BEAM ANALYSIS BY YIELD LINE THEORY
BEAM
ANALYSIS BY YIELD LINE THEORY
1. DATA
l = 3m
g = 6
KN / m
q = 8
KN / m
fck = 20
N/ mm2
fy = 415
N/mm2
2.
CROSS SECTION DIMENSIONS:
D =
Span/12
D =
250mm
Adopt clear
cover 30 mm
d = 220
mm
D = 250
mm
b = 230
mm
3. LOAD CALCULATION:
Self weight of
beam = 0.23
0.25 25
= 1.437 KN / m
Dead Load = 6 KN / m
Finish load = 0.75 KN / m
Total Dead
load(g) = 8.187 KN/m
Live load(q) = 8 KN / m
4. BENDING MOMENT & SHEAR FORCE:
Referring
BM & SF coefficients negative moment
at interior support.
MU
(-ve) = 1.5 =1.5 = 23.05 KNm
Positive Bm @ centre of span
MU (+ve) =
1.5
= 1.5 = 20.01 KNm
Maximum
shear force at the support section
VU = 1.5
0.6×L× (g+q)
= 1.5 0.6 3 (8.187 +
8)
VU = 43:
704 KN
5. LIMITING MOMENT OF THE RESISTANCE:
MU lim = 0.138
fc k bd2
= (0.138
20 230 2002)
10-6
= 25.39 KNm
MU
< MU Lim; The section is
under reinforced.
6. TO GET AST:
Ast
= 282.513 mm2 (Provide 2
bars of 16mm )
Ast
= ,,Ast=402.123mm2
7. SHEAR REINFORCEMENT:
= N/mm2
45
Refer
table 19 (IS-456) page No. 79
N/mm2
Hence
shear reinforcement are required
Balance
shear(VUS) =
VUS =
27.05 KN
Using
8mm two legged
stirrups the spacing is
SV =
=
SV = 266.955mm
8. CHECK FOR DEFLECTION CONTROL:
At
centre of span:
=
Modification
factor Kt=1.0 neglecting bars in compression side KC=1.0
& KF=1
(L/d)
Max = (L/d) basic Kt KF
= 26 1 1= 26
(L/d) Actual = = 15<26
Hence
deflection control is satisfied
9. DESIGN OF SHEAR REINFORCEMENT USING SP16
DESIGN TABLES:
VUS
= 27.05 KN , d= 200mm
Compute
parameter (VUS/d)
VUS/d =
VUS/d
= 1.35 KN/m
Spacing
of 8mm of two legged
stirrups, SV = 230 mm
2. IS CODE DESIGN ANALYSIS
SLAB
(OFFICE ROOM)
1. DATAS:
Size of Room =
6m×6.23m
Live load = 2KN/M2
W,c,Load =
0.8 KN/M2
Wall thickness =
230mm
Concrete grade =
M20
Steel grade = Fe415
2.DESIGN OF SLAB:
Longer span/Shorter span=Ly/Lx
6.23/6=1.03<2
Hence it is two way slab
3.DEPTH REQUIRED FOR STIFFNESS:
Clear span For Longer span=6.23m
Clear span For shorter span=6m
Basic value for l/d for simply
supported beam=32×0.8=28
d=Span/28×MF (Assume M.F=1.0)
=6000/28×1.0
d
=214.28mm
Assume 8mm ø rods & clear
cover 20 mm
D=d+8/2+20
=214.28+4+20=238.28mm
(D=240mm) & (d=220mm)
4.EFFECTIVE
SPAN:
Eff.span=clear span+d
Lx=6000+220=6220mm & Ly=6230+220=6450mm
5.LOAD CALCULATIONS:
Self weight of slab =0.22×25 = 5.5KN/M2
Weight of W.C = 0.8KN/M2
Live Load =2KN/M2
Total Load =8.3KN/M2
Factored Load(Wu) =8.3×1.5
(Wu) =12.45KN/M2
6.FACTORED BENDIND MOMENT:
MX=αX WLX2
MY=αY WLY2
LY/ LX=6.45/6.22=1.036m
Refer IS code book From Table 27 ,Page No 91
αX=0.065, αY=0.061
MX=αX WLX2
=0.065×12.45×6.222
Mx=31.308KNM
MY=αY WLY2
=0.061×12.45×6.222
My=29.381KNM
7.DEPTH REQUIRED:
M.R=Qbd2
29.38×106=2.76×1000× d2
, d=103.17mm
Hence it is safe
8.AREA OF STEEL REINFORCEMENT (Mx DIRECTION):
M.R=0.87fy Ast d(1- fy Ast/bd fck)
31.308×106=0.87×415×
Ast ×220(1-415×Ast/1000×220×20)
31.308×106=79.43×103Ast-7.491Ast2,Ast=410.01mm2(provided)
Area of minimum Reinforcement
Astmin=(0.12/100)bd
= (0.12/100)×1000×220=264 mm2
Area of maximum Reinforcement
Astmax=0.04bd
=0.04×1000×220=8800mm2
Astmin < AstProvided < Astmax
264mm2 <410.01mm2<8800mm2
SPACING
Assume 8 mm ø rods
Spacing=(ast/Astprovided)×1000
=(50.26/410.01)×1000 =122.55mm
Say 130mm
SPACING
LIMIT
1.3d=3×220=660mm
2.450mm
Provided 8mm ø rods @ 130mmc/c
9.AREA
OF STEEL REINFORCEMENT (My DIRECTION):
M.R=0.87fy Ast d(1- fy Ast/bd fck)
29.381×106=39.71×103Ast-7.491Ast2
Ast=383.78 mm2(provided)
Area of minimum Reinforcement
Astmin= (0.12/100)bd
= (0.12/100)×1000×220
Astmin = 264mm2
Area of maximum Reinforcement
Astmax=0.04bd =8800mm2 ,
Astmin <
AstProvided < Astmax
264 mm2 <383.78mm2<8800mm2
Hence it is safe
SPACING
Assume 8 mm ø rods
Spacing=(ast/Astprovided)×1000
=(50.26/383.78)×1000=130.93mm Say
130mm
SPACING LIMIT
1.3d=3×220=660mm
2.450mm
Provided 8mm ø rods @ 130mmc/c
10.MIDDLE STRIP & EDGE STRIP:
Shorter span
Middle Strip for Shorter’s
span=3/4 = 3/4×6.22 = 4.66m
Edge strip=1/8l = 1/8×6.22=0.777m
Longer span
Middle Strip for Shorter’s
span=1/8 ly =
1/8×6.45=0.806m
Size of Torsion reinforcement
mesh=ly/5 = 6.45/5=1.29m
Torsion reinforcement =3/4×AstProvided
=3/4×410.01=307.05mm
Provide 8mmø rods for
reinforcement
Spacing =(50.26/307.05)×1000
=163.74mm Say 170mm
Provide 8mmø rods @ 170mmc/c
11.CHECK FOR SHEAR:
Nominal shear stress <
permissible shear stress
Ú‚v
=K Ú‚c
Ú‚v=Vu/bd
Vu=38.71KN , Ú‚v=0.175N/mm2,
Permissible shear stress (K Ú‚c)
100Ast/bd=0,035N/mm2 , Ú‚c=0.28
N/mm2 , Ú‚v< Ú‚c
Hence it is safe
12.CHECK
FOR DEFLECTION:
FS=O.58 fy
Fs=240.7N/mm2 [M.F=2, BV=20]
Effective depth=Span/(BV×MF)
=77.75mm
77.75mm<220mm
Hence
it is safe.
Table : 3 Comparision between
yeiled line and is code design
analysis (Size of room = 6m X 6.23m)
SL.NO
|
YIELEDLINE ANALYSIS
|
IS CODE DESIGN ANALYSIS
|
1.
|
Total
Depth(D)=200mm
Effective
Depth(d)=175mm
|
Total
Depth(D)=220mm
Effective
Depth(d)=200mm
|
2.
|
Total
Load=7.75KN/m²
Ultimate
Load=11.63KN/m²
|
Total
Load=8.3KN/m²
Ultimate
Load=12.45KN/m²
|
3.
|
Area
of Steel reinforcement=360mm²
|
Area
of Steel reinforcement=410mm²
|
4.
|
Provide
10mmø bars @ 210mm c/c
|
Provide
8mmø bars @130mm c/c
|
BEAM
DESIGN
1.DATA
L=3000mm
Dead load=6KN/m
Live Load=8KN/m
Finished
load=0.75KN/m
Wall Support=230Mm
Concrete grade=M20, Steel grade=fe415
2.EFFECTIVE SPAN OF BEAM:
Effective Span =c/c of support
= 3230 mm
3.SIZE OF BEAM:
Effective depth=Span/12
=269.16mm , Say 270mm
Breadth=1/2d =135mm
Clear cover=25mm,Use 16 mm ø
D=303 mm, Say 310 mm
D=310mm
d = 277mm, Say 280mm
d=280mm
4. LOAD CALCULATIONS:
Self Weight =1× 0.31× 0.135× 25=1.046KN/m
Live Load = 8KN/m
Dead Load =6KN/m
Finished load =0.75KN/m
T0tal Load =15.796KN/m
Factored Load =23.694KN/m, Factored BM=Wl2/8 = 30.89KN/m
5. TO GET AST:
Ast(Required) = 388.33mm2
Numbers
of rods = 2nos
Provide 2nos of 16mmø rods
6. CHECK
FOR MINIMUM & MAXIMUM AST:
ASTmin=(0.85/Fy)
bd =77.421mm2
ASTmax=0.04 bd=1512mm2
7.CHECK FOR SHEAR:
Nominal shear stress <
permissible shear stress
Ú‚v
=K Ú‚c Ú‚ & v=Vu/bd
Vu=38.26KN , Ú‚v=1.012N/mm2
Permissible shear stress (K Ú‚c)
100Ast/bd=1.026N/mm2
Refer From
Table 19 for I.S. 456
Ú‚c
= 0.625, Ú‚cmax=2.8
N/mm2
Ú‚ v > Ú‚c >
Ú‚cmax
K
Ú‚c=1×0.625=0.652N/mm2
8.SHEAR REINFORCEMENT:
Vus =Vu
- Ú‚cbd
Vus =14.625× 103 N
Provided 8mmø rods @ 2 legged Stirrups
1) Sv= ( 0.87 Fy Asv d/ Vus)
=694.41mm, Say 690 mm
2)
Sv= ( 0.8 7 Fy Asv / bd)
=96.01mm
< Say 100mm
3) 0.75 d= 210mm
4) 450mm
Provide 8mmø rods & 2
Legged Stirrups @ 100mm c/c
9.CHECK FOR DEFLECTION:
FS=O.58 fy
Fs=240.7N/mm2 [M.F=2, BV=20]
Effective depth=Span/(BV×MF)
=80.75mm
80.75mm < 280mm
Hence it is safe
1 comment
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