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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