BEAM ANALYSIS BY YIELD LINE THEORY

By
Advertisement Flower

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:
MXX WLX2
MYY 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
MXX WLX2
      =0.065×12.45×6.222
Mx=31.308KNM
MYY 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
ڂv =K ڂc
ڂv=Vu/bd
Vu=38.71KN ,  ڂv=0.175N/mm2, Permissible shear stress (K ڂc)
100Ast/bd=0,035N/mm2     ,     ڂc=0.28 N/mm2   , ڂv< ڂc    
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
ڂv =K ڂc  ڂ    & v=Vu/bd
Vu=38.26KN ,      ڂv=1.012N/mm2
Permissible shear stress (K ڂc)
100Ast/bd=1.026N/mm2
Refer  From  Table 19 for I.S. 456
ڂc  = 0.625,  ڂcmax=2.8 N/mm2
ڂ v      > ڂc    >    ڂcmax
K ڂc=1×0.625=0.652N/mm2
8.SHEAR REINFORCEMENT:
Vus  =Vu  -  ڂcbd
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
Post a Comment
Powered by Blogger.
Don't Forget To Join US Our Community
×
Widget