Shear Wall System

Distribution by Displacement of the Diaphragm due to Story Shear F

a- The lateral sway (deflection) of a cantilever wall ““ due to the shear force of the wall “Qi,disp”

is given as:

= Qi,disp H³ (1)

3EI

Where:

I is the strongest inertia of the wall

E is the concrete modulus of elasticity

H is the height from the fixity level to the floor lever

b- The lateral sway of the whole floor “” due to story shear “F” is given as:

= F H³ (2)

3EΣIi

Where ΣIi represents the summation of the walls inertia

For a diaphragm behavior of the floor slab:

=

Therefore: Qi,disp= F Ii (3)

ΣIi

Distribution by Rotation of the Diaphragm due to F*e

.

The shear due to the torsion moment “F*e” generates different displacements “i,rot“ of the walls

proportional to their distance to the C.O.I:

Q1,rot = 3EI1 1,rot , Q2,rot = 3EI2 2,rot , Q3,rot = 3EI3 3,rot (4)

H³ H³ H³

Let 3E = K → Q1,rot =K I1 1,rot , Q2,rot =K I2 2,rot , Q3,rot =K I3 3,rot (5)

H³

And since 1,rot = 2,rot = 3,rot (6)

x1 x2 x3

→ Q2,rot = KI2 1,rot x2 , Q3,rot = KI3 1,rot x3 (7) x1 x1

The equilibrium of diaphragm system leads to:

F*e = Q1,rot x1+ Q2,rot x2 + Q3,rot x3 (8)

F*e = KI1 1,rot x1 + KI2 1,rot x²2 + KI3 1,rot 1 x²3

x1 x1

F*e = K1,rot (I1 x²1 + I2 x²2 + I3 x²3)

x1

Substituting 1,rot by Q1,rot -equation(5)- →Fe = Q1,rot ΣIi x²i

KI1 I1 x1

→ Q1,rot = Fe I1 x1

ΣIi x²i

→ Qi,rot = Fe Ii xi (9)

ΣIi x²i

Where the floor slab includes walls in both X and Y direction, analogous analysis leads to:

→ Qi,rot = Fe Ii xi (10)

Σ (Ixi x²i+ Iyi y²i)

The total shear force of a wall “I” is therefore:

Qi, = Qi,disp + Qi,rot = F Ii + Fe Ii xi = F Ii 1 + e xi (11)

ΣIi Σ(Ixi x²i+ Iyi y²i) ΣIi Σ(Ixi x²i+ Iyi y²i)