Design of staircase, STAAD Design, Drawing,

1. AT GROUND FLOOR:
PLAN AT R.L. 499.180 m
DESIGN OF STAIRCASE
1

2.  Stair Room size = 7.6 m x 4.0 m (as shown in Figure above)
 Providing Tread (T) = 0.30 m and Riser (R) = 0.144 m
 Considering concrete grade M25 and steel grade Fe 415
fck = 25 N/mm2 and fy = 415 N/mm2
 Floor to floor height = 4.62 m = 4620 mm
 Live Load= 5 kN/m2
 Floor Finish = 1 kN/m2
 No of Risers =
4620
144
= 32.083 Nos. ≈ 32 Nos.
 No of risers in each flight =
32
2
= 16 Nos.
 Therefore, Going = 07 x 300 = 2100 mm (as shown in Figure above)
 Width of Landing = 1200 mm and width of stair = 2000 mm.
 Providing 150 mm thick waist slab. (Considering 1 m width of slab)
 Effective depth = 150-20-(16/2) = 122 mm
2
Data taken:

3. Load on Going:
 Self weight of waist slab = 25 x D x Sec ø
 Sec ø =
𝑅2+𝑇2
𝑇
=
1442+3002
300
= 1.109
 Self weight = 25 x 0.150 x 1.109 = 4.160 𝑘𝑁/ 𝑚≈ 4.2𝑘𝑁/ 𝑚
 Weight of step = 25 x
𝑅
2
= 1.80 𝑘𝑁/ 𝑚
 Floor finish = 1𝑘𝑁/ 𝑚2
 Live Load = 5 𝑘𝑁/ 𝑚2
 Total load = 12.00 𝑘𝑁/ 𝑚
 𝑊𝑢= 12.00 x 1.5 = 18.00 𝑘𝑁/ 𝑚 (Load acting on Going as shown in Figure below)
3
Load Calculations:

4. Load on Landing:
 Weight of slab = 25 x 0.150 = 3.75 𝑘𝑁/ 𝑚
 Floor finish = 1𝑘𝑁/ 𝑚
 Live Load = 5 𝑘𝑁/ 𝑚
 Total load = 9.75 𝑘𝑁/ 𝑚
 𝑊𝑢= 9.75 x 1.5 = 14.625 𝑘𝑁/ 𝑚 (Load acting on Landing as shown in Figure below)
4
Load Calculations:

5. 5
GROUND FLOOR STAIR CASE:
LOADING ON STAIRCASE
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM

6. 6
Area of steel calculation:
For Ground Floor:
For Mu = 50.366 kNm say 50.4 kNm
Ast =
0.5𝑓𝑐𝑘
𝑓𝑦
1 − 1 −
4.6𝑀𝑢
𝑓𝑐𝑘𝑏𝑑2 b.d
Ast = 1418.596 𝑚𝑚2
Using 16 mm diameter bars
Spacing =
1000×
𝜋×162
4
1418.596
= 141.733 mm (Say 140 mm c/c
< 3d or 300 mm)
Astmin =
0.12
100
x 1000 x 150 = 180 𝑚𝑚2
Astprov = 1000
×
201
.
062
140
= 1436.160 𝑚𝑚2
Ptprov=
100× 1436.160
1000×122
= 1.177 %
Check for shear:
For Pt =1.177 %
𝜏 𝑢𝑐=0.682 N/𝑚𝑚2
(From IS 456: 2000, Table
no 19 Pg no.72)
𝑣 𝑢𝑐=
0.682 ×1000×122
1000
= 83.204 kN > 67.669 kN. Hence OK
Distribution steel:
Astreqd = 0.12%bD =
0.12
1000
× 1000 × 150 =
180𝑚𝑚2
Using 8 mm dia bar
Spacing=1000×
50
180
= 277.78mm say 200 mm c/c
< 5d or 450mm.

7. 7
FIRST FLOOR STAIR CASE:

8. 8
 Stair Room size = 7.6 m x 4.0 m (as shown
in Figure above)
 Providing Tread (T) = 0.30 m and Riser (R)
= 0.144 m
 Considering concrete grade M25 and steel
grade Fe 415
𝑓𝑐𝑘 = 25 𝑁/ 𝑚𝑚2and 𝑓𝑦 = 415 𝑁/ 𝑚𝑚2
 Floor to floor height = 4.38 m = 4380 mm
 Live Load= 5 𝑘𝑁/ 𝑚2
 Floor Finish = 1 𝑘𝑁/ 𝑚2
 No of Risers =
4380
146
= 30.00 Nos.
 No of risers in each flight =
30
2
= 15 Nos.
 The going is shown in Figure above.
Data taken:
 Width of Landing = 1300 mm and width of
stair = 2000 mm.
 Providing 150 mm thick waist slab.
(Considering 1 m width of slab)
 Effective depth = 150-20-(16/2) = 122 mm

9. 9
Load on Going:
 Self weight of waist slab = 25 x D x Sec ø
 Sec ø =
𝑅2+𝑇2
𝑇
=
1462+3002
300
= 1.112
 Self weight = 25 x 0.150 x 1.112 = 4.171 𝑘𝑁/ 𝑚≈ 4.2𝑘𝑁/ 𝑚
 Weight of step = 25 x
𝑅
2
= 1.80 𝑘𝑁/ 𝑚
 Floor finish = 1𝑘𝑁/ 𝑚2
 Live Load = 5 𝑘𝑁/ 𝑚2
 Total load = 12.00 𝑘𝑁/ 𝑚
 𝑊𝑢= 12.00 x 1.5 = 18.00 𝑘𝑁/ 𝑚 (Load acting on Going as shown in Figure below)
Load Calculations:

10. 10
 Weight of slab = 25 x 0.150 = 3.75 𝑘𝑁/ 𝑚
 Floor finish = 1𝑘𝑁/ 𝑚
 Live Load = 5 𝑘𝑁/ 𝑚
 Total load = 9.75 𝑘𝑁/ 𝑚
 𝑊𝑢= 9.75 x 1.5 = 14.625 𝑘𝑁/ 𝑚 (Load acting on Landing as shown in Figure below)
Load on Landing:

11. 11
FIRST FLOOR STAIR CASE:
LOADING ON STAIRCASE
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM

12. 12
Area of steel calculation:
For First Floor:
For Mu = 45.82 kNm say 45.9 kNm
Ast =
0.5𝑓𝑐𝑘
𝑓𝑦
1 − 1 −
4.6𝑀𝑢
𝑓𝑐𝑘𝑏𝑑2 b.d
Ast = 1257.842 𝑚𝑚2
Using 16 mm diameter bars
Spacing =
1000×
𝜋×162
4
1257.842
= 159.847 mm (Say 140 mm c/c
< 3d or 300 mm)
Astmin =
0.12
100
x 1000 x 150 = 180 𝑚𝑚2
Astprov = 1000
×
201
.
062
140
= 1436.160 𝑚𝑚2
Ptprov=
100× 1436.160
1000×122
= 1.177 %
Check for shear:
For Pt =1.177 %
τuc = 0.682 N/mm2
(From IS 456: 2000, Table
no 19 Pg no.72)
vuc =
0.682 ×1000×122
1000
= 83.204 kN > 64.497 kN. Hence OK
Distribution steel:
Astreqd = 0.12%bD =
0.12
1000
× 1000 × 150
= 180mm2
Using 8 mm dia bar
Spacing=1000×
50
180
= 277.78mm say 200 mm c/c
< 5d or 450mm.

13. 13
SECOND FLOOR STAIR CASE:

14.  Stair Room size = 7.6 m x 4.0 m (as shown in Figure above)
 Providing Tread (T) = 0.30 m and Riser (R) = 0.144 m
 Considering concrete grade M25 and steel grade Fe 415
fck = 25 N/mm2and fy = 415 N/mm2
 Floor to floor height = 4.82 m = 4820 mm
 Live Load= 5 kN/m2
 Floor Finish = 1 kN/m2
 No of Risers =
4830
142
= 34.014 Nos. ≈ 34 Nos.
 No of risers in each flight =
34
2
= 17 Nos.
 The going is shown in Figure above.
 Width of Landing = 1100 mm and width of stair = 2000 mm.
 Providing 150 mm thick waist slab. (Considering 1 m width of slab)
 Effective depth = 150-20-(16/2) = 122 mm
14
Data taken:

15. 15
Load on Going:
 Self weight of waist slab = 25 x D x Sec ø
 Sec ø =
R2+T2
T
=
1422+3002
300
= 1.106
 Self weight = 25 x 0.150 x 1.106 = 4.149 kN/m≈ 4.2kN/m
 Weight of step = 25 x
R
2
= 1.80 kN/m
 Floor finish = 1kN/m2
 Live Load = 5 kN/m2
 Total load = 12.00 kN/m
 Wu= 12.00 x 1.5 = 18.00 kN/m (Load acting on Going as shown in Figure below)
Load Calculations:

16. 16
 Weight of slab = 25 x 0.150 = 3.75 kN/m
 Floor finish = 1kN/m
 Live Load = 5 kN/m
 Total load = 9.75 kN/m
 Wu= 9.75 x 1.5 = 14.625 kN/m (Load acting on Landing as shown in Figure below)
Load on Landing:

17. 17
SECOND FLOOR STAIR CASE:
LOADING ON STAIRCASE
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM

18. 18
Area of steel calculation:
For Second Floor:
For Mu = 55.568 kNm say 55.6 kNm
Ast =
0.5fck
fy
1 − 1 −
4.6Mu
fckbd2 b.d
Ast = 1619.961 mm2
Using 16 mm diameter bars
Spacing =
1000×
π×162
4
1619.961
= 124.116 mm (Say 120 mm c/c
< 3d or 300 mm)
Astmin =
0.12
100
x 1000 x 150 = 180 mm2
Astprov = 1000
×
201
.
062
120
= 1675.517 mm2
Ptprov=
100× 1675.517
1000×122
= 1.373 %
Check for shear:
For Pt =1.373 %
τuc=0.720 N/mm2
(From IS 456: 2000, Table
no 19 Pg no.72)
vuc=
0.720 ×1000×122
1000
= 87.801 kN > 70.709 kN. Hence OK
Distribution steel;
Astreqd = 0.12% bD =
0.12
1000
× 1000 × 150 =
180mm2
Using 8 mm dia bar
Spacing=1000×
50
180
= 277.78mm say 200 mm c/c
< 5d or 450 mm.

19. 19

20. 20

21. 21

22. 22

23. 23

24. 24