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CLT strip, shell against frame

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Results.

ExampleReferenceRecordsLargest deviation
Five-layer CLT, L/t = 15Independent calculation10.0277%
Five-layer CLT, L/t = 25Independent calculation10.0340%
Five-layer CLT, L/t = 40Independent calculation10.0369%
2D plane stress model, homogeneous beam, L/t = 15Published source10.0012%
2D plane stress model, five-layer CLT, L/t = 15Independent calculation10.495%
2D plane stress model, five-layer CLT, L/t = 40Independent calculation10.0511%
Unsymmetric layup, frame axisIndependent calculation10.282%
Cantilever strip, 0/90 CLT layupClosed form10%
Timber creep, kdef=0.8k_{def} = 0.8Published source10%

Problem. A CLT strip can be modelled as a laminated shell or as a frame with a transformed section. The tests check that the two models agree, and that they diverge for a lay-up with angled plies.

Reference document. The reference is a two-dimensional plane stress model built ply by ply in the X-Z plane. That model is itself checked against the Timoshenko closed form for a homogeneous beam. For creep the reference is Eurocode 5.

The reference model is checked against Timoshenko beam theory, which gives the bending and shear deflection together.

δ=5wL4384 EI+wL28 GAv\delta = \dfrac{5wL^4}{384\,EI} + \dfrac{wL^2}{8\,GA_v}

Creep is applied to the modulus of elasticity with the Eurocode 5 relation. The reduction affects the stiffness and leaves the mass unchanged.

Emean,fin=Emean1+kdefE_{mean,fin} = \dfrac{E_{mean}}{1 + k_{def}}

Numerical model. The shell strip is solved on three meshes and the frame strip with four shear-deformable elements. The reference model's result is the Richardson extrapolation of two meshes.

Five-layer CLT, L/t = 15

Parameters. L = 1.5 · t = 0.1 · layup = 0/90/0/90/0

Independent calculation. Internal consistency, the same layup solved through two element families

QuantityReferenceComputedDeviation
Shell deflection as a ratio of the frame deflection, 64 elements10.9997230.0277%

Note. On the three meshes the ratio is 0.99557, 0.99889 and 0.99972.

Five-layer CLT, L/t = 25

Parameters. L = 2.5 · t = 0.1 · layup = 0/90/0/90/0

Independent calculation. Internal consistency, the same layup solved through two element families

QuantityReferenceComputedDeviation
Shell deflection as a ratio of the frame deflection, 64 elements10.999660.0340%

Note. On the three meshes the ratio is 0.99455, 0.99864 and 0.99966.

Five-layer CLT, L/t = 40

Parameters. L = 4 · t = 0.1 · layup = 0/90/0/90/0

Independent calculation. Internal consistency, the same layup solved through two element families

QuantityReferenceComputedDeviation
Shell deflection as a ratio of the frame deflection, 64 elements10.9996310.0369%

Note. On the three meshes the ratio is 0.99409, 0.99852 and 0.99963.

2D plane stress model, homogeneous beam, L/t = 15

Parameters. mesh = 192×4 and 384×8, Richardson extrapolation

Published source. Timoshenko beam theory

5wL4/384EI+wL2/8GAv5wL^4/384EI + wL^2/8GA_v
QuantityReferenceComputedDeviation
Mid-span deflection0.0007977660.0007977560.0012%

Note. The drawing shows the coarse mesh.

2D plane stress model, five-layer CLT, L/t = 15

Parameters. mesh = 192×4 and 384×8, Richardson extrapolation · L = 1.5

Independent calculation. A two-dimensional plane-stress model, orthotropic layer by layer

QuantityReferenceComputedDeviation
Mid-span deflection0.001214470.001208460.495%

Note. The difference lies between beam theory and the two-dimensional model and grows at short spans. The drawing shows the coarse mesh.

2D plane stress model, five-layer CLT, L/t = 40

Parameters. mesh = 192×4 and 384×8, Richardson extrapolation · L = 4

Independent calculation. A two-dimensional plane-stress model, orthotropic layer by layer

QuantityReferenceComputedDeviation
Mid-span deflection0.04606150.0460380.0511%

Note. The difference lies between beam theory and the two-dimensional model and grows at short spans. The drawing shows the coarse mesh.

Unsymmetric layup, frame axis

Parameters. layup = 0/90/0/90 · z₀ = -0.01172

Independent calculation. A 2D plane stress model built ply by ply

QuantityReferenceComputedDeviation
Mid-span deflection, on the elastic centroidal axis0.005889050.005905630.282%

Cantilever strip, 0/90 CLT layup

Parameters. layup = 0/90 CLT · D₁₆/D₁₁ = 0 · L = 2

Closed form. With no angled layer present, the bending-twisting coupling must be exactly zero

QuantityReferenceComputedDeviation
Twist gradient at the tip as a ratio of the bending gradient01.004×10−141.004 \times 10^{-14}0%

Note. The CLT lay-up has no angled plies, so the two models do not diverge.

Timber creep, kdef=0.8k_{def} = 0.8

Parameters. kdefk_{def} = 0.8

Published source. The Eurocode 5 creep relation

Emean,fin=Emean/(1+kdef)E_{mean,fin} = E_{mean}/(1 + k_{def})
QuantityReferenceComputedDeviation
Long-term deflection as a ratio of the instantaneous one1.81.80%