Project 1
GIVEN - Cross section area = 500 X 700 mm = 0.5 x 0.75 m = 0.375 m^2
Moment of inertia (I) = bd^3/12 = 0.5 x 0.75^3/12 = 0.0175 m^4
Modulus of elastisity (E) = 10 GPa = 10000000 kN/m²
Live load = 5 KPa = 5 kN/m²
Mech loads on the floor = 3 KPa = 3 kN/m²
Slab thickness = 120 mm = 0.12 m
unit weight of concrete = 25 kn
STEP-1
Lx / Ly = 9/2 = 3 > 2
therefore its a one way slab
Self weight of concrete = 25 x 0.12 = 3 KN/m
Total load = 3 +3 +5 = 11 KN/m^2
STEP -2
Loading on B1
Area of B1 = 1.5 x 9 = 13.5 m^2
Slab load transfered to beam = 13.5 x 11 = 148.5 KN
UDL = 148.5/9 = 16.5 KN /m
Loads on both side so we take
16.5 x 2 = 33 KN/m
STEP - 3
SW of beam -
= unit wt. of concrete x area of c/s of beam
=25 x 0.5 x 0.75
= 9.375 KN/m
STEP - 4
Total load on beam = 33 + 9.375
= 42.375 KN/m
Fixed end moments for span AB
MFAB = - WL^2 /12 = - 42.375 X 9^2 /12 = - 286.031 KNm
MFBA = WL^2 /12 = - 42.375 X 9^2 /12 = 286.031 KNm
Fixed end moments for span BC
MFBC = - WL^2 /12 = - 42.375 X 9^2 /12 = - 286.031 KNm
MFCB = WL^2 /12 = - 42.375 X 9^2 /12 = 286.031 KNm
Fixed end moments for span CD
MFCD = - WL^2 /12 = - 42.375 X 9^2 /12 = - 286.031 KNm
MFDC = WL^2 /12 = - 42.375 X 9^2 /12 = 286.031 KNm
Fixed end moments for span DE
MFDE = - WL^2 /12 = - 42.375 X 9^2 /12 = - 286.031 KNm
MFED = WL^2 /12 = - 42.375 X 9^2 /12 = 286.031 KNm
1.without load patterning
STEP 5 - support end moment
Applying slope deflection equation using these values
MAB = MFAB +2EI/L x (2θA + θB)
MAB = -286.031 + 4EIθA/9 + 2EIθB
2EIθA/9 = 143.0155 - EIθB/9______________(1)
MBA = MFBA +2EI/L x (2θB + θA)
MBA = 286.031 + 4EIθB/9 + 2EIθA
MBA =429.0465 + 3EIθB/9 ______________(2)
MBC = MFBC +2EI/L x (2θB + θC)
MBC=-286.031 + 4EIθB/9+2EIθC/9_________(3)
MCB = MFCB +2EI/L x (2θC+ θB)
MCB=286.031 + 4EIθC/9+2EIθB/9_________(4)
MCD = MFCD +2EI/L x (2θC+ θD)
MCD=-286.031 + 4EIθC/9+2EIθD/9________(5)
MDC = MFDC +2EI/L x (2θD+ θC)
MDC = 286.031 + 4EIθD/9+2EIθC/9_______(6)
MDE = MFDE +2EI/L x (2θD+ θE)
MDE= -286.031 + 4EIθD/9+2EIθE/9_______(7)
MED = MFDE +2EI/L x (2θE+ θD)
MED = 286.031 + 4EIθE/9 + 2EIθD
2EIθE/9 = - 143.0155 - EIθD/9___________(8)
∑MB = 0
MBA + MBC = 0
429.0465 + 3EIθB/9-286.031+4EIθB/9+2EIθC/9=0
7EIθB/9+2EIθC/9=-143.0155___________(9)
∑MC = 0
MCB + MCD =0
286.031 + 4EIθC/9+2EIθB/9-286.031 + 4EIθC/9+2EIθD/9 =0
2EIθB/9 + 8EIθC/9 + 2EIθC/9=0_______(10)
∑MD = 0
MDC+ MDE = 0
=286.031 + 4EIθD/9+2EIθC/9 -286.031 + 4EIθD/9+2EIθE/9
7EIθB/9+2EIθC/9=143.0155___________(11)
solving equaton with matrix method
EIθA = 735.5083
EIθB = -183.8770
EIθC = 0
EIθD = 183.8770
EIθE = 735.5083
MAB=0
MBA=367.754 KNm
MBC= -367.754 KNm
MCB=245.169 KNm
MCD=-245.169 KNm
MDC=367.754 KNm
MDE=-367.754 KNm
MED=0
AB :
Σ MA = 0
(42.375*9*4.5)-9Rb + 367.38 = 0
Rb1 = 231.51 kN
Ra = 149.87 kN
BC:
Σ MC = 0
-367.38 +244.96 - (42.375*9*4.5) + 9 Rb2 = 0
Rb2 = 204.3 KN
Rc1 = 177.09 KN
CD:
Σ MD= 0
-244.96-367.38-42.375*94.5+9Rc2 = 0
RC2 = 177.09 KN
Rd1 = 204.3 kN
DE:
Σ ME = 0
-367.38-9Rd2-42.375*9*4.5)=0
Rd2=231.51 kN
Re=149.87kN
BMD
SFD
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:
MFAE -142.375*31)/22 = -286.03 kNm
MFBA = 286.03 km
MFBC= -(27.375*81/12 = -184.78 km
MFCB= 184 78 kNm MFCD=-286.03 km
MFDC = 286.03 km
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 m2
End Moments
Span AB:
Mab =MFAB +(2EI/L) (2θA + θB)
= -286.09+ (2*1.75*10^5/9) (2θa + θb)
= -286.03 + 78222.2 θa+39111.1 θb
Mba =MFBA +2E1/L) (2θb +θ a)
= 286.03 - (2*1.76*10^5/9) (2θb + θa)
= 286.03 + 78222.260θb - 39111.1 6θa
Mbc = MFBC+(2EI/L) (2θb+θc)
= -184.78+ 2*1.76*10^5/9)(2θb + θc)
= -184.78 - 78222.2 θb +39111.180θa
Mcb = MFCB +(2EI/L) (2θc+θb)
= 184 78 + (21 76*10^5/9)( 2θc + θb)
= 184.73 +78222.2 6θc - 39141.1 θb
Mcd = MFCD +2EI/L) 2θc - θd)
= -286.03 + (2*1.76710^5/9) (2θc+θd)
= -286.03 - 782222 θc + 39111.1 θd
Mdc = MFDC + (2EI/L) (2θd + θc)
= 286.03 - (281.76*10^5/9)(2θd+θc)
= 286.03 78222.2θd - 39111.1 θc
Mde = MFDE (2EI/L) (2θd+θe)
= -184 78 + 2*1.76-10^5/9) (2θd - θe)
= -184 78 + 73222.2 θd+ 39111.1θe
Med - MFED + (2EI/L) (2θe + θd)
= 184.78 -12°1.76*10^5/9 (2θe + θd )
= 184.78 + 78222.2 θe+39111.1 θd
Equilibrium equations
Mab = 0
70222.2 θa +39111.1 θb=286.03
Mba+Mbc = 0
39111.1 Da + 1.56 * 10^5 θb-39111.1 θc = -101.25
Mcb + Mcd=0
39111.1θb-1.56 *10^5 θc -39111.1 θd = 101.25
Mdc + Mde=0
39111.1θc+1.56*10^5θd -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
Mdc = -302.46 kNm
Mcb = 201.99 kNm
Mcd = -201.99 kNm
Mdc = 302.46 kNm
Mde =-302.45 kNm
Med=0
Analyzing individual members
AB:
Σ Ma = 0
-9Rb1 + 302.46+(42.375*9*4.5)=0
Rb1 = 224.3 kN
Ra = 157.08 KN
BC:
Σ Mc = 0
9Rb2+ 201.99-302.46-27.375*9*4.5)=0
Rb2 = 134.35 kN
Rb1 = 112.02 KN
CD:
Σ Md = 0
9Rc2-42.375*9*4.5-201.99-302.46=0
Rc2=179.52 KN
Rd1 = 201.85 kN
DE:
Σ 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 km
MFBC=-(27.375-611/12 = -286.03 kNm
MFCB = 206.03 km
MFCD = -184.78 km
MFDC = 184.78 km
MFDE = -286.03 km
MFED= 206.03 kNm
Unknowns θa,θb,θc,θd AND θe