Question: Design typical slab S1 and beam B3 . All the beams carry 230 mm thick brick wall load of 3m height. Assume L.L.$= 3 kN/m^2 .$ Floor finish $= 1 kN/m^2$.
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Use M20 grade of concrete and Fe415 grade of steel. Show all the reinforcement details with neat sketch.

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

Marks : 40

Year : MAY 2012

mumbai university ddrcs • 792 views
 modified 3.2 years ago  • written 3.2 years ago by Aksh_31 • 680
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$d=\dfrac {\text {short eff. span}}{S/D\times ratio \times M.F}=\dfrac {3500}{26\times1.4}=96.15 \\ ∴D=125 mm \\ d=125-20-(10/2) =100 mm$

Load calc:- Assume 1m ⅹ1m panel.

$D.L.=D\times 25=0.125\times 25 = 3.125 kN/m^2 \\ F.F. = 1 kN/m^2 \\ L.L =3 kN/m^2 \\ = 7.225 kN/m^2$

Factored load(w) $=1.5 ×7.225 \\ = 10.83 kN/m^2$

For span $S_1=wl_{x^2} =10.83×3.52 = 132.66$

As the slab design section facing long than coefficients from Pg 91 table 26 long span coefficients.

$ly/lx=6/3.5 = 1.71$

Distribution steel:

$Ast_{min} =150 mm^2$

Provide 8 mm Φ bar

Spacing $=\dfrac {1000\times50}{150}=333.3 \approx 300 mm$

Provide 8 mm Φ bar 300 mm c/c

r/f detail of slab S1

Beams:

Triangular udl= $\dfrac {wlx}3$

Trapezoidal udl= $\dfrac {wlx}2 (1-\dfrac 1{3\beta^2})$

All internal beams (300×500) mm

Self wt. (int.) $=0.3×0.5×25×1.5 \\ = 5.63×kN/m$

No wall load as the beam is internal.

Beam B3= slab load $=16.79+20.8 \\ = 37.59 kN/m$

Self wt.$= 5.63 kN/m \\ = 43.22 \approx 44 kN/m.$

For 111.37 kN

$M_u(kNm) = 111.37 kN \\ b=300, D=500, d=450 mm, M_{20} , Fe_{415} \\ M_{umax} = 0.318fck\times l\times d^2 \\ =0.138 \times20 \times300 \times450^2 \\ = 167.67 kNm \\ M_{umax} \gt M_u$

Hence for singly reinforced section

$Ast_r =\dfrac {0.5\times20\times300\times450}{415}[1-\sqrt{\dfrac {4.6\times111.37\times10^6}{20\times300\times450^2}}] \\ Ast_r=779.11 mm^2$

Provide 4-12 mm Φ steel bar

$Ast_p=804 mm^2 \\ Pt.= \dfrac {100\times Astp}{b\times d}=\dfrac {100\times804}{300\times450}=0.59 \\ Z_{uc} \text {(by interpolation) Pg. 73 of IS 456:2000 }=0.508 \\ V_{uc} = z_{uc} bd=0.508\times300\times450 = 65.58 kN \\ V_{umin} = 0.4bd \times 300\times450=54 kN \\ V_{uc}+ V_{umni} =122.58 kN \\ V_{UD} = 99 \lt V_{uc}+ V+_{umni}$

Hence provide minimum shear r/f

Assume 2 LG 8mm Φ stirrups

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

Spacing

$S_1= \dfrac {0.87fy \times asvd }{V_{umin}}=\dfrac {0.87\times415\times100\times450}{54\times10^3}=300.87 mm \\ S_1=300.87 mm \\ S_2=0.75d =337.5 mm \\ S_3=300 mm$

Provide 8 mm Φ 2LG stirrups @ 300mm c/c

For 198 kNm

$M_u \gt M_{umax } (167.67 kNm)$

Hence provide doubly reinforced section

$M_{u1}=167.67 kNm \\ M_{u2} =198 -167.67=30.33 kNm \\ M_{u1} = T_{U1} \times L_{a1} = 0.87fy Ast_1(d-0.42 x_{umax}) \\ 167.67 \times 10^6 = 0.87 \times 415 \times Ast_1 \times (450-0.42\times 0.48 \times 450) \\ Ast_1=1292.57 mm^2 \\ M_{u2} = T_{U2} \times L_{a2} = 0.87fy\space Ast_2(d-d_c) \\ 30.33 \times 10^6=0.87\times 415\times Ast_2(450-50) \\ Ast_2=210.01 mm^2 \\ Ast = Ast_1+Ast_2=1292.57+210.01 \\ =1502.58 mm^2$

Provide 5-20 mm Φ

$M_{u2} = C_{U2} \times L_{a2}=(f_{sc}-f_{cc}) Asc(d-d_c) \\ f_{sc}=350 N/mm^2 …………\text {assume } \\ f_{cc} = 0.446fck=0.446×20=8.92 N/mm^2 \\ 30.33\times 10^6=(350-8.92)Asc(450-50) \\ Asc=222.3 mm^2$

Provide 2-12 mm Φ.

$Pt.= \dfrac {100\times Astp}{b\times d }=\dfrac {100\times1570}{300\times 450}=1.16\% \\ Z_{uc} \text {(by interpolation) Pg. 73 of IS 456:2000 } =0.652 \\ V_{uc} = z_{uc}bd = 0.652\times 300\times 450 \\ = 88.02 kN \\ V_{umin} =54 kN \\ V_{uc} + V_{umni} =142.02 kN \lt V_{UD}=165 kN$

Hence design and provide shear r/f.

$V_{us} = V_{UD} - V_{UC} \\ =165-142.02 = 22.98 kN \\ S_1=\dfrac {0.87fyasvd}{Vus}=\dfrac {0.87\times415\times100\times450}{22.98\times10^3} \\ =707.01 mm \\ S_2=0.75d=337.5 mm \\ S_3=300 mm$

Provide 8 mm Φ 300 mm c/c