design of concrete stairs
DESCRIPTION
DESIGN OF CONCRETE STAIRS NSCPTRANSCRIPT
DESIGN OF CONCRETE STAIRS
LOWER;
Given:
r = 0.35mm.
t = 0.50mm.
fcβ = 28MPa
Fy = 414MPa
Use:
12mm Ρ RSB main bars
10mm Ρ RSB temperature bars
π =π³
ππ(π. π +
πππ
πππ)
=3662
20(0.4 +
414
700)
= 181.53ππ. π ππ¦ 200ππ.
ππ. ππ πππππ =ππ
πππΈπ
=0.35(0.50)
2(6)(23.54)
= 12.358πΎπ/π
ππ. ππ ππππ =π.πβππ+ππ
π(πΈπ)
=0.2β0.352+0.502
0.50(23.54)
= 5.747πΎπ/π
ππ. ππ ππππ πππ = πΈπh
= 23.54(0.32)
= 7.53πΎπ/π
DL= 12.358+5.747+7.53
=25.635KN/m
LL=2KPa --- from the Design Criteria
πΎπ = π. ππ«π³ + π. ππ³π³
= 39.289 πΎπ/π
Considering 1m strip, (b=1000mm)
ππ’ = 39.289(1)
= 39.289 πΎπ/π
π΄π =πΎππ³π
π
=(39.289)(3.662)2)
8
= 65.859 πΎπ β π.
Effective depth, d
π = π β ππππ ππππππππ β π. ππ π
= 200 β 20 β 0.5(16)
= 172mm.
Solve for π π’
πΉπ =π΄π
β ππ π
=65.859π₯106
0.9(1000)(172)2)
= 2.473
Solve for Ο
π =π.ππππ
β²
ππβ² ((π β βπ β
π(πΉπ)
π.ππππβ² )
= 0.00632
ππππ. =π.π
ππ
= 0.0034
ππππ = π. ππππ
= 0.022
Use π = π. πππππ
Required main bar spacing
π¨π = πππ
= 0.00632(1000)(172)
= 1087.04ππ2
Using 16mm Ρ main bars
π¨π =π
π(πππ)
= 201.06ππ2
πππππππ =π¨π
π¨π(ππππ)
=201.06
1087.04(1000)
= 184.960ππ. π ππ¦ 170ππ.
a.) π1 = 170ππ
b.) 3β = 3(200) = 600ππ. c.) 450ππ
Therefore, Use 16mm. Ρ main bars
SPCD. @ 170mm.
Temperature bars
ππ» = π. ππππ πππ
ππ
= 0.0018
π¨π = ππ»ππ
= 0.0018(1000)(200)
= 360ππ2
Using 10mm Ρ RSB
π¨π =π
π(πππ)
= 78.54ππ2
πππππππ =π¨π
π¨π(ππππ)
=78.54
360(1000)
= 218.167ππ. π ππ¦ 200ππ
a.) π2 = 200ππ
b.) 3β = 3(200) = 450ππ. c.) 450ππ.
Therefore, use 10mm.Ρ temperature
RSB SPCD. @ 200mm. O.C
DESIGN OF CONCRETE STAIRS
UPPER;
Given:
r = 0.37mm.
t = 0.50mm.
fcβ = 28MPa
Fy = 414MPa
Use:
12mm Ρ RSB main bars
10mm Ρ RSB temperature bars
π =π³
ππ(π. π +
πππ
πππ)
=2100
20(0.4 +
414
700)
= 104.25ππ. π ππ¦ 130ππ.
ππ. ππ πππππ =ππ
πππΈπ
=0.37(0.50)
2(4)(23.54)
= 8.71πΎπ/π
ππ. ππ ππππ =π‘βππ+ππ
π(πΈπ)
=0.13β0.372+0.502
0.50(23.54)
= 3.807πΎπ/π
ππ. ππ ππππ πππ = πΈπh
= 23.54(0.32)
= 7.53πΎπ/π
DL= 8.71+3.807+7.53
=20.047KN/m
LL=2KPa --- from the Design Criteria
πΎπ = π. ππ«π³ + π. ππ³π³
= 31.466 πΎπ/π
Considering 1m strip, (b=1000mm)
ππ’ = 31.466(1)
= 31.466 πΎπ/π
π΄π =πΎππ³π
π
=(31.466)(2.1)2)
8
= 17.346 πΎπ β π.
Effective depth, d
π = π β ππππ ππππππππ β π. ππ π
= 130 β 20 β 0.5(16)
= 102mm.
Solve for π π’
πΉπ =π΄π
β ππ π
=17.346π₯106
0.9(1000)(102)2)
= 1.852
Solve for Ο
π =π.ππππ
β²
ππβ² ((π β βπ β
π(πΉπ)
π.ππππβ² )
= 0.00466
ππππ. =π.π
ππ = 0.0034
ππππ = π. ππππ
= 0.022
Use π = π. πππππ
Required main bar spacing
π¨π = πππ
= 0.00466(1000)(102)
= 475.32ππ2
Using 16mm Ρ main bars
π¨π =π
π(πππ)
= 201.06ππ2
πππππππ =π¨π
π¨π(ππππ)
=201.06
475.32(1000)
= 422.99ππ. π ππ¦ 400ππ.
d.) π1 = 400ππ
e.) 3β = 3(130) = 390ππ. f.) 450ππ
Therefore, Use 16mm. Ρ main bars SPCD.
@ 390mm.
Temperature bars
ππ» = π. ππππ πππ
ππ
= 0.0018
π¨π = ππ»ππ
= 0.0018(1000)(130)
= 234ππ2
Using 10mm Ρ RSB
π¨π =π
π(πππ)
= 78.54ππ2
πππππππ =π¨π
π¨π(ππππ)
=78.54
234(1000)
= 335.64ππ. π ππ¦ 300ππ
d.) π2 = 300ππ
e.) 5β = 5(200) = 650ππ. f.) 450ππ.
Therefore, use 10mm.Ρ temperature
RSB SPCD. @ 300mm. O.C
DESIGN OF CONCRETE STAIRS
UPPER
Given:
r = 0.4mm.
t = 0.5mm.
fcβ = 28MPa
Fy = 414MPa
Use:
12mm Ρ RSB main bars
10mm Ρ RSB temperature bars
π =π³
ππ(π. π +
πππ
πππ)
=1200
20(0.4 +
414
700)
= 59.486ππ. π ππ¦ 70ππ.
ππ. ππ πππππ =ππ
πππΈπ
=0.4(0.5)
2(3)(23.54)
= 7.062πΎπ/π
ππ. ππ ππππ =π‘βππ + ππ
π(πΈπ)
=0.07β0.42+0.52
0.5(23.54)
= 2.110πΎπ/π
DL= 7.062+2.110
=9.172KN/m
LL=2KPa --- from the Design Criteria
πΎπ = π. ππ«π³ + π. ππ³π³
= 16.241 πΎπ/π
Considering 1m strip, (b=1000mm)
ππ’ = 16.241(1)
= 16.241 πΎπ/π
π΄π =πΎππ³π
π
=(16.241)(1.22)
8
= 2.924 πΎπ β π.
Effective depth, d
π = π β ππππ ππππππππ β π. ππ π
= 70 β 20 β 0.5(16)
= 42mm.
Solve for π π’
πΉπ =π΄π
β ππ π
=2.924π₯106
0.9(1000)(42)2)
= 1.842
Solve for Ο
π =π.ππππ
β²
ππβ² ((π β βπ β
π(πΉπ)
π.ππππβ² )
= 0.00464
ππππ. =π.π
ππ
= 0.0034
ππππ = π. ππππ
= 0.022
Use π = π. πππππ
Required main bar spacing
π¨π = πππ
= 0.00464(1000)(72)
= 334.08ππ2
Using 16mm Ρ main bars
π΄π =π
4(162)
= 201.06ππ2
πππππππ =π¨π
π¨π(ππππ)
=201.06
334.08(1000)
= 601.832π. π ππ¦ 580ππ.
g.) π1 = 580ππ
h.) 3β = 3(70) = 210ππ. i.) 450ππ
Therefore, Use 16mm. Ρ main bars SPCD.
@ 210mm.
Temperature bars
ππ» = π. ππππ πππ
ππ
= 0.0018
π¨π = ππ»ππ
= 0.0018(1000)(70)
= 126ππ2
Using 10mm Ρ RSB
π¨π =π
π(πππ)
= 78.54ππ2
πππππππ =π¨π
π¨π(ππππ)
=78.54
126(1000)
= 623.33ππ. π ππ¦ 600ππ
g.) π2 = 600ππ
h.) 5β = 5(70) = 350ππ. i.) 450ππ.
Therefore, use 10mm.Ρ temperature
RSB SPCD. @ 350mm. O.C
DESIGN OF CONCRETE STAIRS
LOWER;
Given:
r = 0.35mm.
t = 0.50mm.
fcβ = 28MPa
Fy = 414MPa
Use:
12mm Ρ RSB main bars
10mm Ρ RSB temperature bars
π =π³
ππ(π. π +
πππ
πππ)
=4300
20(0.4 +
414
700)
=213.157ππ. π ππ¦ 220ππ.
ππ. ππ πππππ =ππ
πππΈπ
=0.35(0.5)
2(7)(23.54)
= 14.418πΎπ/π
ππ. ππ ππππ =π‘βππ+ππ
π(πΈπ)
=0.22β0.352+0.502
0.50(23.54)
= 6.321πΎπ/π
DL= 9.416+14.418+6.321
=30.155KN/m
LL=2KPa --- from the Design Criteria
πΎπ = π. ππ«π³ + π. ππ³π³
= 45.617 πΎπ/π
Considering 1m strip, (b=1000mm)
ππ’ = 45.617(1)
= 45.617 πΎπ/π
π΄π =πΎππ³π
π
=(45.617)(403)2)
8
= 105.425 πΎπ β π.
Effective depth, d
π = π β ππππ ππππππππ β π. ππ π
= 220 β 20 β 0.5(16)
= 102mm.
Solve for π π’
πΉπ =π΄π
β ππ π
=105.425π₯106
0.9(1000)(192)2)
= 3.178
Solve for Ο
π =π.ππππ
β²
ππβ² ((π β βπ β
π(πΉπ)
π.ππππβ² )
= 0.0083
ππππ. =π.π
ππ
= 0.0034
ππππ = π. ππππ
= 0.022
Use π = π. πππππ
Required main bar spacing
π¨π = πππ
= 0.0083(1000)(192)
= 1593.6ππ2
Using 16mm Ρ main bars
π΄π =π
4(162) = 201.06ππ2
πππππππ =π¨π
π¨π(ππππ)
=201.06
1593.6(1000)
= 126.167ππ. π ππ¦ 120ππ.
j.) π1 = 120ππ
k.) 5β = 5(130) = 6600ππ. l.) πππππ
Therefore, Use 16mm. Ρ main bars SPCD.
@ 1200mm.
Temperature bars
ππ» = π. ππππ πππ
ππ
= 0.0018
π¨π = ππ»ππ
= 0.0018(1000)(220)
= 396ππ2
Using 10mm Ρ RSB
π¨π =π
π(πππ)
= 78.54ππ2
πππππππ =π¨π
π¨π(ππππ)
=78.54
234(1000)
= 198.33ππ. π ππ¦ 180ππ
j.) π2 = 180ππ
k.) 5β = 5(220) = 660ππ. l.) 450ππ.
Therefore, use 10mm.Ρ temperature
RSB SPCD. @ 180mm. O.C
DESIGN OF CONCRETE STAIRS
STAIRS IN THE BLEACHERS
Given:
r = 0.3mm.
t = 0.3mm.
fcβ = 28MPa
Fy = 414MPa
Use:
12mm Ρ RSB main bars
10mm Ρ RSB temperature bars
π =π³
ππ(π. π +
πππ
πππ)
=2121
20(0.4 +
414
700)
= 105.141ππ. π ππ¦ 120ππ.
ππ. ππ πππππ =ππ
πππΈπ
=0.3(0.3)
2(5)(23.54)
= 5.296πΎπ/π
ππ. ππ ππππ =π‘βππ + ππ
π(πΈπ)
=0.12β0.32 + 0.32
0.3(23.54)
= 3.995πΎπ/π
DL= 5.296+3.995
=9.291 KN/m
LL=2KPa --- from the Design Criteria
ππ’ = 1.4π·πΏ + 1.7πΏπΏ
= 16.407 πΎπ/π
Considering 1m strip, (b=1000mm)
ππ’ = 16.407(1)
= 16.407πΎπ/π
π΄π =πΎππ³π
π
=(16.407)(2.121)2)
8
= 9.226 πΎπ β π.
Effective depth, d
π = π β ππππ ππππππππ β π. ππ π
= 120 β 20 β 0.5(16)
= 92mm.
Solve for π π’
πΉπ =π΄π
β ππ π
=9.226π₯106
0.9(1000)(92)2)
= 1.211
Solve for Ο
π =π.ππππ
β²
ππβ² ((π β βπ β
π(πΉπ)
π.ππππβ² )
= 0.0030
ππππ. =π.π
ππ
= 0.0034
ππππ = π. ππππ
= 0.022
Use π = π. πππ
Required main bar spacing
π¨π = πππ
= 0.0034(1000)(92)
= 312.8ππ2
Using 16mm Ρ main bars
π¨π =π
π(πππ) = 201.06ππ2
πππππππ =π¨π
π¨π(ππππ)
=201.06
312.8(1000)
= 642.774ππ. π ππ¦ 6000ππ.
m.) π1 = 600ππ
n.) 3β = 3(120) = 360ππ. o.) 450ππ
Therefore, Use 16mm. Ρ main bars SPCD.
@ 360mm.
Temperature bars
ππ» = π. ππππ πππ
ππ
= 0.0018
π¨π = ππ»ππ
= 0.0018(1000)(120)
= 216ππ2
Using 10mm Ρ RSB
π¨π =π
π(πππ)
= 78.54ππ2
πππππππ =π¨π
π¨π(ππππ)
=78.54
216(1000)
= 363.611ππ. π ππ¦ 350ππ
m.) π2 = 350ππ
n.) 5β = 5(120) = 1600ππ. o.) 450ππ.
Therefore, use 10mm.Ρ temperature
RSB SPCD. @ 160mm. O.C
DESIGN OF CONCRETE STAIRS
Given:
r = 0.6mm.
t = 0.75mm.
fcβ = 28MPa
Fy = 414MPa
Use:
12mm Ρ RSB main bars
10mm Ρ RSB temperature bars
π =π³
ππ(π. π +
πππ
πππ)
=6000
20(0.4 +
414
700)
=297.43ππ. π ππ¦ 320ππ.
ππ. ππ πππππ =ππ
πππΈπ
=0.6(0.75)
2(7)(23.54)
= 37.075 πΎπ/π
ππ. ππ ππππ =π‘βππ+ππ
π(πΈπ)
=0.32β0.62+0.752
0.75(23.54)
= 9.647 πΎπ/π
ππ. ππ ππππ πππ = πΈπh
= 23.54(0.4)
= 9.416πΎπ/π
DL= 37.075+9.416+9.647
=56.138 KN/m
LL=2KPa --- from the Design Criteria
πΎπ = π. ππ«π³ + π. ππ³π³
= 81.993 πΎπ/π
Considering 1m strip, (b=1000mm)
ππ’ = 81.993(1)
= 81.993 πΎπ/π
π΄π =πΎππ³π
π
=(81.993)(6)2)
8
= 368.968 πΎπ β π.
Effective depth, d
π = π β ππππ ππππππππ β π. ππ π
= 320 β 20 β 0.5(16)
= 102mm.
Solve for π π’
πΉπ =π΄π
β ππ π
=368.968π₯106
0.9(1000)(102)2)
= 4.808
Solve for Ο
π =π.ππππ
β²
ππβ² ((π β βπ β
π(πΉπ)
π.ππππβ² )
= 0.013
ππππ. =π.π
ππ
= 0.0034
ππππ = π. ππππ
= 0.022
Use π = π. πππ
Required main bar spacing
π¨π = πππ
= 0.013(1000)(292) = 3796ππ2
Using 16mm Ρ main bars
π΄π =π
4(162)
= 201.06ππ2
πππππππ =π¨π
π¨π(ππππ)
=201.06
3796(1000)
= 52.966ππ. π ππ¦ 50ππ.
p.) π1 = 50ππ
q.) 3β = 3(320) = 960ππ. r.) 450ππ
Therefore, Use 16mm. Ρ main bars SPCD.
@ 50mm.
Temperature bars
ππ» = π. ππππ πππ
ππ
= 0.0018
π¨π = ππ»ππ
= 0.0018(1000)(320)
= 576ππ2
Using 10mm Ρ RSB
π΄π =π
4(102)
= 78.54ππ2
πππππππ =π¨π
π¨π(ππππ)
=78.54
576(1000)
= 136.354ππ. π ππ¦ 120ππ
p.) π2 = 120ππ
q.) 5β = 5(320) = 1600ππ. r.) 450ππ.
Therefore, use 10mm.Ρ temperature
RSB SPCD. @ 120mm. O.C