Page 63 - Advanced Design Examples of Seismic Retrofit of Structures

Example of a Two-Story Unreinforced Masonry Building Chapter 2 55

W Live ¼ W Live 1 + W Live 2
W Live ¼ 283,768 + 109,108 ¼ 392,875 kg

W ¼ 832,167 + 925,586 + 25% 392, 875Þ ¼ 1855Ton
ð
According to Eq. (2.1), the total design shear force is:
Q E ¼ 1:41 1 1 0:825 1855 ¼ 2156Ton

2.7.2 Vertical Distribution of Demand Forces
Once the total seismic demand of the building is determined, this force should
be distributed between different stories. Based on Section 3.3.3.3 of Code 360,
the seismic force applied at any floor level i shall be determined in accordance
with parts 2.7.2.1 and 2.7.2.2 [10].

2.7.2.1 Deformation-Controlled Demand Force
For deformation-controlled structural elements, the seismic force applied at any
floor level i, F UDi , is determined based on Eq. (2.8):
W i h k i
F UDi ¼ X n k Q E (2.8)
W j h
j¼1 j
where:
Q E ¼pseudo-lateral force based on Eq. (2.1); and
k ¼vertical distribution factor;
k ¼2.0 for T 2.5s, 1.0 for T 0.5s, and 0.5T +0.75 for 0.5< T <2.5s;
W j ¼portion of the effective seismic weight W located on or assigned to
level j;
W i ¼portion of the effective seismic weight W located on or assigned to
level i;
h j ¼height from the base to level j; and
h i ¼height from the base to level i.
For the floor and the first stories, we have:
1,012,585 3
F UD1 ¼ 2,156,407 ¼ 784Ton
1,012,585 3 + 843,386 6:3
843,386 6:3
F UD2 ¼ 2,156,407 ¼ 1372Ton
1,012,585 3 + 843,386 6:3

2.7.2.2 Force-Controlled Demand Force
For force-controlled structural elements, the seismic force applied at any floor
level i, F UFi , is determined based on Eq. (2.8):