Question: A building having floor to floor height as 3.2 m is to be provided a dog legged staircase. The stair hall is 3m x 5m. Grade of concrete is M20 and steel Fe415.
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a) Draw a dog legged staircase.

b) Draw the plan showing flight details, mid-landings etc. Draw reinforcement details in a flight.

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

ddrcs mumbai university • 136 views
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modified 8 weeks ago by gravatar for Yashbeer Yashbeer160 written 5 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Given:

height = 3.2m

dimension = $3m \times 5m$

M20 Fe415

Solution:

1] Design Constant

$f_{ck} = 20 N/mm^2$

$f_y = 415 N/mm^2$


2] Design Parameter

Height = 3.2 m

Each flight = 1600 mm

$\therefore$ provide rise = 10 each 160mm

$\therefore$ treads = 9 each 250mm

$\therefore$ Plan is

enter image description here


3] effective span and depth

Less for $I^{st} = 2.25 + 1.375 = 3.625 m$

Less for $II^{nd} = 5m$

$d = \frac{Less}{B_v mf}$ ($B_v = 20$ fixed and $mf = 1.35$ for % of steal = 0.4%)

$\therefore d = \frac{5000}{20 \times 1.35} = 185.18 mm$

assume $d = 200 mm \quad D = 225 mm$ (cover 25mm)


4] Load calculations:

(R = Riser = 0.16m, T = Tread = 0.25m)

i) On goeing

$\begin{aligned} S.W.S &= 25 \times \frac{0.225 \sqrt{0.16^2 + 0.25^2}}{0.25} \bigg( \frac{250 \sqrt{R^2 + T^2}}{T} \bigg) \\ &= 6.678 \end{aligned}$

$S.W. Step = 2\big( \frac{25R}{2} \big) \quad LL = 3 \quad FF = 1$

Total load = 12.678

UL = 19.017 kN/m

II) On slab / landing (25 X D)

$SW = 25 \times 0.225 = 5.625 \quad LL = 3 \quad FF=$

$TL = 9.625 kN/m$

$\therefore$ Total ultimate = 14.437


5] BM calculations

Case 1:

enter image description here

$R_A+R_B=62.64$

$R_A = 33.274 \ KN$

$R_B = 29.366 \ KN$

enter image description here

$\therefore \quad M_{max} =29.11 \mathrm{kNm}$

Case 2:

enter image description here

$\therefore \quad M_{max} = 55.1 \mathrm{kNm}$


6] Check for depth

$d^2 = \frac{55.1 \times 10^6}{0.138 \times 20 \times 1000} \left( \frac{M}{0.138 fck \ b} \right)$

$d = 141.3 mm \lt d_{pro} = 200 mm \therefore Safe$


7] Calculation of Reinforcement

For Fight (I)

$M = 29.11 KNm$

$Ast =\frac{0.5 \times 20}{415}\left[1- \sqrt{ 1 - \frac{4.6 \times 29.11 \times 10^{6}}{20 \times 1000 \times 200^{2}}} \right] 1000 \times 200$

$Ast = 421.79 mm^2$

$\therefore$ provide 10mm $\pi$ @ 185mm c/c

$\text{Ast provided} = 424.54 mm^2$

For flight (II)

$M = 55.1 KNm$

Ast = $835.93 mm^2$

$\therefore$ provide 12mm $\phi$ @ 130mm c/c

$\text{Ast provided} = 870 mm^2$

Distribution steel:

$0.12 \% bd = 240 mm^2$

8mm $\phi$ @ 205 mm c/c


8] Check for deflection

$\text{% of steel} = \frac{870}{1000 \times 200} \times 100 \left( \frac{\text{Ast provided}}{bd} \right)$

$0.435 \not{\lt} 0.4$

Detail check

$f_s = 231.274 \quad mf = 1.3$

$d = \frac{5000}{20 \times 1.3} = 192 \lt 200 \therefore Safe$


9] Check for shear

Consider 19.017 through out span

$Vu_{max} = \frac{wl}{2} = \frac{19.017 \times 5}{2} = 47.54 KN$

$\tau_v = 0.2377 \quad (Vu_{max} \times b \times d)$

For D = 225mm $\quad$ K = 1.15 from IScode

$\tau_e = K \tau_c$

$1.15 \times 0.28 = 0.322 \quad \therefore Safe$


10] Development

$L_d = \frac{0.87 \times 1115 \times 12}{4 \times 1.2 \times 1.6} = 564.14 (\frac{0.87 fy \phi}{4 \tau_D d})$

$ = \frac{1.3 \times 27.55 \times 10^6}{117.54 \times 1.^3} + 100 = 853.4 \quad \therefore Safe$

enter image description here

enter image description here

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modified 8 weeks ago  • written 8 weeks ago by gravatar for Yashbeer Yashbeer160
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