Appendix C

Simple one-storey structure comprising four columns whose tops are connected by a rigid slab-diaphragm

In general, when a horizontal force H acts on a storey level, all the points of the slab including the column tops move in accordance with the same rules due to the in-plane rigidity of the slab.

These rules induce the diaphragm to develop a parallel (translational) displacement by δ_xo, δ_yo and a rotation θ_z about the a point C_T(x_CT, y_CT) , the centre of stiffness, in the xC_T y coordinate system, which has as origin the point C_T and is rotated with respect to the initial coordinate system X0Y by an angle a.

 

 

Χ0Υ     initial coordinate system
x’0y’     auxiliary coordinate system
xC_Ty    principal coordinate system

ζ= x·cosφ+y·sinφ η=-x·sinφ+y·cosφ   x=ζ·cosφ-η·sinφ y=ζ·sinφ+η·cosφ   x-vector: ζ=x·cosφ & η=-x·sinφ   y-vector: ζ=y·sinφ & η=y·cosφ

Relationships of axis transformation due to rotation

When a horizontal force H_x acts on the C_T along the x direction, the 3 following equilibrium equations should apply:

·         The sum of forces acting in x direction is be equal to H_x, i.e. H_x=Σ(V_xxoi)                        (i)

·         The sum of forces acting in y direction is equal to zero, i.e. Σ(V_xyoi)=0.0                         (ii)

·         The sum of moments V_xxoi and V_xyoi  about point C_T is equal to zero,      
           i.e. Σ(V_xxoi·y_i-V_xyoi·x_i)=0                                                                                                      (iii)

where V_xyoi=δ_xo·K_xyi with K_xyi=1/2·(K_ζi-K_ηi)·sin2φ’_i   and φ’_i=φ_i-a

 Expression (i) yields H_x=Σ(δ_xo·K_xxi)=δ_xo·Σ(K_xxi) →  where  

Because of the identity sin (w₁-w₂)=sinw₁·cosw₂-cosw₁·sinw₂, K_xyi=1/2·(K_ζi-K_ηi)·(sin2φ_i·cos2a-cos2φ_i·sin2a) equation (ii) becomes Σ(K_xyi)=0 → Σ[(K_ζi-K_ηi)·sin2φi·cos2a]=Σ[(K_ζi-K_ηi)cos2φ_i·sin2a] → cos2a·Σ[(K_ζi-K_ηi)·sin2φ_i]=sin2a·Σ[(K_ζi-K_ηi)·cos2φ_i] →

                                                                (3)

The angle a of the principal system is calculated from this expression. However, the coordinates x_CT, y_CT of C_T are still unknown. It is possible, though, to work on the auxiliary system x’0y’ which is parallel to the xC_Ty and has the advantage that its stiffnesses are the same with the ones of the final principal system xC_Ty. Therefore, coordinates X,Y are transferred to the x’,y’.

Example:

Calculation of angle a of the principal system :

Using the same values for K_ζ, K_η previously calculated and since 2φ₁=2φ₂=0˚, 2φ₃=60˚ and 2φ₄=90˚:

(3) → tan2a=Σ[(K_ζi-K_ηi)·sin2φ_i]/Σ[(K_ζi-K_ηi)·cos2φ_i] →

tan2a=[(31.1-31.1)·10⁶·sin0˚ +(31.1-31.1)·10⁶·sin0˚+(186.6-26.2)·10⁶·sin60˚+ (19.7-78.7)·10⁶·sin90˚]/

/[( 31.1-31.1)·10⁶·cos0˚+(31.1-31.1)·10⁶·cos0˚ + (186.6-26.2) ·10⁶·cos60˚+ (19.7-78.7)·10⁶·cos90˚]= =[0+0+138.9-59.0]/[0+0+80.2+0.0]=79.9/80.2=0.996 → 2a=44.88˚ → a=22.44˚

Therefore the angle of the principal system is a=22.44˚

Coordinate system transformation: From the initial system X0Y we are transferred to the auxiliary x’0y’, which is parallel to the principal system xC_Ty, therefore angle of the initial system is a=22.4˚:

 sina=0.3817, cosa=0.9243

C₁: from C₁(0,0) and φ₁=0˚ becomes C₁(0,0) φ’₁=-22.44˚

C₂: from C₂(6.0,0) and φ₂=0˚ becomes C₂(6.0·0.924=5.544, -6.0·0.3817=-2.290), φ’₂=-22.44˚

C₃: from C₃(0.0,5.0) and φ₃=30˚  becomes C3(5.0·0.3817=1.9085, 5.0·0.924=4.62), φ’₃=30˚-22.44˚ =7.56˚ → C₃(5.544, -2.290), φ’₃=7.56˚

C₄: from C₄(6.0, 5.0) and φ₄=45˚  becomes C₄(6.0·0.924+5.0·0.3817=7,4542, -6.0·0.3817+5.0·0.9243=2.331), φ’₄=45˚-22.44˚ =22.56˚ → C4(7.454, -2.331), φ’₄=22.56˚

C_M: from C_M(3.0, 2.5) becomes C_M(3.0·0.924+2.5·0.3817=3.727, -3.0·0.3817+2.5·0.9243=1.166)

 

Working on the auxiliary coordinate system x’0y’

x_i=x’_i-x’_CT and y_i=y’_i-y’_CT  therefore equation (iii) yields:

δ_xo·Σ((y’_i-y_o)·K_xxi-(x’_i- x’_CT)·K_xyi)=0 → Σ(y’_i·K_xxi)-y_o·Σ(K_xxi)-Σ(x’_i·K_xyi)+ x’_CT·Σ(K_xyi)=0 →

Σ((x’_i- x’_CT)·K_yyi-(y’_i- y’_CT)·K_xyi)=0 → -Σ(y’_i·K_xyi)+ y’_CT ·Σ(K_xyi)+Σ(x’_i·K_yyi)- x’_CT·Σ(K_yyi)=0.

Since Σ(K_xyi)=0, Σ(y’_i·K_xxi)- y’_CT·Σ(K_xxi)-Σ(x’_i·K_xyi)=0 → 

-Σ(y’_i·K_xyi)+Σ(x’_i·K_yyi)- x’_CT·Σ(K_yyi)=0 → 

  

 

Accordingly, taking into account that K_yxi=K_xyi, the corresponding expressions are:

 where ,       

Therefore the final relationships are:

,   where                             (4)

,   where                            (5)

Example:

Columns stiffnesses in the principal system xC_Ty:

Since C₁, C₂ have a square cross section → K_xx1=K_yy1=K_xx2=K_yy2=31.1·10⁶ N/m, K_xy1=K_xy2=0.0

C₃: φ’₃=7.56˚ (a) → K_xx3= K_ζ3·cos²φ₃+K_η3·sin²φ₃ =[186.6·0.9913²+26.2·0.1316²]·10⁶=183.82·10⁶ N/m

(b) →  K_xy3=(1/2)·(K_ζ3-K_η3)·sin2φ₃=0.5·(186.6-26.2)· 10⁶·0.261=20.93·10⁶ N/m

(d) → K_yy3= K_ζ3·sin²φ_i+K_η3·cos²φ₃ =[186.6·0.1316²+26.2·0.9913²] ·10⁶ N/m =28.98·10⁶ N/m

C4: φ’₄=22.56˚ (a) → K_xx4=K_ζ4·cos²φ₄+K_η4·sin²φ₄=[19.7·0.9235²+78.7·0,3837²] ·10⁶=28.39·10⁶ N/m

(b) →  K_xy4=(1/2)·(K_ζ4-K_η4)·sin2φ₄=0.5·(19.7-78.7)·10⁶·0.709=-20.92·10⁶ N/m

(d) → K_yy4= K_ζ4·sin²φ_i+K_η4·cos²φ₃ =[19.7·0,3837²+78.7·0.9235²] ·10⁶=70.02·10⁶ N/m

Total stiffnesses of the diaphragm and structural moments of the stiffnesses with respect to  C_T :

K_xx=Σ(K_xxi)=( 31.1+31.1+183.82+28.39) ·10⁶=274.41·10⁶ N/m

K_yy=Σ(K_yyi)= (31.1+31.1+28.98+70.02) ·10⁶=161.2·10⁶ N/m

Σ(x’_i·K_yyi)=[0.0·31.1+5.544·31.1+1.9085·28.98+7.4542·70.02]·10⁶ =749.55·10⁶ N

Σ(x’_i·K_xyi)=[0.0·0+5.544·0+1.9085·20.93+7.4542·(-20.93)]·10⁶=-116.07·10⁶ N

Σ(y’_i·K_xxi)=[0.0·31.1 -2.29·31.1 +4.62·183.82+2.331·28.39]·10⁶=844.21·10⁶ N

Σ(y’_i·K_xyi)= [0.0·0.0 -2.29·0.0 +4.62·20.93+2.331· (-20.92)]·10⁶=47.93·10⁶ N

x’_CK=[Σ(x’_i·K_yyi)-Σ(y’_i·K_xyi)]/Σ(K_yyi)=( 749.55-47.93)/161.2=4.353 m

y’_CK=[Σ(y’_i·K_xxi)-Σ(x’_i·K_xyi)]/Σ(K_xxi)=( 844.21+116.07)/274.41=3.500 m

 

Displacements due to rotation from a moment M applied in CT

Displacements of a random point i of the diaphragm due to rotation by θz δi=ri·θz δxi =-δi·sinωi sinωi =yi /ri δxi =-ri·θz·yi/ri=-yi·θz δyi=xi·θz

Consider the principal coordinate system xC_Ty

The deformation induced by the external moment M applied in C_T , is examined. In order to examine this deformation a transfer is needed from the auxiliary system x’0y’ to the principal system xC_Ty by a simple parallel translation.  By also transferring the centre of mass in the principal system as well, the structural eccentricities e_ox, e_oy of C_M with respect to C_T, derive from the expressions:

 ,  , ,                                        (6)

The deformation of the diaphragm is a rotation θ_z about C_T. Rotation θ_z of diaphragm induces a displacement δ_i at the top of each column i with coordinates x_i,y_i with respect to the coordinate system having as origin C_T. If the distance of point i from C_T is r_i, the two components of the (infinitesimal) deformation δ_i are: δ_xi=-θ_z·y_i and δ_yi= θ_z·x_i

Example:

Coordinate transfer in the principal system and calculation of the structural eccentricities :

Expressions (6) yield:

C₁(0.0-4.353=-4.35, 0.0-3.500=-3.50) φ’₁=-22.44˚                K_xx=31.1·10⁶, K_yy=31.1·10⁶, K_xy=0.0

C₂(5.544-4.353=1.19, -2.290-3.500=-5.79), φ’₂=-22.44˚                   K_xx=31.1·10⁶, K_yy=31.1·10⁶, K_xy=0.0

C₃(1.9085-4.353=-2.44, 4.62-3.500=1.12), φ’₃=7.56˚                        K_xx=183.82, K_yy=28.98, K_xy=20.93

C₄(7,4542-4.353=3.10, 2.3312-3.500=-1.17), φ’₄=22.56˚      K_xx=28.39, K_yy=70.02, K_xy=-20.92

C_M(3.727-4.353=-0.626, 1.166-3.500=-2.334) and obviously C_T(0,0)

e_ox=-0.626 m, e_oy=-2.334 m

 

Displacements δ_xi, δ_yi incuce shear forces V_xi and V_yi in each column:

V_xi=K_xxi·δ_xi= K_xxi·(-θ_z·y_i ) → V_xi= -θ_z·K_xxi·y_i and            V_yi=K_yyi·δ_yi=k_yyi·θ_z·x_i →            V_yi=θ_z·k_yyi·x_i

The resultant of the moments of all shear forces V_xi, V_yi with respect to the C_T should be equal to the external moment M_CT i.e.:

M_CT=Σ[-V_xi·y_i + V_yi·x_i+K_zi] → M_CT= θ_z·Σ[K_xxi·y_i² + K_yyi·x_i²+K_zi]  →

  where                                                        (7)

Example:

Calculation of diaphragm torsional stiffness:

Κ_θ=Σ(K_xxi·y_i²+K_yyi·x_i²+Kzi)=Σ[31.1·3.5²+31.1·4.35²+31.1·5.79²+31.1·1.19²+183.82·1.12²+ 28.98·2.44²+28.39·1.17²+70.02·3.10²]·10⁶ N·m= =[381.0+588.5+1042.6+44.0+230.6+172.5+38.9+672.9] ·10⁶ N·m → Κ_θ=3174·10⁶ N·m

The quantity Κ_θ=Σ(K_xxi·y_i² + K_yyi·x_i²+K_zi) is called torsional stiffness of the diaphragm, measured in units of moment e.g. N·m, by analogy with the quantities K_x=Σ(K_xi), K_y=Σ(K_yi) which are called lateral stiffnesses of the diaphragm in direction x and y respectively and measured in N/m. The torsional stiffness of the column itself is usually insignificant and can be ignored (§5.4.3.4).

Definitions:

Lateral stiffness Κ_jk of diaphragm denotes the force in j direction required to cause displacement of the diaphragm by one unit in the k direction.

Torsional stiffness Κ_θ of diaphragm denotes the moment required to cause rotation of the diaphragm by one unit.

  

Question: Create a simple idealised equivalent structural system with the same seismic behaviour as the actual structural system.

Solution: Four idealised columns E₁, E₂ and E₃, E₄ are placed symmetrically with respect to the centre C_T and to the axes x and y, i.e. all four idealised columns have x and y coordinates with the same absolute values. Each idealised column is assumed to have a stiffness equal to K_x=1/4·Σ(K_xi) and K_y=1/4·Σ(K_yi).

This system satisfies the two conditions of the actual system, concerning the translations of all diaphragm columns since:

Stiffness by x: 4·1/4·Σ(K_xi)=Σ(K_xi) and stiffness by y: 4·(1/4)·Σ(K_yi)=Σ(K_yi).

To satisfy the third condition, the torsional stiffness of the idealised system should be

K_θ,eq=[K_x1·y²+K_x2·y²+K_y1·x²+K_y2·x²]=Σ(K_xxi)·y²+Σ(K_yyi) ·x²

equal to the torsional stiffness of the actual system

K_θ,re=K_θ=Σ(K_xxi·y_i² + K_yyi·x_i²-2K_xyi·x_i·y_i+K_zi),

i.e. should K_θ,eq= K_θ,re à Σ(K_xxi)·y²+Σ(K_yyi)·x²=K_θ à

                            where  and                          (8)

The curve equation (8) is an ellipse with centre C_T and semiaxes r_x, r_y oriented along the principal axes

Example C.6:

Calculation of the torsional stiffness ellipse

r_x=√(K_θ/K_yy)=√(3134·10⁶N·m /161.2·10⁶N/m)=4.41 m

r_y=√(K_θ/K_xx)=√(3134·10⁶ N·m /274.41·10⁶N/m)=3.38 m

Conclusion:

The torsional behaviour of a floor can be described by the torsional stiffness ellipse (C_T, r_x, r_y) which represents the equivalent distribution of the diaphragm stiffness.

The radii r_x, r_y, of the ellipse are called torsional radii.

There are infinite solutions of idealised dual system sets, whose most characteristic is the one with the four idealised columns placed in the four ellipse ends.

Generally, there are infinite solutions with n-tuple diametrically opposed systems, where each idealised column stiffness is equal to 1/n of the total system stiffness.

 

So far, all the calculations depended on the structure geometry and were not affected by the magnitude of the external loading. For instance the centre of stiffness, the structural eccentricities, and the torsional radii, are independent of the seismic force magnitude.

Next, the displacements and stress resultants of the structure, due to external seismic loading H, will be calculated.

The relevant seismic force H is applied at the centre of mass C_M of the diaphragm. This force is resolved in two forces H_x and H_y parallel to the two axes of the principal system. In order to apply the previous analysis, the forces H_x, H_y are transferred to the centre of stiffness C_T together with the moment M_CT according to the expression:

                                                                                                        (9)

Example C.7-1: Given a horizontal force in x direction equal to is H=90.6 kN, carry-out the analysis of the structure.

Calculation of moment M_CT :

The structural eccentricities are e_ox =-0.626 and e_oy= y_CM=-2.334m

For a given force Η=H_X=90.6 kN in the initial system X0Y, the equivalent loading in the principal system (a=22.44^ο) is: H_x=H_X·cosa=90.6·0.924=83.72 kN and H_y=-H_X·sina=-90.6·0.382=-34.64 kN and

Μ_CT=-H_x·e_oy+H_y·e_ox=-83.72kN·(-2.334m)+[-34.64kN·(-0.626m)]=195.4 kNm+21.7kNm=217.1 kNm

 

The following quantities are calculated using the H_x, H_y, M_CT:

·         the displacements  δ_xo, δ_yo and θ_z of C_T using the expressions:

              ,  ,                                                                          (10)

where  , ,  

Example C.7-2:

Quantities K_xx=Σ(K_xxi)= 161.2·10⁶N/m, K_yy=Σ(K_yyi)= 161.2·10⁶ N/m and Κ_θ=3134·10⁶ N·m are already calculated therefore:

δ_xo=H_x/Κ_xx =83.72·10³N/(274.41·10⁶N/m)=0.305 mm,

δ_yo=H_y/Κ_yy =-34.46·10³N/(161.2·10⁶ N/m)=- 0.214 mm and

θ_z=M_CT/K_θ=217.1kNm/(3134·10³kNm)=0.692·10^-4 

·         the displacements δ_xi, δ_yi of the top of each column using the expressions:

,                                                 (11)

·         and the displacements δ_ζi, δ_ηi by transferring δ_xi, δ_yi to the local system of each column using the expressions:

,   where                        (12)

Example C.7-3:

C₁: δ_x1= δ_xo- θ_z·y₁=0.305mm-0.692·10^-4·(-3.50·10³mm)=(0.305+0.242)mm=0.547 mm

δ_y1= δ_yo+ θ_z·x₁=-0.214mm+0.692·10^-4·(-4.35·10³mm)=(-0.214-0.301)mm=-0.515 mm

By transferring to the local system where φ’₁=0.0-22.44°=-22.44°.

δ_ζ1= δ_x1·cosφ’₁+δ_y1·sinφ’₁=0.547·0.924-0.515·(-0.382)=0.505+0.197=0.702 mm

δ_η1=-δ_x1·sinφ’₁+δ_y1·cosφ’₁=-0.547·(-0.382)+(-0.515)·0.924=0.209-0.476=-0.267 mm

C₂: δ_x2= δ_xo- θ_z·y₂=0.305mm-0.692·10^-4·(-5.79·10³mm)=(0.305+0.400)mm=0.705 mm

δ_y2= δ_yo+ θ_z·x₂=-0.214mm+0.692·10^-4·1.19·10³mm=(-0.214+0.082)mm=-0.132  mm

By transferring to the local system where φ’₂=0.0-22.44°=-22.4°.

δ_ζ2= δ_x2·cosφ’₂+δ_y2·sinφ’₂=0.705·0.924+(-0.291)·(-0.132)=0.652+0.038=0.701 mm

δ_η2=-δ_x2·sinφ’₂+δ_y2·cosφ’₂=-0.705·(-0.382)+(-0.132)·0.924=0.269-0.122=0.147 mm

C₃: δ_x3= δ_xo- θ_z·y₃=0.305mm-0.692·10^-4·1.12·10³mm=(0.305-0.078)mm=0.227 mm

δ_y3= δ_yo+ θ_z·x₃=-0.214mm+0.692·10^-4·(-2.44·10³mm)=(-0.214-0.169)mm=-0.383 mm

By transferring to the local system where φ’₃=30.0-22.44°=7.56°.

δ_ζ3= δ_x3·cosφ’₃+δ_y3·sinφ’₃=0.227·0.991+(-0.383)·0.132=0.225-0.050=0.175 mm

δ_η3=-δ_x3·sinφ’₃+δ_y3·cosφ’₃=-0.227·0.132+(-0.383)·0.991=-0.030-0.380=-0.410 mm

C₄: δ_x4= δ_xo- θ_z·y₄=0.305mm-0.692·10^-4·(-1.17·10³mm)=(0.305+0.081)mm=0.386 mm

δ_y4= δ_yo+ θ_z·x₄=-0.214mm+0.692·10^-4·3.10·10³mm=(-0.214+0.215)mm=0.001 mm

By transferring to the local system where φ’₄=45.0-22.44°=22.56°.

δ_ζ4= δ_x4·cosφ’₄+δ_y4·sinφ’₄=0.386·0.923+0.001·0.384=0.355+0.000=0.355 mm

δ_η4=-δ_x4·sinφ’₄+δ_y4·cosφ’₄=-0.386·0.384+0.001·0.923=-0.148+0.001=-0.147 mm

In the example considered, for seismic action in X direction, the deformation due to rotation at column C₂ gives δ_x2,θ=0.400 mm, higher than the deformation  due to translation δ_xo=0.305 mm. The total displacement is therefore δ_x2=0.305+0.400=0.705 mm.

     Figure C.7: Moment distribution (Mji,1- Mji,2=Vji·h)

·        shear forces and bending moments of each column in its local system based on the relationships

                                                 ,                                                   (13)

                ,                              (14)

            ,  

 

 

Example C.7-4:

Calculation of shear forces and bending moments :

The moment distribution factor is assumed the same in both directions a_ζi=a_ηi=0.50

C₁: V_ζ1=δ_ζ1·K_ζ1=0.702mm·31.1·10⁶ N/m =21.8 kN

V_η1=δ_η1·K_η1=-0.267mm·31.1·10⁶ N/m =-8.3 kN

M_ζ1,1 =V_ζ1·h·0.50=21.8·3.0·0.50=32.7 kNm              M_ζ1,2=- M_ζ1,1=-32.7 kNm

M_η1,1=V_η1·h·0.50=-8.3·3.0·0.50=-12.5 kNm              M_ηι,2=- M_η1,1=12.5 kNm

C₂: V_ζ2=δ_ζ2·K_ζ2=0.701mm·31.1·10⁶ N/m =21.8 kN

V_η2=δ_η2·K_η2=0.147mm·31.1·10⁶ N/m =4.6 kN

M_ζ2,1=V_ζ2·h·0.50=21.8·3.0·0.50=32.7 kNm               M_ζ2,2=- M_ζ2,1=-32.7 kNm

M_η2,1=V_η2·h·0.50=4.6·3.0·0.50=6.9 kNm                  M_η2,2=- M_η2,1=-6.9 kNm

C₃: V_ζ3=δ_ζ3·K_ζ3=0.175mm·186.6·10⁶ N/m =32.7 kN

V_η3=δ_η3·K_η3=-0.410mm·26.2·10⁶ N/m =-10.8 kN

M_ζ3,1=V_ζ3·h·0.50=32.7·3.0·0.50=49.1 kNm               M_ζ3,2=- M_ζ3,1=-49.1 kNm

M_η3,1=V_η3·h·0.50=-10.8·3.0·0.50=-16.2 kNm                        M_η3,2=- M_η3,1=16.2 kNm

C₄: V_ζ4=δ_ζ4·K_ζ4=0.355mm·19.7·10⁶ N/m =7.0 kN

V_η4=δ_η4·K_η2=-0.147mm·78.7·10⁶ N/m =-11.6 kN

M_ζ4=V_ζ4·h·0.50=7.0·3.0·0.50=10.5 kNm                   M_ζ4,2=- M_ζ4,1=-10.5 kNm

M_η4,1=V_η4·h·0.50=-11.6·0·0.50=-17.4 kNm               M_η4,2=- M_η4,1=17.4 kNm

Frame and model of a single-storey space frame

The analysis of the one-storey space frame illustrated in the figure, under horizontal seismic force H=90.6 kN is performed by means of four methods: (i) analysis using manual calculations, assuming fixed-ended columns, (ii) analysis using the Excel file, assuming fixed-ended columns, (iii) analysis using the Excel file, assuming columns with k=6, (iv) assuming using software, assuming actual beam and column torsional stiffnesses.

(i)            Analysis using manual calculations, assuming fixed columns (k=12)

This method was described via the example presented to all previous paragraphs of this appendix.

(ii)          Analysis using the Excel file, assuming fixed-ended columns (k=12)

The related <diaphragm_general.xls> spreadsheet is used assuming fixed-ended columns.

The results are identical to those of the practical calculations.

The torsional stiffness ellipse and the equivalent columns of the structure are drafted at the end of the spreadsheet.

(iii)         Analysis using the Excel file, assuming columns with k=6

The centre of stiffness, the torsional radii and the stress resultants derived from this method are exactly the same as those of the previous methods, with the exception of deformations, whose their relative values, however, remain constant 12/6=2.00.

(iv)         Analysis using software, assuming actual beam and column torsional stiffnesses

In order to define the diaphragmatic function of a building’s storey, with actual stiffness and not with assumption of fixed function of its columns, it is necessary to use the appropriate software. Structural analysis software is necessary for one storey, as well as multistorey buildings, even in case of rectangular columns in parallel arrangement.

The method which describes the diaphragmatic behaviour of each floor is presented in Appendix D and this example is analysed as example 1.