Dead loads;
At the roof;
PRD = 92.3psf(11.8' + 112.5psf(5.5' = 1,708plf (24.91KN/m)
)
)
At the floors;
PFD = 95.3psf(11.8' + 112.5psf(11' = 2,362plf (34.45KN/m)
)
)
Live loads; (Live loads are unreducable due to the small tributary area)
At the roof;
PRL = 20psf(11.8' = 236plf (3.44KN/m)
)
At the floors;
PFL = 40psf(11.8' = 472plf (6.88KN/m)
)
Confirm that walls are structural;
Per FEMA 302 Section 9.1.1.13, walls with Pu > 0.35Po shall not be considered to contribute to the
calculated strength of the structure for resisting earthquake induced forces. By inspection, load case 4a
governs; U = 1.386D + QE + 0.5L;
Pu = [1.386(1.71+2(2.36))klf + 0.5(0.236+2(0.472))klf]29.5'= 280k (1.25MN)
The nominal strength of the wall is given by;
Po = 0.85fc' (A g - A st ) + f y A st
(EQ. 10-2 ACI 318-95)
where; Ag = 23.5' 12"/1' 9" = 2,538-in2 (1.64X106 mm2)
(
)
Ast = lwtρv = 23.5' 12"/1' 9"(0.0037) = 9.39-in2 (6.06X103 mm2)
(
)
fc'= 4,000psi (46.9MPa)
Po = 0.85(4,000psi)(2,538 - in2 - 9.39 - in2 ) + 60,000psi(9.39 - in2 ) = 9,161k (40.75MN)
∴ Pu = 280k << 3,206k = 0.35Po (1.25MN << 14.26MN)
Therefore, walls are structural
Determine if boundary elements are required;
This requirement will be checked at the first story level only for this is the worst case condition, and if
boundary elements are not required at that level they will not be required at the levels above.
Per FEMA 302 Section 9.1.1.13, boundary elements are not required if;
(1) Pu ≤0.10A g fc'
and either
Mu
≤1.0
(2)
Vul w
Mu
(3) Vu ≤3A cv f c' , and
≤3.0
or
Vul w
Check condition (1);
By inspection, load case 4a governs; U = 1.386D + QE + 0.5L;
Pu = 280k < 1,015k = 0.1(23.5' (12"/1' 9"(4,000psi)(1k/1000-lb) = 0.1Agfc' (1.25MN < 4.51MN)
)
)
O.K.
Check condition (2);
By inspection, Mu and Vu are the same for all load cases. Therefore;
2,600ft - kips
Mu
=
= 1.06 > 1.0
N.G.
Vul w 104 k (23.5' )
Therefore, check condition 3;
Vu = 104 k < 3(9" )282" 4,000psi = 482 k (0.46MN < 2.14MN)
O.K.
Mu
= 1.06 < 3.0
O.K.
Vul w
Therefore, no boundary elements are required
Determine if wall has adequate capacity for flexural and axial loads combined;
Determine design loads;
Pu = [1.386(1.71+2(2.36))klf + 0.5(2(0.472))klf]23.5'= 221k (0.98MN)
Load case 4a;
Mu = 1.0(2,600-ft-kips) = 2,600-ft-kips (3.53MN-m)
Pu = [0.714(1.71+2(2.36))klf]23.5'= 108k (480.4KN-m)
Load case 5b;
Mu = 1.0(2,600-ft-kips) = 2,600-ft-kips (3.53MN-m)
H2-32
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