L4. Torsion of beams: St. Venant (uniform) torsion, CCSM: chap...

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


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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25

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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

26


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

y{zzzz 27


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


Thanks for today!


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


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


üExamples:

Evaluate: Shear stresses and torsional constant









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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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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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



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

33



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


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



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


Thanks for today!


τs =Tx

2 Ai t∴ Kv =

4 Ai2

l


∴ 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

L4. Torsion of beams: St. Venant (uniform) torsion, CCSM: chap...

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