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Following figure shows the framing plan of a residential building. Floor to floor height is 3.2m. Grade of concrete is M20 and steel Fe415. All columns are 300mm x 300mm in size.

a) Design slab S1

b) Draw the reinforcement details of slab S1

c) Design the beam B1

d) Draw the reinforcement details of beam B1.

Beam B1 is provided with 8mm diameter stirrups @150mm c/c throughout the length.

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

ddrcs mumbai university • 361  views
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A] Design of slab S1

$l_y = 5m$

$l_n =4m$

$\frac {l_y}{l_n} = \frac{5}{4} = 1.25 \lt 2$

$\therefore$ Two way continuous slab

1) Depth Calculation

$d_{read} = \frac{l_x}{\frac{l}{2} \times M.F }$

= $\frac{4000}{26 \times 1.4}$

= $109.89 mm$

$d_{req} \approx 120mm$

Overall Depth

$D = d + c.c + \frac{\phi}{2}$

= $120 +20+\frac{10}{2}$

D = 145 mm

a) s/w of slab = $25 \times D$

= $25 \times 0.145 = 3.625 kN/m^2$

b) Live load = $3 kN/m^2$

c) Floor Finish = $\frac {1.5 kN/m^2}{w = 8.124 kN/m^2}$

$w_u = 1.5 \times 8.125$

$w_u = 12.18 kN.m$

3) Calculation of B.M Coefficient

$\frac{l_y}{l_n} = 1.25$

$\alpha_x = 0.075$ and $\alpha_y= 0.056$

$M_{ux}= \alpha_x \times w_u \times l_n^2$

= $0.075 \times 12.18 \times 4^2$

$M_{ux}= 14.616 kN.m$

$M_{uy}= 0.056 \times 12.18 \times 4^2$

$M_{uy}= 10.91 kN.m$

4) Check for depth

$d = \sqrt{\frac{14.616 \times 10 ^6}{20 \times 1000 \times 0.138}}$

$d = 72.72 mm \lt 120mm$

$\therefore$ safe in depth

5) Calculation of steel

a) Astx = ?

$Astx =\frac{0.5 \times 20 \times 1000 \times 120}{415} [1- \sqrt{1-\frac{4.6 \times 14.616 \times 10^6}{20 \times 1000 \times 120^2}}]$

$Astx = 359.91 mm^2$

$Ast_{min} = \frac{0.12}{100} \times 1000 \times 145$

$Ast_{min} = 174 mm^2$

$Ast_x \gt Ast_{min}$

Use $Astx = 359.91 mm^2$

use 10 mm $\phi$ bar

$Spacing = \frac{\frac{\pi}{4} \times 10^2}{359.91} \times 1000$

= $218.22$

$\approx 200 mm$

Provide 10 mm $\phi$ bar @ 200 mm c/c

b) Asty = ?

$M_{uy} = 10.91 kN.m$

$Astx =\frac{0.5 \times 20 \times 1000 \times 120}{415} [1- \sqrt{1-\frac{4.6 \times 10.91 \times 10^6}{20 \times 1000 \times 120^2}}]$

$Asty = 263.98 mm^2$

$Ast_y \gt Ast_{min}$

use 10 mm $\phi$ bar

$Spacing = \frac{\frac{\pi}{4} \times 10^2}{263.98} \times 1000$

= $297.73$

$\approx 275 mm$

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

6) Check for shear

1) $Vux = 0.076 \times 12.18 \times 4 = 3.65 kN$

2) $Vuy = 0.056 \times 12.18 \times 4 = 2.72 kN$

$Vuc = k \tau_{uc} \ bd$

$k =1.3 for d \ lt 150 mm$

$A_{stp} = \frac{\frac{\pi}{4} \times 10^2}{200} \times 1000 = 392.7 mm^2$

%$pt = \frac{392.7}{1000 \times 120} \times 100$

$\tau_{uc} = 0.396$

$Vuc = k \tau_{uc} bd$

$1.3 \times 0.396 \times 1000 \times 120$

$Vuc = 61.77 kN \gt 3.656 kN$

$\therefore$Safe in shear

7) Check for deflection

$F_s = 0.58 f_y \frac{Ast_{req}}{Ast_d}$

= $0.58 \times 415 \times \frac{359.91}{392.7}$

$F_s =220.6$

$\approx 240 N/mm^2$

%$pt = 0.327$

M.F = 1.5

$d = \frac{4000}{26 \times 1.6} = 102.56 \lt 120mm$

$\therefore$ Safe in deflection

7) Check for development length

$Ld \leq 1.3 \frac{M_1}{V} + l_o$

$M_1 = 14.616 kN.m$

$V = 3.65 kN$

$l_o = 12 d or d$

$l_o = 120 mm$

$\therefore RHS = 5325.66 \ mm$

$LHS = \frac{0.87 f_y}{4 \tau bd}$

= $\frac {0.87 \times}{4 \times 1.2 \times 1.8}$

= 470.11mn

$\therefore RHS \gt LHS$

Safe in development length

B) Beam design $B_1$

a) Load transformed by Slab S1

= $\frac{w_u l_x}{2}[1-\frac{1}{3 \beta^2}]$

= $\frac{12.18 \times 4}{2}[1-\frac{1}{3 \times 1.25^2}]$

= $19.16kW/m$

b) Wall load = $1.5 \times l \times b \times h$

$1.5 \times 20 \times 0.23 \times 3.2$

= $22.08 kN/h$

C) S/W of beam = 10% (a+b)

= 10% (19.16 + 22.08)

= 4.124 kN/m

= $w_u = 45.364 kN/n$

$V_A =V_B = \frac{45.36 \times 5}{2} = 113.41$

midspan AB

$M_u =141.76 kN.m$

$b_f = \frac{l_o}{6} + 6 D_f + bw$

$\frac{0.7 \times 5000}{6} + 6 \times 145 + 230$

$1683.33 mm$

$D = \frac{5000}{10}$

$D = 500 mm$

$d = 500-25 = 475 mm$

Assume $x \leq D_f$

$Mu_r = 0.36 Fc_k x_u b_f (d-0.42 xl)$

=$0.36 \times 20 \times 145 \times 1683.33 (475-0.42 \times 145)$

$Mu_r = 727.73 \gt 14176 kN$

Assumption is correct

$Ast_1 =\frac{0.5 \times 20 \times 230 \times 475}{415} [1- \sqrt{1-\frac{4.6 \times 141.76 \times 10^6}{20 \times 1000 \times 470^2}}]$

$Ast = 1027.54 mm^2$

Use 20 mm $\phi$ bar

No. of Bar = $\frac{1027.54}{\frac{pi}{4} \times 20^2}$

= $3.27$

$\approx 4 Nos$

Provide 4 -20 mm $\phi$ bars

Design of shear reinforcement

$Vu_D = 113.41 kN$

$D Vu_c= \tau_{uc} bd$

$Ast = 4 \times \frac{\pi}{4} \times 20^2 = 1256.64 mm^2$

%$pt = \frac{Ast}{bd} \times 100$

%$pt = 1.16$

$\tau_{uc} 0.652$

$Vu_c = \tau_{uc} bd$

= $0.652 \times 230 \times 475$

$Vu_c = 71.23 kN$

$Vu_s = Vu_D- Vu_c$

= $113.41-71.23$

$Vu_s = 42.18 kN$

$Vu_s = Vu_{sv} = 42.18 kN$

$S_v = \frac{0.87 F_y As_v d}{Vu_{sv}}$

$S_v = \frac{0.87 \times 415 \times 2 \times \frac{\pi}{4} \times 8^2 \times 475 }{42.18 \times 10^3}$

= $402.82$

$\approx 300 mm$

$\therefore$ 2L-8mm $\phi$ 300 mm c/c

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