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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.

enter image description here

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.

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A] Design of slab S1

ly=5m

ln=4m

lyln=54=1.25<2

Two way continuous slab

1) Depth Calculation

dread=lxl2×M.F

= 400026×1.4

= 109.89mm

dreq120mm

Overall Depth

D=d+c.c+ϕ2

= 120+20+102

D = 145 mm

2) Load Calculation

a) s/w of slab = 25×D

= 25×0.145=3.625kN/m2

b) Live load = 3kN/m2

c) Floor Finish = 1.5kN/m2w=8.124kN/m2

wu=1.5×8.125

wu=12.18kN.m

3) Calculation of B.M Coefficient

lyln=1.25

αx=0.075 and αy=0.056

Mux=αx×wu×l2n

= 0.075×12.18×42

Mux=14.616kN.m

Muy=0.056×12.18×42

Muy=10.91kN.m

4) Check for depth

d=14.616×10620×1000×0.138

d=72.72mm<120mm

safe in depth

5) Calculation of steel

a) Astx = ?

Astx=0.5×20×1000×120415[114.6×14.616×10620×1000×1202]

Astx=359.91mm2

Astmin=0.12100×1000×145

Astmin=174mm2

Astx>Astmin

Use Astx=359.91mm2

use 10 mm ϕ bar

Spacing=π4×102359.91×1000

= 218.22

200mm

Provide 10 mm ϕ bar @ 200 mm c/c

b) Asty = ?

Muy=10.91kN.m

Astx=0.5×20×1000×120415[114.6×10.91×10620×1000×1202]

Asty=263.98mm2

Asty>Astmin

use 10 mm ϕ bar

Spacing=π4×102263.98×1000

= 297.73

275mm

Provide 10 mm ϕ bar @ 275 mm c/c

6) Check for shear

1) Vux=0.076×12.18×4=3.65kN

2) Vuy=0.056×12.18×4=2.72kN

Vuc=kτuc bd

k=1.3ford lt150mm

Astp=π4×102200×1000=392.7mm2

%pt=392.71000×120×100

τuc=0.396

Vuc=kτucbd

1.3×0.396×1000×120

Vuc=61.77kN>3.656kN

Safe in shear

7) Check for deflection

Fs=0.58fyAstreqAstd

= 0.58×415×359.91392.7

Fs=220.6

240N/mm2

%pt=0.327

enter image description here

M.F = 1.5

d=400026×1.6=102.56<120mm

Safe in deflection

7) Check for development length

Ld1.3M1V+lo

M1=14.616kN.m

V=3.65kN

lo=12dord

lo=120mm

RHS=5325.66 mm

LHS=0.87fy4τbd

= 0.87×4×1.2×1.8

= 470.11mn

RHS>LHS

Safe in development length

B) Beam design B1

enter image description here

1) Load Calculation

a) Load transformed by Slab S1

= wulx2[113β2]

= 12.18×42[113×1.252]

= 19.16kW/m

b) Wall load = 1.5×l×b×h

1.5×20×0.23×3.2

= 22.08kN/h

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

= 10% (19.16 + 22.08)

= 4.124 kN/m

= wu=45.364kN/n

enter image description here

VA=VB=45.36×52=113.41

enter image description here

midspan AB

Mu=141.76kN.m

enter image description here

bf=lo6+6Df+bw

0.7×50006+6×145+230

1683.33mm

D=500010

D=500mm

d=50025=475mm

Assume xDf

Mur=0.36Fckxubf(d0.42xl)

=0.36×20×145×1683.33(4750.42×145)

Mur=727.73>14176kN

Assumption is correct

Ast1=0.5×20×230×475415[114.6×141.76×10620×1000×4702]

Ast=1027.54mm2

Use 20 mm ϕ bar

No. of Bar = 1027.54pi4×202

= 3.27

4Nos

Provide 4 -20 mm ϕ bars

Design of shear reinforcement

VuD=113.41kN

DVuc=τucbd

Ast=4×π4×202=1256.64mm2

%pt=Astbd×100

%pt=1.16

τuc0.652

Vuc=τucbd

= 0.652×230×475

Vuc=71.23kN

Vus=VuDVuc

= 113.4171.23

Vus=42.18kN

Vus=Vusv=42.18kN

Sv=0.87FyAsvdVusv

Sv=0.87×415×2×π4×82×47542.18×103

= 402.82

300mm

2L-8mm ϕ 300 mm c/c

enter image description here

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