Question: The floor system shown below is subjected to service live load 3 kN/m2 and floor finish load 1 kN/m2. External beams are supporting 230 mm thick wall at height 3.6 m. Adopt M20 and $Fe_{415}$ use LSM.
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Mumbai University > Civil Engineering > Sem 8 > Design and Drawing of Reinforced Concrete Structure

Marks : 40

Year : MAY 2014

mumbai university ddrcs • 848 views
 modified 3.2 years ago  • written 3.2 years ago by Aksh_31 • 680
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Slabs:-

$\dfrac {ly}{lx}=\dfrac {6.1}3= 2.03 m \geq 2$

Hence it is one way slab

$dr=\dfrac {\text {short eff. span}}{\text {S/D ratio * M.F.}} \\ =\dfrac {3000}{26\times 1.4} =82.41 \\ \text {Provide } D=125 mm$

Assume (1mⅹ1m) panel

$D.L.=D \times 25 \\ =0.125 \times 25 = 3.125 kN/m^2 \\ F.F. = 1 kN/m^2 \\ =4.125 kN/m^2 \\ \text {Factored }D.L.=1.5 ×4.125 \\ = 6.187kN/m^2 \\ L.L=3 kN/m^2$

Factored $L.L.=1.5 ⅹ3=4.5 kN/m^2$

For moment coefficient

$w_dl_{x^2}=6.18\times3^2=55.62 \\ w_Ll_{x^2}=4.5\times 3^2=40.5$

Pg. 36 of IS: 456:2000

Shear calculation:-

$w_dL_x=6.18\times 3=18.54 \\ w_LL_x=4.5\times3 =13.5$

r/f detail of slab S1-S2-S1

Distribution steel:

$Ast_{min}= 150 mm^2$

Provide 8 mm Φ bar , spacing $=\dfrac {1000\times 50}{150}=333.3\approx 300 mm$

Provide 8 mm Φ bar 300 mm c/c

Beam

As shown in the figure, Beam B1 forms the shorter sides of all the slabs, therefore it is not carry slab load. Also it is an internal beam, so wall load will not be acting.

Assume internal beam $(300 \times 500) mm \\ Self wt. (I)=0.3 \times 0.5 \times 25 \times1.5 \\ = 5.62 kN/m$

As beam runs from column to column, it is a priority beam , design it as simply supported.

$V_{UD}=\dfrac {wl}2=\dfrac {5.62\times9}2= 25.29 kN \\ M_u=\dfrac {wl^2}8=\dfrac {5.62\times 9^2}8=56.9 kNm \\ b=300 mm, d=450 mm , fck=20, fy=415 \\ M_{umax}=0.138fckld^2 = 0.138 \times 20 \times 300 \times 450^2 \\ =167.67 kNm\gt\gtMU Mu$

Hence it follows singly reinforced section.

$Astr =\dfrac {0.5\times20\times300\times450}{415}(1-\sqrt{1-\dfrac {4.6\times56.9\times0^6}{20\times300\times450^2}})$

Provide 4-12 mm Φbars

$Astp=452 mm^2 \\ Pt= \dfrac {100Astp}{bd}=\dfrac {100\times452}{300\times 450}=0.33 \\Z_{uc}=0.398 (pg. 73 of IS:456) \\ V_{uc}=z_{uc}bd=0.398\times300\times450 \\ =53.73 kN ∴safe \\ V_{umin} =0.4bd=0.4 \times 300\times 450 = 54 kN. \\ V_{uc} +V_{umin}=107.73 kN\gt\gt V_{UD}$

Hence provide minimum share reference Assume 2LG 8mm Φ stirrups

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

Spacing

$S_1= \dfrac {0.87fy \times asvd}{V_{umin}} =\dfrac {0.87\times415\times100\times450}{54\times10^3}=300.87 mm \\ S_2=0.73d=337.5 mm \\ S_3=300 mm$

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