Po = 0.85fc' (A g - A st ) + f y A st
(EQ. 10-2 ACI 318-95)
where; Ag = 29.5' 12"/1' 9" = 3,186-in2 (2.05X106 mm2)
(
)
Ast = lwtρv = 29.5' 12"/1' 9"(0.0037) = 11.8-in2 (7.61X103 mm2)
(
)
fc'= 4,000psi (46.90MPa)
Po = 0.85(4,000psi)(3,186 - in2 - 11.8 - in2 ) + 60,000psi(11.8 - in2 ) = 11,500k (51.15MN)
∴ Pu = 203k << 4,025k = 0.35Po (902.9KN<<18.70MN)
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 = 203k << 1,273k = 0.1(29.5' (12"/1' 9"(4,000psi)(1k/1000-lb) = 0.1Agfc'
O.K.
)
)
(902.9KN<<5.66MN)
Check condition (2);
By inspection, Mu and Vu are the same for all load cases. Therefore;
1,881ft - kips
Mu
=
= 0.85 < 1.0
O.K.
Vul w 75.2 k (29.5' )
Therefore, no boundary elements are required
Determine if wall has adequate capacity for flexural and axial loads combined;
Per FEMA 302 Section 9.1.1.13, walls subject to combined flexural and axial loads shall be designed in
accordance with Sections 10.2 and 10.3 of ACI 318-95 except that Section 10.3.6 of ACI 318-95 and the
nonlinear strain requirements of 10.2.2 do not apply. To satisfy these requirements, an analysis program
entitled ` CACOL'produced by the Portland Cement Association was utilized. This program produces an
P
interaction diagram for the wall cross section and plots the loads acting on the section.
Note: At the time of writing of this problem, the current version of PCACOL program was developed for
ACI 318-89. However, there are no changes between the 95 code and the 89 code which will affect the
results of this shear wall analysis.
Determine design loads;
Pu = [1.386(1.17+2(1.81))klf + 0.5(2(0.24))klf]29.5'= 203k (902.9KN)
Load case 4a;
Mu = 1.0(1,881-ft-kips) = 1,881-ft-kips (2.55MN-m)
Pu = [0.714(1.79+2(1.81))klf]29.5'= 114k (507.1KN)
Load case 5a;
Mu = 1.0(1,881-ft-kips) = 1,881-ft-kips (2.55MN-m)
Figure 9 shows the design interaction diagram (obtained from PCACOL) for the shear wall section. The
section has a 9-in. (220.6mm) thick web reinforced with two curtains of reinforcement each having #4
(~10M) vertical and horizontal bars spaced at 12-in. (308.8mm) on center with fc' = 4,000psi (46.90MPa)
and fy = 60,000psi (413.7MPa). Two design load combinations are listed. The point marked "1" represents
the Pu-Mu combination corresponding to load case 4a, and the point marked "2" represents the Pu-Mu
combination corresponding to load case 5a. This figure shows that the walls have sufficient capacity for
axial and overturning forces.
Therefore, use 2 curtains of #4 (~10M) bars at 12" (308.8mm) o.c. each way
H2-29
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