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Design a suitable doglegged staircase for a public building to re located in a stair room of size $6m \times 3m$. Take floor to floor height 3.6m. Live load $=4kN/m^2$.

Finish load $=1.25kN/m^2.$ Use LSM with concrete and steel grade M20 and Fe415. Also draw:

a) Plan showing flight dimensions, mid landing and position of support.

c) Reinforcement details in flight through sectional elevation.

Mumbai University > Civil Engineering > Sem 8 > Design and Drawing of Reinforced Concrete Structure

Marks : 20

Year : DEC 2015

mumbai university ddrcs • 2.7k  views
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Formulas:

Riser=120 mm to 150 mm

$\therefore$ Maximum no. of riser(R) in one flight=12

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

D.L. of waist slab

$=D\times 25\times \sec\phi\\ \therefore \sec\phi =\dfrac {\sqrt{T^2+R^2}}T\\ \therefore T=Tread \\ R=Riser$

It is a doglegged staircase.

Floar to floar height=3.6m

Height/flight $=3.6/2= 1.8m$

Assume Riser $=150mm \\ Tread=250 mm.$

No. of riser in one flight $=1800/150=12$ Nos.

No. of tread in one flight $=R-1=12-1=11$

Trial depth $=\dfrac {\text {eff-span }}{S/D ratio \times M.F.}$

$S/D ratio=20-\gt$(for simply supported) M.F. (Modification Factor) $=1.4 ( \because Fe 415) \\ d_{reqd} =\dfrac {6300}{20\times 1.4} =225mm$

Assume D=300 mm.

$\therefore$ d=D-clear cover $-(ΙΈ/2)\\ = 300-25-(20/2) \\ D=265 mm.$

a) D.L. of waist slab

$=D\times 25\times \dfrac {\sqrt{R^2+T^2}}T \\ =0.3\times25\times\dfrac {\sqrt{150^2+250^2}}{250} \\ =8.74 kN/m^2$

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

c) $F.F. =1.25 kN/m^2.$

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

Total $= 8.74 +1.875+1.25+4 \\ = 15.86 kN/m^2$

Factored Load $=1.5\times 15.86 \\ =23.8 kN/m^2$

a) D.L. of slab $=D \times 25=0.3 \times 25=7.5 kN/m^2$

b) F.F. = $1.25kN/m^2$

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

$\therefore$ Total $=7.5+1+4=12.5kN/m^2$

$\therefore$ Factored Load $=12.5 \times 1.5 \\=18.75 kN/m^2$

First Flight

Calculations:

$\sum M_c=0\\ V_A\times4.4-62.47\times(1.312+1.775)-33.28\times0.88=0 \\ V_A=50.48 kN \\ \sum f_y=0 \\ V_A+V_B=62.47+33.28 \\ V_B=45.27kN \\ B.M_{max} =50.48\times 2.12 - \dfrac {2.8\times2.12^2}2 =53.53 kNm \\ B.M_B =45.27\times1.775 -\dfrac {18.75\times1.775^2}2=50.81kNm$

Typical Flight:

Calculations:

$\sum M_0 =0 \\ V_A\times 6.3-33.28\times (0.88+2.75+1.755)-65.45\times(1.375+1.775)-33.28\times0.88=0 \\ \therefore V_A=65.92 kN \\ \sum F_y=0 \\ V_A+V_D=33.28 +65.45 +33.28 \\ V_D=66.09 kN\\ B.M._{max} =65.92\times(1.775+1.375)-33.28\times(0.88+1.375)-\dfrac {23.8\times1.375^2}2\\ B.M._{max}=110.1kNm \\ B.M_{boro}=65.92\times1.775- \dfrac {18.75\times1.755^2}2=87.47 kNm$

Data:- b=100 mm , d=265 mm, fck=$20N/mm2 \\ F_y=415 N/mm^2$

Distribution steel: Assume 8 mm and Admin $=360 mm^2 = (\dfrac \pi4 \times8^2)$

Spacing $= \dfrac {1000Asv}{Ast_{min}}=\dfrac {1000\times50}{360}\\ =138.88 mm \\ \approx 130 mm$

Checks:

1)Check for deflection:

For typical flight

$Fs=0.58 fy \dfrac {Astr}{Astp}\times 0.58\times415\times \dfrac {1279.49}{1340} \\ =229.83 N/mm^2 \\ Pt. =0.5 \\M.F.=1.45 \\ \therefore d=\dfrac {6300}{20\times1.45}=217.24 \lt 265 mm \\ \therefore safe$

2)Check for development length

$M_0=M/2=\dfrac {110.1}2=55.05kNm \\ V=66.09kN\\ L_0=\dfrac {bs}2-25+3\times\phi =\dfrac {300}2-25+3\times16=173mm \\ L_d=\dfrac {0.87fy\phi}{4.zbd}=47\phi=47\times16=752 mm \\ L_d \leq \dfrac {1.3M_0}V+L_0 \\ 752 \leq \dfrac {1.3\times55.05\times10^6}{66.09\times10^3}+173=1255.84 \\ \therefore safe$

Reinforcement details:

First flight:

Typical Flight:

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