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A building having floor to floor height as 3.1m is to be provided an open well staircase.

Design open well staircase. Draw the plan showing flight details, mid-landings etc. Draw reinforcement details in a flight. Grade of concrete is M20 and steel Fe415.

staircase reinforcement • 1.3k  views
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Step 1:

Floor to floor height = 3.1m (Open well)

M20 Fe415

$fck = 20 N/mm^2$

$f_y = 415 N/mm^2$

$M - 0.138 \ fck \ bd^2$

Step 2: Preliminary Dimensions

Floor to floor height = 3.1m

height of each fright = $\frac{3.1}{2} = 1.55m$ But in this case we need to design open well staircase

$\therefore$ Assume Riser = 155mm

Total Riser = $\frac{3100}{155} = 20$

Since in open well staircase there are 3 flights

$\therefore$ Total tread = 20 - 3 = 17

$\therefore$ Provide 7 tread in $1^{st}$ flight, 3 tread in $2^{nd}$ flight, 7 tread in $3^{rd}$ flight

Assume $tread = 250mm$

Assume width of the landing to be 1m (effective)

Step 3: Effective span and depth

Less for $I^{st}$ and $III^{rd}$ flight

Less = 1.75 + 1 = 2.75 m

Less of $II^{nd}$ flight

Less = 1 + 1 0.75 = 2.75 m

$d = \frac{less}{B_v \times mf}$ assume % of steel = 0.4%, mf = 1.15, $Bv$ = 20

$\frac{2750}{20 \times 1.15} = 119.6$

$\therefore$ Provide d = 130 mm

effective cover = 25 mm

$D_{overall} = 155 mm$

Step 4: Load calculations

On going portion

\begin{aligned} \text{1) Self wt. of slab } &= \frac{25D \sqrt{R^2 + T^2}}{T} \\ &= \frac{25 \times 0.155 \times \sqrt{0.155^2 + 0.25^2} }{0.25} \\ &= 4.56 KN/m \\ \text{} \\ \text{2) Self wt of step } &= \frac{25 R}{2} \\ &= \frac{25 \times 0.155}{2} \\ &= 1.94 KN/m \\ \text{} \\ \text{3)} LL = 3 KN/m & \quad P = 1 KN/m \end{aligned}

Total load = 10.5 KN/m

Ultimate load = 16.75 KN/m

1) Self wt. of slab = $25 \times D$ = 3.9 KN/m

2) LL = 3 KN/m $\quad$ FF = 1 KN/m

Total = 7.9 KN/m

Ultimate = 11.85 KN/m

As per clause 33.2 page 63 on half load has to be considered in open well staircase

Step 5: Pending Moment

For $I^{st}$ and $III^{rd}$ flight

$R_A + R_B = (5.925 \times 1) + (15.75 \times 1.75) = 33.5 KN$

Moment of B

[email protected] = 0 = $[5.925 \times 1 \times (0.5+1.75) ] + \frac{15.75 \times 1.75^2}{2} - R_A \times 2.75$

$R_A = 3.62 KN \quad R_B = 19.88 KN$

$x = 1.26 m$

$M_{max} = (19.88 \times 1.26) - (\frac{15.75 \times 1.26^2}{2}) = 12.55 KN/m$

for $II^{nd}$ flight

$R_A + R_B = 23.66$

$R_A = R_B = 11.83$

Max at mid span = $(11.83 \times 1.375) - [5.925 \times (0.5 + \frac{075}{2} )] - (\frac{15.75 \times 0.375^2}{2})$

$M_{max} = 10 KNm$

Step 6: Check for depth

$d = \sqrt{\frac{M}{0.138 \ fck \ b}} = 67.43 mm \lt d_{provided} \quad \therefore \ Safe$

Step 7: Ast calculation

Since moment is almost same provide Ast for Max moment

$Ast = \left( \frac{0.5 \times 20}{4.5} \right) \left[ 1 - \sqrt{1 - \frac{4.6 \times 12.55 \times 10^6}{20 \times 1000 \times 130^2}} \right] 1000 \times 130$

$Ast = 280 mm^2$

$\therefore$ Provide #10mm @ 275mm c/c

Distribution Steel

$Ast_{min} = \frac{0.12}{100} bD = 186 mm^2$

provide #10mm @ 265 mm c/c

Step 8: Check for deflection

% of Steel provided = $\frac{285.6}{1000 \times 130} \times 100 = 0.22 % \lt 0.4$

$\therefore$ Safe

Step 9: Check for Shear

$V_{max} = \frac{15.75 \times 2.75}{2} = 21.66 KN$

$V_{UC} = \tau_c \ bd \ K = 1.15 \times 0.28 \times 1000 \times 130 = 41.86 KN$

$\therefore$ Safe

Step 10: Check for development length

$L_d \lt \frac{1.3 M_1}{V} + L_D$

$L_d = \frac{\phi \ f_y \ 0.87}{4 \tau_{bd}} = \frac{10 \times 415 \times 0.87}{4 \times 1.6 \times 1.2} = 470.12 mm$

$M_1 = \frac{12.55}{2} = 6.275 KNm$

$V = 4.66 KN$

$\therefore \frac{1.3 \times 6.275 \times 10^6}{4.66 \times 10^3} = 476 mm$

$\therefore$ Safe

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This reinforcement details are wrong