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For the floor plan as shown below, the slabs are subjected to L.L of $4 kN/m^2$ Thickness of walls may be considered as 200 mm & height of wall is 3.5 m.

Design slab S1 Beams SB1 & MB1 Draw reinforcement details for slab and beams. Adopt M20 and Fe415 Use LSM.

Mumbai University > Civil Engineering > Sem 8 > Design and Drawing of Reinforced Concrete Structure

Marks : 40

Year : DEC 2013

mumbai university ddrcs • 745  views
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$\dfrac {l_y}{l_x}=\dfrac 5{2.5}= 2 \geq 2 $

Hence it is a one way slab

$d_{reqd}= \dfrac {\text {short eff. span}} {\text {S/D ratio * M.F.}} \\ \therefore d=\dfrac {2500}{26\times 1.4} =68.68 \approx 75 mm \\ D=d+20+(10/2) = 75+20+(10/2) \\ = 100 mm$

Load calculation:

Assume (1mⅹ1m) panel

$D.L.=D\times 25 = 0.1 \times 25=2.5 kN/m^2 \\ F.F. = 1 kN/m^2 \\ = 3.5 kN/m^2 $

Factored $D.L.=1.5 ×3.5 \\ = 5.25 kN/m^2 \\ L.L=4 kN/m^2$

Factored $L.L.=1.5 \times 4=6 kN/m^2 $

For moment coefficient

$w_dl_{x^2}=5.25\times 2.5^2=32.81 \\ w_Ll_{x^2}=6\times 2.5^2=37.5$

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For shear coefficients:

$w_dl_{x}=5.25\times 2.5=13.12 \\ w_Ll_{x}=6\times 2.5=15$

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Distribution steel:

$Ast_{min} =120 mm^2$ Provide 8mm Φ bar

Spacing $=\dfrac {1000\times 50}{120}=416.66 \approx 400 mm c/c$

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Slab load= D.L.+L.L.=5.25 +6 =11.25 kN

No wall as the beams are internal.

Assume internal beam dimension.(300ⅹ500)mm

Self wt.(I)$=(0.3\times0.5\times25) \times1.5 \\ = 5.62 kN/m.$

Beam SB1

Slab load: $2.5\times 5 \times11.25=140.62 kN $

Slab load/m $140.62/5 = 28.12 kN$

Self wt $ = 5.62 \\ = 33.74 \approx 34 kN/m.$

Provide 34 kN/m throughout.

$M_u=\dfrac {wl^2}8=956.25 kNm \\ V_{ud}=\dfrac {wl}2=255 kNm \\ At \space \space x=7.5 , V_{ud} = 255 kNm \\ At\space \space x=5m ,V_{ud}=? \\ \dfrac {7.5}5=\dfrac {255}x \\ x=\dfrac {255\times5}{7.5} =170 kN \\ l=300 mm , D=500 mm, dc=50 mm \\ ∴d=450 mm \\ M_{umax} =0.138fckld^2 = 0.138\times20\times300\times450^2 \\ =167.67 kN_m\lt\ltM_U $

Hence provide doubly reinforced section

$M_{U2} =956.25-167.67=788.58 kNm \\ M_{U1}=T_{U1} × L_{a1} \\ = 0.87 fy Ast_1 (d-0.42 x_{umax}) \\ 167.67×10^6=0.87×415×Ast_1(450-0.42×0.48×450) \\ Ast_1=1292.57 mm^2 \\ M_{U2} = T_{U2} × L_{a2}=0.87 fyAst_2 (d-d_c) \\ 788.58×10^6=0.87×415×(450-50) \\ Ast_2=5460.32 mm^2 \\ Ast= Ast_1+Ast_2=5460.32+1292.57 \\ = 6752.89 mm^2 $

Provide 6-48 Φ bar

$\dfrac {d_c}d=\dfrac {50}{450}=0.11 $

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$f_{cc}=0.446 f_ck = 0.446×20 \\ = 8.92 N/mm^2 \\ M_{U2} = C_{U2}×L_{a2}\\ 788.58×106=(fsc-fcc)Asc(d-dc) \\ = (350.8-8.92)Asc(450-50) \\ ∴Asc=5766.49 mm^2$

Provide 5-40 mm Φ bar

Shear reinforcement:-

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$\dfrac {V_{UD}}{7.05}=\dfrac {255}{7.5} \\ V_{UD}=239.7 kN \\ Zuc=\dfrac {V_{UD}}{ld}=\dfrac {239.7\times10^3}{300\times450}=1.77 \lt 2.8 N/mm^2 \therefore safe \\ Pt=\dfrac {100Astp}{bd}=\dfrac {100\times6\dfrac \pi4\times 40^2}{300\times450}=5.58\% \\ Z_{uc} (pg.73:IS:456:2000)=0.82 \\ V_{uc} = z_ubd=0.82\times300\times 450=110.7 kN \\ V_{umin}=0.4\times300\times450 = 54 kN \\ V_{uc}+V_{umin} = 164.7 kN\ltVUD$

Hence design and provide shear r/f

$V_{US}=239.7-164.7 = 75 kN $

Provide 8 mm Φ 2 LG stirrups

$S_1=\dfrac {0.87fyasvd}{Vus}=\dfrac {0.87\times415\times100\times450}{75\times10^3}\\ =216.33 mm \\ S_2 = 0.75 \times 450 = 337.5 mm \\ S_3=300mm$

Provide 8 mm Φ @ 200 mm c/c.

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Beam MB1

• Beam MB1 will not contain slab load as it is a shorter side of slab

• Self wt. = 5.62 kN/m

Point load = 170 kN

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$M_U=123.81 kNm \\ M_{umax} =0.138fckld^2 = 0.138\times20\times300\times450^2 \\ =167.67 kNm\gtM_U $

Design a singly reinforced section.

B=300 mm, d=450 mm, fck=20 N/mm2 , fy=415 N/mm2

$Astr =\dfrac{0.5\times20\times300\times450}{415}(1-\sqrt{1-\dfrac {4.6\times123.8\times10^6}{20\times300\times450^2}}) \\ = 881.98 mm^2 \\ Ast_{min}=\dfrac {0.85bd}{fy}=283.01mm^2 \lt Astr $

Provide 3-12 mm Φ

$Astp=942 mm^2 \\ Pt= \dfrac {100Astp}{bd}=\dfrac {100\times942}{300\times450}=0.69\% \\ Z_{uc} (pg. 73 of IS:456:2000)=0.54 \\ V_{uc} = z_{uc}bd=0.54\times300\times450 \\ =72.9 kN \\ V_{umin} = 0.4bd=0.4\times300\times450 = 54 kN. \\ V_{uc} + V_{umin}=126.9 kN\gt V_{UD}$

Hence provide minimum share reference

provide 8mm Φ2LG stirrups

$asv=2\times\dfrac \pi4\times 8^2=100 mm^2 $

Spacing

$S_1= \dfrac {0.87fv \times asvd}{V_{umin}}=\dfrac {0.87\times415\times 100\times450}{54\times10^3}=300.87 mm \\ S_2=0.75d=337.5 mm \\ S_3=300 mm$

provide 8mmΦ 2LG stirrups @300 mm c/c throughout.

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