Question: Design a doglegged staircase for a office building as shown in figure given in Also show arrangement of flights giving details. Draw reinforcement details for both flights.
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Use M20 grade concrete and Fe 415 steel.

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

Marks : 20

Year : MAY 2015

mumbai university ddrcs • 3.3k views
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modified 3.2 years ago  • written 3.2 years ago by gravatar for Aksh_31 Aksh_31630
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enter image description here

Trial depth= $\dfrac {\text {effspan}}{S/D\space ratio \times M.F.} \\ S/D \space ratio= 20, M.F.=1.4 \\ d_{reqd}=\dfrac {5400}{20\times1.4}=192.85 mm $

Assume D=250 mm

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

Load calculation on step portion:

a) D.L. of waist slab $=D\times25\times \dfrac {\sqrt{R^2+T^2}}T \\ =0.25\times25\times\dfrac {\sqrt{150^2+300^2}}{300} \\ =7kN/m^2$

b) D.L. of step= $R/2\times25=\dfrac {0.150}2\times25=1.875 kN/m^2$

c) $F.F=1.5 kN/m^2$

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

Total $=7+1.875+1.5+4 \\ =14.37 kN/m^2$

Factored Load $=14.37 \times 1.5 \\ = 21.56 kN/m^2.$

Load calculation on landing portion:

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

b) $F.F. = 1.5 kN/m^2 $

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

Total $=6.25+1.5+4=11.75 kN/m^2$

Factored Load $=11.75 \times 1.5=17.62 kN/m^2.$

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Calculations:

$\sum^{[email protected]}=0 \\ V_A \times 4.05-61.44\times (1.425+1.2) -21.14\times (0.6)=0 \\ V_A=42.95 kN \\ \sum F_y\\ V_A+V_C=61.44+21.14 \\ V_C=39.63 kN \\ B.M._{max}=42.95\times1.99 - \dfrac {21.56\times1.99^2}2 \\ B.M._{max}=42.78 kNm \\ M_{@b}=39.63\times1.2-21.14\times0.6 \\ M_{@B}=34.26kNm$

Typical Flight:

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Calculations:

$ƸM_D=0 \\ V_A×5.4-21.14×(0.6+3+1.2)-64.68×(1.5+1.2)-21.14×(0.6)=0 \\ V_A=53.28 kN \\ ƸFy=0 \\ V_A+V_B=21.14+64.48+21.14 \\ =53.28 kN \\M_{@B } \space or\space D=0 \\ =53.48 × 1.2 – 21.14 × 0.6 = 51.49 kNm \\ BM_{MAX}=0 \\ =53.48 × (1.2 + 1.5) – 21.14 × (0.6+1.5) - \dfrac {21.56\times3^2}2 \\ = 75.75 kNm.$

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

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Distribution steel

Assume 8 mm ɸ & Astmin = 300 mm2

Spacing $=\dfrac {1000Asv}{Ast_{min}}=\dfrac {1000\times50}{300}=1666.66 \approx 150 mm$

Provide $8 mm ɸ @ 150$ mm c/c.

Checks:

1) Check For deflection: (For typical Flight)

$fs= 0.58 fy \dfrac {Astr}{astp}=0.58\times415\times\dfrac {1091.25}{1130} \\ = 232.44 \\ Pt. = 0.52 \\ M.F. = ?$

From Graph at pg. 73 of IS. 456:2000

$M.F.=1.48 \\ d=\dfrac {5400}{20\times1.48}=182.44 \lt 215 mm \\ \therefore safe$

2) Check for development length:

$M_0=M/2=75.75/2 =37.87 kNm \\ V-53.48 kN \\ L_0=\dfrac {bs}2 -25+3\phi =\dfrac {230}2-25 +3 \times 12 \\ =126mm \\ L_d=\dfrac {0.87fy\phi}{4Z_{bd}}=47\phi=47\times12=564 mm \\ L_d \leq \dfrac {1.3M_0}V +L_0 \\ 564 \leq \dfrac {1.3\times37.87\times10^6}{53.48\times10^3}+126=1046.54 \\ \therefore safe $

Reinforcements details

First flight:

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Typical flight:

enter image description here

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written 3.2 years ago by gravatar for Aksh_31 Aksh_31630
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