DIAPHRAGMATIC BEHAVIOUR OF ONE-STOREY SPACE FRAME – GENERAL CASE

C.1 Subject description

Figure C.1-1: Simple one-storey structure comprising four columns whose tops are connected by a rigid slab-diaphragm.

Figure C.1-1: 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 [1] [*]Note Hereafter, the term “column” implies both columns and walls. 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 rotationθ_z about the a point C_T(x_CT, y_CT), the centre of stiffness, in the xC_Ty 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.

To better understand this chapter it is useful to study first §5.4.

Figure C.1-2

Figure C.1-2

Χ 0Υ initial coordinate system
x’0y’ auxiliary coordinate system
xC_Ty principal coordinate system

Figure C.1-3: Parallel translation in both directions and rotation of the diaphragm, due to force H

Figure C.1-3: Parallel translation in both directions and rotation of the diaphragm, due to force H

C.2 Axis transformation

It is useful to recall the relationship between the two sets of components for the vector v(x,y) transferred from one coordinate system x0y to another ζ0η rotated by an angle φ with respect to the original system. This vector can be either a displacement vector, a force vector, or defining the location of a point.

Figure C.2: Relationships of axis transformation due to rotation Figure C.2: Relationships of axis transformation due to rotation

ζ = x · cos φ +y · sin φ η =-x · sin φ +y · cos φ x= ζ · cos φ - η · sin φ y= ζ · sin φ + η · cos φ x-vector: ζ =x · cos φ & η =-x · sin φ y-vector: ζ =y · sin φ & η =y · cos φ

Application of the above relationships on the stiffnesses of a column with cross-section is inclined by an angle φ_i with respect to the initial system x0y:

The displacement of a column top i by δ_xi is equivalent to the displacements δ_{ζ i}, δ_ηi in the principal system ζ, η of column i:

δ_{ζ i} = δ _xi · cos φ _iand δ_ηi=-δ_xi·sinφ _i

Displacements δ_ζi and δ_ηi will result in the corresponding shear forces:

V _{ζ i} =K _{ζ i} · δ _xi · cos φ _iand V_ηi=-K_ηi·δ _xi·sinφ_i.

The equivalent shear forces V_xxi and V_xyi [2] [*]Note The symbol V_jk denotes shear force in k direction due to displacement in j direction, i.e. VXY means shear force in Υ direction due to displacement in X direction. derive from the transfer of shear forces V_ζi and V_ηi to the system x0y:

V_xxi=V _{ζ i} · cos φ _i -V _{η i} · sin φ _i =(K _{ζ i} · δ _xi · cos φ _i ) · cos φ _i -(-K _{η i} · δ _xi · sin φ _i ) · sin φ _i =K _{ζ i} · δ _xi · cos² φ _i +K _{η i} · δ _xi · sin² φ _i →

V_xxi = δ _xi · ( K _{ζ i} · cos² φ _i +K _{η i} · sin² φ _i )

V_xyi=V _{ζ i} · sin φ _i +V _{η i} · cos φ _i =(K _{ζ i} · δ _xi · cos φ _i ) · sin φ _i +(-K _{η i} · δ _xi · sin φ _i ) · cos φ _i =K _{ζ i} · δ _xi · (1/2)sin2 φ _i -K _{η i} · δ _xi · (1/2) · sin2 φ _i →
V_xyi= δ _xi · (1/2) · ( K _{ζ i} -K _{η i} ) · sin2 φ _i

In a simpler form, these two expressions, as a function of the normalised stiffness coefficients, of column i in direction x, K_xxi and K_xyi, are:

V_xxoi= δ _xi × K_xxiwhere K_xxi= K_ζi×cos²φ_i +K_ηi×sin²φ_i (a)

V_xyoi= δ _xi × K_xyiwhere K_xyi=(1/2)×(K_ζi-K_{η i})×sin2φ_i (b)

Quantities K_xxi and K_xyi are called normalised stiffnesses of column i in direction x.

In a similar way, the expressions as a function of the normalised stiffness coefficients K_yxi and K_yyi of column in direction y are:

V_yxoi= δ _yi × K_yxiwhere K_yxi=(1/2)×(K_ζi-K_{η i})×sin2φ_i= K_xyi (c)

V_yyoi= δ _yi × K_yyiwhere K_yyi=K_ζi×sin²φ_i +K_ηi×cos²φ_i (d)

Notes:

• In every coordinate system, for each column i always K_xyi=K_yxi.
• Stiffness coefficients K_xxi, K_xyi and K_yyi are identical in all systems parallel to x0y, therefore to x’0y’ as well.
• In rectangular columns the angle φ is irrelevant since K_ζ=K_η therefore (a) and (d) always give K_xx=K_yy =K_ζ=K_η and (b) always gives K_xy=0.

Examples:

Column stiffnesses :

Stiffness K_ij of each column i in direction j is Κ _ji =(k_ji × EI _ji / h ³)×k_ji,vα (§4.1.5).The columns are assumed fixed-ended → k_ji=12 and the shear effect is ignored [3] [*]NoteIf in the accompanying software , in project 4Columns, the option is selected to take into account works from shears, the factors of stiffness reduction kva have value 0.966 in the C1,C2 and the stronger value in column C3 by x: kva,x3=0.875. → k_ji,vα=1.0 (§5.1.1). Therefore, the column stiffness derives from the expression Κ_ji=12×EI_ji/h³ . Also, the column torsional stiffness is assumed as K_zi=0.

The columns of the example have the same E and h, therefore their stiffness can be assumed as Κ _ji =C × I_ji where C=12E/h³

As elasticity modulus is assumed Ε =32.80 GPa corresponding to concrete C30/37 (§6.1) therefore C=12E/h³=12 × 32.80 × GPa/(3.0³m³)=14.58 × 10⁹N/m⁵ .

Column stiffnesses with respect to their local coordinate systems ζ , η :

C ₁ , C ₂ 400/400 K _ζ1 = K _1η = K _2ζ = K _2η =14.58·10⁹ N / m ⁵ × 0.400 × 0.400³/12 × m ⁴ =31.1 × 10⁶ N / m ,

C ₃ 800/300 K _3ζ =14.58 × 10⁹ × 0.300 × 0.800³/12=186.6 × 10⁶ N / m , K _3η =14.58 × 10⁹·0.800 × 0.300³/12=26.2 × 10⁶ N / m

C ₄ 300/600 K _4ζ =14.58 × 10⁹ × 0.600 × 0.300³/12=19.7 × 10⁶ N / m , K _4η =14.58 × 10⁹ × 0.300 × 0.600³/12=78.7 × 10⁶ N / m

The diaphragmatic behaviour may be considered as superposition of three cases:

(a) parallel translation of the diaphragm in X direction due to horizontal force component H_X,

(b) parallel translation of the diaphragm in Y direction due to horizontal force component H_Y,

(c) rotation of the diaphragm due to moment M_CT applied at the centre of stiffness C_T

The horizontal seismic forces are applied at each point with mass, while the resultant force is applied at the centre of mass C_M.

In case the direction of the force H passes through the point C_T as well as C_M the moment has zero value and therefore the diaphragm develops zero rotation.

C.3 Case 1: Translation of centre of stiffness C_T along x direction by δ_xο

Figure C.3

Figure C.3

• 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)

Expression (i) yields H_x=Σ(δ_xo× K_xxi)=δ_xo×Σ(K_xxi) →

(3) [4] [*]NoteExpressions (1) and (2), are described in §5.4.2.

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

Figure C.4

Figure C.4

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

Example:

Columns stiffnesses in the principal system xC_Ty [6] [*]Note These stiffnesses are the same in the auxiliary system x’0y’ also, since the principal system xCTy and the auxiliary system x’0y’ are parallel between them. :

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)·sin2 φ₃=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

C.5 Case 3: Rotation of the diaphragm by an angle θ_z about C_T

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 alsotransferring 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:

The deformation of the diaphragm is a rotation θ_z about C_T. Rotation θ_zof 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 ifrom 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] →

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 ofmoment 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.

C.6 Torsional stiffness ellipse, torsional radii and equivalent system

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 isassumed 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 → Σ(K_xxi)·y²+Σ(K_yyi)·x²=K_θ →

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.

Notes :

C.7 Superposition of the three cases

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 forcesH_x and H_y parallel to the two axes of the principal system. In order to apply the previous analysis, the forcesH_x, H_y are transferred to the centre of stiffness C_T together with the moment M_CT according to the expression:

Example C.7-1: Given a horizontal force in x direction equal to is H=90.6 kN [7] [*]NoteThis force correspond to a seismic acceleration applied in the position of mass a=0.20g, therefore Η=Σ(mi)×0.20g= 45.3×103kgr×0.20×10m/sec2=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:

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

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_{ζ 1} = δ _x1 × cos φ ’₁+ δ _y1 × sin φ ’₁=0.547 ×0.924-0.515×(-0.382)=0.505+0.197=0.702 mm

δ_{η 1} =- δ _x1 × sin φ ’₁+ δ _y1 × cos φ ’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_{ζ 2} = δ _x2 × cos φ ’₂+ δ _y2 × sin φ ’₂=0.705 ×0.924+(-0.291)×(-0.132)=0.652+0.038=0.701 mm

δ_{η 2} =- δ _x2 × sin φ ’₂+ δ _y2 × cos φ ’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_{ζ 3} = δ _x3 × cos φ ’₃+ δ _y3 × sin φ ’₃=0.227 ×0.991+(-0.383)×0.132=0.225-0.050=0.175 mm

δ_{η 3} =- δ _x3 × sin φ ’₃+ δ _y3 × cos φ ’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_{ζ 4} = δ _x4 × cos φ ’₄+ δ _y4 × sin φ ’₄=0.386 ×0.923+0.001×0.384=0.355+0.000=0.355 mm

δ_{η 4} =- δ _x4 × sin φ ’₄+ δ _y4 · cos φ ’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.

Example C.7-4:

Calculation of shear forces and bending moments :

The moment distribution factor is assumed the same in both directions a _{ζ i} =a _{η i} =0.50 [8][*]NoteMji,1-Mji,2=Vji·h (where h is the column height) regardless of the actual stiffness of the columns and the fixity factors at their top and base. a_ji=0.50 means equal distribution of the moment at the top and at the base of the column.

C₁: V _{ζ 1} = δ_{ζ 1} · K _{ζ 1} =0.702mm · 31.1 · 10⁶ N/m =21.8 kN

V _{η 1} = δ_{η 1} · K _{η 1} =-0.267mm · 31.1 · 10⁶ N/m =-8.3 kN

M _{ζ 1,1} =V _{ζ 1} · h · 0.50=21.8·3.0·0.50=32.7 kNm M_ζ1,2=- M_ζ1,1 =-32.7 kNm

M _{η 1,1} =V _{η 1} · h · 0.50=-8.3·3.0·0.50=-12.5 kNm M_ηι,2=- M_η1,1 =12.5 kNm

C₂: V _{ζ 2} = δ_{ζ 2} · K _{ζ 2} =0.701mm · 31.1 · 10⁶ N/m =21.8 kN

V _{η 2} = δ_{η 2} · K _{η 2} =0.147mm · 31.1 · 10⁶ N/m =4.6 kN

M _{ζ 2,1} =V _{ζ 2} · h · 0.50=21.8·3.0·0.50=32.7 kNm M_ζ2,2=- M_ζ2,1 =-32.7 kNm

M _{η 2,1} =V _{η 2} · h · 0.50=4.6·3.0·0.50=6.9 kNm M_η2,2=- M_η2,1 =-6.9 kNm

C₃: V _{ζ 3} = δ_{ζ 3} · K _{ζ 3} =0.175mm · 186.6 · 10⁶ N/m =32.7 kN

V _{η 3} = δ_{η 3} · K _{η 3} =-0.410mm · 26.2 · 10⁶ N/m =-10.8 kN

M _{ζ 3,1} =V _{ζ 3} · h · 0.50=32.7·3.0·0.50=49.1 kNm M_ζ3,2=- M_ζ3,1 =-49.1 kNm

M _{η 3,1} =V _{η 3} · h · 0.50=-10.8·3.0·0.50=-16.2 kNm M_η3,2=- M_η3,1 =16.2 kNm

C₄: V _{ζ 4} = δ_{ζ 4} · K _{ζ 4} =0.355mm · 19.7 · 10⁶ N/m =7.0 kN

V _{η 4} = δ_{η 4} · K _{η 2} =-0.147mm · 78.7 · 10⁶ N/m =-11.6 kN

M _{ζ 4} =V _{ζ 4} · h · 0.50=7.0·3.0·0.50=10.5 kNm M_ζ4,2=- M_ζ4,1 =-10.5 kNm

M _{η 4,1} =V _{η 4} · h · 0.50=-11.6·0·0.50=-17.4 kNm M_η4,2=- M_η4,1 =17.4 kNm

C.8 Work Methodology

Summarising the theory for the calculation of displacements and stress resultants of a floor diaphragm the following computational procedure can be formulated. This procedure has also been applied in the related file <diaphragm_general.xls> [9] [*]NoteThere is also the related file supporting the special case of parallel rectangular columns described in the § 5.4. . This Excel file may be used to verify the calculations of this and the following examples, but also of any other floor diaphragm.

GEOMETRIC DATA OF THE DIAPHRAGM (independent of the loading)

1. Calculation of column stiffnesses K _{ζ i} , K _{η i} , K_zi and angle φ _i [10][*]Note Angle φi in the local system ζ,η is met only in elements of unsymmetrical cross-section e.g. L, unequal T, Z, etc. in the local system.

2. Calculation of angle a of the main system using expression (3)

3. Calculation of column angles φ ’_i in the principal system x,y using the expression φ ’_i= φ _i -a

4. Calculation of column stiffnesses K_xxi, K_xyi, K_yyi in the principal system x,y using expressions (a), (b), (c)

5. Calculation of each column centre of gravity and centre of mass (x’_CM, y’_CM) using the relationships for the transformation from the principal system X0Y to the auxiliary system x’0y’.

6. Calculation of diaphragm total stiffnesses using the expressions K_x= Σ (K_xxi), K_y= Σ (K_yyi). Calculation of the quantities Σ (x’_i · K_yyi), Σ (y’_i · K_xyi), Σ (y’_i · K_xxi), Σ (x’_i · K_xyi) and the coordinates of the centre of stiffness C_T(x’_CT, y’_CT) using the expressions (4) and (5)

7. Transfer of the coordinates x’, y’ of the diaphragm C_M and each column centre of gravity in the principal system. Calculation of the structural eccentricities e_ox, e_oy using the expressions (6’).

8. Calculation of torsional stiffness K _θ using the expression (7)

9. Calculation of torsional radii r_x, r_y using the expressions (8)

Up to this point the calculations are independent of the external loading. Next, the stress distribution and the elastic deformations of the diaphragm elements, due to the external loading, are calculated.

FLOOR DISPLACEMENTS AND STRESS RESULTANTS (dependent on the loading)

10. Transfer of horizontal forces H_x, H_y and moment M_CT to the C_T using the expression (9).

11. Calculation of displacements δ _xo , δ _yo and θ _z of the C_T using the expressions (10).

12. Calculation of the displacements δ _xi , δ _yi of each column top using the expressions (11).

13. Calculation of the corresponding displacements δ_{ζ i} , δ_{η i} in the local system of each column using the expressions (12)

14. Calculation of principal shear forces V _{ζ i} , V _{η i} of each column using the expressions (13)

15. Calculation of principal bending moments M _{ζ i,1} , M _{ζ i,2} & M _{η i,1} , M _{η i,2} using the expressions (14)

C.9 Expressions relating the initial system X0Y and the principal system xC_Ty

Seismic force Η _X =H acting in X direction (H_Υ=0) in the initial system X0Y:

The equivalent forces H_x, H_y in the principal system xC_Ty are: H_x=H·cosa and H_y=-H·sina and the equivalent deformations of C_T, δ_x, δ_y, in the principal system are: δ_x = δ_XXο·cosa + δ_ΧΥο·sina and δ_y=- δ_{XX ο}·sina +δ_ΧΥο·cosa. H_x=K_xx ·δ _x and H_y=K_yy·δ_y therefore:

H · cosa = K_xx · ( δ _{XX ο} · cosa + δ_ΧΥο · sina) → H=K_xx·( δ_XXο+tana· δ_ΧΥο) (i)

-H · sina = K_yy · (- δ _{XX ο} · sina + δ_ΧΥο · cosa) →-tana·H=K_yy·(-tana·δ_{XX ο}+δ_ΧΥο) (ii)

Seismic force Η _Y =H acting in Y direction (H_X=0) in the initial system X0Y:

The equivalent forces H_x, H_y in the principal system xC_Ty are: H_x=H·sina and H_y=H·cosa and the equivalent deformations of C_T, δ_x, δ_y,in the principal system are:

δ _x = δ_ΧΥο · cosa+ δ _YYo · sina and δ_y =-δ_YXo·sina+ δ _YYo·cosa.

H_x=K_xx · δ _x and H_y=K_yy·δ_y due to δ _YXo = δ_ΧΥο yields:

H · sina=K_xx · ( δ_ΧΥο · cosa+ δ _YYo · sina) → tana·H=K_xx·( δ_ΧΥο+tana ·δ_YYo) (iii)

H · cosa=K_yy · (- δ_ΧΥο · sina+ δ _YYo · cosa) → H=K_yy·(-tana·δ_ΧΥο+δ _YYo) (iv)

(iv) → K_yy=H/(δ_YYo - tana·δ_ΧΥο)

(i) → K_xx=H/(δ_XXο + tana·δ_ΧΥο)

(iii) → tana·H= H/( δ_XXο+tana·δ_ΧΥο)·( δ_ΧΥο+tana·δ _YYo) →

tana · ( δ _{XX ο} +tana · δ_ΧΥο )= δ_ΧΥο +tana · δ _YYo ) → tana·δ_XXο +tan²a· δ_ΧΥο = δ_ΧΥο +tana·δ_YYo →

δ_ΧΥο · (1- tan²a)=tana · ( δ _{XX ο} - δ_ΧΥο )→ δ_ΧΥο=[tana/(1- tan²a)]·( δ_{XX ο} - δ_YYo) →

tan2a = 2 δ_ΧΥο / ( δ _{XX ο} - δ _YYo ) (C.9.1)

K_xx=H/( δ _{XX ο} +tana · δ_ΧΥο ) (C.9.2)

K_yy=H/( δ _YYo -tana · δ_ΧΥο ) (C.9.3)

Κ_θ =H · (Y_CT-Y_CM)/ θ _xz (C.9.4)

r_x= √(K _θ /K_yy) (C.9.5)

r_y= √(K _θ /K_xx) (C.9.6)

C.10 One-storey space frame with rectangular columns in random arrangement

The method which describes the diaphragmatic behaviour of each floor is presented in Appendix D. Also relevant examples are presented verifying the validity of the algorithms of the general method.

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.

The names, loadings, coordinates and dimensions of slabs and columns are shown in figure of §C.1. The cross-section of beams is 250/500. The concrete class is C30/37 (Ε=32.80 GPa).

C.10.1 Analysis using manual calculations, assuming fixed columns (k=12)

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

C.10.2 Analysis using the Excel file, assuming fixed-ended columns (k=12)

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

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

C.10.3 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.

C.10.4 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.