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

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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