Seismic base shear of a masonry building

Step 1: Fundamental period

$$T = \frac{0.09\,h}{\sqrt{d}} = \frac{0.09 \times 6}{\sqrt{8}} = 0.19\ \text{s}$$

For medium soil with $0.1 \le T \le 0.55$ s, the spectral value $S_a/g = 2.5$ (plateau).

Step 2: Design horizontal coefficient

$$A_h = \frac{0.24}{2}\times\frac{1.0}{1.5}\times 2.5 = 0.12 \times 0.667 \times 2.5 = 0.20$$

Step 3: Base shear

Seismic weight $W = 450 + 600 = 1050$ kN.

$$V_B = A_h W = 0.20 \times 1050 = 210\ \text{kN}$$

Step 4: Distribution of base shear

$$Q_i = V_B \frac{W_i h_i^2}{\sum W_j h_j^2}$$

        Q2 = 157.5 kN
   --->  =================  roof (6 m)
   |
        Q1 = 52.5 kN
   --->  =================  floor (3 m)
   |
   ^ base shear VB = 210 kN

Remarks

• The design force is 20 percent of the weight. NBC 105 uses a similar approach: base shear = $C_d(T)\,W$, where $C_d(T)$ depends on zone factor, soil type, importance factor, ductility and overstrength factors.
• The shear is then shared among the walls in the direction of force in proportion to their stiffness (rigid diaphragm) or their tributary area (flexible diaphragm), and each wall is checked for shear and overturning.

Answer: $T = 0.19$ s, $A_h = 0.20$, base shear $V_B = 210$ kN; lateral forces: 157.5 kN at roof and 52.5 kN at first floor.