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)