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Clark-Nielsen (Wood-Armer for in-plane forces)
The Wood-Armer
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 in-plane force triad consisting
of two in plane forces per unit length (Nx, Ny) and an in-plane
shear force per unit length (Nxy). These equations are
generally known as the Clark-Nielsen 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
Clark-Nielsen calculation may be carried out by
following the procedure below. All calculations proceed after the determination of the
in-plane stress field Nx, Ny, Nxy. For an elastic section subject only to in-plane 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*(cotø-cscø),-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)/2-SQRT(((NxT-NyT)/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 Clark-Nielsen
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 Clark-Nielsen 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 Clark-Nielsen 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
|
Clark-Nielsen 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
Clark-Nielsen
calculations for this example are satisfactory.
Other Wood-Armer
related topics
|