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  1. Home/
  2. SAI HARSHITA M M/
  3. Project 1

Project 1

PROJECT 1 Given, Concrete floor thickness = 120 mm = 0.12 m S.W of floor = 0.12*25 = 3 kPa Floor finish and mech loads = 3 kPa Live load = 5 kPa Total load = 11 kPa S.W of beam = 0.5*0.75*25 = 9.375 kN/m Load on the indicated beam  = (11*1.5)*2 =33 kN/m Adding S.w of beam  = 33 + 9.375 = 42.375 kN/m Analyzing…

    • SAI HARSHITA M M

      updated on 10 Feb 2022

    PROJECT 1

    Given,

    Concrete floor thickness = 120 mm = 0.12 m

    S.W of floor = 0.12*25 = 3 kPa

    Floor finish and mech loads = 3 kPa

    Live load = 5 kPa

    Total load = 11 kPa

    S.W of beam = 0.5*0.75*25 = 9.375 kN/m

    Load on the indicated beam  = (11*1.5)*2

    =33 kN/m

    Adding S.w of beam  = 33 + 9.375 = 42.375 kN/m

    • Analyzing without considering load patterning

    • Solving by slope deflection method

      Fixed End Moments:

      MFAB = -(42.375*81)/12 = -286.03 kNm

      MFBA = 286.03 kNm

      MFBC=  -286.03 kNm

      MFCB = 286.03 kNm

      MFCD = -286.03 kNm

      MFDC = 286.03 kNm

      MFDE = -286.03 kNm

      MFED = 286.03 kNm

      Unknowns : θa, θb, θc, θd, θe

      E = 10 GPa = 10 * 10^6 kPa

      = 1 *10^7 kPa

      I = (0.5*0.75^3)/12 = 0.0176 m4

      EI = 1.76*10^5 kNm2

       

      End Moments:

      Span AB :

      M ab = MFAB + (2EI/L)(2θa + θb)

      =  -286.03 + (2*1.76*10^5/9)( 2θa  + θb )

      = -286.03 + 78222.2 θa + 39111.1 θb

      M ba = MFBA + (2EI/L)(2θb + θa )

      =  286.03 + (2*1.76*10^5/9)( 2θb  + θa )

      = 286.03 + 78222.2 θb + 39111.1 θa

      M bc = MFBC + (2EI/L)(2θb + θc)

      =  -286.03 + (2*1.76*10^5/9)( 2θb  + θc )

      = -286.03 + 78222.2 θb + 39111.1 θc

      M cb = MFCB + (2EI/L)(2θc + θb)

      =  286.03 + (2*1.76*10^5/9)( 2θc  + θb )

      = 286.03 + 78222.2 θc + 39111.1 θb

      M cd = MFCD + (2EI/L)(2θc + θd)

      =  -286.03 + (2*1.76*10^5/9)( 2θc  + θd )

      = -286.03 + 78222.2 θc + 39111.1 θd

      M dc = MFDC + (2EI/L)(2θd + θc )

      =  286.03 + (2*1.76*10^5/9)( 2θd  + θc )

      = 286.03 + 78222.2 θd + 39111.1 θc

      M de = MFDE + (2EI/L)(2θd + θe)

      =  -286.03 + (2*1.76*10^5/9)( 2θd  + θe )

      = -286.03 + 78222.2 θd + 39111.1 θe

      M ed = MFED + (2EI/L)(2θe + θd )

      =  286.03 + (2*1.76*10^5/9)( 2θe  + θd )

      = 286.03 + 78222.2 θe + 39111.1 θd

       

      Equilibrium equations:

      Mab = 0

       78222.2 θa + 39111.1 θb= 286.03

      Mba + Mbc = 0

      (286.03 + 78222.2 θb + 39111.1 θa) + (-286.03 + 78222.2 θb + 39111.1 θc) = 0

      39111.1 θa + 1.56 * 10^5 θb +39111.1 θc = 0

      Mcb + Mcd = 0

      39111.1 θb + 1.56 * 10^5 θc +39111.1 θd = 0

      Mdc + Mde = 0

      39111.1 θc + 1.56 * 10^5 θd +39111.1 θe = 0

      Med = 0

      39111.1 θd + 78222.2 θe = -286.03

      Solving the unknowns,

      θa = 0.00418

      θb = -0.00105

      θc = 0

      θd = 0.00105

      θe = -0.00418

      On Substituting,

      Mab = 0

      Mba = 367.38 kNm

      Mbc = -367.38 kNm

      Mcb = 244.96  kNm

      Mcd = -244.96 kNm

      Mdc = 367.38  kNm

      Mde = -367.38  kNm

      Med = 0

      Analyzing individual members,

      AB:

      ∑Ma = 0

      (42.375*9*4.5)-9Rb + 367.38 = 0

      Rb1 = 231.51 kN

      Ra = 149.87 kN

      BC:

      ∑Mc = 0

      -367.38 + 244.96 – (42.375*9*4.5) + 9 Rb2 = 0

      Rb2 = 204.3 kN

      Rc1 = 177.09 kN

      CD:

      ∑Md = 0

      -244.96+367.38-(42.375*9*4.5)+9Rc2 = 0

      Rc2 = 177.09 kN

      Rd1 = 204.3 kN

      DE:

      ∑Me = 0

      -367.38+9Rd2-(42.375*9*4.5)=0

      Rd2=231.51kN

      Re=149.87kN

      • Max positive and max negative moment in each span (considering load patterning)

    •  

      To get maximum positive moment in AB, AB and CD has to be loaded. Similarly to get maximum positive moment in CD, AB and CD has to be loaded with Live load including SW and DL. The remaining spans has to be loaded with self weight and DL alone

       

       

      S.W of floor =  3 kPa

      Mech services and Floor finishes = 3 kPa

      Total load = 6 kPa

      S.W of beam = 9.375 kN/m

      Load on the beam  = (6*1.5*2) = 18 kN/m

      Adding S.W = 18 + 9.375 = 27.375 kN/m

      Live Load on spans AB and CD = ( 5*1.5*2) = 15 kN/m

      Total load on AB and CD = 15 + 27.375 = 42.375 kN/m

      Solving by slope deflection method,

      Fixed End Moments:

      MFAB = -(42.375*81)/12 = -286.03 kNm

      MFBA = 286.03 kNm

      MFBC=  -(27.375*81)/12= -184.78 kNm

      MFCB = 184.78 kNm

      MFCD = -286.03 kNm

      MFDC = 286.03 kNm

      MFDE = -184.78 kNm

      MFED = 184.78 kNm

      Unknowns : θa, θb, θc, θd, θe

      E = 10 GPa = 10 * 10^6 kPa

      = 1 *10^7 kPa

      I = (0.5*0.75^3)/12 = 0.0176 m4

      EI = 1.76*10^5 kNm2

       

      End Moments:

      Span AB :

      M ab = MFAB + (2EI/L)(2θa + θb)

      =  -286.03 + (2*1.76*10^5/9)( 2θa  + θb )

      = -286.03 + 78222.2 θa + 39111.1 θb

      M ba = MFBA + (2EI/L)(2θb + θa )

      =  286.03 + (2*1.76*10^5/9)( 2θb  + θa )

      = 286.03 + 78222.2 θb + 39111.1 θa

      M bc = MFBC + (2EI/L)(2θb + θc)

      =  -184.78 + (2*1.76*10^5/9)( 2θb  + θc )

      = -184.78 + 78222.2 θb + 39111.1 θc

      M cb = MFCB + (2EI/L)(2θc + θb)

      =  184.78 + (2*1.76*10^5/9)( 2θc  + θb )

      = 184.78 + 78222.2 θc + 39111.1 θb

      M cd = MFCD + (2EI/L)(2θc + θd)

      =  -286.03 + (2*1.76*10^5/9)( 2θc  + θd )

      = -286.03 + 78222.2 θc + 39111.1 θd

      M dc = MFDC + (2EI/L)(2θd + θc )

      =  286.03 + (2*1.76*10^5/9)( 2θd  + θc )

      = 286.03 + 78222.2 θd + 39111.1 θc

      M de = MFDE + (2EI/L)(2θd + θe)

      =  -184.78 + (2*1.76*10^5/9)( 2θd  + θe )

      = -184.78 + 78222.2 θd + 39111.1 θe

      M ed = MFED + (2EI/L)(2θe + θd )

      =  184.78 + (2*1.76*10^5/9)( 2θe  + θd )

      = 184.78 + 78222.2 θe + 39111.1 θd

       

      Equilibrium equations:

      Mab = 0

       78222.2 θa + 39111.1 θb= 286.03

      Mba + Mbc = 0

      39111.1 θa + 1.56 * 10^5 θb +39111.1 θc = -101.25

      Mcb + Mcd = 0

      39111.1 θb + 1.56 * 10^5 θc +39111.1 θd = 101.25

      Mdc + Mde = 0

      39111.1 θc + 1.56 * 10^5 θd +39111.1 θe = -101.25

      Med = 0

      39111.1 θd + 78222.2 θe = -184.78

      Solving the unknowns,

      θa = 0.00474

      θb = -0.00216

      θc = 0.00130

      θd = -0.00044

      θe = -0.00214

      On Substituting,

      Mab = 0

      Mba = 302.46 kNm

      Mbc = -302.46 kNm

      Mcb = 201.99 kNm

      Mcd = -201.99 kNm

      Mdc = 302.46 kNm

      Mde = -302.46  kNm

      Med = 0

      Analyzing individual members,

      AB:

      ∑Ma = 0

      -9Rb1 + 302.46+(42.375*9*4.5)=0

      Rb1 = 224.3 kN

      Ra = 157.08 kN

      BC:

      ∑Mc = 0

      9Rb2 + 201.99-302.46-(27.375*9*4.5)=0

      Rb2 = 134.35 kN

      Rc1 = 112.02 kN

      CD:

      ∑Md = 0

      9Rc2 –(42.375*9*4.5)-201.99-302.46=0

      Rc2=179.52 kN

      Rd1 = 201.85 kN

      DE:

      ∑Me = 0

      9Rd2 –(27.375*9*4.5)-302.46=0

      Rd2 = 156.79 kN

      Re = 89.58 kN

      To get max positive and negative moment in span BC (as well as DE), BC and DE have to be loaded with LL

      Solving by slope deflection method,

      Fixed End Moments:

      MFAB = -(42.375*81)/12 = -184.78 kNm

      MFBA = 184.78 kNm

      MFBC=  -(27.375*81)/12= -286.03 kNm

      MFCB = 286.03 kNm

      MFCD = -184.78 kNm

      MFDC = 184.78 kNm

      MFDE = -286.03 kNm

      MFED = 286.03 kNm

      Unknowns : θa, θb, θc, θd, θe

      E = 10 GPa = 10 * 10^6 kPa

      = 1 *10^7 kPa

      I = (0.5*0.75^3)/12 = 0.0176 m4

      EI = 1.76*10^5 kNm2

       

      End Moments:

      Span AB :

      M ab = MFAB + (2EI/L)(2θa + θb)

      =  -184.78 + (2*1.76*10^5/9)( 2θa  + θb )

      = -184.78 + 78222.2 θa + 39111.1 θb

      M ba = MFBA + (2EI/L)(2θb + θa )

      =  184.78 + (2*1.76*10^5/9)( 2θb  + θa )

      = 184.78 + 78222.2 θb + 39111.1 θa

      M bc = MFBC + (2EI/L)(2θb + θc)

      =  -286.03 + (2*1.76*10^5/9)( 2θb  + θc )

      = -286.03+ 78222.2 θb + 39111.1 θc

      M cb = MFCB + (2EI/L)(2θc + θb)

      =  286.03 + (2*1.76*10^5/9)( 2θc  + θb )

      = 286.03+ 78222.2 θc + 39111.1 θb

      M cd = MFCD + (2EI/L)(2θc + θd)

      =  -184.78 + (2*1.76*10^5/9)( 2θc  + θd )

      = -184.78+ 78222.2 θc + 39111.1 θd

      M dc = MFDC + (2EI/L)(2θd + θc )

      =  184.78 + (2*1.76*10^5/9)( 2θd  + θc )

      = 184.78+ 78222.2 θd + 39111.1 θc

      M de = MFDE + (2EI/L)(2θd + θe)

      =  -286.03 + (2*1.76*10^5/9)( 2θd  + θe )

      = -286.03+ 78222.2 θd + 39111.1 θe

      M ed = MFED + (2EI/L)(2θe + θd )

      =  286.03 + (2*1.76*10^5/9)( 2θe  + θd )

      = 286.03+ 78222.2 θe + 39111.1 θd

       

      Equilibrium equations:

      Mab = 0

       78222.2 θa + 39111.1 θb= 184.78

      Mba + Mbc = 0

      39111.1 θa + 1.56 * 10^5 θb +39111.1 θc = 101.25

      Mcb + Mcd = 0

      39111.1 θb + 1.56 * 10^5 θc +39111.1 θd = -101.25

      Mdc + Mde = 0

      39111.1 θc + 1.56 * 10^5 θd +39111.1 θe = 101.25

      Med = 0

      39111.1 θd + 78222.2 θe = -286.03

      Solving the unknowns,

      θa = 0.00214

      θb = 0.00044

      θc = -0.00130

      θd = 0.00216

      θe = -0.00474

      On Substituting,

      Mab = 0

      Mba = 302.46 kNm

      Mbc = -302.46 kNm

      Mcb = 201.99 kNm

      Mcd = -201.99 kNm

      Mdc = 302.46 kNm

      Mde = -302.46  kNm

      Med = 0

      Analyzing individual members,

      AB:

      ∑Mb = 0

      9Ra + 302.46-(27.375*9*4.5)=0

      Rb1 = 156.79 kN

      Ra = 89.58 kN

      BC:

      ∑Mc = 0

      9Rb2 + 201.99-302.46-(42.375*9*4.5)=0

      Rb2 = 201.85 kN

      Rc1 = 179.52 kN

      CD:

      ∑Md = 0

      9Rc2 –(27.375*9*4.5)-201.99-302.46=0

      Rc2=112.02 kN

      Rd1 = 134.35 kN

      DE:

      ∑Me = 0

      9Rd2 –(42.375*9*4.5)-302.46=0

      Rd2 = 224.3 kN

      Re = 157.08 kN

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