l4. torsion of beams: st. venant (uniform) torsion, ccsm: chap...
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Material and Computational Mechanics Group 1
L4. Torsion of beams: St. Venant (uniform) torsion, CCSM: chap 10.1, 10.2.1-2
àCharacteristics of torsion problem
üAngle of twist - warping
üKinematics of torsion
à St. Venant and Vlasov torsion
à St. Venant torsion
üKinematics
ü Stress - strain relation (Hooke's law)
ü Equilibrium
ü Prandtl's stress function
üGoverning equations
Material and Computational Mechanics Group 2
àCharacteristics of torsion problem
üAngle of twist - warping
üKinematics of torsion
Assume rigid cross section and twist-rotation about the SC (TC): From geometry
ur=rθ@xD, w =ur cos@αDv = −ur sin@αD, y =rcos@αD
z =rsin@αD=⇒w@x, yD=y θ@xD, v@x, zD= −zθ@xDNote!
x =JyzN; x̂ =J−z
yNwithx⋅x̂ =0 ⇒u=Jv
wN= θ@xD J−z
yN= θ@xD x̂
The induced strain state now becomes:
εx= ∂ux∂x
; εy = ∂v∂y
:=0; εz = ∂w∂z
:=0
γxy =∂ux∂y
+∂v∂x
=∂ux∂y
−zθ
γxz = ∂ux∂z
+ ∂w∂x
= ∂ux∂z
+y θ
γyz =∂v∂z
+∂w∂y
= − θ@xD+θ@xD:=0
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Material and Computational Mechanics Group 3
à St. Venant and Vlasov torsion
Consider characteristics of torsion depending on the kinematics:
1) St. Venant (uniform) torsion is obtained for cross sections who preserve their shapealong the beam during torsion
γyz = εy = εz = εx = ∂ ux∂x
= 0
γxy = ∂ux∂y − z θ ≠ 0
γxz = ∂ux∂z + y θ ≠ 0
= or J γxy
γxzN =
ikjjjj ∂ux∂y
∂ux∂z
y{zzzz + θ J −zy
N or γ = ∇ ux + θ x̂
Note! the gradient operator ∇ is associated with postion x, i.e.
∇ ⋅ x = 2 and ∇ ⋅ x̂ = 0
2) Vlasov (nonform) torsion is obtained for cross sections who change their shape alongthe beam during torsion
γyz = εy = εz = 0 ; γ = ∇ ux + θ x̂ := 0
εx = ∂ ux∂ x
≠ 0
Note! Vlasov torsion will be considered later on. Cf. examples of cross sections
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à St. Venant torsion
üKinematics
Let us restate the kinematics as: only out-of-plane shear deformations are assumed tooccur
γxy = ∂ ux∂y
− z θ ≠ 0, γxz = ∂ ux∂ z
+ y θ ≠ 0 or γ = ∇ux + θ x̂
ü Stress - strain relation (Hooke's law)
Note! due to kinematics
γyz = εy = εz = εx = 0 ⇒ τyz = σy = σz = σx = 0
From Hooke's law we have
τ = G γ or J τxy
τxzN = G J γxy
γxzN with G =
iso
E2 H1 + νL = Shear modulus
Stress tensor w.r.t Cartesian basis, cf. fig. x,
σ =ikjjjj 0
τxy
τxz
τxy
00
τxz
00
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Material and Computational Mechanics Group 5
ü Equilibrium
Equilibrium requires
∇¯
⋅ σ + q = 0 ⇒ 9 ∂τxy∂y + ∂τxz
∂z + qx = ∇ ⋅ τ + qx = 0
∂τxy∂x + qy = 0
∂τxz∂x + qz = 0
Assume: qx = 0fl Equilibrium condition
∇ ⋅ τ = 0
ü Prandtl's stress function
Introduce Prandtl's stress function φ = φ@y, zD so that
τ = −∇ˆ
φ with ∇ˆ
⋅ x̂ = 2 and ∇ˆ
⋅ x = 0 ⇒
∇ ⋅ τ = −∇ ⋅ ∇ˆ
φ =. .. = 0 ⇒ ∇ ⋅ τ := 0 with qx = 0
Consider also
∇ˆ
⋅ τ =
−∇ˆ
⋅ ∇ˆ
φ = −∆φ = G ∇ˆ
⋅ γ = G ∇ˆ
⋅ H∇ ux + θ x̂L = G H∇ˆ ⋅ ∇ ux + θ ∇ˆ
⋅ x̂L = 2 G θ ⇒
∴ ∆φ + 2 G θ = 0
fl Given θ , we may compute φ w.r.t boundary conditons!!! fl τ = −∇ˆ
φ
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How do we determine the boundary conditions?
fl no shear stress is allowed on the boundaries of the cross section, cf. fig. x fl considertraction vector w.r.t outward normal
t =ikjjjj 0
τxy
τxz
τxy
00
τxz
00
y{zzzz ikjjjj nx = 0
nynz
y{zzzz =ikjjjj τxy ny + τxz nz
00
y{zzzz ⇒
τ̄ = ny τxy + nz τxz = n ⋅ τ = 0 ∀ x ∈ Γ
Interpretation:
n ⋅ τ = −n ⋅ ∇ˆ
φ =. .. = n̂ ⋅ ∇φ = dφds
= 0 ⇒ φ = C ∀ x ∈ Γ
Note!
n ⋅ ∇ˆ
φ = −ny ∂ φ∂z
+ nz ∂ φ∂y
= −J−nz ∂ φ∂y
+ ny ∂ φ∂z
N = −n̂ ⋅ ∇φ
Note! C = 0 ∀ x ∈ Γ normally chosen for outer boundary of the cross section.
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üGoverning equations
Model completion fl
Consider momentum balance relation HMx + d MxL − Mx + q dx = 0d Mxdx
+ q = 0
fl Governing equations
Mx = GKv θ Hconstitutive + kinematicsLMx + q = 0 HequilibriumL = HGKv θ L + q = 0
BC: Geometric type θ = θ̄Force type Mx = M
¯x
Material and Computational Mechanics Group 8
Thanks for today!
Material and Computational Mechanics Group 9
L5. Torsion of beams: St. Venant (uniform) torsion, CCSM: chap 10.2.3-10.2.6
ü Torque due to St. Venant torsion
ü Solid cross sections: Stress analysis
ü Examples:
Evaluate: Shear stresses and torsional constant
Example: Thin rectangular cross section
Example: Open thin-walled cross section
Example: Thin, curved cross section
Example: Closed thin-walled cross section
Example: Closed thin-walled channel cross section
Example: Stiffened rectangular cross section
Material and Computational Mechanics Group 10
ü Torque due to St. Venant torsion
Torque defined as the stress resultant, c.f. fig.,
Mx = ‡ Hτxy H−zL + τxz HyLL A = ‡ τ ⋅ x̂ A = −‡ ∇ˆ
φ ⋅ x̂ A = −‡ ∇φ ⋅ x A
Make use of the divergence theorem:
−Mx = ‡ ∇φ ⋅ x A = ‡ ∇ ⋅ HφxL A − ‡ φ H∇ ⋅ xL A = φ ‡Γ
n ⋅ x Γ − 2 ‡ φ A
Assume: Solid cross section with φ = 0 along (outer) boundary fl
Mx = 2 ‡ φ A
Note! Torque basically area trapped under stress function φ !
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ü Solid cross sections: Stress analysis
Introduce normalized stress function via
φ = φ0 2 G θ with φ0 = φ0@y, zD ⇒ ∴ ∆φ0 + 1 = 0
Mx = 2 ‡ φ A = G Kv θ with Kv = 4 ‡ φ0 A = torsional constant
Given torque Mx fl Computational steps
1) Solve normalized stress function
∆φ0 + 1 = 0 in with φ0 := 0 along outer boundary
2) Evaluate the torsional stiffness
Kv = 4 ‡ φ0 A ⇒ G θ = MxKv
3) Compute the shear stresses
τ = −∇ˆ
φ ; τxy = 2 G θ ∂ φ0
∂z= 2 Mx
Kv ∂ φ0
∂ z; τxz = −2 G θ ∂ φ0
∂ y= −2 Mx
Kv ∂ φ0
∂ y
Material and Computational Mechanics Group 12
üExamples:
Evaluate: Shear stresses and torsional constant
Example: Thin rectangular cross section
Example: Thin, curved cross section
Example: Open thin-walled cross section
Example: Closed thin-walled cross section
Example: Closed thin-walled channel cross section
Example: Stiffened rectangular cross section
Material and Computational Mechanics Group 13
Example: Thin rectangular cross section
Note! we have in this case
b p t ⇒ φ0@y, zD ≈ φ0@zD ⇒
1) Solve normalized stress functions, consider
∆φ0 + 1 =d2 φ0
d z2+ 1 = 0 ⇒ φ0@zD = −
12
z2 + a1 z + a2
BCfl 8φ0@− t2 D = 0, φ0@ t
2 D = 0< fl a1 → 0, a2 → t2
8 fl
φ0@zD = 18
Ht2 − 4 z2L2) Determine torsional constant via integration
Kv = 4 ‡ φ0 = 4 ‡− t
2
t2
φ0@zD b z =b t3
3
3) Evaluate the shear stresses
τxy = 2 MxKv
∂ φ0
∂z= − 2 Mx
Kv z , τxz = − 2 Mx
Kv
∂ φ0
∂ y= 0
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Example: Open thin-walled cross section
Consider open thin-walled cross section consisting of rectangular portions!
Note! assumption bi p ti ⇒ φ0 i@yi, ziD ≈ φ0@ziD ⇒ "uncoupled components"
BCi : 8φ0 i@− ti2 D = 0, φ0 i@ ti
2 D = 0< fl
φ0 i@zD = 18
Hti2 − 4 z2L- Torsional constant becomes
Kv = 4 ‡ φ0 = 4 ‚i=1
N ‡ φ0 i = 13
‚i=1
N
bi ti3
- Evaluation of shear stresses
τxyi =2 MxKv
∂ φ0 i
∂ z= −
2 MxKv
z , τxzi = 0
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Material and Computational Mechanics Group 15
Example: Thin, curved cross section
Note! we have in this case
L p t ⇒ φ0@s, nD ≈ φ0@nD ⇒ φ0@nD =18
Ht@sD2 − 4 n2L- Determine torsional constant via integration
Kv = 4 ‡ φ0 = 4 ‡s1
s2‡− t
2
t2 18
Ht@sD2 − 4 n2L n s = ‡s1
s2t@sD3
3 s
- Evaluate the shear stresses; First note!
τxs = n̂ ⋅ τ = −n̂ ⋅ ∇ˆ
φ =. .. = −n ⋅ ∇ φ = n̄ ⋅ ∇φ =dφdn
⇒
⇒ τxs =2 MxKv
dφ0
dn= −
2 MxKv
n
Material and Computational Mechanics Group 16
Example: Closed thin-walled cross section
Consider (in this case) the more general situation:
Mx = 2 ‡ φ − ‡Γi
φ n ⋅ x Γ = 2 ‡ φ − φi ‡Γi
n ⋅ x Γ =9φi ‡Γi
n ⋅ x Γ = −φi ‡Γi
n̄ ⋅ x Γ = −φi ‡ ∇ ⋅ x = −φi 2 Ai= =
2 ‡ φ + 2 φi Ai
Note again! τs = n̂ ⋅ τ = −n̂ ⋅ ∇ˆ
φ = −n ⋅ ∇φ = n̄ ⋅ ∇ φ = dφdn
Evaluate φ@s, nD ; Assume 1) φ@s, nD = φ@nD , 2) Linear var. in thickness dir. fl
φA− t2E = 0 at Γ , φA t
2E = φi at Γi ⇒ φ@nD = I 1
2+ n
tM φi
Tangential shear stress from normal derivative
τs = dφdn
⇒ φi = t τs
Conclusion! Constant φi= constant shear flow, φi = t τs .Note! Evaluation of the torque simplified as
Mx = 2 ‡ φ + 2 φi Ai ≈ 2 φi Ai = 2 t τs Ai
fl well known shear stress formula for closed thin-walled cross sections
τs = Mx2 Ai t
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Closed thin-walled cross section (cont'd) Express the torsional constant fl formulate normalized stress function
Mx = 2 ‡ φ + 2 Ai φi ≈ G H4 Ai φ0 iL ϕ = G Kv ϕ with Kv = 4 Ai φ0 i
Consider normalized stress function: φ0 i = φi2 G ϕ = t τs
2 G ϕ = ?
fl Evaluate the shear stress in terms of kinematics! fl
τs = G γs, γs = ∂ux∂ s
+ ∂ us∂x
= ∂ux∂s
+ R@sD θ
t τs = G t γs = 1G
t I ∂ ux∂s
+ R@sD θ M t τs dst
= G Hdux + R@sD θ dsLUtlizing t τs = C = constant shear flow :
® t τs 1t
ds = t τs l = G J® dux + ϕ ® R@sD dsN = 2 G θ Ai
with
l = ® dst@sD = normalized segment length;
Carefully note! ® dux = 0; 2 Ai = ® R@sD ds
fl Conclusion
φ0 i = t τs
2 G θ= Ai
l⇒ ∴ Kv =
4 Ai2
l
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Example: Closed thin-walled channel cross section
Consider again the special situation
Mx ≈ 2 ‚i=1
N
φi Ai = 2 Hφ1 A1 + φ2 A2L = 2 Ht1 τs1 A1 + t2 τs2 A2LNote! φi = ti τsi = constant shear flow,
Kinematics, cf. closed thin-walled cross section, + integration:
t1 τs1 dst1
= G Hdux + R@sD θ dsL ⇒ φ1 l1 = G J‡ABCD
dux + θ ‡ABCD
R@sD sNt2 τs2 ds
t2= G Hdux + R@sD θ dsL ⇒ φ2 l2 = G J‡
DEFA
dux + θ ‡DEFA
R@sD dsNt3 τs3
dst3
= G Hdux + R@sD θ dsL ⇒ Hφ2 − φ1L l3 = G J‡AD
dux + θ ‡AD
R@sD dsNNote ! Normalized segment lenghts
l1 = ‡ABCD
1t1
ds , l2 = ‡DEFA
1t2
ds , l3 = ‡AD
1t3
ds
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Cont'd, From conditions
®A1
dux = ®A2
dux = 0
form integration curves
‡ABCD
• s − ‡AD
•ds = ®A1
• s, ‡DEFA
• s + ‡AD
•ds = ®A2
• s
leading to
φ1 l1 − Hφ2 − φ1L l3 = G θ ®A1
R@sD s = G θ 2 A1
φ2 l2 + Hφ2 − φ1L l3 = G θ ®A2
R@sD ds = G θ 2 A2
The shear flows may now be established as
φ1 = A2 l3 + A1 Hl2 + l3Ll2 l3 + l1 Hl2 + l3L 2 G θ , φ2 = A1 l3 + A2 Hl1 + l3L
l2 l3 + l1 Hl2 + l3L 2 G θ
The torque may now be expressed as fl
Mx = 2 Hφ1 A1 + φ2 A2L =2 HA22 l1 + A1
2 l2 + HA1 + A2L2 l3Ll2 l3 + l1 Hl2 + l3L 2 G θ
Material and Computational Mechanics Group 20
Cont'd fl Shear stress formulae for the channel cross section problem
τs1 = HA1 l2 + A1 l3 + A2 l3L2 D t1
Mx
τs2 = HA2 l1 + A1 l3 + A2 l3L2 D t2
Mx
τs3 = A2 l1 − A1 l22 D t3
Mx
with
D = A22 l1 + A1
2 l2 + HA1 + A2L2 l3Mx = G
ikjjjj4 ‚i=1
N
φ0 i Aiy{zzzz θ = G Kv θ with Kv = 4 ‚
i=1
N
φ0 i Ai
Express the torsional constant fl formulate normalized stress function
Consider normalized stress function defined via:
φ01 = A2 l3 + A1 Hl2 + l3Ll2 l3 + l1 Hl2 + l3L , φ02 = A1 l3 + A2 Hl1 + l3L
l2 l3 + l1 Hl2 + l3Lfl Conclusion
∴ Kv = 4 A22 l1 + A1
2 l2 + HA1 + A2L2 l3l2 l3 + l1 Hl2 + l3L
Material and Computational Mechanics Group 21
Example: Stiffened rectangular cross section
Consider cross sectional properties:
A1 → α b h, A2 → H1 − αL b h
l1 → ht
+ 2 α bt, l2 → h
t+ 2 H1 − αL b
t, l3 → h
t, t1 → t, t2 → t, t3 → t
fl Torsional constant
Kv = 8 b2 h2 t Hb H1 − αL α + h H1 − α + α2LL4 b h + 3 h2 + 4 b2 H1 − αL α
Plot 8b = h = 1, t = 1 ê 20< versus the parameter a :
Figure 1
0.2 0.4 0.6 0.8 1
0.051
0.052
0.053
0.054
0.055
0.056
0.057
Note! Increase of approximately 14% may be achieved if the internal web is placed close toeither of the external webs.
Kv@αD
α
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Thanks for today!
Material and Computational Mechanics Group 23
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τs =Tx
2 Ai t∴ Kv =
4 Ai2
l
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∴ Kv = 4 A22 l1 + A1
2 l2 + HA1 + A2L2 l3l2 l3 + l1 Hl2 + l3L
τs1 = HA1 l2 + A1 l3 + A2 l3L2 D t1
Tx
τs2 = HA2 l1 + A1 l3 + A2 l3L2 D t2
Tx
τs3 = A2 l1 − A1 l22 D t3
Tx