Slab Design
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CARMEL B. SABADO BSCE-5
CE-162 2nd Excel Program
PROF. GERONIDES P. ANCOG SEPT. 6, 2009
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)
****====DESIGN OF ONE-WAY SLAB====****
4.6 m
GIVEN: Concrete,fc' =
4.6 m
27.5 Mpa 414 Mpa 23.5 kN/m3 4.8 kPa 4.6 m 10 mm Ø 8 mm Ø
Steel,fy = Unit weight of concrete,wc = Slab live load = clear span,L = main bar diameter = temp. bar diameter =
TYPE OF SLAB SECTION:
1) 2) 3) 4)
3
L L L L
h = L / 28 ;
L 28 165
h = say h =
/ / / /
20 24 28 10
; ; ; ;
simply supported one end continuous two end continuous cantilever
TWO END CONTINUOUS
=
164.29 mm
mm
COMPUTE LOADS: DEAD LOAD(DL)
self weight = wch assume floor finish assume floor finish = ywc total DL
= = = =
3.88 Kpa 25 mm 0.59 Kpa 4.47 KPa
service load
=
4.8 Kpa
=
13.04 Kpa
LIVE LOAD(LL)
wu = 1.2DL + 1.6LL COMPUTE MOMENTS:
M = cwL2 L(m) 4.6 4.6 4.6
coefficients, c
0.11 0.07 0.04
w(kPa) 13.04 13.04 13.04
(kN.m/m) 30.65 19.71 11.5
COMPUTE FOR ρ: 1.4/ fy √¯fc'/ 4fy use ρmin
=
0 0
=
0
=
.85 fc' β1600 ρmax =.75 = fy(600 + fy)
.85 fc' β1600 = fy(600 + fy)
ρmax =.75
0.02
main bar reinforcement = temp bar reinforcement =
78.54 50.27
COMPUTE THE REQUIRED As AND SPACING OF BAR REINFORCEMENTS interior support at mid-span exterior support
Mu = d = h - (20+Ø/2) = Ru = Mu/(.9bd2) ρ= use ρ As = ρbd s = 1000Ao/As
30.65 120 2.37 0.01 0.01 724.3 108.44
= = = = = = =
19.71 120 1.52 0 0 456.08 172.21
11.5 120 0.89 0 0 380 206.68
(kN-m/m) (mm2)
(mm2)
TEMPERATURE BAR SPACING REQUIREMENT: fy < 414MPa
ρtemp.
=
0
fy = 414MPa
ρtemp.
=
0
ρtemp.
=
fy > 414MPa MINIMUM RHO FOR TEMPERATURE BARS:
ρmin ρtemp. As,temp =ρtempbd temp bar spacing
= =
200/fy
=
0.0018
=
0
=
Err:502
0.0018(414)/fy
0.48
(mm2) (mm2)
SLAB DETAILS FOR BENDING OF REINFORCEMENTS fixed bar spacing at mid-span, s 150 = As(required) 456.08 = As(actual) = 1000Ao/s(fixed) 523.6 design is ok!!!=) at exterior support Ap = As(actual)2/3 = 349.07 As(required) 380 = design is not safe!!!=( provide extra bars,n = 1 Ap = Ap + n(Ao) 427.61 = design is ok!!!=) at interior support Ap = As(actual)4/3 = 698.13 As(required) 724.3 = design is not safe!!!=( provide extra bars, n = 2 Ap = Ap + n(Ao) 855.21 = design is ok!!!=)
Slab Details Scheme 1: Bending of reinforcements 10 mm
Ø bent 2/3 spaced @
150 mm 1 -extra top bar is needed
2
1 -extra top bar is needed
8 10 mmØ temp bars @ Err:502 mm
-extra top bars are needed
165 mm
mm Ø bent 2/3 spaced @ 150 mm
2 -extra top bars are needed
Slab Details Scheme 2: Cutting of reinforcements
10 mmØ 190 oc
10 mmØ 380 oc
10 mmØ 190 mm spacing
10 190 oc
8 mmØ Err:502 mm spacing
8 mmØ Err:502 oc
10 mmØ 190 mm spacing mmØ 95 mm spacing
10
CARMEL B. SABADO BSCE-5
CE-162 PROF. GERONIDES P. ANCOG 2nd Excel Program SEPT. 6, 2009
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)
****====DESIGN OF TWO-WAY SLAB====****
7.7 m
6.2 m
GIVEN: fc ' =
Estimate slab thickness based on code minimum thickness requirement. Trial depth, h: 20.7 MPa 414 MPa 23.6 kN/m³ 12 mmφ 6.65 kPa 6.2 m
fy = wc = bar size = Live load = LA = LB =
h=
2( A + B ) 180
=
154.44 mm
say
160 mm
7.7 m
Slab load: DL: slab: wch =
3.78 kPa 0.59 kPa 4.37 kPa 6.65 kPa 17.42 kPa
finish, assume 25mm cement finish 0.025wc = TOTAL
LL: Ultimate load: wu = 1.4DL + 1.7LL = Slab aspect ratio, m: L m = A= LB
=
0.81
Positve moments:
Negative moments:
Case 4: Ma, pos DL Ma, pos LL Mau, pos = 1.2DL + 1.6LL
coeff, c 0.04 0.05
Mb, pos DL Mb, posLL Mbu, pos = 1.2DL + 1.6LL
0.02 0.02
w 4.37 6.65 4.37 6.65
M=cwL²
L 6.2 6.2
6.545 12.270
7.7 7.7
27.486 4.142 7.886 17.587
At continuous edge: Ma, neg Mb,neg
coeff, c 0.07 0.03
wu
M=cwL²
L
17.42
6.2
47.536 kN-m/m
17.42
7.7
29.948 kN-m/m
At discontinuous edge: M = 1/3(Mpos) Ma, neg Mb,neg
9.162 kN-m/m 5.862 kN-m/m
Design of middle strip in the short direction: h Mu = Mmax = Trial d = h - (20 + φ/2) =
Ru =
M u /φ bd 2
= =
d 47.536 Kn-m 134 mm
=
2.94
=
0
.85 f c ' β1 .003E s = fy .003E s + f y
0.02
1.4 fc ' , f y 4 f y
ρmin =
min
ρmax = .75
ρ=
2ωRu .85 f c ' 1 2 Ru 1 − 1 − = 1 − 1 − ω f y f y .85 f c '
Ao =
π 4
Db
0.01 ok, use this rho!!!=)
2
As = ρbd s=
=
1000 Ao As
=
113.1 mm2
=
1048.63 mm2
=
107.85 mm oc
At mid-span: Mu Ru p As s
= = = = =
27.486 Kn-m 1.7 0 ok, use this rho!!!=) 580.06 mm2 194.98 mm oc
At discontinuous end: Mu Ru p As s
= = = = =
9.162 Kn-m 0.57 0 use pmin!!!=( 368.15 mm2 307.2 mm oc
Design od middle strip in the long direction: h Mu dL = h - (20 + 1.5φ) Ru p Ao As s
= = = = = = =
29.948 Kn-m 122 mm 2.24 0.01 ok, use this rho!!!=) 113.1 mm2 776.56 mm2 145.64 mm oc
At mid-span: Mu Ru p
= = =
17.587 Kn-m 1.31 0 ok, use this rho!!!=)
d
As s
= =
402.52 mm2 280.98 mm oc
At discontinuous end: Mu Ru p As s
= = = = =
5.862 Kn-m 0.44 0 use pmin!!!=( 335.19 mm2 337.42 mm oc
Check for shear: Total load on panel, Wt = LALBwu Shear per m of long beam, Case 4: C A
vA = Shear concrete:
= =
C AWT 2 LB d
Vuc = φvc
fc ' 6
bd
831.51 Kn 0.71
=
38.34 Kn/m
=
134 mm
=
91.45 shear
is okay!!=)
SLAB DETAILS FOR BENDING OF REINFORCEMENTS
Short Direction: fixed bar spacing at mid-span, s
= = =
150 mm oc 753.98 >As required at mid-span, ok 580.06
Ap = As(actual)2/3 As(required)
= =
502.65 >As required at exterior support,ok 368.15
Ap = As(actual)4/3 As(required) provide extra bars, n Ap = Ap + n(Ao)
= = = =
1005.31As required at interior support,ok
As(actual) = 1000Ao/s(fixed) As(required) at exterior support
at interior support
Long Direction: fixed bar spacing at mid-span, s
= = =
220 mm oc 514.08 >As required at mid-span, ok 402.52
Ap = As(actual)2/3 As(required)
= =
342.72 >As required at exterior support,ok 335.19
Ap = As(actual)4/3 As(required) provide extra bars, n Ap = Ap + n(Ao)
= = = =
685.44As required at interior support,ok
As(actual) = 1000Ao/s(fixed) As(required) at exterior support
column strip
LB/4
at interior support
provide 2 extra top bars
LB/2
column strip middle strip
12mmφ @ 220 oc bent 2/3 provide 2 extra top bars
LB/4
LB/4
12mmφ @150 oc bent 2/3
LA/4
LA/2
LA/4
column strip middle strip
LA/4
column strip column strip
Slab thickness = 160 mm
12φ bot bars @ 250
12φ top bars @ 150 column strip
LB/4
12φ top bars @ 105
12φ bot bars @ 175
column strip 12φ top bars @ 245
12φ top bars @ 245
LB/4
LB/2
LB/4
Slab Details Scheme 2: Cutting of Reinforcements
LA/4
LA/2
column strip middle strip
LA/4
LA/4
column strip column strip
Slab thickness = 160 mm
>As required at exterior support,ok
>As required at interior support,ok
>As required at exterior support,ok
>As required at interior support,ok
CE-162 2nd Excel Program
PROF. GERONIDES P. ANCOG SEPT. 6, 2009
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)
****====DESIGN OF ONE-WAY SLAB====****
4.6 m
GIVEN: Concrete,fc' =
4.6 m
27.5 Mpa 414 Mpa 23.5 kN/m3 4.8 kPa 4.6 m 10 mm Ø 8 mm Ø
Steel,fy = Unit weight of concrete,wc = Slab live load = clear span,L = main bar diameter = temp. bar diameter =
TYPE OF SLAB SECTION:
1) 2) 3) 4)
3
L L L L
h = L / 28 ;
L 28 165
h = say h =
/ / / /
20 24 28 10
; ; ; ;
simply supported one end continuous two end continuous cantilever
TWO END CONTINUOUS
=
164.29 mm
mm
COMPUTE LOADS: DEAD LOAD(DL)
self weight = wch assume floor finish assume floor finish = ywc total DL
= = = =
3.88 Kpa 25 mm 0.59 Kpa 4.47 KPa
service load
=
4.8 Kpa
=
13.04 Kpa
LIVE LOAD(LL)
wu = 1.2DL + 1.6LL COMPUTE MOMENTS:
M = cwL2 L(m) 4.6 4.6 4.6
coefficients, c
0.11 0.07 0.04
w(kPa) 13.04 13.04 13.04
(kN.m/m) 30.65 19.71 11.5
COMPUTE FOR ρ: 1.4/ fy √¯fc'/ 4fy use ρmin
=
0 0
=
0
=
.85 fc' β1600 ρmax =.75 = fy(600 + fy)
.85 fc' β1600 = fy(600 + fy)
ρmax =.75
0.02
main bar reinforcement = temp bar reinforcement =
78.54 50.27
COMPUTE THE REQUIRED As AND SPACING OF BAR REINFORCEMENTS interior support at mid-span exterior support
Mu = d = h - (20+Ø/2) = Ru = Mu/(.9bd2) ρ= use ρ As = ρbd s = 1000Ao/As
30.65 120 2.37 0.01 0.01 724.3 108.44
= = = = = = =
19.71 120 1.52 0 0 456.08 172.21
11.5 120 0.89 0 0 380 206.68
(kN-m/m) (mm2)
(mm2)
TEMPERATURE BAR SPACING REQUIREMENT: fy < 414MPa
ρtemp.
=
0
fy = 414MPa
ρtemp.
=
0
ρtemp.
=
fy > 414MPa MINIMUM RHO FOR TEMPERATURE BARS:
ρmin ρtemp. As,temp =ρtempbd temp bar spacing
= =
200/fy
=
0.0018
=
0
=
Err:502
0.0018(414)/fy
0.48
(mm2) (mm2)
SLAB DETAILS FOR BENDING OF REINFORCEMENTS fixed bar spacing at mid-span, s 150 = As(required) 456.08 = As(actual) = 1000Ao/s(fixed) 523.6 design is ok!!!=) at exterior support Ap = As(actual)2/3 = 349.07 As(required) 380 = design is not safe!!!=( provide extra bars,n = 1 Ap = Ap + n(Ao) 427.61 = design is ok!!!=) at interior support Ap = As(actual)4/3 = 698.13 As(required) 724.3 = design is not safe!!!=( provide extra bars, n = 2 Ap = Ap + n(Ao) 855.21 = design is ok!!!=)
Slab Details Scheme 1: Bending of reinforcements 10 mm
Ø bent 2/3 spaced @
150 mm 1 -extra top bar is needed
2
1 -extra top bar is needed
8 10 mmØ temp bars @ Err:502 mm
-extra top bars are needed
165 mm
mm Ø bent 2/3 spaced @ 150 mm
2 -extra top bars are needed
Slab Details Scheme 2: Cutting of reinforcements
10 mmØ 190 oc
10 mmØ 380 oc
10 mmØ 190 mm spacing
10 190 oc
8 mmØ Err:502 mm spacing
8 mmØ Err:502 oc
10 mmØ 190 mm spacing mmØ 95 mm spacing
10
CARMEL B. SABADO BSCE-5
CE-162 PROF. GERONIDES P. ANCOG 2nd Excel Program SEPT. 6, 2009
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)
****====DESIGN OF TWO-WAY SLAB====****
7.7 m
6.2 m
GIVEN: fc ' =
Estimate slab thickness based on code minimum thickness requirement. Trial depth, h: 20.7 MPa 414 MPa 23.6 kN/m³ 12 mmφ 6.65 kPa 6.2 m
fy = wc = bar size = Live load = LA = LB =
h=
2( A + B ) 180
=
154.44 mm
say
160 mm
7.7 m
Slab load: DL: slab: wch =
3.78 kPa 0.59 kPa 4.37 kPa 6.65 kPa 17.42 kPa
finish, assume 25mm cement finish 0.025wc = TOTAL
LL: Ultimate load: wu = 1.4DL + 1.7LL = Slab aspect ratio, m: L m = A= LB
=
0.81
Positve moments:
Negative moments:
Case 4: Ma, pos DL Ma, pos LL Mau, pos = 1.2DL + 1.6LL
coeff, c 0.04 0.05
Mb, pos DL Mb, posLL Mbu, pos = 1.2DL + 1.6LL
0.02 0.02
w 4.37 6.65 4.37 6.65
M=cwL²
L 6.2 6.2
6.545 12.270
7.7 7.7
27.486 4.142 7.886 17.587
At continuous edge: Ma, neg Mb,neg
coeff, c 0.07 0.03
wu
M=cwL²
L
17.42
6.2
47.536 kN-m/m
17.42
7.7
29.948 kN-m/m
At discontinuous edge: M = 1/3(Mpos) Ma, neg Mb,neg
9.162 kN-m/m 5.862 kN-m/m
Design of middle strip in the short direction: h Mu = Mmax = Trial d = h - (20 + φ/2) =
Ru =
M u /φ bd 2
= =
d 47.536 Kn-m 134 mm
=
2.94
=
0
.85 f c ' β1 .003E s = fy .003E s + f y
0.02
1.4 fc ' , f y 4 f y
ρmin =
min
ρmax = .75
ρ=
2ωRu .85 f c ' 1 2 Ru 1 − 1 − = 1 − 1 − ω f y f y .85 f c '
Ao =
π 4
Db
0.01 ok, use this rho!!!=)
2
As = ρbd s=
=
1000 Ao As
=
113.1 mm2
=
1048.63 mm2
=
107.85 mm oc
At mid-span: Mu Ru p As s
= = = = =
27.486 Kn-m 1.7 0 ok, use this rho!!!=) 580.06 mm2 194.98 mm oc
At discontinuous end: Mu Ru p As s
= = = = =
9.162 Kn-m 0.57 0 use pmin!!!=( 368.15 mm2 307.2 mm oc
Design od middle strip in the long direction: h Mu dL = h - (20 + 1.5φ) Ru p Ao As s
= = = = = = =
29.948 Kn-m 122 mm 2.24 0.01 ok, use this rho!!!=) 113.1 mm2 776.56 mm2 145.64 mm oc
At mid-span: Mu Ru p
= = =
17.587 Kn-m 1.31 0 ok, use this rho!!!=)
d
As s
= =
402.52 mm2 280.98 mm oc
At discontinuous end: Mu Ru p As s
= = = = =
5.862 Kn-m 0.44 0 use pmin!!!=( 335.19 mm2 337.42 mm oc
Check for shear: Total load on panel, Wt = LALBwu Shear per m of long beam, Case 4: C A
vA = Shear concrete:
= =
C AWT 2 LB d
Vuc = φvc
fc ' 6
bd
831.51 Kn 0.71
=
38.34 Kn/m
=
134 mm
=
91.45 shear
is okay!!=)
SLAB DETAILS FOR BENDING OF REINFORCEMENTS
Short Direction: fixed bar spacing at mid-span, s
= = =
150 mm oc 753.98 >As required at mid-span, ok 580.06
Ap = As(actual)2/3 As(required)
= =
502.65 >As required at exterior support,ok 368.15
Ap = As(actual)4/3 As(required) provide extra bars, n Ap = Ap + n(Ao)
= = = =
1005.31
As(actual) = 1000Ao/s(fixed) As(required) at exterior support
at interior support
Long Direction: fixed bar spacing at mid-span, s
= = =
220 mm oc 514.08 >As required at mid-span, ok 402.52
Ap = As(actual)2/3 As(required)
= =
342.72 >As required at exterior support,ok 335.19
Ap = As(actual)4/3 As(required) provide extra bars, n Ap = Ap + n(Ao)
= = = =
685.44
As(actual) = 1000Ao/s(fixed) As(required) at exterior support
column strip
LB/4
at interior support
provide 2 extra top bars
LB/2
column strip middle strip
12mmφ @ 220 oc bent 2/3 provide 2 extra top bars
LB/4
LB/4
12mmφ @150 oc bent 2/3
LA/4
LA/2
LA/4
column strip middle strip
LA/4
column strip column strip
Slab thickness = 160 mm
12φ bot bars @ 250
12φ top bars @ 150 column strip
LB/4
12φ top bars @ 105
12φ bot bars @ 175
column strip 12φ top bars @ 245
12φ top bars @ 245
LB/4
LB/2
LB/4
Slab Details Scheme 2: Cutting of Reinforcements
LA/4
LA/2
column strip middle strip
LA/4
LA/4
column strip column strip
Slab thickness = 160 mm
>As required at exterior support,ok
>As required at interior support,ok
>As required at exterior support,ok
>As required at interior support,ok
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