Design of staircase ppt

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Information about Design of staircase ppt

Published on December 12, 2016

Author: VibhanshuSingh5

Source: slideshare.net

1. DESIGN OF STAIRCASE

2. Diagram showing support condition

3. Data and calculations •Height between floor=2.79 m •Tread=250 mm •Riser=155 mm •Width of staircase=1000 mm =1 m •Number of riser =2790/155 =18 nos.

4. Continued… •So giving First flight: 8 tread and 9 riser Second flight: 8 tread and 9 riser •Height of landing = 9155 =1395 mm =1.395m

5. Continued… • Then effective span will be provided as =going span +x+y; l=2+x+y l=2+0.45+0.3675 l=2.82 m (since: x<1m and y<1m.)

6. Based upon IS 456:2000 code specification • Minimum reinforcement for HYSD bar =0.12% • For pt =0.12%. • Modification factor for tension reinforcement =1.25 • Taking (span/depth) ratio 32. • l/d =32. • d=2820/32 =88.125 mm • Clear cover = 15 mm; • Steel diameter =10 mm.

7. Continued.. • Total depth of waist =88.125+15+10/2 =108.125 mm • Provide total depth = 110 mm. • B= 0.1552 + 0.252 = 0.294m

8. LOAD CALCULATION: Load on landing: Live load =3 KN/m2. Finishing load=1 KN/m2. Dead load =250.11=2.75 KN/m2 Total load on landing = 6.75 KN/m2.(not factored) Load on going: Live load =3 KN/m2. Finishing load=1 KN/m2 Dead load = (WB+ 0.5RT)25/T ; (refer diag.) =(0.1100.294+0.50.1550.25)25/0.25; =5.1715 KN/m2; Total load on going =9.1715 KN/m2 (not factored)

9. Loading diagram

10.  MB=0  RA2.82 - 10.1250.37(0.370.5+2.45) - 13.75721.45 - 10.1250.50.452=0 RA=18.0112 KN RA+RB = 35.8165 KN  RB=17.805 KN MX=17.805X – 10.1250.37(X-0.370.5) -13.757(X-0.37)2 0.5 dM/dx =0 x=1.39166 m hence MX=13.073 KNm/m Mu/bd2 =(13.07106)/(1000852)=1.809

11. Using SP 16 From table 1; for M20 and fy=415 N/mm2 Pt =0.565 % Pt =

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