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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:

  • Top:

NxT=Nx/2, NyT=Ny/2, -NxyT=-Nxy/2

  • Bottom: 

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

  • T1: Determine the generalized Clark-Nielsen forces: 

NxT1=NxT+2*NxyT*cotÝ+NyT*(cotÝ)^2+ABS((NxyT+NyT*cotÝ)/sinÝ) 

NÝT1=NyT/((sinÝ)^2)+ABS((NxyT+NyT*cotÝ)/sinÝ)

  • T2: Suppose NxT1<0; top layer is in X-direction compression.

NxT2=0 

NÝT2=(1/(sinÝ)^2)*(NyT+ABS((NxyT+NyT*cotÝ)^2/(NxT+2*NxyT*cotÝ+NyT*(cotÝ)^2)))

  • T3: Suppose NÝT1<0; top layer is in Ý-direction compression.

NxT3=NxT+2*NxyT*cotÝ+NyT*(cotÝ)^2+ABS((NxyT+NyT*cotÝ)^2/NyT) 

NÝT3=0 

  • T4: Select the appropriate values from the above for output. Note that if NxT1<0 and NÝT1<0, then no top reinforcement is required.

Nx(T)=IF(NÝT1<0,IF(NxT3<0,0,NxT3),IF(NxT1<0,0,NxT1)) 

NÝ(T)=IF(NxT1<0,IF(NÝT2<0,0,NÝT2),IF(NÝT1<0,0,NÝT1))

Bottom reinforcement

  • Since NxT=NxB, NyT=NyB, NxyT=NxyB in this case, the top and bottom layer equations are identical.

Concrete forces (top)

  • F1: Determine the generalized Clark-Nielsen concrete force.

FcT1=IF(FcTI<0,-2*FcTI*(cotÝ-cscÝ),-2*FcTI*(cotÝ+cscÝ)) 

where FcTI=NxyT+NyT*cotÝ 

  • F2: Suppose NxT1<0; top layer is in X-direction compression.

FcT2=((NxT+NxyT*cotÝ)^2+(NxyT+NyT*cotÝ)^2)/(NxT+2*NxyT*cotÝ+NyT*(cotÝ)^2)

  • F3: Suppose NÝT1<0; top layer is in Ý-direction compression.

FcT3=NyT+(NxyT^2/NyT)

  • F4: Suppose Nx(T)=0 and NÝ(T)=0, there is no cracking and "concrete force parallel to the crack" has no meaning. A reasonable value to give is the second (most compressive) principle stress: 

FcTP=(NxT+NyT)/2-SQRT(((NxT-NyT)/2)^2+NxyT^2)

  • F5: Select the appropriate values from the above for output. 

Fc(T)=IF(AND(NÝ(T)=0,Nx(T)=0),FcTP,IF(NÝT1<0,FcT3,IF(NxT1<0,FcT2,FcT1)))

Concrete forces (bottom)

  • Since NxT=NxB, NyT=NyB, NxyT=NxyB in this case, the top and bottom layer equations are identical. 

Note that for orthogonal reinforcement, Ý=90į.

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.


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