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