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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.

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 • 3.9k  views
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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

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

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

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

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:

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

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:

Typical flight:

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