Solid Mechanics 2020/2021 - ULisboa · Solid Mechanics 2020/2021 Class 17 Torsion of thin-walled...
Transcript of Solid Mechanics 2020/2021 - ULisboa · Solid Mechanics 2020/2021 Class 17 Torsion of thin-walled...
Solid Mechanics2020/2021
Class 17
Torsion of thin-walled open sections
Examples
Torsion of thin-walled closed sections, single and multicellular
Examples
November 24, 2020
Recap: Statical Linear Theory of Elasticity
σij = Eijkl ekl, if isotropic σij = E / (1 + ν) [eij + ν / (1 – 2 ν) ekk δij],
eij = ½ (∂ui / ∂xj + ∂uj / ∂xi)
∂σij / ∂xj + bi = 0.
We illustrate the theory by considering the problem of torsion
Recap: Torsion of a Non-Circular Cylinder
Warping function ψ(x1, x2) or the
Proposed displacement field is:
u1(x1, x2, x3)= – α x2 x3
u2(x1, x2, x3) = α x1 x3,
u3(x1, x2, x3) = α ψ(x1, x2)
α is the angle of twist per unit length,
∂2ψ/∂x12 + ∂2ψ/∂x2
2 = 0 in S,
∂ψ/∂n = x2 n1 – x1 n2 on c.
Determination of the stress distribution over a cross section of a bar in torsion consists in finding either the:
Stress function φ(x1, x2)
σ31 = ∂φ / ∂x2
σ32 = - ∂φ / ∂x1
G is the shear modulus.
∂2φ/∂x12 + ∂2φ/∂x2
2 = - 2 G α in S,φ = 0 on c.
Recap: Torsion of thin-walled open sections
When the rectangular cross section is narrow, b >> t, an approximate solution exists.
Performing the rescaling
x1 = b ξ1, - ½ ≤ ξ1 ≤ ½ , x2 = t ξ2, - ½ ≤ ξ2 ≤ ½
and using
∂/∂x1 = 1/b ∂/∂ξ1, ∂/∂x2 = 1/t ∂/∂ξ2
The equation for the stress function
∂2φ(x1, x2) / ∂x12 + ∂2φ(x1, x2) / ∂x2
2 = - 2 G α in S
φ(x1, x2) = 0 on c,
becomes, with ε = t / b
ε2 ∂2φ/∂ξ12 + ∂2φ/∂ξ2
2 = - 2 G α t2.
Recap: Torsion of thin-walled open sections
To lowest order in ε = t / b << 1,
∂2φ(x1, x2) / ∂x12 + ∂2φ(x1, x2) / ∂x2
2 = - 2 G α in S
φ(x1, x2) = 0 on the boundary c,
reduces to
∂2φ/∂ξ22 = - 2 G α t2, - ½ < ξ2 < ½
φ = 0 on ξ2 ± ½,
whose solution is
φ = G α t2 (1/4 - ξ22).
Back to the physical variables,
φ = G α (t2/4 - x22).
Recap: Torsion of thin-walled open sections
φ = G α (t2/4 - x22).
Compute the stress and the torsional rigidity from
σ31 = ∂φ / ∂x2 and κ = T / α = 2 / α ∫S φ(x1, x2).
σ31 = - 2 G α x2, κ = G (b t3 / 3), J = b t3 / 3
which confirms the formula for the torsional rigidity of the rectangular cross-section with b / t = ∞.
In terms of the couple T
σ31 = - 2 T /(b t3 / 3) x2
τmax = T /(b t2 / 3).
Recap: Torsion of thin-walled open sections
Contrast the
approximate solution
with the exact solution
Recall:
T = ∫S (x1 σ32 - x2 σ31) dS,
∫S x1 σ32 = - ∫S x2 σ31) dS = T /2
Torsion of thin-walled open sections: different shapes
J = b t3/3, J = (b1 t13 + b2 t2
3)/3, J = (b1 t13 + 2 b2 t2
3)/3,
J = (b1 t13 + 2 b2 t2
3)/3, J = b t3/3.
The corners marked A have zero stress. Those marked B have a large stress concentration depending on the radius of the fillet.
Torsion of thin-walled open sections: different shapes
J = b t3/3, J = (b1 t13 + b2 t2
3)/3, J = (b1 t13 + 2 b2 t2
3)/3,
J = (b1 t13 + 2 b2 t2
3)/3, J = b t3/3.
For a section with n elements J = 1/3 Σn bi ti3
The twisting moment in each element is Ti = Ji T/J
The maximum stress in each element is τmaxi = Ti ti / Ji = T ti / J.
The maximum stress on the cross-section is τmax = T tmax / J.
Torsion of thin-walled single cell closed sections
Torsion of a hollow circular section
The torsional rigidity is
κ = T / α = G IP = G π (a4 – b4) / 2
and the stress
τ = σθz = G α r = T r / IP
Note that the stress is almost constant across the cross-section thickness.
Torsion of thin-walled single cell closed sections
For a non-circular hollow cylindrical bar in torsion – cylinder with thin-walled closed section – it is possible to develop approximate formulas.
The thin tubular section of variable thickness t(s) follows a midline c.
Assume that the shear stress τ is constantacross the thickness of the wall.
Torsion of thin-walled single cell closed sections
If the wall thickness is not constant, consider the equilibrium of a part of the wall.
Summing forces in the vertical x3
direction,
τ1 t1 L = τ2 t2 L
We conclude that the product τ(s) t(s) is constant along the section.
Define the constant shear flow q as
q = τ(s) t(s) = constant.
Torsion of thin-walled closed/ open sections
Stress distribution in open sections.
Stress distribution in closed sections.
Torsion of thin-walled single cell closed sections
The total couple T associated with the shear stresses (constant across the cross-section thickness) is
∫S x x σ(x) n dS = ∫c x x τ(s) t(s) ta(s) ds
= T e3.
where ta(s) is the tangent to c(s).
From ta = (dx1/ds, dx2/ds, 0),
x x τ (s) t(s) ta
= τ(s) t(s) (x1 dx2/ds – x2 dx1/ds) e3.
Torsion of thin-walled single cell closed sections
T e3 = ∫c x x τ (s) t(s) ta (s) ds
= ∫c τ (s) t(s) (x1 dx2/ds – x2 dx1/ds) ds e3
= ∫c τ (s) t(s) (x1 dx2 – x2 dx1) ds e3
= q ∫Ω (1 + 1) dS = 2 q Ω e3
where Ω is the area enclosed by the midline c.
Torsion of thin-walled single cell closed sections
Consider a more physical deduction:
the total couple T associated with the shear flow is
T = ∫c r(s) q ds = q ∫c r(s) ds,
r(s) being the distance between the origin O and the direction tangent to the midline c at s.
The product r(s) ds is twice the elemental triangle area, r(s) being the triangle height and ds its base, and
∫c r(s) ds = 2 Ω,
where Ω is the area enclosed by the midline c.
Torsion of thin-walled single cell closed sections
The total couple T associated with the shear flow is
T = ∫c r(s) q ds = q ∫c r(s) ds
∫c r(s) ds = 2 Ω
where Ω is the area enclosed by the midline c.
T = 2 q Ω = 2 τ(s) t(s) Ω
τ(s) = T / (2 t(s) Ω).
Torsion of thin-walled single cell closed sections
T = 2 q Ω = 2 τ(s) t(s) Ω
τ(s) = T / (2 t(s) Ω).
Using the formula
∫c τs ds = 2 G Ω α,
∫c τs ds = ∫c T/(2 t(s) Ω) ds
= T/(2 Ω) ∫c 1 /t(s) ds,
obtain the torsional rigidity,
T = 4 G Ω2 α / ∫c 1 /t(s) ds = G J α
J = 4 Ω2 / ∫c 1 /t(s) ds
Example: sections with equal perimeter
The torsional rigidity is T/α = 4 G Ω2 / ∫c 1 /t(s) ds, J = 4 Ω2 / ∫c 1 /t(s) ds
where Ω is the area enclosed by the midline c.
ε = t / R J = ⅓ 2 π R4 ε3
J = 2 π R4 ε, J = 2 π R4 ε3
Example: sections with equal enclosed area
The torsional stiffness is T = 4 G Ω2 α / ∫c 1 /t(s) ds, J = 4 Ω2 / ∫c 1 /t(s) ds
where Ω is the area enclosed by the midline c.
J = a3 t, J = ¾ a3 t
Torsion of thin-walled multicellular closed sections
In the multicellular case the shear flow is no longer constant along the cross-section wall.
At a junction we will have, from equilibrium in the x3 direction, L being the cylinder length,
q1 L + q3 L = q2 L or q1 + q3 = q2.
Torsion of thin-walled multicellular closed sections
The total couple T associated with the shear stresses is
T e3 = ∫S x x σ(x) n dS
= ∫c1 x x q (s) ta (s) ds + ∫c2 x x q (s) ta (s) ds + ∫c3 x x q (s) ta (s) ds
= q1 ∫c1 x x ta (s) ds + q2 ∫c2 x x ta (s) ds + q3 ∫c3 x x ta (s) ds
= 2 (q1 ΩABCOA + q2 ΩOCDAO + q3 ΩACOA) e3
Torsion of thin-walled multicellular closed sections
The total couple T associated with the shear stresses is
T = 2 (q1 ΩABCOA + q2 ΩOCDAO + q3 ΩACOA)
But q3 = q2 - q1
= 2 (q1 ΩABCOA + q2 ΩOCDAO + (q2 - q1) ΩACOA)
= 2 (q1 ΩABCA + q2 ΩACDA)
Torsion of thin-walled multicellular closed sections
For a section with n interconnected cells
T = 2 ∑i=1n qi Ωi.
But the problem now is statically indeterminate.
We need compatibility equations.
The equation for a single tube will be used and we force the twist of all the tubes to be the same:
From ∫c τs ds = ∫c q / t(s) ds = 2 G Ω α,
∫ci q / t(s) ds = 2 Gi Ωi α.
Torsion of thin-walled multicellular closed sections
Continuing with the example, q3 = q2 - q1
∫ABC q1/t(s) ds - ∫CA q3/t(s) ds = 2 G1 Ω1 α.
∫CDA q2/t(s) ds + ∫AC q3/t(s) ds = 2 G2 Ω2 α.
The twist in both tubes is the same
∫ABC q1 / t(s) ds - ∫CA (q2 - q1) / t(s) ds
= ∫ABCA q1 / t(s) ds - ∫CA q2 / t(s) ds
= q1 ∫ABCA 1 / t(s) ds – q2 ∫CA 1 / t(s) ds = 2 G1 Ω1 α.
∫CDA q2 / t(s) ds + ∫AC (q2 - q1) / t(s) ds
= ∫CDAC q2 / t(s) ds - ∫AC q1 / t(s) ds
= q2 ∫CDAC 1 / t(s) ds - q1 ∫AC 1 / t(s) ds = 2 G2 Ω2 α.
Torsion of thin-walled multicellular closed sections - Example
Stiffness T = 28 /11 G a3 t α
Strength τmax = 12 / 11 G a α = 3 T / (7 a2 t)