Figure 10 shows the design interaction diagram (obtained from PCACOL) for the shear wall section. As in
Figure 9 only two design load combinations are listed. The point marked "1" represents the PuMu
combination corresponding to load case 4a, and the point marked "2" represents the PuMu 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
(2) Longitudinal moment frame design: use fc'= 4,000psi (46.90MPa), fy = 60ksi (413.7MPa)
Per paragraph 74.f, special moment frames (SMF' ) are frames conforming to the requirements of
s
Sections 21.1 through 21.5 of ACI 31895 in addition to the ACI 31895 requirements for ordinary moment
frames (OMF' ).
s
Determine design loads;
The RISA2D model, used previously to determine the relative rigidity of the frames, was reanalyzed for
lateral loading. The resultant forces acting on the frame at each floor level, determined in section 3d of this
problem solution, were distributed over the length of the structure at each floor level. Note that there are no
live loads acting on the frame, and the dead loads are due only to the frames self weight, weight of
windows, and nonstructural infill walls. Cracked section properties in accordance with ACI 31895
Section 10.11.1 were used.
Note: The RISA2D program does not distribute loads over rigid end offsets. Therefore, distributed loads
acting within the rigid end offsets were added as point loads on the accompanying node.
Dead loads;
At the roof ; wRD = distributed load due to weight of beam and parapet
wRD = {16"(18")+27"(12")}150pcf(1k/1000lb)/144in2/ft2 = 0.638klf (9.30KN/m)
PRD = point load due to tributary weight of column and parapet over rigid end offsets
PRD = {18"(24")}150pcf(3.46' (1k/1000lb)/144in2/ft2+0.638klf(2' =2.83k
)
)
(12.6KN)
Note: The tributary 3.46ft (1.06m) column height was calculated in the seismic weights section of this
problem.
At the floor; wFD = distributed load due to weight of beam + infill + windows
wFD = {18"(24")}150pcf(1k/1000lb)/144in2/ft2+...
... +{3' 40psf)+6' 8psf)}(1k/1000lb) = 0.618klf (9.01KN/m)
(
(
Note: The tributary 6ft glass height was calculated in the seismic weights section of this problem.
PFD = point load due to tributary weight of column and over rigid end offsets
PFD = {18"(24")}150pcf(11' (1k/1000lb)/144in2/ft2 = 4.95k (22.02KN)
)
The earthquake story forces were uniformly distributed over a length equal to the length of the frame minus
the length of the rigid end offsets.
(QE)R = 121.5k/(141' 12(1' ) = 0.941klf (13.72KN/m)
At the roof;

)
rd
k
At the 3 floor; (QE)3 = 100 /(141' 12(1' ) = 0.775klf (11.30KN/m)

)
nd
k
At the 2 floor; (QE)2 = 50.8 /(141' 12(1' ) = 0.394klf (5.75KN/m)

)
Design beams;
Check that axial loading may be ignored for beam design per Section 21.3.1.1 of ACI 31895;
From the analysis output, the largest compressive axial load in a floor beam is 11.6k (51.6KN) (due to load
combination 5a), and the largest compressive axial load in a roof beam is 16.8k (74.73KN) (due to load
combination 4a).
A g f c' 16" (18" )4ksi
= 115k > 16.8k (511.5KN > 74.73KN)
=
O.K.
At the roof;
10
10
H233