Clark-Nielsen (Wood Armer for in plane forces)

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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*cotų+NyT*(cotų)^2+ABS((NxyT+NyT*cotų)/sinų)

NųT1=NyT/((sinų)^2)+ABS((NxyT+NyT*cotų)/sinų)

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

NxT2=0

NųT2=(1/(sinų)^2)*(NyT+ABS((NxyT+NyT*cotų)^2/(NxT+2*NxyT*cotų+NyT*(cotų)^2)))

T3: Suppose NųT1<0; top layer is in ų-direction compression.

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

NųT3=0

T4: Select the appropriate values from the above for output. Note that if NxT1<0 and NųT1<0, then no top reinforcement is required.

Nx(T)=IF(NųT1<0,IF(NxT3<0,0,NxT3),IF(NxT1<0,0,NxT1))

Nų(T)=IF(NxT1<0,IF(NųT2<0,0,NųT2),IF(NųT1<0,0,NųT1))

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*(cotų-cscų),-2*FcTI*(cotų+cscų))

where FcTI=NxyT+NyT*cotų

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

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

F3: Suppose NųT1<0; top layer is in ų-direction compression.

FcT3=NyT+(NxyT^2/NyT)

F4: Suppose Nx(T)=0 and Nų(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(Nų(T)=0,Nx(T)=0),FcTP,IF(NųT1<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, ų=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 (LUSAS mdl file)

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.