Structural Concrete - Hassoun

662 Chapter 17 Design of Two-Way Slabs
and
For an exterior column,
and
J c = 2dx3
3
J c = d ( )
x
3
2 3 + x2 y
+ xd3
6
(17.29)
A c = d(2x + y) (17.30)
(
)
+(2x + y)dx 2 1 − 2dx 2 x 1 − yd3
12
(17.31)
where x, x 1 ,andy are as shown in Fig. 17.26. The maximum shear stress, v 1 = V u /A c + M v C/J c ,
must be less than φ(4 √ f ′ c); otherwise, shear reinforcement should be provided.
Example 17.8
Determine the moments at the exterior and interior columns in the long direction of the flat plate in
Example 17.4.
Solution
1. Find the exterior column moment. From Examples Example 17.4 and Example 17.5,
q Du =(136.5)(1.2) =0.16 ksf
0.5q Lu = 0.5 ×(1.6 × 100) =80 psf
(
l 2 = l ′ 2 = 20 ft l n = l′ n = 22.33 ft 1 + 1
)
= 1.87
α ec
The unbalanced moment to be transferred to the exterior column using Eq. 17.22b is
M u = 0.08
1.87 [(0.16 + 0.08)(20)(22.33)2 − 0] =102 K ⋅ ft
If Eq. 17.22a is used, M u = 168 K ⋅ ft, which is a conservative value.
2. At an interior support, the slab stiffness on both sides of the column must be used to compute α ec :
α ec =
K ec
∑ (Ks + K b )
(Eq. 17.21)
From Example 17.5, K ec = 233 E c , K s = 202.5 E c ,andK b = 0. Therefore,
α ec =
233E c
= 0.58
(2)202.5E c
(
1 + 1
)
= 1 + 1
α ec 0.58 = 2.72
From Eq. 17.22b, the unbalanced moment at an interior support is
M u = 0.08
2.72 [(0.16 + 0.08)(20)(22.33)2 − 0.16(20)(22.33) 2 ]=23 K ⋅ ft
If Eq. 17.22a is used, M u = 42 K ⋅ ft, which is a conservative value.