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ClarkNielsen (WoodArmer for inplane forces)
The WoodArmer
equations
resolve bending and twisting moments for the purpose of RC slab
design. Similar equations may be derived to resolve the
forces required to resist an inplane force triad consisting
of two in plane forces per unit length (Nx, Ny) and an inplane
shear force per unit length (Nxy). These equations are
generally known as the ClarkNielsen equations.
For a full explanation and derivation of the formulae, the
reader is referred to either:
 "Concrete slabs: analysis and design" L.A Clark
and R.J Cope (Elsevier Applied Science)
 "Concrete Bridge Design to BS5400" L.A Clark
(Construction Press) Chapter 5 (section entitled
"Reinforced Concrete Plates") and Appendix A
By reference to
"Concrete Bridge Design to BS5400" Appendix A, the
ClarkNielsen calculation may be carried out by
following the procedure below. All calculations proceed after the determination of the
inplane stress field Nx, Ny, Nxy. For an elastic section subject only to inplane forces, the
stress field is:
NxT=Nx/2, NyT=Ny/2, NxyT=Nxy/2
NxB=Nx/2, NyB=Ny/2, NxyB=Nxy/2
The
sign convention for Nxy differs between LUSAS and "Concrete
Bridge Design to BS5400", and so the sign for Nxy is
reversed. The equations are described in a form that can be
translated directly to a spreadsheet format:
Top
reinforcement
NxT1=NxT+2*NxyT*cotø+NyT*(cotø)^2+ABS((NxyT+NyT*cotø)/sinø)
NøT1=NyT/((sinø)^2)+ABS((NxyT+NyT*cotø)/sinø)
NxT2=0
NøT2=(1/(sinø)^2)*(NyT+ABS((NxyT+NyT*cotø)^2/(NxT+2*NxyT*cotø+NyT*(cotø)^2)))
NxT3=NxT+2*NxyT*cotø+NyT*(cotø)^2+ABS((NxyT+NyT*cotø)^2/NyT)
NøT3=0
Nx(T)=IF(NøT1<0,IF(NxT3<0,0,NxT3),IF(NxT1<0,0,NxT1))
Nø(T)=IF(NxT1<0,IF(NøT2<0,0,NøT2),IF(NøT1<0,0,NøT1))
Bottom
reinforcement
Concrete
forces (top)
FcT1=IF(FcTI<0,2*FcTI*(cotøcscø),2*FcTI*(cotø+cscø))
where
FcTI=NxyT+NyT*cotø
FcT2=((NxT+NxyT*cotø)^2+(NxyT+NyT*cotø)^2)/(NxT+2*NxyT*cotø+NyT*(cotø)^2)
FcT3=NyT+(NxyT^2/NyT)
FcTP=(NxT+NyT)/2SQRT(((NxTNyT)/2)^2+NxyT^2)
Fc(T)=IF(AND(Nø(T)=0,Nx(T)=0),FcTP,IF(NøT1<0,FcT3,IF(NxT1<0,FcT2,FcT1)))
Concrete
forces (bottom)
Note
that for orthogonal reinforcement, ø=90°.
Simple example
A simple example may be used to demonstrate the ClarkNielsen
calculation for orthogonally placed (minimised area)
reinforcement. The example used here is a a shell subjected to an
unsymmetrical tensile load; a thin rectangular slab fixed in
translation at one edge and with a point load applied to an
opposing corner. No
bending field (Mx, My, Mxy) will be generated, while a stress
field will be generated where various combinations of positive
and negative stresses for Nx, Ny and Nxy exist, enabling the
verification of the ClarkNielsen calculations for a variety of
conditions. The
calculations for Top and Bottom face reinforcement will produce
the same result and so a single hand calculation will verify
both results.
 Rectangular
surface, plan dimensions length 16 units, width 10 units
 Mesh attributes: Any quadrilateral shell element (QSI4
elements used in subsequent calcs) regular mesh of element size 1 unit
 Geometric attributes: thickness 0.2 units
 Material attributes: E=1E6, poissons ratio=0.3
 Supports:
fixed
in translation on bottom edge
 Loading attributes: Structural
load, Concentrated in Y direction 100 units total
Download
Clark Nielsen example model (for LUSAS Version 21)
Download
Clark Nielsen example model (for earlier LUSAS versions)
Stress field from LUSAS Modeller, extracted at 4 nodes for
calculations to be checked explicitly:
Node

29

161

129

183

Nx

87.673

4.765

2.804

11.192

Ny

29.469

47.202

7.529

0.574

Nxy

31.530

9.839

0.321

4.694

Calculation of ClarkNielsen stresses by hand, determined
from the stress field (Nx, Ny, Nxy) using the procedure
explained above.
Download
spreadsheet calculations (MSExcel format)
Node

29

161

129

183

Nx(T)=Nx(B)

0.000

7.302

1.563

0.000

Ny(T)=Ny(B)

0.000

28.520

3.925

0.698

Fc(T)=Fc(B)

50.740

9.839

0.321

6.581

ClarkNielsen stresses from LUSAS
Modeller, extracted at the
same 4 nodes for comparison to the hand calculations undertaken:
Node

29

161

129

183

Nx(B)

0.000

7.302

1.563

0.000

Ny(B)

0.000

28.520

3.925

0.698

Fc(B)

50.740

9.839

0.321

6.581

By inspection the results tabulated above agree closely with
those derived by hand calculation and demonstrate that the
ClarkNielsen
calculations for this example are satisfactory.
Other WoodArmer
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