Walls on grid lines 2 through 8;
One design will be produced for the worst case and used for all walls on grid lines 2 through 8. By
inspection, the worst case situation occurs on either grid line 2 or 8 between grid lines C and D because
these walls carry the largest shear, largest overturning moment, and smallest axial load (the walls moment
capacity increases with an increase in axial load).
Determine design loads;
The following diagram shows the maximum shear and overturning moment at each level;
1-ft = 0.30m, 1-kip = 4.448KN
Therefore, at the base of the wall Vmax = 104-kips (462.6KN), and Mmax = 2,600-ft-kips (3.53MN-m).
Design for shear;
By inspection, only minimum reinforcement will be required.
Try 2 curtains of #4 (~10M) bar at 12-in. (308.8mm) on center each way as used for walls on grid lines 1
and 9;
Check capacity;
Vu ≤φ n
with φ = 0.6
V
(EQ. 11-1 ACI 318-95)
Since hw/lw = 33'26.5'= 1.25 < 2.0 in accordance with Section 21.6.5.3 of ACI 318-95;
/
Vn = A cv (α c f c' + ρ nf y )
(EQ. 21-7 ACI 318-95)
where; Acv = 23.5' 9")(12"/1' = 2,538-in2 (1.64X106 mm2)
(
)
fc'= 4,000psi (46.90MPa)
αc = 3.0 for hw/lw = 1.3 < 1.5
ρn = 0.0037
fy = 60ksi (413.7MPa)
Note: The more conservative Equation 21-6 could have been used due to the light loading.
0.6(2,538 - in2 )
e
j
3 4,000psi + 0.0037(60,000psi) = 627 k (2.79MN)
∴ φ n=
V
1000 - kips / lb
φ n = 627 k > 104 k = Vu (2.79MN > 462.6KN)
O.K.
V
Design for flexural and axial loads;
Determine axial loads acting at each story;
Note: The tributary width of diaphragm is 23.5-ft(1/2) = 11.8-ft. (3.60m), and the story to story height is
11-ft (3.36m).
H2-31
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