and the second flights starts from mid landing beam to the floor beam of upper storey. Use concrete M20 and steel of grade Fe415. Adopt LSM. Draw plan and sketch reinforcement details for both flights.

It is a dog-legged staircase

Floor to Floor height=3 m

Height/flight$=3/2=1.5 m$

Assume Riser$=150mm$

No. of riser in one flight$=1500/150=10$ Nos.

No. of tread in one flight$=R-1=9$

Trial depth$=\dfrac {\text {eff.span}}{S/D\space ratio \times M.F.}\\ S/D ratio=20, M.F. = 1.4 \\ d_{reqd}=\dfrac {5300}{20\times 1.4}=189.28 $

Assume D=250 mm.

d=D-clear cover $-ɸ/2. \\ =250-25-20/2=215 mm.$

Load calculation on step portion:

a) D.L. of waist slab $=D× 25 × \dfrac {\sqrt{R^2+T^2}}T \\ =0.25×25×\dfrac {\sqrt{150^2+250^2}}{250} \\ =7.28 kN/m^2 $

b) D.L. of step $= R/2 × 25 \\ = (0.15/2) × 25= 1.875 kN/m^2$

c) $F.F. = 1 kN/m^2$ Assumed

d) $L.L=4 kN/m^2$ Assumed

Total $= 7.28 + 1.875+1+4 \\ =14.15 kN/m^2 $

Factored Load $=1.5 × 14.15=21.23 kN/m^2$

Load calculation on landing portion:

a) D.L. of slab $=D ×25= 0.25 25=6.25 kN/m^2$

b) $F.F. = 1 kN/m^2$

c) $L.L.=4 kN/m^2$

Total $= 6.25+1+4=11.25kN/m^2$

Factored Load $=11.25 × 1.5 \\ =16.88 kN/m^2$

First Flight:

Calculations:

$\sum M_{@C}=0\\ V_A× 3.65-45.11×(1.06+1.525)-25.74×(0.76)=0 \\ V_A=37.3 kN \\ \sum Fy=0 \\ V_A+V_C =45.11+25.74 \\ V_C=33.55kN \\ B.M_{MAX}=37.3×1.75-\dfrac {21.23×1.75^2}2 \\ =32.76kNm \\ B.M\gt_{@B}=33.55×1.525-25.74×0.76 \\ =31.6 kNm.$

Typical Flight

Calculations:

$\sum M_{@D}=0 \\ V_A ×5.3-25.74×(0.76+2.25+1.525)- 47.76 × (1.125+1.525)-25.74 (0.76)=0 \\ V_A =49.59 kN \\ \sum Fy=0 \\ V_A+V_D=25.74+25.74+47.76 \\ V_D=49.65 kN$

$B.M_{max}=49.59 × (1.525+1.125)-25.74 × (0.76+1.125) – \dfrac {21.23×1.125^2}2\\ = 69.45 kNm \\ B.M_{@Barc}=49.59 ×1.525-25.74 × 0.76 \\ = 56.06 kNm$

Data:- $b=1000 mm, d=215 mm, fck=20 N/mm^2\\ Fy=415 N/mm^2$

Distribution steel:

Assume $8 mm ɸ$ & Astmin$= 300 mm^2$

Spacing $=\dfrac {1000Asv}{Ast_{min}}=\dfrac {1000×50}{300}=166.66mm \approx 150 mm $

Provide $8mm ɸ @ 150$ mm c/c.

Checks:

1) Check for deflection: (For typical flight)

$Fs=0.58 × fy \dfrac {Astr}{Astp}=0.58×415× \dfrac {989.64}{1027.27} \\ = 231.88 N/mm^2 \\Pt=0.47 , M.F.=?$

From graph at pg. 38 of IS:456:2000

$M.F. =1.44 \\ d= \dfrac {5300}{20×1.44}=184.02 mm \lt 215 mm\\ \therefore safe$

2) Check for development length:-

$M_0 = M/2=69.45/2=34.72 kNm \\ V= 49.65 kN. \\ L_0 \dfrac {bs}2 -25 + 3\phi =\dfrac {300}2 -25+ 3 × 12= 161 mm \\ L_d= \dfrac {0.87fy\phi}{4Z_{bd}}=47\phi =47×12=564 mm \\ L_d \leq \dfrac {1.3M_0}V + L_0 \\ 564 \leq \dfrac {1.3×34.72×10^6}{49.65×10^3}+161 \\ 564 \leq 1070.08 mm \\ \therefore safe $

Reinforcement details:

First Flight:

Typical Flight:-