Laminate generator

Results.

Example	Reference	Records	Largest deviation
Homogeneous section, layered shear integral	Published source	1	0%
Five-layer CLT, along the longitudinal layer direction	Published source	1	0%
CLT strip section, mass consistency	Closed form	1	0%
Homogeneous section, layer stress	Closed form	1	0%
Homogeneous section, transverse shear distribution	Closed form	1	0%
Equilibrium identities, CLT section	Closed form	1	0%
Equilibrium identities, homogeneous section	Closed form	1	0%
Five-layer CLT, continuity of the transverse shear	Closed form	1	0%

Problem. The generator turns a ply lay-up into section stiffness. The shear integral, the beam formula and the equilibrium identities are tested on a homogeneous section, and the rolling shear in the cross plies on CLT.

Reference document. Classical lamination theory is taken from Jones, Mechanics of Composite Materials. The transverse shear stiffness is built from energy equivalence. For a homogeneous rectangle the shear factor must come out as five sixths.

In classical lamination theory the membrane, coupling and bending matrices are built by summing the contributions of the layers.

A = \sum \bar{Q}_k t_k,\qquad B = \sum \bar{Q}_k t_k \bar{z}_k,\qquad D = \sum \bar{Q}_k \left(t_k \bar{z}_k^2 + \dfrac{t_k^3}{12}\right)

The transverse shear stiffness of a layered section is computed from this relation. $S(z)$ is the first moment of area above the level.

GA = \dfrac{(EI)^2}{\displaystyle\int \dfrac{S(z)^2}{G(z)}\,dz}

The integral of the layer stresses through the thickness must equal the section forces and moments.

\int \sigma_{xx}\,dz = N_{xx},\qquad \int \sigma_{xx}\,z\,dz = M_{xx},\qquad \int \tau_{xz}\,dz = Q_x

Numerical model. There is no finite element model, and the compared quantities are read directly from the section matrices and ply stresses. The CLT lay-up is five 20 mm plies, 0/90/0/90/0.

Homogeneous section, layered shear integral

Parameters. t = 0.18 · layer = 3

Published source. The Timoshenko shear correction factor, rectangular section

\kappa = 5/6

Quantity	Reference	Computed	Deviation
Shear stiffness $GA$ , per unit width	576,923	576,923	0%

Note. A homogeneous section divided into three equal plies is used.

Five-layer CLT, along the longitudinal layer direction

Parameters. layup = 0/90/0/90/0 · layer thickness = 0.02

Published source. Classical lamination theory

D = \sum \bar{Q}_k\,(t z^2 + t^3/12)

Quantity	Reference	Computed	Deviation
Bending stiffness $D_{11}$	765.413	765.413	0%

Note. With the Poisson ratio taken as zero, $\bar{Q}_{11} = E$ . The cross layers enter with the transverse modulus.

CLT strip section, mass consistency

Parameters. layup = 0/90/0/90/0

Closed form. Conservation of mass

\sum \rho_k t_k b

Quantity	Reference	Computed	Deviation
Equivalent density times the transformed area	0.42	0.42	0%

Homogeneous section, layer stress

Parameters. t = 0.2 · N = 250 · M = 30

Closed form. The beam formula

N/A + Mc/I

Quantity	Reference	Computed	Deviation
Stress in the topmost fibre	5,750	5,750	0%

Note. A homogeneous section divided into four equal layers is used.

Homogeneous section, transverse shear distribution

Parameters. t = 0.2 · Q = 40

Closed form. The maximum shear stress in a rectangular section

\tau_{max} = 3V/2A

Quantity	Reference	Computed	Deviation
Shear stress at the mid-plane	300	300	0%

Note. The value comes out exactly zero at the faces.

Equilibrium identities, CLT section

Parameters. layup = CLT

Closed form. An equilibrium equation independent of the constitutive relation

Quantity	Reference	Computed	Deviation
$\int \sigma\,dz$ as a ratio of $N_{xx}$	120	120	0%

Equilibrium identities, homogeneous section

Parameters. layup = homogeneous

Closed form. An equilibrium equation independent of the constitutive relation

Quantity	Reference	Computed	Deviation
$\int \sigma\,dz$ as a ratio of $N_{xx}$	120	120	0%

Five-layer CLT, continuity of the transverse shear

Parameters. Q = 30 · layup = 0/90/0/90/0

Closed form. Equilibrium requires the shear stress to be continuous across the interface

Quantity	Reference	Computed	Deviation
Jump at the layer interface	0	$1.137 \times 10^{-13}$	0%

Note. The peak rolling shear is 366.4 in the cross ply and 388.9 in the longitudinal ply, and it is zero at the faces.

Homogeneous section, layered shear integral#

Five-layer CLT, along the longitudinal layer direction#

CLT strip section, mass consistency#

Homogeneous section, layer stress#

Homogeneous section, transverse shear distribution#

Equilibrium identities, CLT section#

Equilibrium identities, homogeneous section#

Five-layer CLT, continuity of the transverse shear#