A 3 Story Medical building essential facility with ordinary

Problem 11.5 A 3 story medical building (essential facility) with ordinary reinforced masonry shear walls and ordinary reinforced concrete moment frames is to be constructed on a stiff soil base in Charleston, South carolina (80° longitude, 33 N latitude). Determine the lateral seismic forces at each floor

Solution

Solution:-

Design seismic base shear = A_h W

Fundamental natural period = 0.075 (h)^0.75

                                          T_a = 0.075 x (11.25)^0.75

                                               = 0.46 second

Where h is in meter

Design horizontal seismic coefficient

A_h = Z/2 *I/R *Sa/g

    = 0.36/2 *1.5/3 *2.17 = 0.1953

Where Z = zone factor = 0.36 (for South Carolina )

For medical building( I) = 1.5

R =Response reduction factor =3

Sa/g = 1/ Ta = 1/ (0.46) = 2.17

V_B =A_h* W

       = 0.1953 * 44 *10⁵ =8.59 *10⁵ lb

W = seismic weight of building

Design lateral seismic force at first floor

Q₁ =     V_B * W₁ * h₁²/(W₁h₁² +W₂h₂² + W₃h₃² )

     = 8.59 *10⁵ * 17 *10⁵*12²/(17*10⁵* 12² + 15 * 10⁵ * 24² + 12 * 10⁵* 36²)

     = 7.81 * 10⁴ lb      Answer

Design lateral force at second floor

Q₂ =8.59 *10⁵ *15 *10⁵ * 24²/ ( 17 *10⁵ *12² + 15 *10⁵ * 24² + 12 * 10⁵ *36² )

Q₂ =2.78 * 10⁵ lb Answer

lateral seismic force At third floor

Q₃ = 8.59 * 10⁵ * 12 * 10⁵ * 36² / (17 *10⁵ *12² + 15 * 10⁵ *24² + 12 * 10⁵ * 36²)

Q₃ = 5.01 * 10⁵ lb Answer