'' leads to =>
Given
−Eη''⋅Iz − Eζ''⋅Iyz = −M z
solving two equations two unknowns
−Eη''⋅Iyz − Eζ''⋅Iy = M y
 Iyz⋅My + Mz⋅Iy 
 −I 2 + I ⋅I 
yz
y z 
Find( Eη'' , Eζ'') → 
 −Iyz⋅Mz − My⋅Iz 


 −Iyz2 + Iy⋅Iz


Eη'' :=
(Iyz⋅My + Mz⋅Iy)
Eζ'' :=
 −I 2 + I ⋅II 
y z
 yz
(−Iyz⋅Mz − My⋅Iz)
 −I 2 + I ⋅I 
y z
 yz
reversed signs
now substituting back into the relation for axial stress ( σ := −E⋅ η''⋅y − E⋅ζ''⋅ z ) =>
σ := −Eη''⋅ y − Eζ''⋅ z
σ→
(
− Iyz⋅M y + M z⋅Iy
2
)
⋅y −
−Iyz + Iy ⋅Iz
or ....... rearranging in
terms of My and Mz =>
−Iyz⋅M z − M y ⋅Iz
2
⋅z
−Iyz + Iy ⋅Iz
σ :=
My
(−Iy⋅y + Iyz⋅z)⋅Mz + (Iz⋅z − Iyz⋅y)⋅M
 −I 2 + I ⋅I 
y z
 yz
6
notes_12_bending_wo_twist.mcd
as a check on this development let:
Iyz := 0
and
σ→
−M z
Iz
σ :=
⋅y +
My
Iy
⋅z
(
− Iyz⋅M y + M z⋅Iy
 −I 2 + I ⋅I 
y z
 yz
integration of s
q1( x) = 0
(
(−Iyz⋅Mz − My⋅Iz)
 −I 2 + I ⋅I 
y z
 yz
⋅zz
d
d 
q +  σ ⋅t = 0 (the equilibrium relationship above) along s leads to :
ds
dx 
where q1( x) is f(x) 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:
 d σ  ⋅t ds

dx 
0
⋅y −
which matches our previous understanding on bending
c) Shear Stress
⌠
q( s, x) = q1( x) − 

⌡
)
)
(
)
−Iyz⋅M z − M y ⋅Iz 
d  − Iyz⋅M y + M z⋅Iy
d

σ =
⋅y −
⋅z
dx  
2
2
dx




  −Iyz + Iy ⋅Iz
 −Iyz + Iy ⋅Iz 
Vz+dVz/dx*dx
Vz
where only My and Mz are x
dependent and
d
d
M z = Vy and M y = −Vz =>
dx
dx
y
x
My+dMy/dx*dx
My
Vy
Mz
z
M z( x) := Vy ⋅x
d 
dx 

M y ( x) := −V
Vz⋅x
(
− Iyz⋅M y ( x) + M z( x)⋅Iy
 −I 2 + I ⋅I 
y z
 yz
)
⋅y −
Vy+dVy/dx*dx
x
Mz+dMz/dx*dx
Iyz := Iyz
(−Iyz⋅Mz(x) − My( x)⋅Iz)
 −I 2 + I ⋅I 
y z
 yz

Iyz⋅Vz − Vy ⋅Iy


−Iyz + Iy ⋅Iz
⋅z →
2
⋅y −
−Iyz⋅Vy + Vz⋅Iz
2
⋅z
−Iyz + Iy ⋅Iz
copy and substitute
⌠
q( s, x) = −

⌡
s
0
q( s, x) =
s
⌠
 d σ  ⋅t ds = −  ( Iyz⋅Vz − Vy⋅Iy ) ⋅y − ( −Iyz⋅Vy + Vz⋅Iz) ⋅z ⋅t ds


dx 
 2

 −I 2 + I ⋅I  
   −Iyz + Iy ⋅Iz
y z
 yz

⌡
0
−1
2
−Iyz + Iy ⋅Iz

⌠
⌠

⋅ Iyz⋅Vz − Vy ⋅Iy ⋅ y ⋅ t ds − −Iyz⋅Vy + Vz⋅Iz ⋅ z⋅ t ds


⌡
⌡

(
)
s
(
0
7
)
s
0

notes_12_bending_wo_twist.mcd
if we designate the integrals which are the static moments of the cross section area: Qy and
Qz:
s
s
⌠
Qz =  y ⋅ t ds
⌡
⌠
Qy =  z⋅ t ds
⌡
0
−1
q( s, x) :=
2
−Iyz + Iy ⋅Iz
0
(
q( s, x) collect , Vy , Vz →
q( s, x) :=
)
(
−1
2
−Iyz + Iy ⋅Iz
(
)
⋅ −Iy ⋅Qz + Iyz⋅Qy ⋅Vy −
or rearranging as we
did for axial stress
1
2
−Iyz + Iy ⋅Iz
(
)
⋅ Iyz⋅Qz − Iz⋅Qy ⋅Vz
−1
⋅  ( Iyz⋅Qz − Iz⋅Qy ) ⋅Vz + ( −Iy ⋅Qz + Iyz⋅Qy ) ⋅V
Vy

 −I 2 + I ⋅I  
yz
y
z


or if the axes are principal (Ixy
= 0)
q( s, x) :=
)
⋅  Iyz⋅Vz − Vy ⋅Iy ⋅Qz − −Iyz⋅Vy + Vz⋅Iz ⋅Q
Qy


Iyz := 0
−1
⋅  ( Iyz⋅Qz − Iz⋅Qy ) ⋅Vz + ( −Iy ⋅Qz + Iyz⋅Qy ) ⋅V
Vy

 −I 2 + I ⋅I  
yz
y
z


Vy
Vz
q( s, x) simplify collect , Qy →
⋅Qy + Qz⋅
Iz
Iy
q( s, x) :=
 Qy⋅Vz Qz⋅Vy 
+

Iz
Iy


Iyz := 0
d) Shear Center
the above relationships apply for bending without twist i.e. when
⌠
 σ dA = 0
⌡
q→
−1
2
−Izy + Iy ⋅Iz
⌠
 q⋅ h ds = 0
p

⌡
(
)
(
possible only if lateral loads pass
through P
)
⋅  Izy⋅Qz − Iz⋅Qy ⋅Vz + −Iy ⋅Qz + Izy⋅Qy ⋅Vy


8
from above (reset in hidden
area
notes_12_bending_wo_twist.mcd
this point P was designated (by Maillart in 1921, see page 106 Kollbrunner) as the shear
center. In the centroidal coordinate system, this is located at yD and zD. the second condition
applies for a shear force V. The center of action must pass through P. Divide Q into
components Vy and Vz
y
thus from (Qy), the moment the moment
equilibrium wrt C (in centroidal coordinates)
is:
q
⌠
 q V ⋅h ds + V ⋅z = 0
y D
y c

⌡
(
)
Vy
⌠
 q V ⋅h ds = −V ⋅z
y D
y c

⌡
(
)
( )
q Vy
set
reset
−1
2
−Izy + Iy ⋅Iz
q :=
(
zD =
⋅  ( Iyz⋅Qz − Iz⋅Qy ) ⋅Vz + ( −Iy ⋅Qz + Iyz⋅Qy ) ⋅V
Vy

 −I 2 + I ⋅I  
yz
y
z


)
⌠

2
 −I + I ⋅I   
⌡
y z
 zy
−Vy
substitute
−1
( )
−Vy ⋅zD =
−1
⋅ −Iy ⋅Qz + Izy⋅Qy ⋅Vy
⌠
⌠
 q V ⋅h ds = −V ⋅z = 
y D 
y c

⌡

⌡
 −I 2 + I ⋅I 
y z
 zy
(
(
moment from V.z
)
⋅ −Iy ⋅Qz + Izy⋅Qy ⋅Vy ⋅hc ds

−Iy ⋅Qz + Izy⋅Qy ⋅hc ds

)
⌠

⋅   −Iy ⋅Qz + Izy⋅Qy ⋅hc ds =

 −I 2 + I ⋅I   
⌡

y z
 zy
1
yD,zD
z
is q with Vz set to 0 and hc is
perpendicular distance from centroid to
line of action (i.e. y(s) and z(s)).
Vz := 0
q→
Vz
hC
(
)
⌠
⌠
−Iy ⋅  Qz⋅hc ds + Izy⋅  Qy ⋅hc ds


⌡
⌡
⌠
 q V ⋅h ds − V ⋅y = 0
z D
z c

⌡
( )
9
or ...

2
−Izy + Iy ⋅Iz
⌠
 q V ⋅h ds = V ⋅y
z D
z c

⌡
( )
notes_12_bending_wo_twist.mcd
( )
is q with Vy set to 0 and hc is perpendicular distance from centroid to line of action
(i.e. y(s) and z(s)).
q Vz
reset
reset
set
Vz := Vz
−1
q→
2
−Izy + Iy ⋅Iz
(
)
( )
yD =
q :=
⋅ Izy⋅Qz − Iz⋅Qy ⋅Vz
⌠
⌠
 q V ⋅h ds = V ⋅y = 
z D 
z c

⌡

⌡
⌠
Vz⋅y D = 


⌡
Vy := 0
−1
 −I 2 + I ⋅I 
y z
 zy
−1
 −I 2 + I ⋅I 
y z
 zy
(
−1
⋅  ( Iyz⋅Qz − Iz⋅Q
Qy ) ⋅Vz + ( −Iy ⋅Qz + Iyz⋅Qy ) ⋅Vy

 −I 2 + I ⋅I  
yz
y
z


substitute
(
)
⋅ Izy⋅Qz − Iz⋅Qy ⋅Vz⋅hc ds
)
⋅ Izy⋅Qz − Iz⋅Qy ⋅Vz⋅hc ds =
 ⌠

⌠
⋅  Izy⋅  Qz⋅hc ds − Iz⋅  Qy ⋅hc ds

 −I 2 + I ⋅I   
⌡
⌡

y
z
zy


−Vz
 ⌠

⌠
⋅  Izy⋅  Qz⋅hc ds − Iz⋅  Qy ⋅hc ds

 −I 2 + I ⋅I   
⌡
⌡

y z
 zy
−1
for principal axes (Iyz = 0):
Izy := 0
 ⌠

⌠
y D :=
⋅  Izy⋅  Qz⋅hc ds − Iz⋅  Qy ⋅hc ds

 −I 2 + I ⋅I   
⌡
⌡

zy
y
z


−1
yD =
⌠
 Q ⋅h ds
y c

⌡
zD :=
zD =
Iy
10
⌠
⌠
−Iy ⋅  Qz⋅hc ds + Izy⋅  Qy ⋅hc ds


⌡
⌡
2
−Izy + Iy ⋅IIz
⌠
− Qz⋅hc ds

⌡
Iz
notes_12_bending_wo_twist.mcd