Question: Design a dog legged staircase for an office building in a room which measures 3m x 6m (clear dimensions or inside dimensions).
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Floor to floor height is 3.5m. The building is liable to overcrowding as it’s a public building. Stairs are supported on brick walls 230 mm thick at the ends of landings (i.e. landing spans parallel to stairs). Provide M20 concrete & Fe415 steel. Fixing the dimensions of risers & treads for the human comfort is expected. Carry out the necessary checks. Show the reinforcement details. Use limit state method.

Civil Eng > Sem 8 > Design and Drawing of Reinforced Concrete structure

ddrcs mumbai university • 283 views
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modified 5 months ago by gravatar for Yashbeer Yashbeer170 written 8 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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1. Given:

Hall dimension: $3m \times 6m$

Floor to Floor height: $3.5m$

Brick wall: $230mm$

M20 concrete

Fe 415 Steel

Assume: Floor finish = 1 and Liveload = $3 kN/m$

enter image description here

Floor to floor height = $3.5m$

Floor to floor height / Flight = $\frac{3.5}{2} = 1.75 = 1750 mm$

Assume Riser $=160 \mathrm{mm}$

No. of riser $=\frac{1710}{160}=10.93 \approx 11 \text{ Nos.}$

$T=R-1 = 11-1 = 10 \text{ Nos.}$

Assume Tread T $=250 \mathrm{mm}$

$\begin{aligned} \text{Going} &= \text{No. of tread} \times \text{Dimension of Tread} \\ &= 10 \times 250 \\ &= 2.5 mm \end{aligned}$

$\begin{aligned} \text { width of landing } &= 5000-2500 \\ &=120 \mathrm{mm} \end{aligned}$


2. Calculation of effective length

Flight 1 -

$\begin{aligned} L_{eff} &= 2500 + 1250 + \frac{230}{2} - \frac{250}{2} \\ &= 3740 mm \\ &= 3.74 m \end{aligned}$

Flight 2 -

$\begin{aligned} L_{eff} &= 1250 + 2500 + 1250 + \frac{230}{2} + \frac{230}{2} \\ &= 5230 mm \\ &= 5.23 m \end{aligned}$


3. Depth Calculation

$\begin{aligned} d_{req} &= \frac{L_{eff}}{\left( \frac{L}{d}\right)_b \times M.F}\\ &= \frac{5230}{20 \times 1.4} \\ &= 186.78 mm \sim 200 mm \end{aligned}$

Overall depth

$D = d + c.c. + \frac{d}{2} = 200 + 20 + \frac{b}{2} = 225 mm$


4. Load calculation

a) Flight 1 -

$\begin{aligned} \text{s/w of slab} &= 25 \times D \times sec \ d \\ &= 25 \times D \times \frac{\sqrt{R^2 + T^2}}{T} \\ &=25 \times 0.225 \times \frac{\sqrt{160^{2}+250^{2}}}{280} \\ &= 6.68 KN/m^2 \end{aligned}$

$\begin{aligned} \text { weight of steps } &=\frac{R}{2} \times 25 \\ &= \frac{0.160}{2} \times 25 \\ &= 2 KN/m^2 \end{aligned}$

$\text{Floor Finish} = 1 KN/m^2 \text{ ...Assume}$

$\text{Live load} = 8 KN/m^2 \text{ ...Assume}$

$T.L = 12.68 KN/m^2$

$\begin{aligned} w_u = 1.5 \times 12.68 = 19.02 KN/m \text{ ...Assume } L_m \text{ of strip} \end{aligned}$

b) Mid landing

$\begin{aligned} \text{Self wt of slab} &= 25 \times D \\ &= 25 \times 0.225 \\ &= 5.625 KN/m^2 \\ FF &= 1 KN/m^2 \\ LL &= 3 KN/m^2 \\ TL &= 9.6 KN/m^2 \end{aligned}$

$w_u = 1.5 \times 9.625 = 14.13 KN/m \text{ ... Assuming } L_m \text{ of strip}$

enter image description here

$\sum M_a A = 0$

$19.02+\frac{2.375^{2}}{2}+14.43 \times 1.365 \times \left( 2.375 + \frac{1.365}{2} \right) - V_B \times 3.74 = 0 $

$V_B = 30.45 KN$


$\sum F_y = 0$

$v_{A}-19.02 \times 2.375-14.43 \times 1.365+30.45=0$

$V_{A} = 34.42 KN$


S.F.D -

$A_{L} = 0$

$A_R=34.42 \mathrm{KN}$

$C = 34.42 - 19.02 \times 2.375 = -10.77 \mathrm{KN}$

$B_{L} = -10.77-14.43 \times 1.365=-30.45 \mathrm{KN}$

$B_{R} = -30.45+30.45=0$

$\frac{34.42}{\eta}-\frac{10.77}{2.375-\eta}$


B.M.D -

$M_A = 0$

$M_C = 30.45 \times 1.365-14.43 \times \frac{1.365}{2} \times \frac{1.365}{2} = 28.12 \ KN$

$M_D = 34.42 \times 1.81 - 19.02 \times \frac{1.81^2}{2} = 31.14 KNm$

$$V_u = 34.42 KN$$

$$M_u = 31.14 KNm$$


Check for depth

$\begin{aligned} Mu_{max} &= Ru_{max} \ bd^2 \\ d &= \sqrt{\frac{Mu_{max}}{Ru_{max} \ b}} \\ &= \sqrt{\frac{31.14 \times 10^6}{0.138 \times 20 \times 1000}} \\ &= 106.21 mm \lt 200 mm \\ \therefore \quad & \text{ Safe in depth} \end{aligned}$


Calculation of steel

a) Main Steel

$Ast = \frac{0.5 \ fck \ bd}{fy} \left[ 1 - \sqrt{1 - \frac{4.6 M_y}{fck \ db^2}} \right]$

$Ast = \frac{0.5 \times 20 \times 1000 \times 200}{415} \left[ 1 - \sqrt{1 - \frac{4.6 \times 31.14 \times 10^6}{20 \times 1000 \times 200^2}} \right]$

$Ast = 452.72 mm^2$

$\begin{aligned} Ast_{min} &= 0.12\% bD \\ &= \frac{0.12}{100} \times 1000 \times 225 \\ &= 270 mm^2 \end{aligned}$

$Ast_{min} = Ast$

$\text{Provide Ast = 452.72} mm^2$

$Use \ 12mm \phi$

$Spacing = \frac{\pi / 4 \times 12^2}{452.72} \times 1000 = 249.81$

$\therefore$ Use 12mm d @ 225 mm c/c

$Ast_p = \frac{\pi / 4 \times 12^2}{225} \times 1000 = 502.65 mm^2$


b) Distribution steel

$Ast_{min} = 270 mm^2$

Use 10mm $\phi$

$Spacing = \frac{\pi / 4 \times 10^2}{270} \times 1000 = 290.78 \sim 275 mm$

Use 10 mm $\phi$ @ 275 mm c/c

Check for shear

$\begin{aligned} V_{ue} &= \tau \ bd \\ pt\% &= \frac{Ast_p}{bd} \times 100 \\ &= \frac{502.65}{1000 \times 200} \times 100 \\ &= 0.25 \end{aligned}$

$$ 0.25 - 0.36 N/mm^2 $$

$$ \tau_{uc} = 0.36 N/mm^2 $$

$\begin{aligned} V_{uc} &= \tau_{uc} \times b \times d \\ &= 0.36 \times 1000 \times 200 \\ &= 72 KN \gt V_{UD} \\ \therefore \quad & \text{Safe in shear} \end{aligned}$

B] Design of Flight 2

enter image description here

$R_E=R_F=\frac{19.43 \times 1.365+19.02 \times 2.5+14.43 \times 1.365}{2}$

$R_E=R_F= 43.47 KN$


S.F.D -

$E_L = 0$

$E_R = 43.47 KN$

$G = 43.47 - 14.43 \times 1.365 = 23.775 \ KN$

$H = 23.775 - 19.02 \times 2.5 = -23.775 \ KN$

$F_L = -23.775 - 14.43 \times 1.365 = -43.47 \ KN$

$F_R = -43.47 + 43.47 = 0 \ KN $


B.M.D -

enter image description here

$M_E = 0$

$M_F = 43.47 \times (1.365 + 1.25) - 14.43 \times (1.365) \times (\frac{1.365}{2} + 1.25) - 19.02 \times \frac{1.25^2}{2} = 60.75 KNm$

$MF = 0$

Check for depth

$M_u = M_I = 60.75 KNm$

$d = \sqrt{\frac{60.75 \times 10^6}{0.138 \times 20 \times 1000}}$

$d = 148.36 mm \gt 200 mm$

$\therefore \quad Safe in depth$

Calculation of steel

a) Main Steel

$Ast = \frac{0.5 \times 20 \times 1000 \times 200}{415} \left[ 1- \sqrt{1 - \frac{4.6 \times 60.75 \times 10^6}{20 \times 1000 \times 200^2} }\right]$

$Ast = 931.78 mm^2$

$\begin{aligned} Ast_{min} &= 0.12\% bD \times (\text{Tor Steel}) \\ &= \frac{0.12}{100} \times 1000 \times 225 \\ &= 270 m^2 \\ Ast &\gt Ast_{min} \\ Provide \ Ast &= 931.78 mm^2 \\ \end{aligned}$

Use 12mm $\phi$ bar

$Spacing = \frac{\pi / 4 \times 12^2}{831.78} \times 1000 = 121.37$

Use 12mm $\phi$ @ 110 mm c/c

$\therefore Asp = \frac{\pi / 4 \times 12^2}{110} \times 1000 = 1028.16 mm^2$

b) Distribution steel

$Ast_{min} = 270 mm^2$

Use 10mm $\phi$

$Spacing = \frac{\pi / 4 \times 10^2}{270} \times 1000 = 290.98 \sim 275 mm$

Provide 10mm $\phi$ @ 275 mm c/c

Check for shear

$V_{uc} = \tau_{uc} bd$

$\begin{aligned} Pt \% &=\frac{Ast_P}{b a} \times 100 \\ &= \frac{1028.15}{1000 \times 200} \times 100 \\ &= 0.51 \end{aligned}$

$\tau_{uc} = 0.48 N/mm^2$

$\begin{aligned} V_{uc} &=0.488 \times 1000 \times 200 \\ &=97.6 \mathrm{kN} \gt \mathrm{V} \end{aligned}$

$\therefore \text{ Safe in shear}$

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modified 5 months ago  • written 5 months ago by gravatar for Yashbeer Yashbeer170
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