Example 5.4.3.6 (the simple structure used so far in the theory)

(a) GEOMETRIC DATA OF THE DIAPHRAGM (independent of the loading)

Torsional stiffness calculation

The stiffness K_ij of each column i in direction j is:

Κ _ji =k_ji · ( EI _ji / h ³ ) · k_{ji , v α} (§4.1.5)

where

k_ji=12 (columns are assumed to be fixed-ended)

k_{ji , v α} =1.0 (§4.1.1) (the shear effect is not taken into account [*]Note If in the related file (in sheet 4Columns) the shear effect is taken into account, the stiffness reduction factors kva are equal to 0.966 for C1,C2 and the more important value of k_va,x3=0.875 in column C3 by x . )

Therefore, Κ _ji =12 · EI _ji / h ³ .

Moreover, we assume that all columns have the same torsional stiffness K_zi=0.

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

Modulus of Elasticity is taken as Ε =32.80 GPa, corresponding to concrete class C30/37 (volume C §1.1). Therefore, C=12E/h³=12 · 32.80 · GPa/(3.0³m³)=14.58 ·10⁹N/m⁵.

C₁, C₂ 400/400:

K_1x=K_1y=K_2x=K_2y=14.58 · 10⁹N/m⁵ · 0.400 · 0.400³/12 · m⁴=31.1 ·10⁶ N/m

C₃ 800/300:

K_3x=14.58 ·10⁹·0.300·0.800³/12=186.6·10⁶ N/m

K_3y=14.58 ·10⁹·0.800·0.300³/12=26.2·10⁶ N/m

C₄ 300/600:

K_4x=14.58 ·10⁹·0.600·0.300³/12=19.7·10⁶ N/m

K_4y=14.58 ·10⁹·0.300·0.600³/12=78.7·10⁶ N/m

Therefore based on the expressions (3):

Calculation of the Centre of Stiffness and the Lateral Stiffnesses : expressions (4') and (5')

K_x= Σ (K_xi)=(31.1+31.1+186.6+19.7) ·10⁶ N/m=268.5·10⁶ N/m

K_y= Σ (K_yi)=(31.1+31.1+26.2+78.7) ·10⁶ N/m=167.1·10⁶ N/m

Χ _CT =Σ( X_i · K_yi )/ K_y =[(0.0·31.1+6.0·31.1+0.0·26.2+6.0·78.7)·10⁶]/(167.1· 10⁶)=658.8/167.1=3.94 m

Υ _CT = Σ (Y_i · K_xi)/K_x=[(0.0·31.1+0.0·31.1+5.0·186.6+5.0·19.7)·10⁶]/(268.5· 10⁶)=1031.5/268.5=3.84 m

Coordinate transfer to the principal system xC_Ty : expressions (6')

C₁(-3.94,-3.84), C₂(2.06,-3.84), C₃(-3.94,1.16), C₄(2.06,1.16), C_M(-0.94,-1.34), C _T(0.0,0.0)

e_ox=-0.94 m, e_oy=-1.34 m

Calculation of the diaphragm's torsional stiffness : expression (7')

K _θ = Σ [K_xi · y_i²+K_yi · x_i²+K_zi]=

(31.1·3.84²+31.1·3.94²+31.1·3.84²+31.1·2.06²+186.6·1.16²+26.2·3.94²+19.7·1.16²+78.7·2.06²)· ·10⁶N/m·m²=(1194.8+1355.4)·10⁶Nm=2550·10⁶ Nm [*]Note The torsional stiffnesses K_z of columns are equal to: K_z1=K_z2=0.5E × I_d/h=0.5×32.80×10⁹Pa×0.1404×0.4⁴m⁴/(3.0m)=20× 10⁶ N × m K_z3= 0.5E × n × b³ × d/h=0.5×32.80×10⁹Pa×0.251×0.3³m³×0.8m/(3.0m)=30× 10⁶ Nm K_z4= 0.5E × n × b³ × d/h=0.5×32.80×10⁹Pa×0.229×0.3³m³×0.6m/(3.0m)=20× 10⁶ Nm, therefore K_θ=(2550+90)×10⁶ Nm=2640×10⁶ Nm, i.e. an increase of the floor torsional stiffness by 3.5%.

Diaphragm's torsional radii : expressions (8')

r_x=√(K _θ /K_y)=√[2550 · 10⁶Nm/(167.1 · 10⁶N/m)]=3.91 m

r_y=√(K _θ /K_x)=√[2550 · 10⁶Nm/(268.5 · 10⁶N/m)]=3.08 m

(b) DISPLACEMENTS AND STRESS RESULTANTS (independent of the loading)

Given the seismic force H=90.6 kN [*]Note This force corresponds to seismic acceleration (applied at mass position) a=0.20g,therefore Η=Σ(mi)·0.20g=45.3·10³kgr×0.20×10m/sec²=90.6 kN. and the interstorey height h=3.0 m, calculate the displacements of the centre of stiffness C_T and those of each column, the shear forces carried by each column and the corresponding bending moments for the following three loading cases:

(A) H_x=90.6 kN, H_y=0, (B) H_x=0 kN, H_y=90.6 kN and ( Γ ) H_x=90.6 kN, H_y=-27.2 kN

Case Α : H_x=90.6 kN, H_y=0.0

Seismic moment at the centre of stiffness : expression (9')

M_CT=-H_x · e_oy+H_y · e_ox=-90.6kN · (-1.34m)=121.4 kNm

Displacements of the centre of stiffness : expressions (10')

δ _xo =H_x/K_x=90.6 · 10³N/(268.5 · 10⁶N/m)=0.337 · 10^-3m=0.337 m

δ _yo =0 [*]Note In the general case of inclined columns, displacement by y (δ_xyo) is also developed, in addition to the displacement by x (δ_xxo), from the corresponding loading (see appendix C). In a system of parallel orthogonal columns, as this one, the secondary displacements δ_xyo, and δ_yxo, are zero.

θ _z =M_CT/K _θ =121.4 · 10³Nm/( 2550 · 10⁶Nm)=47.6 ·10^-6

Displacements of columns : (expressions 11')

δ _xi = δ _xo + δ _{x θ i} = δ _xo - θ _z · y_i

δ _yi =0.0+ θ _z · x_i= θ _z ·x_i

To compare the quantities, the displacements due to rotation are calculated separately.

Column C₁:

δ _{x θ 1} =- θ _z · y₁=-47.6 ·10^-6·(-3.84·10³mm)=0.183 mm

δ _{y θ 1} = θ _z · x₁=-0.188 mm

Summarising,

C₁: δ _{x θ 1} = 0.183 mm, δ _{y θ 1} =-0.188 mm, δ _x1 =0.337+0.183=0.520 mm, δ _y1 =-0.188 mm

Likewise, for the rest of the columns:

C₂: δ _{x θ 2} = 0.183 mm, δ _{y θ 2} = 0.098 mm, δ _x2 =0.337+0.183=0.520 mm, δ _y2 = 0.098 mm

C₃: δ _{x θ 3} =-0.055 mm, δ _{y θ 3} =-0.188 mm, δ _x3 =0.337-0.055=0.282 mm, δ _y3 =-0.188 mm

C₄: δ _{x θ 4} =-0.055 mm, δ _{y θ 4} = 0.098 mm, δ _x4 =0.337-0.055=0.282 mm, δ _y4 = 0.098 mm

Seismic shear forces : expressions (13')

C₁: V_x1= δ _x1 · K_x1=0.520·10^-3m·31.1·10⁶N/m=16.17 kN, V_y1=δ_y1 ·K_y1=-0.188·31.1=-5.85 kN

C₂: V_x2= δ _x2 · K_x2=0.520·31.1= 16.17 kN, V_y2=δ_y2·K_y2= 0.098· 31.1= 3.05 kN

C₃: V_x3= δ _x3 · K_x3=0.282·186.6=52.62 kN, V_y3=δ_y3·K_y3=-0.188· 26.2=-4.93 kN

C₄: V_x4= δ _x4 · K_x4=0.282·19.7= 5.56 kN, V_y4=δ_y4·K_y4= 0.098· 78.7= 7.71 kN

Verification: Σ (Vxi)=90.52 kN≈90.6 kN and Σ (Vyi)=-0.02≈0.0 as expected

Seismic bending moments : expressions (14')

It is assumed (for all columns) a_xi=a_yi=0.50, and therefore (1-a_xi)=(1-a_yi)=0.50.

C₁: M_x1,1=0.5 ·16.17·3.0=24.3 kNm, M_x1,2=-24.3 kNm,

M_y1,1=0.5 · (-5.85) ·3.0=-8.8 kNm, M_y1,2= 8.8 kNm

C₂: M_x2,1=0.5 ·16.17·3.0=24.3 kNm, M_x2,2=-24.3 kNm,

M_y2,1=0.5 ·3.05·3.0= 4.6 kNm, M_y2,2= -4.6 kNm

C₃: M_x3,1=0.5 ·52.62·3.0=78.9 kNm, M_x3,2=-78.9 kNm,

M_y3,1=0.5 ·(-4.93)·3.0=-7.4 kNm, M_y3,2= 7.4 kNm

C₄: M_x4,1=0.5 ·5.56·3.0= 8.3 kNm, M_x4,2= -8.3 kNm,

M_y4,1=0.5 ·7.71·3.0= 11.6 kNm, M_y4,2=-11.6 kNm

Case B: H_x=0.0 kN, H_y=90.6

Seismic moment at the centre of stiffness : expression (9')

M_CT=-H_x·e_oy+H_y·e_ox=90.6kN ·(-0.94m)=-85.2 kNm

Displacements of the centre of stiffness : expressions (10')

δ _xo =0

δ _yo =H_y/K_y=90.6 ·10³N/(167.1·10⁶N/m)=0.542 mm

θ _z =M_CT/K _θ =-85.2 ·10³Nm/(2550·10⁶Nm)=-33.4·10^-6

Only the quantities for column C₃ are calculated here. The results for the remaining columns can be derived anyway from the related file <diaphragm ortho.xls>.

Displacements of columns (C₃) : expressions (11')

δ _{x θ 3} =- θ _z · y₃=-(-33.4) ·10^-6·1.16·10³mm=0.039 mm

δ _{y θ 3} = θ _z · x₃=(-33.4) ·(-3.94·10³mm)=0.132 mm

δ _{x θ 3} =0.039 mm, δ _{y θ 3} =0.132 mm, δ _x3 =0.039 mm, δ _y3 =0.542+0.132=0.674 mm

Seismic shear forces (C₃) : expressions (13')

V_x3= δ _x3 · K_x3=0.039 ·186.6=7.28 kN

V_y3= δ _y3 · K_y3=0.674 ·26.2=17.66 kN

Seismic bending moments (C₃) : expressions (14')

M_x3,1=0.5 ·7.28·3.0=10.9 kNm

M_x3,2=-10.9 kNm

M_y3,1=0.5 ·17.66·3.0=26.5 kNm

M_y3,2=-26.5 kNm

Case C: H_x=90.6 kN, H_y=-27.2 kN [*]Note The seismic regulations impose consideration of seismic loading in one direction combined with a percentage of seismic loading in the other e.g. 30%. All seismic loadings should be taken into account with both positive and negative signs.

Seismic moment in the center of stiffness : expression (9')

M_CT=-H_x · e_oy+H_y · e_ox=-90.6 ·(-1.34)+(-27.2)·(-0.94m)=121.4+25.6=147.0 kNm

Displacements of the centre of stiffness : expressions (10')

δ _xo =H_x/K_x=90.6/268.5mm=0.337 mm

δ _yo =H_y/K_y=-27.2/167.1=-0.163 mm

θ _z =M_CT/K _θ =147 ·10³Nm/(2550·10⁶Nm)=57.65·10^-6

Next, only the data of column C₃ will be calculated. The other results can be derived, from the related file <diaphragm ortho.xls>.

Displacements of columns (C₃) : expressions (11')

δ _{x θ 3} =- θ _z · y₃=-57.65 ·10^-6·1.16·10³mm=-0.067 mm

δ _{y θ 3} = θ _z · x₃=57.65 ·(-3.94·10³mm)=-0.227 mm

δ _{x θ 3} =-0.067 mm, δ _{y θ 3} =-0.227 mm, δ _x3 =0.337-0.067=0.270 mm, δ _y3 =-0.163-0.227=-0.390 mm

Seismic shear forces (C₃) : expressions (13')

V_x3= δ _x3 ·K_x3=0.270 ·186.6=50.38 kN

V_y3= δ _y3 ·K_y3=-0.390 ·26.2=-10.22 kN

Seismic bending moments (C₃) : expressions (14')

M_x3,1=0.5 ·50.38·3.0=75.6 kNm

M_x3,2=-75.6 kNm

M_y3,1=0.5 ·(-10.22)·3.0=-15.3 kNm

M_{y 3,2} =15.3 kNm

Notes

• The analysis of such problems is practically performed by a table format. In this respect, using the related Excel files, one can define his own structures and examine the effect of various factors e.g. the elasticity modulus of the columns, the stiffness factors, the cross-sectional dimensions, etc.
• For seismic action in the x direction, the displacement of the centre of stiffness C_T is 0.337 mm, while the diaphragm maximum displacement due to rotation is 0.520 mm, i.e. 54% greater than the average value. For seismic action in y direction, the C_T displacement is 0.520 mm while the diaphragm maximum displacement due to rotation is 0.674 mm, i.e. 30%, greater than the average value.
• The actual displacement is much greater than the maximum of 0.674 mm which is indicative and idealised. This is due to the fact that the actual stiffness [EC8 §4.3.1(6),(7)] due to cracking should be considered reduced by an order of 50%, therefore resulting to a double displacement 2·0.674=1.35 mm. If all columns are considered fixed-ended simultaneously, the stiffness factor would be taken equal to k=3 instead of k=12 and due to the previous reduction by 50% k=1.5. Therefore in accordance with this stiffness factor, the maximum elastic displacement should be 12/1.5=8 times greater, i.e. δ=8·0.0.674=5.4 mm (results can be verified introducing the changes in the Excel file).
• For the calculation of the actual displacement, EC8 imposes the multiplication of the elastic displacement with the behaviour factor q. Using an average value of q=3.50 the previous displacement would be 3.5·5.4=19mm. Obviously considering the small loads of an one-storey building of these dimensions, the size of its columns is very large, and for this reason the displacement is relatively small. In case of columns of half cross-sectional dimensions e.g. 400/400 instead of 200/200, the displacement of the diaphragm would be (400/200)⁴=16 times greater i.e. δ_yo=16·19=304 mm. This displacement is extremely large and prohibited from other provisions of EC8. In case that this building would be a multistorey one, then, the need for strong columns would be even more obvious.