DESIGN OF CONCRETE STAIRS

LOWER;

Given:

r = 0.35mm.

t = 0.50mm.

fc’ = 28MPa

Fy = 414MPa

Use:

12mm ф RSB main bars

10mm ф RSB temperature bars

𝒉 =𝑳

𝟐𝟎(𝟎. 𝟒 +

𝟒𝟏𝟒

𝟕𝟎𝟎)

=3662

20(0.4 +

414

700)

= 181.53𝑚𝑚. 𝑠𝑎𝑦 200𝑚𝑚.

𝒘𝒕. 𝒐𝒇 𝒔𝒕𝒆𝒑𝒔 =𝒓𝒕

𝟐𝒏𝜸𝒄

=0.35(0.50)

2(6)(23.54)

= 12.358𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒔𝒍𝒂𝒃 =𝟎.𝟐√𝒓𝟐+𝒕𝟐

𝒕(𝜸𝒄)

=0.2√0.352+0.502

0.50(23.54)

= 5.747𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒍𝒂𝒏𝒅𝒊𝒏𝒈 = 𝜸𝒄h

= 23.54(0.32)

= 7.53𝐾𝑁/𝑚

DL= 12.358+5.747+7.53

=25.635KN/m

LL=2KPa --- from the Design Criteria

𝑾𝒖 = 𝟏. 𝟒𝑫𝑳 + 𝟏. 𝟕𝑳𝑳

= 39.289 𝐾𝑁/𝑚

Considering 1m strip, (b=1000mm)

𝑊𝑢 = 39.289(1)

= 39.289 𝐾𝑁/𝑚

𝑴𝒖 =𝑾𝒖𝑳𝟐

𝟖

=(39.289)(3.662)2)

8

= 65.859 𝐾𝑁 − 𝑚.

Effective depth, d

𝒅 = 𝒉 − 𝒔𝒍𝒂𝒃 𝒄𝒐𝒗𝒆𝒓𝒊𝒏𝒈 − 𝟎. 𝟓𝒅𝒃

= 200 − 20 − 0.5(16)

= 172mm.

Solve for 𝑅𝑢

𝑹𝒖 =𝑴𝒖

∅𝒃𝒅𝟐

=65.859𝑥106

0.9(1000)(172)2)

= 2.473

Solve for ρ

𝝆 =𝟎.𝟖𝟓𝒇𝒄

′

𝒇𝒄′ ((𝟏 − √𝟏 −

𝟐(𝑹𝒖)

𝟎.𝟖𝟓𝒇𝒄′ )

= 0.00632

𝝆𝒎𝒊𝒏. =𝟏.𝟒

𝒇𝒚

= 0.0034

𝝆𝒎𝒂𝒙 = 𝟎. 𝟕𝟓𝝆𝒃

= 0.022

Use 𝝆 = 𝟎. 𝟎𝟎𝟔𝟑𝟐

Required main bar spacing

𝑨𝒔 = 𝝆𝒃𝒅

= 0.00632(1000)(172)

= 1087.04𝑚𝑚2

Using 16mm ф main bars

𝑨𝒃 =𝝅

𝟒(𝟏𝟔𝟐)

= 201.06𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=201.06

1087.04(1000)

= 184.960𝑚𝑚. 𝑠𝑎𝑦 170𝑚𝑚.

a.) 𝑆1 = 170𝑚𝑚

b.) 3ℎ = 3(200) = 600𝑚𝑚. c.) 450𝑚𝑚

Therefore, Use 16mm. ф main bars

SPCD. @ 170mm.

Temperature bars

𝝆𝑻 = 𝟎. 𝟎𝟎𝟏𝟖 𝟒𝟏𝟓

𝒇𝒚

= 0.0018

𝑨𝒔 = 𝝆𝑻𝒃𝒉

= 0.0018(1000)(200)

= 360𝑚𝑚2

Using 10mm ф RSB

𝑨𝒃 =𝝅

𝟒(𝟏𝟎𝟐)

= 78.54𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=78.54

360(1000)

= 218.167𝑚𝑚. 𝑠𝑎𝑦 200𝑚𝑚

a.) 𝑆2 = 200𝑚𝑚

b.) 3ℎ = 3(200) = 450𝑚𝑚. c.) 450𝑚𝑚.

Therefore, use 10mm.ф temperature

RSB SPCD. @ 200mm. O.C

DESIGN OF CONCRETE STAIRS

UPPER;

Given:

r = 0.37mm.

t = 0.50mm.

fc’ = 28MPa

Fy = 414MPa

Use:

12mm ф RSB main bars

10mm ф RSB temperature bars

𝒉 =𝑳

𝟐𝟎(𝟎. 𝟒 +

𝟒𝟏𝟒

𝟕𝟎𝟎)

=2100

20(0.4 +

414

700)

= 104.25𝑚𝑚. 𝑠𝑎𝑦 130𝑚𝑚.

𝒘𝒕. 𝒐𝒇 𝒔𝒕𝒆𝒑𝒔 =𝒓𝒕

𝟐𝒏𝜸𝒄

=0.37(0.50)

2(4)(23.54)

= 8.71𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒔𝒍𝒂𝒃 =𝐡√𝒓𝟐+𝒕𝟐

𝒕(𝜸𝒄)

=0.13√0.372+0.502

0.50(23.54)

= 3.807𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒍𝒂𝒏𝒅𝒊𝒏𝒈 = 𝜸𝒄h

= 23.54(0.32)

= 7.53𝐾𝑁/𝑚

DL= 8.71+3.807+7.53

=20.047KN/m

LL=2KPa --- from the Design Criteria

𝑾𝒖 = 𝟏. 𝟒𝑫𝑳 + 𝟏. 𝟕𝑳𝑳

= 31.466 𝐾𝑁/𝑚

Considering 1m strip, (b=1000mm)

𝑊𝑢 = 31.466(1)

= 31.466 𝐾𝑁/𝑚

𝑴𝒖 =𝑾𝒖𝑳𝟐

𝟖

=(31.466)(2.1)2)

8

= 17.346 𝐾𝑁 − 𝑚.

Effective depth, d

𝒅 = 𝒉 − 𝒔𝒍𝒂𝒃 𝒄𝒐𝒗𝒆𝒓𝒊𝒏𝒈 − 𝟎. 𝟓𝒅𝒃

= 130 − 20 − 0.5(16)

= 102mm.

Solve for 𝑅𝑢

𝑹𝒖 =𝑴𝒖

∅𝒃𝒅𝟐

=17.346𝑥106

0.9(1000)(102)2)

= 1.852

Solve for ρ

𝝆 =𝟎.𝟖𝟓𝒇𝒄

′

𝒇𝒄′ ((𝟏 − √𝟏 −

𝟐(𝑹𝒖)

𝟎.𝟖𝟓𝒇𝒄′ )

= 0.00466

𝝆𝒎𝒊𝒏. =𝟏.𝟒

𝒇𝒚 = 0.0034

𝝆𝒎𝒂𝒙 = 𝟎. 𝟕𝟓𝝆𝒃

= 0.022

Use 𝝆 = 𝟎. 𝟎𝟎𝟒𝟔𝟔

Required main bar spacing

𝑨𝒔 = 𝝆𝒃𝒅

= 0.00466(1000)(102)

= 475.32𝑚𝑚2

Using 16mm ф main bars

𝑨𝒃 =𝝅

𝟒(𝟏𝟔𝟐)

= 201.06𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=201.06

475.32(1000)

= 422.99𝑚𝑚. 𝑠𝑎𝑦 400𝑚𝑚.

d.) 𝑆1 = 400𝑚𝑚

e.) 3ℎ = 3(130) = 390𝑚𝑚. f.) 450𝑚𝑚

Therefore, Use 16mm. ф main bars SPCD.

@ 390mm.

Temperature bars

𝝆𝑻 = 𝟎. 𝟎𝟎𝟏𝟖 𝟒𝟏𝟓

𝒇𝒚

= 0.0018

𝑨𝒔 = 𝝆𝑻𝒃𝒉

= 0.0018(1000)(130)

= 234𝑚𝑚2

Using 10mm ф RSB

𝑨𝒃 =𝝅

𝟒(𝟏𝟎𝟐)

= 78.54𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=78.54

234(1000)

= 335.64𝑚𝑚. 𝑠𝑎𝑦 300𝑚𝑚

d.) 𝑆2 = 300𝑚𝑚

e.) 5ℎ = 5(200) = 650𝑚𝑚. f.) 450𝑚𝑚.

Therefore, use 10mm.ф temperature

RSB SPCD. @ 300mm. O.C

DESIGN OF CONCRETE STAIRS

UPPER

Given:

r = 0.4mm.

t = 0.5mm.

fc’ = 28MPa

Fy = 414MPa

Use:

12mm ф RSB main bars

10mm ф RSB temperature bars

𝒉 =𝑳

𝟐𝟎(𝟎. 𝟒 +

𝟒𝟏𝟒

𝟕𝟎𝟎)

=1200

20(0.4 +

414

700)

= 59.486𝑚𝑚. 𝑠𝑎𝑦 70𝑚𝑚.

𝒘𝒕. 𝒐𝒇 𝒔𝒕𝒆𝒑𝒔 =𝒓𝒕

𝟐𝒏𝜸𝒄

=0.4(0.5)

2(3)(23.54)

= 7.062𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒔𝒍𝒂𝒃 =𝐡√𝒓𝟐 + 𝒕𝟐

𝒕(𝜸𝒄)

=0.07√0.42+0.52

0.5(23.54)

= 2.110𝐾𝑁/𝑚

DL= 7.062+2.110

=9.172KN/m

LL=2KPa --- from the Design Criteria

𝑾𝒖 = 𝟏. 𝟒𝑫𝑳 + 𝟏. 𝟕𝑳𝑳

= 16.241 𝐾𝑁/𝑚

Considering 1m strip, (b=1000mm)

𝑊𝑢 = 16.241(1)

= 16.241 𝐾𝑁/𝑚

𝑴𝒖 =𝑾𝒖𝑳𝟐

𝟖

=(16.241)(1.22)

8

= 2.924 𝐾𝑁 − 𝑚.

Effective depth, d

𝒅 = 𝒉 − 𝒔𝒍𝒂𝒃 𝒄𝒐𝒗𝒆𝒓𝒊𝒏𝒈 − 𝟎. 𝟓𝒅𝒃

= 70 − 20 − 0.5(16)

= 42mm.

Solve for 𝑅𝑢

𝑹𝒖 =𝑴𝒖

∅𝒃𝒅𝟐

=2.924𝑥106

0.9(1000)(42)2)

= 1.842

Solve for ρ

𝝆 =𝟎.𝟖𝟓𝒇𝒄

′

𝒇𝒄′ ((𝟏 − √𝟏 −

𝟐(𝑹𝒖)

𝟎.𝟖𝟓𝒇𝒄′ )

= 0.00464

𝝆𝒎𝒊𝒏. =𝟏.𝟒

𝒇𝒚

= 0.0034

𝝆𝒎𝒂𝒙 = 𝟎. 𝟕𝟓𝝆𝒃

= 0.022

Use 𝝆 = 𝟎. 𝟎𝟎𝟒𝟔𝟒

Required main bar spacing

𝑨𝒔 = 𝝆𝒃𝒅

= 0.00464(1000)(72)

= 334.08𝑚𝑚2

Using 16mm ф main bars

𝐴𝑏 =𝜋

4(162)

= 201.06𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=201.06

334.08(1000)

= 601.832𝑚. 𝑠𝑎𝑦 580𝑚𝑚.

g.) 𝑆1 = 580𝑚𝑚

h.) 3ℎ = 3(70) = 210𝑚𝑚. i.) 450𝑚𝑚

Therefore, Use 16mm. ф main bars SPCD.

@ 210mm.

Temperature bars

𝝆𝑻 = 𝟎. 𝟎𝟎𝟏𝟖 𝟒𝟏𝟓

𝒇𝒚

= 0.0018

𝑨𝒔 = 𝝆𝑻𝒃𝒉

= 0.0018(1000)(70)

= 126𝑚𝑚2

Using 10mm ф RSB

𝑨𝒃 =𝝅

𝟒(𝟏𝟎𝟐)

= 78.54𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=78.54

126(1000)

= 623.33𝑚𝑚. 𝑠𝑎𝑦 600𝑚𝑚

g.) 𝑆2 = 600𝑚𝑚

h.) 5ℎ = 5(70) = 350𝑚𝑚. i.) 450𝑚𝑚.

Therefore, use 10mm.ф temperature

RSB SPCD. @ 350mm. O.C

DESIGN OF CONCRETE STAIRS

LOWER;

Given:

r = 0.35mm.

t = 0.50mm.

fc’ = 28MPa

Fy = 414MPa

Use:

12mm ф RSB main bars

10mm ф RSB temperature bars

𝒉 =𝑳

𝟐𝟎(𝟎. 𝟒 +

𝟒𝟏𝟒

𝟕𝟎𝟎)

=4300

20(0.4 +

414

700)

=213.157𝑚𝑚. 𝑠𝑎𝑦 220𝑚𝑚.

𝒘𝒕. 𝒐𝒇 𝒔𝒕𝒆𝒑𝒔 =𝒓𝒕

𝟐𝒏𝜸𝒄

=0.35(0.5)

2(7)(23.54)

= 14.418𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒔𝒍𝒂𝒃 =𝐡√𝒓𝟐+𝒕𝟐

𝒕(𝜸𝒄)

=0.22√0.352+0.502

0.50(23.54)

= 6.321𝐾𝑁/𝑚

DL= 9.416+14.418+6.321

=30.155KN/m

LL=2KPa --- from the Design Criteria

𝑾𝒖 = 𝟏. 𝟒𝑫𝑳 + 𝟏. 𝟕𝑳𝑳

= 45.617 𝐾𝑁/𝑚

Considering 1m strip, (b=1000mm)

𝑊𝑢 = 45.617(1)

= 45.617 𝐾𝑁/𝑚

𝑴𝒖 =𝑾𝒖𝑳𝟐

𝟖

=(45.617)(403)2)

8

= 105.425 𝐾𝑁 − 𝑚.

Effective depth, d

𝒅 = 𝒉 − 𝒔𝒍𝒂𝒃 𝒄𝒐𝒗𝒆𝒓𝒊𝒏𝒈 − 𝟎. 𝟓𝒅𝒃

= 220 − 20 − 0.5(16)

= 102mm.

Solve for 𝑅𝑢

𝑹𝒖 =𝑴𝒖

∅𝒃𝒅𝟐

=105.425𝑥106

0.9(1000)(192)2)

= 3.178

Solve for ρ

𝝆 =𝟎.𝟖𝟓𝒇𝒄

′

𝒇𝒄′ ((𝟏 − √𝟏 −

𝟐(𝑹𝒖)

𝟎.𝟖𝟓𝒇𝒄′ )

= 0.0083

𝝆𝒎𝒊𝒏. =𝟏.𝟒

𝒇𝒚

= 0.0034

𝝆𝒎𝒂𝒙 = 𝟎. 𝟕𝟓𝝆𝒃

= 0.022

Use 𝝆 = 𝟎. 𝟎𝟎𝟒𝟔𝟔

Required main bar spacing

𝑨𝒔 = 𝝆𝒃𝒅

= 0.0083(1000)(192)

= 1593.6𝑚𝑚2

Using 16mm ф main bars

𝐴𝑏 =𝜋

4(162) = 201.06𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=201.06

1593.6(1000)

= 126.167𝑚𝑚. 𝑠𝑎𝑦 120𝑚𝑚.

j.) 𝑆1 = 120𝑚𝑚

k.) 5ℎ = 5(130) = 6600𝑚𝑚. l.) 𝟒𝟓𝟎𝒎𝒎

Therefore, Use 16mm. ф main bars SPCD.

@ 1200mm.

Temperature bars

𝝆𝑻 = 𝟎. 𝟎𝟎𝟏𝟖 𝟒𝟏𝟓

𝒇𝒚

= 0.0018

𝑨𝒔 = 𝝆𝑻𝒃𝒉

= 0.0018(1000)(220)

= 396𝑚𝑚2

Using 10mm ф RSB

𝑨𝒃 =𝝅

𝟒(𝟏𝟎𝟐)

= 78.54𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=78.54

234(1000)

= 198.33𝑚𝑚. 𝑠𝑎𝑦 180𝑚𝑚

j.) 𝑆2 = 180𝑚𝑚

k.) 5ℎ = 5(220) = 660𝑚𝑚. l.) 450𝑚𝑚.

Therefore, use 10mm.ф temperature

RSB SPCD. @ 180mm. O.C

DESIGN OF CONCRETE STAIRS

STAIRS IN THE BLEACHERS

Given:

r = 0.3mm.

t = 0.3mm.

fc’ = 28MPa

Fy = 414MPa

Use:

12mm ф RSB main bars

10mm ф RSB temperature bars

𝒉 =𝑳

𝟐𝟎(𝟎. 𝟒 +

𝟒𝟏𝟒

𝟕𝟎𝟎)

=2121

20(0.4 +

414

700)

= 105.141𝑚𝑚. 𝑠𝑎𝑦 120𝑚𝑚.

𝒘𝒕. 𝒐𝒇 𝒔𝒕𝒆𝒑𝒔 =𝒓𝒕

𝟐𝒏𝜸𝒄

=0.3(0.3)

2(5)(23.54)

= 5.296𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒔𝒍𝒂𝒃 =𝐡√𝒓𝟐 + 𝒕𝟐

𝒕(𝜸𝒄)

=0.12√0.32 + 0.32

0.3(23.54)

= 3.995𝐾𝑁/𝑚

DL= 5.296+3.995

=9.291 KN/m

LL=2KPa --- from the Design Criteria

𝑊𝑢 = 1.4𝐷𝐿 + 1.7𝐿𝐿

= 16.407 𝐾𝑁/𝑚

Considering 1m strip, (b=1000mm)

𝑊𝑢 = 16.407(1)

= 16.407𝐾𝑁/𝑚

𝑴𝒖 =𝑾𝒖𝑳𝟐

𝟖

=(16.407)(2.121)2)

8

= 9.226 𝐾𝑁 − 𝑚.

Effective depth, d

𝒅 = 𝒉 − 𝒔𝒍𝒂𝒃 𝒄𝒐𝒗𝒆𝒓𝒊𝒏𝒈 − 𝟎. 𝟓𝒅𝒃

= 120 − 20 − 0.5(16)

= 92mm.

Solve for 𝑅𝑢

𝑹𝒖 =𝑴𝒖

∅𝒃𝒅𝟐

=9.226𝑥106

0.9(1000)(92)2)

= 1.211

Solve for ρ

𝝆 =𝟎.𝟖𝟓𝒇𝒄

′

𝒇𝒄′ ((𝟏 − √𝟏 −

𝟐(𝑹𝒖)

𝟎.𝟖𝟓𝒇𝒄′ )

= 0.0030

𝝆𝒎𝒊𝒏. =𝟏.𝟒

𝒇𝒚

= 0.0034

𝝆𝒎𝒂𝒙 = 𝟎. 𝟕𝟓𝝆𝒃

= 0.022

Use 𝝆 = 𝟎. 𝟎𝟏𝟑

Required main bar spacing

𝑨𝒔 = 𝝆𝒃𝒅

= 0.0034(1000)(92)

= 312.8𝑚𝑚2

Using 16mm ф main bars

𝑨𝒃 =𝝅

𝟒(𝟏𝟔𝟐) = 201.06𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=201.06

312.8(1000)

= 642.774𝑚𝑚. 𝑠𝑎𝑦 6000𝑚𝑚.

m.) 𝑆1 = 600𝑚𝑚

n.) 3ℎ = 3(120) = 360𝑚𝑚. o.) 450𝑚𝑚

Therefore, Use 16mm. ф main bars SPCD.

@ 360mm.

Temperature bars

𝝆𝑻 = 𝟎. 𝟎𝟎𝟏𝟖 𝟒𝟏𝟓

𝒇𝒚

= 0.0018

𝑨𝒔 = 𝝆𝑻𝒃𝒉

= 0.0018(1000)(120)

= 216𝑚𝑚2

Using 10mm ф RSB

𝑨𝒃 =𝝅

𝟒(𝟏𝟎𝟐)

= 78.54𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=78.54

216(1000)

= 363.611𝑚𝑚. 𝑠𝑎𝑦 350𝑚𝑚

m.) 𝑆2 = 350𝑚𝑚

n.) 5ℎ = 5(120) = 1600𝑚𝑚. o.) 450𝑚𝑚.

Therefore, use 10mm.ф temperature

RSB SPCD. @ 160mm. O.C

DESIGN OF CONCRETE STAIRS

Given:

r = 0.6mm.

t = 0.75mm.

fc’ = 28MPa

Fy = 414MPa

Use:

12mm ф RSB main bars

10mm ф RSB temperature bars

𝒉 =𝑳

𝟐𝟎(𝟎. 𝟒 +

𝟒𝟏𝟒

𝟕𝟎𝟎)

=6000

20(0.4 +

414

700)

=297.43𝑚𝑚. 𝑠𝑎𝑦 320𝑚𝑚.

𝒘𝒕. 𝒐𝒇 𝒔𝒕𝒆𝒑𝒔 =𝒓𝒕

𝟐𝒏𝜸𝒄

=0.6(0.75)

2(7)(23.54)

= 37.075 𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒔𝒍𝒂𝒃 =𝐡√𝒓𝟐+𝒕𝟐

𝒕(𝜸𝒄)

=0.32√0.62+0.752

0.75(23.54)

= 9.647 𝐾𝑁/𝑚

𝒘𝒕. 𝒐𝒇 𝒍𝒂𝒏𝒅𝒊𝒏𝒈 = 𝜸𝒄h

= 23.54(0.4)

= 9.416𝐾𝑁/𝑚

DL= 37.075+9.416+9.647

=56.138 KN/m

LL=2KPa --- from the Design Criteria

𝑾𝒖 = 𝟏. 𝟒𝑫𝑳 + 𝟏. 𝟕𝑳𝑳

= 81.993 𝐾𝑁/𝑚

Considering 1m strip, (b=1000mm)

𝑊𝑢 = 81.993(1)

= 81.993 𝐾𝑁/𝑚

𝑴𝒖 =𝑾𝒖𝑳𝟐

𝟖

=(81.993)(6)2)

8

= 368.968 𝐾𝑁 − 𝑚.

Effective depth, d

𝒅 = 𝒉 − 𝒔𝒍𝒂𝒃 𝒄𝒐𝒗𝒆𝒓𝒊𝒏𝒈 − 𝟎. 𝟓𝒅𝒃

= 320 − 20 − 0.5(16)

= 102mm.

Solve for 𝑅𝑢

𝑹𝒖 =𝑴𝒖

∅𝒃𝒅𝟐

=368.968𝑥106

0.9(1000)(102)2)

= 4.808

Solve for ρ

𝝆 =𝟎.𝟖𝟓𝒇𝒄

′

𝒇𝒄′ ((𝟏 − √𝟏 −

𝟐(𝑹𝒖)

𝟎.𝟖𝟓𝒇𝒄′ )

= 0.013

𝝆𝒎𝒊𝒏. =𝟏.𝟒

𝒇𝒚

= 0.0034

𝝆𝒎𝒂𝒙 = 𝟎. 𝟕𝟓𝝆𝒃

= 0.022

Use 𝝆 = 𝟎. 𝟎𝟏𝟑

Required main bar spacing

𝑨𝒔 = 𝝆𝒃𝒅

= 0.013(1000)(292) = 3796𝑚𝑚2

Using 16mm ф main bars

𝐴𝑏 =𝜋

4(162)

= 201.06𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=201.06

3796(1000)

= 52.966𝑚𝑚. 𝑠𝑎𝑦 50𝑚𝑚.

p.) 𝑆1 = 50𝑚𝑚

q.) 3ℎ = 3(320) = 960𝑚𝑚. r.) 450𝑚𝑚

Therefore, Use 16mm. ф main bars SPCD.

@ 50mm.

Temperature bars

𝝆𝑻 = 𝟎. 𝟎𝟎𝟏𝟖 𝟒𝟏𝟓

𝒇𝒚

= 0.0018

𝑨𝒔 = 𝝆𝑻𝒃𝒉

= 0.0018(1000)(320)

= 576𝑚𝑚2

Using 10mm ф RSB

𝐴𝑏 =𝜋

4(102)

= 78.54𝑚𝑚2

𝒔𝒑𝒂𝒄𝒊𝒏𝒈 =𝑨𝒃

𝑨𝒔(𝟏𝟎𝟎𝟎)

=78.54

576(1000)

= 136.354𝑚𝑚. 𝑠𝑎𝑦 120𝑚𝑚

p.) 𝑆2 = 120𝑚𝑚

q.) 5ℎ = 5(320) = 1600𝑚𝑚. r.) 450𝑚𝑚.

Therefore, use 10mm.ф temperature

RSB SPCD. @ 120mm. O.C