Structural Concrete - Hassoun

17.10 Transfer of Unbalanced Moments to Columns 665
3. For transfer by shear at exterior column, the critical section is located at a distance d/2 from the
face of the column (Fig. 17.28).
q u = 330 psf
(
V u = 0.33 20 × 12.83 − 23.95
12 × 27.9 )
= 83.1K
12
Locate the centroid of the critical section by taking moments about AB:
(
2 23.95 × 23.95 )
=(2 × 23.95 + 27.9)x
2
l
Therefore, x l = 7.6 in. The area of the critical section A c is 2(23.95 × 7.9) + (27.9 × 7.9) =
599 in. 2 Calculate J c = I x + I y for the two equal areas (7.9 × 23.95) with sides parallel to the
direction of moment and the area (7.9 × 27.9) perpendicular to the direction of moment, all about
the axis through CD.
J c = I y + I x = ∑ ( )
bh 3
12 + Ax2
= 2
[7.9 (23.95)3
12
( ) 23.95 2
]
+(7.9 × 23.95) − 7.6
2
[ 27.9
+(1)
12 (7.9)3] +[(27.9 × 7.9)(7.6) 2 ]=39,208 in. 4
or by using Eq. 17.31 for an exterior column. Calculate the maximum and minimum nominal
shear stresses using Eq. 17.27:
v max = V u
+ M v c = 83,100
A c J c 599 + 40.2(12,000)(7.6) = 232 psi
39,208
v min = 138 psi
Allowable v c = φ4 √ √
f c ′ = 0.75 × 4 4000 = 190 psi
Shear stress is greater than the allowable v c , so increase the slab thickness f c ′ or use shear reinforcement.
4. Check the concentration of reinforcement at the exterior column; that is, check that the flexural
capacity of the section is adequate to transfer the negative moment into the exterior column.
The critical area of the slab extends 1.5h on either side of the column, giving an area
(20 + 3 × 9) = 47 in. wide and 9 in. deep. The total moment in the 120-in.-wide column strip is
107 K ⋅ ft, as calculated ) in Example 17.4 (step 5). The moment in a width, c 2 + 3h = 47 in., is
equal to 107 = 41.9K⋅ ft.
(
47
120
If equal spacing in the column strip is used, then the additional reinforcement within the 47-in.
width will be needed for a moment equal to M f − 41.9 = 66 − 41.9 = 24.1 K ⋅ ft. The required
A s = 0.73 in. 2 and four no. 4 bars (A s = 0.8 in. 2 ) may be used. An alternative solution is to arrange
the reinforcement within the column strip to increase the reinforcement within a width of 47 in.
The amount of steel needed within this width should be enough to resist a moment of 0.6 times
the negative moment in the column strip, or 0.6 × 107 = 64.2 K ⋅ ft.
Assume a = 1.0 in. Then
A s =
A s =
M u
φf y (d − a∕2)
64.2(12)
= 1.93 in.2
0.9 × 60(7.9 − 0.5)