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Design S-2
Two Way Slab, Coefficient Method 2 Case 1
f'c = 3000 psi, fy = 60000 psi, β₁ = 0.85
L = 240", S = 192", h = 5.5", d = 4.3", m = 0.80
Loading: (Take 1 ft strip)
Slab (Self)
69 lbs/ft
Dead Load
40 lbs/ft
Live Load
100 lbs/ft
Short Direction: Continuous edge
Rn =
42880.2
0.9(12)(4.3)2
= 214.73 psi
ρ=
0.85(3000)
60000
[1 − √1 −
200
60000
= 0.00333 < 0.00374 (use ρ)
= 290.8 lbs/ft
Solve for Flexure
Short Direction: Midspan
Moments at Short Direction
Rn =
0.048(290.8)(192)2
12
= 42880.20 in lbs, Continuous edge (CE)
0.036(290.8)(192)2
ρ=
2(161.05)
0.85(3000)
]
Long Direction: Continuous edge
Rn =
0.025(290.8)(192)2
29480.14
0.9(12)(4.3)2
= 147.63 psi
12
= 22333.44 in lbs, Midspan
ρ=
Check for Slab Thickness
2(147.63)
0.85(3000)
[1 − √1 −
]
60000
0.85(3000)
= 0.00254
0.75(0.85)(3000)(0.85)(87000)
(60000)(87000 + 60000)
200
60000
= 0.00333 > 0.00254 (use ρmin )
ρmin =
= 0.01604
(0.01604)(60)
]
3
= 780.24 psi
Long Direction: Midspan
Rn =
Mumax = 0.9(780.24)(12)(4.3)2
22333.44
0.9(12)(4.3)2
= 111.84 psi
= 155807.69 in lbs > Mu (ok)
Solve for Steel Ratio ρ
page 1 of 2 (S-2)
60000
[1 − √1 −
200
60000
= 0.00333 > 0.00277 (use ρmin )
0.033(290.8)(192)2
Rnmax = (0.01604)(60000)[1 − 0.59
0.85(3000)
ρmin =
12
= 29480.14 in lbs, Continuous edge (CE)
ρmax =
0.9(12)(4.3)2
= 161.05 psi
Moments at Long Direction
+Mu =
32160.15
= 0.00277
12
= 32160.15 in lbs, Midspan
−Mu =
]
ρmin =
Wu = 1.2(69 + 40) + 1.6(100)
+Mu =
0.85(3000)
= 0.00374
Factored Load
−Mu =
2(214.73)
powered by eStructural
ρ=
0.85(3000)
60000
[1 − √1 −
2(111.84)
0.85(3000)
]
VDL =
200
=
60000
= 0.00333 > 0.00191 (use ρmin )
Design for Steel
VDL =
As = 0.00374(12)(4.3)
2
= 0.19298 in
= 1830.09 lbs
Long Direction
VDL =
As = 0.00333(12)(4.3)
VLL =
0.17183
= 13.97 "; use 13.5 "
Long Direction Bars: 1/2"Ø, Grade 60
Continuous edge
As = 0.00333(12)(4.3)
= 0.17183 in2
(12)(0.20)
0.17183
= 13.97 "; use 13.5 "
Midspan
As = 0.00333(12)(4.3)
= 0.17183 in2
Sp =
100(192)
3(12)
Vu = 1.2(581.33) + 1.6(533.33)
= 1550.92 lbs
Test for Shear Strength
1830.09
< Vc = 2√3000(12)(4.3)
0.75
2440.12 lbs < 5652.50 lbs (ok)
Vn =
Recommendation
f 'c = 3000 psi, fy = 60000 psi
Panel Dimension (L x S): 20' x 16'
Thickness (h): 5.5"
Short Direction Bars: Grade 60
Top Bars 4' at CE: 1/2"Ø @ 12" O.C.
Bottom Bars: 1/2"Ø @ 13.5" O.C.
Long Direction Bars: Grade 60
Top Bars 5' at CE: 1/2"Ø @ 13.5" O.C.
Bottom Bars: 1/2"Ø @ 13.5" O.C.
Remark:
C.E.: Continuous Edge
(12)(0.20)
0.17183
= 13.97 "; use 13.5 "
Solve for Shear Forces
Short Direction
page 2 of 2 (S-2)
3(12)
= 533.33 lbs
= 0.17183 in2
(12)(0.20)
109(192)
= 581.33 lbs
Midspan
Sp =
3(12)(2)
= 629.33 lbs
(12)(0.20)
0.19298
= 12.44 "; use 12.0 "
Sp =
100(192)(3 − 0.802 )
Vu = 1.2(685.97) + 1.6(629.33)
Short Direction Bars: 1/2"Ø, Grade 60
Continuous edge
Sp =
3(12)(2)
= 685.97 lbs
= 0.00191
ρmin
109(192)(3 − 0.802 )
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