Structural Concrete - Hassoun

17.10 Transfer of Unbalanced Moments to Columns 663
Example 17.9
For the flat plate in Example 17.4, calculate the shear stresses in the slab at the critical sections due to
unbalanced moments and shearing forces at an interior and exterior column. Check the concentration of
reinforcement and torsional requirements at the exterior column. Use f c ′ = 4ksiandf y = 60 ksi.
Solution
1. The unbalanced moment at the interior support is M u = 23 K ⋅ ft (Example 17.8), where γ f = 0.6
(because c 1 = c 2 = 20 in.). The moment to be transferred by flexure is
The moment to be transferred by shear is
M f = γ f M u = 0.6 × 23 = 13.8K⋅ ft
M v = 23 − 13.8 = 9.2K⋅ ft
Alternatively, moments calculated from Eq. 17.22a may be used producing higher shear stresses.
Using d = 7.9 in. (Example 17.4),
[ ( ) 27.9 2
]
V u = 0.33 20 × 24 − = 156.6K
12
From Fig. 17.27,
A c = 4(27.9)(7.9) =882 in.
J c = d ( )
x
3
2 3 + x2 y + xd3
6
= 7.9
[ (27.9)
3
]
+(27.9) 2 (27.9)
2 3
v max = 156,600
882
= 177 + 13 = 190 psi
v min = 177 − 13 = 164 psi
+ 9.2(12,000)(27.9∕2)
114,670
+ 27.9
6 (7.9)3 = 114,670 in. 4
Figure 17.27
moment.
Example 17.9: Shear stresses at interior column due to unbalanced