design of two way rigid slab - example 10.1 of book by reinforced concrete structures _ design...
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8/3/2019 Design of two way rigid slab - Example 10.1 of Book By Reinforced Concrete Structures _ Design according to CSA
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322 Chapter 10
Step 8.
a) Compute the cross-sectional area ofthe steel reinforcement required in thecolumn and middle strips for negativemoments (computed in Step 6).
b) Compute the minimum reinforcement toensure structural integrity in the eventof shear failure as well as the cornerreinforcement, if applicable.
c) Select bar diameter and spacing, takinginto account minimum requirements andpaying particular attention to simplicity
of placement between strips (consider theiron worker-rodmen and site inspector).
Draw up a computation table.
See Figure 9.6 (Chapter 9) for cornerreinforcement.
See Equation 10.39 for the minimumarea of reinforcement required forstructural integrity.
As,min = 0.002Ag (Equation 10.38)
See Table 10.4 for the maximumallowable spacing.
Use Figure 10.14 for curtailment oftop and bottom reinforcement aswell as bar anchoring.
Note. If slab thickness cannot be increased, drop panels or capitals can be economicalsolutions. If transverse reinforcement proves necessary, refer to Section 10.3.6c.
10.5 Examples
Example 10.1 Design of a Two-Way Slabon Non-Rigid Beams
Problem Statement
Consider a two-way slab supported on beams, as illustrated in Figure X10.1A. The clearheight between oors is 4 m. The columns are 300 mm 300 mm. All the beams are300 mm (bw) by 550 mm (htotal). In addition to its self-weight, the slab supports a deadload of 1 kPa and a live load of 4.8 kPa.
Design the slab along centreline B and centreline 2, including reinforcements and theirspacing.
Given: fc = 30 MPa;fy = 400 MPa; normal-density concrete (unit weight = 24 kN/m3;
bc(slab) = 25 mm.
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8/3/2019 Design of two way rigid slab - Example 10.1 of Book By Reinforced Concrete Structures _ Design according to CSA
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Two-Way Slabs: Direct-Design Method 323
Figure X10.1A Example X10.1
Solution
Step 1. Determine whether the direct-design method can be used
a)L
L
b
a
= =7500
5500
1 36. < 2.0 OK
b) 11
1
1
1
= =
EI
EI
I
I
b
s
b
s
22
2
2
2
= =
EI
EI
I
I
b
s
b
s
SinceIb1 =Ib2, then:
1
2
2
1
1
2
= =
I
I
l
l
s
s
Therefore:
1 2
2
2 1
2
1 2
2
2 1
2
2
2
7 500 5 50 0
5 50 0 7 50 0
l
l
l l
l l= =
= 00 73.
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8/3/2019 Design of two way rigid slab - Example 10.1 of Book By Reinforced Concrete Structures _ Design according to CSA
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324 Chapter 10
0 2 0 73 5 01 2
2
2 1
2. . .< = =96 9 34 2. .kN kN OK