torsion thin rectangle

8/2/2019 Torsion Thin Rectangle

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Torsion of a Thin Rectangular Section

y

z

1, 0 y zn n= =1, 0

y zn n= − =

0, 1 y zn n= =

0, 1 y zn n= = −

( )

( )

2 2

2 2

1

2

1

2

d y z

n ds

d y z z

y dz

ψ

ψ

∂= +

∂

∂ = + =∂

ds dz=

ds dy= −

ds dz= −

ds dy=

( )

( )

2 2

2 2

1

2

1

2

d y z

n ds

d y z y

z dy

ψ

ψ

∂= +

∂∂

= + = −∂ −

( )

( )

2 2

2 2

1

2

1

2

d y z

n ds

d y z z

y dz

z y

ψ

ψ

ψ

∂= +

∂∂

= + = −−∂ −

∂=

∂

( )

( )

2 2

2 2

1

2

1

2

d y z

n ds

d

y z y z dy

ψ

ψ

∂= +

∂∂

= + =−∂



z

z yψ ∂ =∂

z yψ ∂ =∂

t

2 22

2 20

y zψ ψ ψ ∂ ∂∇ = + =

∂ ∂

yzψ =

xy

xz

G z y

G y z

ψ σ φ

ψ σ φ

⎛ ⎞∂′= −⎜ ⎟∂⎝ ⎠

∂⎛ ⎞′= +⎜ ⎟∂⎝ ⎠

22

0

xz

eff

xy

TyG y

J σ φ

σ

′= =

=

xzσ 2 2

32

4 43

eff

A

z

A

J z y dA y z

bt y dA I

ψ ψ ⎧ ⎫⎛ ⎞∂ ∂⎪ ⎪⎛ ⎞= − + +⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠⎪ ⎪⎩ ⎭

= = =

∫

∫ 3

3eff

bt J =

eff T GJ φ ′=

y

max

eff

Tt

J

τ =



Out of plane warping

>> z= linspace(-5,5, 20);

>> y =linspace(-0.5,0.5,10);>> [zz, yy] = meshgrid(z,y);>> ux=zz.*yy;

>> mesh(zz,yy,ux)>> axis equal>> view (50,20)

nowarping

no warping

xu yzφ ′=



a( ) J eff = 1

3bt 3

b( ) J eff =1

3b1t 1

3 +1

3b2t 2

3

c( ) J eff =1

3b1t 13

+1

3 b2t 23

+1

3 b3t 33

The results we obtained for the torsion of a thin rectangle can also be

used, with some qualifications, for other thin open sections such as shown in the

figure below

For example, the effective area moments for the cross sections shown can be calculated as



Also, the maximum shear stress formula can still be applied as

τ max = Tt max

J eff

t max

is the largest thickness of the cross section. However, this maximum shear stress occurs on the

outer edges of the thickest section and does not account for the stress concentrations that occur at re-

entrant corners such as those marked with aC

in Fig. 1. At such locations, the stresses depend on thelocal radius of curvature of the corner and may be considerably larger than the value predicted from Eq.

(1). Such stress concentrations can be taken into account by finding either numerically or experimentally

a stress concentration factor, K , for each re-entrant corner and then examining all high stress points and

choosing the one with the highest stress, i.e.

τ max = K Tt

J eff

⎡⎣⎢ ⎤

⎦⎥

max

where



0

0.5

1

1.5

2

2.5

3

0.5 1.0 1.5

r

t

maxτ τ

open section

r

t

maxτ τ

/ r t

maxK τ

τ =

closed section

Stress Concentrations



Centerline warping of thin open sections

r ⊥

ds

d Ω

xu φ ω ′= −

0r dsω ω ⊥= +∫ sectorial area function

O

center of twist

0

1

2

2 2

d r ds

d r ds

d

ω

ω ω

⊥

⊥

=

Ω =

= Ω = Ω +∫

T

since we are taking T as

positive counterclockwise,ω is positive if the area isswept out in acounterclockwise manner

ω + ω −



To locate the center of twist, O, we must have

0

0

y dA

z dA

ω

ω

=

=

∫∫

y and z are measured from the centroidof the cross section

To fix we can specify0ω

dA oω =∫

An ω satisfying all three of the above conditions is called a principalsectorial area function, ωp

x pu φ ω ′= −

torsion thin rectangle

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