Problem Statement:
Use the ACI 318 Direct Design Method to design an interior bay of a flat plate slab system of
multi bay building. The Dimensions of an interior bay are shown in Figure 1.
Figure 1. Interior Bay Dimensions
Design Data
Column Size - 24” x 24”
Materials
• Concrete: normal weight (150 pcf), f’c = 4,000 psi
• Welded Wire Reinforcement (WWR) , fy = 80,000 psi
Loads
• Superimposed dead loads = 30 psf
• Live load = 50 psf
Slab Thickness
Longest clear span ln = 24 – (24/12) = 22.0 ft
Minimum thickness h per ACI Table 9.5(c) = ln/30 = 8.8 in. Use h 9.0 in.
qu =1.2(9/12*150 + 30) + 1.6(50) = 251.0 psf
Design for Flexure
Check if the Direct Design Method of ACI Sect. 13.6 can be utilized to compute the bending moments due to the gravity loads:
• 3 continuous spans in one direction, more than 3 in the other O.K.
• Rectangular panels with long-to-short span ratio = 24/20 = 1.2 < 2 O.K.
• Successive span lengths in each direction are equal O.K.
• No offset columns O.K.
• L/D = 50/(112.5 + 30) = 0.35 < 2 O.K.
Since all requirements are satisfied, the Direct Design Method can be used.
Short Direction (20 ft)
l2 = 24 ft and ln = 20’-2’= 18.0 ft
Factored moment
Mo = =
Division of the total panel moment Mo into negative and positive moments, and then column and middle strip moments, involves the direct application of the moment coefficients in the following Table 1.Middle
Strip
Span Location
D8 at 6 in.
Table 2. Required Bay Reinforcement.
Long Direction (24 ft)
l2 = 20 ft and ln = 24-2= 22.0 ft
Factored moment per span.
Mo = =
Division of the total panel moment Mo into negative and positive moments, and then column and
middle strip moments, involves the direct application of the moment coefficients in the following
Table 3.
212.10 106.05
159.08 63.63
53.03 42.42
Total moment
Slab Moment Int. Neg Int. Pos.
Column Strip
Middle Strip
Table 3 Moment Distribution.
Use the ACI 318 Direct Design Method to design an interior bay of a flat plate slab system of
multi bay building. The Dimensions of an interior bay are shown in Figure 1.
Figure 1. Interior Bay Dimensions
Design Data
Column Size - 24” x 24”
Materials
• Concrete: normal weight (150 pcf), f’c = 4,000 psi
• Welded Wire Reinforcement (WWR) , fy = 80,000 psi
Loads
• Superimposed dead loads = 30 psf
• Live load = 50 psf
Slab Thickness
Longest clear span ln = 24 – (24/12) = 22.0 ft
Minimum thickness h per ACI Table 9.5(c) = ln/30 = 8.8 in. Use h 9.0 in.
qu =1.2(9/12*150 + 30) + 1.6(50) = 251.0 psf
Design for Flexure
Check if the Direct Design Method of ACI Sect. 13.6 can be utilized to compute the bending moments due to the gravity loads:
• 3 continuous spans in one direction, more than 3 in the other O.K.
• Rectangular panels with long-to-short span ratio = 24/20 = 1.2 < 2 O.K.
• Successive span lengths in each direction are equal O.K.
• No offset columns O.K.
• L/D = 50/(112.5 + 30) = 0.35 < 2 O.K.
Since all requirements are satisfied, the Direct Design Method can be used.
Short Direction (20 ft)
l2 = 24 ft and ln = 20’-2’= 18.0 ft
Factored moment
Mo = =
Division of the total panel moment Mo into negative and positive moments, and then column and middle strip moments, involves the direct application of the moment coefficients in the following Table 1.Middle
Strip
Span Location
D8 at 6 in.
Table 2. Required Bay Reinforcement.
Long Direction (24 ft)
l2 = 20 ft and ln = 24-2= 22.0 ft
Factored moment per span.
Mo = =
Division of the total panel moment Mo into negative and positive moments, and then column and
middle strip moments, involves the direct application of the moment coefficients in the following
Table 3.
212.10 106.05
159.08 63.63
53.03 42.42
Total moment
Slab Moment Int. Neg Int. Pos.
Column Strip
Middle Strip
Table 3 Moment Distribution.
No comments:
Post a Comment