Question: The floor system shown below is subjected to live load of $3kN/m^2$ and floor finish load of $1 kN|m^2$ In addition to self-weight.
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All the external beams support 230 mm thick brick wall of height 3.3m and internal beams carry 115 mm thick brick wall of same height of 3.3 m. Use LSM and adopt concrete M20, Steel Fe415 to design

a) slab $s_1$ and $s_2$

b) Beam $B_2 – B_3$ and $B_4$

also sketch the reinforcement details:

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

Marks : 40

Year : DEC 2015

mumbai university ddrcs • 912 views
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modified 3.0 years ago  • written 3.1 years ago by gravatar for Aksh_31 Aksh_31630
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Section X-X:

$\dfrac {l_y}{l_x}=\dfrac 53=1.67 \lt 2 $

Hence it is two way slab.

Trial depth(d)= $\dfrac {\text {short eff span}}{\text {S/D ratio * M.F.}}$

S/D ratio=20 (as slab in simply supported)

$d=\dfrac {3000}{20\times 1.4}=107.14mm $

Assume D=150 mm

$∴d=150-20-10/2 \\ d=125 mm$

Load calculations: Assume $1m \times 1m$ pamel

$D.L.=D\times 25 = 0.12\times25 = 3 \\ .75 kNm \\ F.F.= = 1 kN/m \\ L.L.= =3 kN/m\\ =7.75 kN/m$

Factored load(w) $=1.5\times 7.75=11.62 \approx 1 kN/m $

As it is simply supported

$V_{UD}= \dfrac {wl^2}2=\dfrac {12\times5}2=30kN \\ M_u=\dfrac {wl^2}8=\dfrac {12\times5^2}8=37.5 kNm \\ b=1000 mm, d=125 mm , fck=20, fy=415 \\ M_{umax } =0.138fckld^2 = 0.138\times 20\times 1000\times 125^2 \\ =43.12 kN \gt M_U ∴safe \\ Astr=\dfrac {0.5\times 20\times 1000\times 125}{415} (1-\sqrt{1-\dfrac {4.6\times37.5 \times 10^6}{20\times 1000\times 125^2}}) \\ Astr=996 mm^2 \\ Asr_{min}=(0.12bD/100)=180 mm^2 \lt Astr $

Provide $16 mm Φ$ bars

Spacing $=\dfrac {100\times201}{996} = 201.8 \approx 200 mm \\ Astp= \dfrac {1000\times 201}{200} =1005 mm^2 \\ Pt= \dfrac {100\times 1005}{1000\times 125}=0.8\% \\ Z_{uc}(pg. 73 of IS:456)=0.572. \\ V_{uc}=kz_{uc} bd=1.3\times 0.572\times 1000\times 125 \\ =92.95 kN \gt VUD ∴safe$

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Dist. Steel:- $Ast_{min} =180 mm^2 , 8 mm Φ$ bars

Spacing $=\dfrac {1000\times 50}{180}=277.7\approx 250 mm $

r/f detail of slab along X-X

Section Y-Y:

It is also a two way slab

Trial depth:- dr = $\dfrac {\text {short eff span}}{\text {S/D ratio * M.F.}} \\ dr=\dfrac {3000}{26\times 1.4}$

(26:- Now it is continues with S3)

$dr = 82.41 mm $

Assume D=125 mm

$∴d=125-20-(10/2)=100 mm$

Load calculations: Assume 1m ⅹ1m pamel

$D.L.=D\times 25 = 0.125 \times 25 = 3.125 kNm\\ F.F.= = 1 kN/m \\ L.L.= =3 kN/m \\ =7.125 kN/m $

Factored load(w) $=1.5 \times 7.125=10.68 \approx 11 kN/m $

∴for both span :- $wlx^2 =99$

From pg. 91, of IS 456:2000 coefficients.

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Dist. Steel:- $Ast_{min} =150 mm^2 , 8 mm Φ$ bar

Spacing= $\dfrac {1000\times 50}{180}=333.33\approx 300 mm $

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All external beams (300ⅹ600)mm

All Internal beams=(300ⅹ500)mm

Self weight(ext.)=(0.3ⅹ0.6ⅹ25)×1.5

=6.75 kN/m

Self wt.(int)= (0.3ⅹ0.5ⅹ25) ⅹ1.5

= 5.63 kN/m

Wall load =(0.23ⅹ3.6ⅹ20)ⅹ1.5

=24.84 kN/m2 25 kN/m.

Beam B2:- slab load = 9 kN/m.

B4 Self wt.(ext)=6.75 kN/m

Wall load = 25

$= 40.75 \approx 41 kN/m $

Beam B3:-

slab load = 40 kN/m.

Self wt.(int)=5.63 kN/m

$= 45.63 \approx 46 kN/m $

By moment distribution method

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Provide minimum shear r/f for both cases.

Provide 8mm Φ 2 LG stirrups

$S_1=\dfrac {0.87 f_y asvd}{V_{umun}}=\dfrac {0.87\times 415\times 100 \times 450}{54 \times 10^3} \\ =300.87 mm \\ S_2=0.75 \times d=0.75 \times 450=337.5 mm \\ S_3=300 mm $

Provide 8mm Φ@ 300 mm c/c.

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modified 3.0 years ago  • written 3.0 years ago by gravatar for Aksh_31 Aksh_31630
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