Project 1
SOLUTION:
1). Given
Concrete Slab Thickness = 0.12 m
Dead Load of Slab = Thickness x Unit Wt. of Concrete
= 0.12 x 25
= 3 KN/m2 = 3 kpa.
Floor Finish and Mech loads on the Floor = 3 kPa.
The Live load on the Floor = 5 kPa.
Total Load = 11 kpa.
Beam c/s 500 mm x 750 mm = 0.5 x 0.75 m
Dead Load of Beam = C/S x Unit Wt. of Concrete
= 0.5 x 0.75 x 25
= 9.375 KNm
Load on Beam = (11x1.5) + (11x1.5)
= 33 + 9.375 = 42.375 KN/m
Considering the Beam
Solving by the Slope Deflection Method:
Step-1: Fixed End Moments (FEM)
AB:`M^Fab`=`(-WL^2)/12`=- 286.03 KNm
`M^Fba`=`(WL^2)/12`= 286.03 KNm.
BC:`M^Fbc`=`(-WL^2)/12`=- 286.03 KNm
`M^Fcb`=`(WL^2)/12`= 286.03 KNm.
CD:`M^Fcd`=`(-WL^2)/12`=- 286.03 KNm
`M^Fdc`=`(WL^2)/12`= 286.03 KNm.
DE:`M^Fde`=`(-WL^2)/12`=- 286.03 KNm
`M^Fed`=`(WL^2)/12`= 286.03 KNm.
Step-2: Identify the Unknowns
θa, θb, θc, θd & θe.
Step-3: End Moments Equation
where,
E = 10 GPa = `10^7`Kpa
I = `(bd^3)/12` = (0.5x0.75)/12 =0.017 `m^4`
EI = 1.75 x`10^5`KN/m2
L = 9m
Mab = `M^Fab`+ `(2EI)/L`(2θa + θb)
= - 286.03 + (2x1.75x`10^5`)/9 (2θa + θb)
= - 286.03 +7.77x`10^4`θa + 3.88x`10^4`θb-----(1)
Mba = `M^Fba`+ `(2EI)/L`(θa + 2θb)
= 286.03 + (2x1.75x`10^5`)/9 (θa + 2θb)
= 286.03 +3.88x`10^4`θa + 7.77x`10^4`θb-----(2)
Mbc = `M^Fbc`+ `(2EI)/L`(2θb + θc)
= - 286.03 + (2x1.75x`10^5`)/9 (2θb + θc)
= - 286.03 +7.77x`10^4`θb + 3.88x`10^4`θc-----(3)
Mcb = `M^Fcb`+ `(2EI)/L`(θb + 2θc)
= 286.03 + (2x1.75x`10^5`)/9 (θb + 2θc)
= 286.03 +3.88x`10^4`θb+ 7.77x`10^4`θc-----(4)
Mcd = `M^Fcd`+ `(2EI)/L`(2θc + θd)
= - 286.03 + (2x1.75x`10^5`)/9 (2θc + θd)
= - 286.03 +7.77x`10^4`θc + 3.88x`10^4`θd-----(5)
Mdc = `M^Fdc `(2EI)/L`(θc+ 2θd)
= 286.03 + (2x1.75x`10^5`)/9 (θc + 2θd)
= 286.03 +3.88x`10^4`θc+ 7.77x`10^4`θd-----(6)
Mde = `M^Fde`+ `(2EI)/L`(2θd + θe)
= - 286.03 + (2x1.75x`10^5`)/9 (2θd + θe)
= - 286.03 +7.77x`10^4`θd+ 3.88x`10^4`θe-----(7)
Med = `M^Fed`+ `(2EI)/L`(θd + 2θe)
= 286.03 + (2x1.75x`10^5`)/9 (θd + 2θe)
= 286.03 +3.88x`10^4`θd + 7.77x`10^4`θe-----(8)
Step-4: Equilibrium Equation
Mab = 0
Mba + Mbc = 0
Mcb + Mcd = 0
Mdc + Mde = 0
Med = 0
Finding Equation Mab = 0
7.77x`10^4`θa + 3.88x`10^4`θb = 286.03-----(1)
Finding Equation Mba + Mbc = 0
3.88x`10^4`θa + 154700 θb + 3.88x`10^4`θc =0-----(2)
Finding Equation Mcb + Mcd = 0
3.88x`10^4`θb +154700 θc + 3.88x`10^4`θd =0-----(3)
Finding Equation Mdc + Mde = 0
3.88x`10^4`θc +154700 θd + 3.88x`10^4`θe =0-----(4)
Finding Equation Med = 0
3.88x`10^4`θd + 7.77x`10^4`θe = -286.03-----(5)
By substituting the above equations 1, 2, 3, 4 we get.
θa = 0.00421
θb = - 0.00106
θc = 0.0000
θd = 0.00106
θe = - 0.00421
By substituting the θa, θb, θc,θd & θe in End Moments equations, we get.
Mab = - 286.03 +7.77x`10^4`θa + 3.88x`10^4`θb
= -286.03 + (77700 x 0.00421) + (38800 x -0.00106)
= - 286.03 + 286
= 0
Mba = 286.03 +3.88x`10^4`θa + 7.77x`10^4`θb
= 286.03 + (38800 x 0.00421) + (77700 x -0.00106)
= 286.03 + 80.98
= 367 KNm
Mbc = - 286.03 +7.77x`10^4`θb + 3.88x`10^4`θc
= -286.03 + (77700 x -0.00106) + (38800 x 0.000)
= - 286.03 - 821
= -367 KNm
Mcb = 286.03 +3.88x`10^4`θb + 7.77x`10^4`θc
= 286.03 + (38800 x -0.00106) + (77700 x 0.000)
= 286.03 - 41.12
= 245 KNm
Mcd = - 286.03 +7.77x`10^4`θc + 3.88x`10^4`θd
= -286.03 + (77700 x 0.000) + (38800 x 0.00106)
= - 286.03 + 41.12
= - 245 KNm
Mdc = 286.03 +3.88x`10^4`θc + 7.77x`10^4`θd
= 286.03 + (38800 x 0.000) + (77700 x 0.00106)
= 286.03 + 82.36
= 368.4 KNm
Mde = - 286.03 +7.77x`10^4`θd + 3.88x`10^4`θe
= -286.03 + (77700 x 0.00106) + (38800 x -0.00421)
= - 286.03 - 81
= -368 KNm
Med = 286.03 +3.88x`10^4`θd + 7.77x`10^4`θe
= 286.03 + (38800 x 0.00106) + (77700 x -0.00421)
= 286.03 - 286
= 0
Span @ AB:
From moment equilibrium @ B
`sumMb=0`
Ra x 9 + 367-( 42.375 x 9 x 4.5) = 0
Ra =149.88 KN
`sumV=0`
Ra + Rb1 = 381.37 KN
Rb1 = 381.37 - Ra(149)
Rb1 = 231.48 KN
Span @ BC:
From moment equilibrium @ C
`sumMc=0`
Rb2 x 9 -367+245 -( 42.375 x 9 x 4.5) = 0
Rb2 =204.22 KN
`sumV=0`
Rb2 + Rc1 = 381.37 KN
Rc1 = 381.37 - Rb2(204.22)
Rc1 = 177.11 KN
Span @ CD:
From moment equilibrium @ D
`sumMd=0`
Rc2 x 9 -245+368 -( 42.375 x 9 x 4.5) = 0
Rc2 =177.04 KN
`sumV=0`
Rc2 + Rd1 = 381.37 KN
Rd1 = 381.37 - Rc2(177.04)
Rd1 = 204.37 KN
Span @ DE:
From moment equilibrium @ D
`sumMe=0`
Rd2 x 9 -368 - ( 42.375 x 9 x 4.5) = 0
Rd2 =231.55 KN
`sumV=0`
Rd2 + Re = 381.37 KN
Re = 381.37 - Rd2(231.55)
Re = 149.81 KN
SHEAR FORCE AND BENDING MOMENT DIAGRAM
2).
Self weight of Floor = 3 kpa
Floor Finish and Mech loads on the Floor = 3 kPa.
Total Load = 6 kpa.
Self weight of Beam = 9.375 KNm
Load on Beam = (6x1.5) + (6x1.5) = 18 KNm
Dead load of Beam = 18 + 9.375 = 27.375 KN/m
The Live load on the Floor = 5 kPa.
The Live load on the Beam = (5 x 1.5) + (5x1.5) = 15 kpa
= 15 + 27.375 = 42.375 KNm.
a). Considering the Beam load patterning to determine maximum positive moment to span AB & CD.
Solving by the Slope Deflection Method:
Step-1: Fixed End Moments (FEM)
AB:`M^Fab`=`(-WL^2)/12`=- 286.03 KNm
`M^Fba`=`(WL^2)/12`= 286.03 KNm.
BC:`M^Fbc`=`(-WL^2)/12`=-184.78 KNm
`M^Fcb`=`(WL^2)/12`= 184.78 KNm.
CD:`M^Fcd`=`(-WL^2)/12`=- 286.03 KNm
`M^Fdc`=`(WL^2)/12`= 286.03 KNm.
DE:`M^Fde`=`(-WL^2)/12`= -184.78 KNm
`M^Fed`=`(WL^2)/12`= 184.78 KNm.
Step-2: Identify the Unknowns
θa, θb, θc, θd & θe.
Step-3: End Moments Equation
where,
E = 10 GPa = `10^7`Kpa
I = `(bd^3)/12` = (0.5x0.75)/12 =0.017 `m^4`
EI = 1.75 x`10^5`KN/m2
L = 9m
Mab = `M^Fab`+ `(2EI)/L`(2θa + θb)
= - 286.03 + (2x1.75x`10^5`)/9 (2θa + θb)
= - 286.03 +7.77x`10^4`θa + 3.88x`10^4`θb-----(1)
Mba = `M^Fba`+ `(2EI)/L`(θa + 2θb)
= 286.03 + (2x1.75x`10^5`)/9 (θa + 2θb)
= 286.03 +3.88x`10^4`θa + 7.77x`10^4`θb-----(2)
Mbc = `M^Fbc`+ `(2EI)/L`(2θb + θc)
= - 184.78 + (2x1.75x`10^5`)/9 (2θb + θc)
= - 184.78 +7.77x`10^4`θb + 3.88x`10^4`θc-----(3)
Mcb = `M^Fcb`+ `(2EI)/L`(θb + 2θc)
= 184.78 + (2x1.75x`10^5`)/9 (θb + 2θc)
= 184.78 +3.88x`10^4`θb+ 7.77x`10^4`θc-----(4)
Mcd = `M^Fcd`+ `(2EI)/L`(2θc + θd)
= - 286.03 + (2x1.75x`10^5`)/9 (2θc + θd)
= - 286.03 +7.77x`10^4`θc + 3.88x`10^4`θd-----(5)
Mdc = `M^Fdc `(2EI)/L`(θc+ 2θd)
= 286.03 + (2x1.75x`10^5`)/9 (θc + 2θd)
= 286.03 +3.88x`10^4`θc+ 7.77x`10^4`θd-----(6)
Mde = `M^Fde`+ `(2EI)/L`(2θd + θe)
= - 184.78 + (2x1.75x`10^5`)/9 (2θd + θe)
= - 184.78 +7.77x`10^4`θd+ 3.88x`10^4`θe-----(7)
Med = `M^Fed`+ `(2EI)/L`(θd + 2θe)
= 184.78 + (2x1.75x`10^5`)/9 (θd + 2θe)
= 184.78 +3.88x`10^4`θd + 7.77x`10^4`θe-----(8)
Step-4: Equilibrium Equation
Mab = 0
Mba + Mbc = 0
Mcb + Mcd = 0
Mdc + Mde = 0
Med = 0
Finding Equation Mab = 0
7.77x`10^4`θa + 3.88x`10^4`θb = 286.03-----(1)
Finding Equation Mba + Mbc = 0
3.88x`10^4`θa + 154700 θb + 3.88x`10^4`θc =-101.25-----(2)
Finding Equation Mcb + Mcd = 0
3.88x`10^4`θb +154700 θc + 3.88x`10^4`θd =101.25-----(3)
Finding Equation Mdc + Mde = 0
3.88x`10^4`θc +154700 θd + 3.88x`10^4`θe =-101.25-----(4)
Finding Equation Med = 0
3.88x`10^4`θd + 7.77x`10^4`θe = -184.78-----(5)
By substituting the above equations 1, 2, 3, 4 we get.
θa = 0.00477
θb = - 0.00218
θc = 0.00131
θd = - 0.00044
θe = - 0.00216
By substituting the θa, θb, θc,θd & θe in End Moments equations, we get.
Mab = - 286.03 +7.77x`10^4`θa + 3.88x`10^4`θb
= -286.03 + (77700 x 0.00477) + (38800 x -0.00218)
= - 286.03 + 286.04
= 0
Mba = 286.03 +3.88x`10^4`θa + 7.77x`10^4`θb
= 286.03 + (38800 x 0.00477) + (77700 x -0.00218)
= 286.03 + 15.69
= 301.72 KNm
Mbc = - 184.78 +7.77x`10^4`θb + 3.88x`10^4`θc
= -184.78 + (77700 x -0.00218) + (38800 x 0.00131)
= - 184.78 - 49.13
= -233.91 KNm
Mcb = 184.78 +3.88x`10^4`θb + 7.77x`10^4`θc
= 184.78 + (38800 x -0.00218) + (77700 x 0.00131)
= 184.78 + 17.20
= 201.98 KNm
Mcd = - 286.03 +7.77x`10^4`θc + 3.88x`10^4`θd
= -286.03 + (77700 x 0.00131) + (38800 x -0.00044)
= - 286.03 + 84.7
= - 201.33 KNm
Mdc = 286.03 +3.88x`10^4`θc + 7.77x`10^4`θd
= 286.03 + (38800 x 0.00131) + (77700 x -0.00044)
= 286.03 + 16.64
= 302.67 KNm
Mde = - 184.78 +7.77x`10^4`θd + 3.88x`10^4`θe
= -184.78 + (77700 x 0.00044) + (38800 x -0.00216)
= - 184.78 - 49.62
= -234.4 KNm
Med = 184.78 +3.88x`10^4`θd + 7.77x`10^4`θe
= 184.78 + (38800 x 0.00044) + (77700 x -0.00216)
= 184.78 - 150.76
= 34.02
Span @ AB:
From moment equilibrium @ B
`sumMb=0`
Ra x 9 + 301.72-( 42.375 x 9 x 4.5) = 0
Ra =157.14 KN
`sumV=0`
Ra + Rb1 = 381.37 KN
Rb1 = 381.37 - Ra(157.14)
Rb1 = 227.23 KN
Span @ BC:
From moment equilibrium @ C
`sumMc=0`
Rb2 x 9 -234+201 -( 27.375 x 9 x 4.5) = 0
Rb2 =126.77 KN
`sumV=0`
Rb2 + Rc1 = 381.37 KN
Rc1 = 381.37 - Rb2(126.77)
Rc1 = 254.6 KN
Span @ CD:
From moment equilibrium @ D
`sumMd=0`
Rc2 x 9 -201+302 -( 42.375 x 9 x 4.5) = 0
Rc2 =179.44 KN
`sumV=0`
Rc2 + Rd1 = 381.37 KN
Rd1 = 381.37 - Rc2(179.44)
Rd1 = 201.93 KN
Span @ DE:
From moment equilibrium @ E
`sumMe=0`
Rd2 x 9 -234 -( 27.375 x 9 x 4.5) = 0
Rd2 =149.11 KN
`sumV=0`
Rd2 + Re = 381.37 KN
Re = 381.37 - Rd2(149.11)
Re = 232.26 KN
SHEAR FORCE AND BENDING MOMENT DIAGRAM
b). Considering the Beam load patterning to determine maximum positive moment to span BC & DE.
Solving by the Slope Deflection Method:
Step-1: Fixed End Moments (FEM)
AB:`M^Fab`=`(-WL^2)/12`=- 184.78 KNm
`M^Fba`=`(WL^2)/12`= 184.78 KNm.
BC:`M^Fbc`=`(-WL^2)/12`=-286.03 KNm
`M^Fcb`=`(WL^2)/12`= 286.03 KNm.
CD:`M^Fcd`=`(-WL^2)/12`=- 184.78 KNm
`M^Fdc`=`(WL^2)/12`= 184.78 KNm.
DE:`M^Fde`=`(-WL^2)/12`= -286.03 KNm
`M^Fed`=`(WL^2)/12`= 286.03 KNm.
Step-2: Identify the Unknowns
θa, θb, θc, θd & θe.
Step-3: End Moments Equation
where,
E = 10 GPa = `10^7`Kpa
I = `(bd^3)/12` = (0.5x0.75)/12 =0.017 `m^4`
EI = 1.75 x`10^5`KN/m2
L = 9m
Mab = `M^Fab`+ `(2EI)/L`(2θa + θb)
= - 184.78 + (2x1.75x`10^5`)/9 (2θa + θb)
= - 184.78 +7.77x`10^4`θa + 3.88x`10^4`θb-----(1)
Mba = `M^Fba`+ `(2EI)/L`(θa + 2θb)
= 184.78 + (2x1.75x`10^5`)/9 (θa + 2θb)
= 184.78 +3.88x`10^4`θa + 7.77x`10^4`θb-----(2)
Mbc = `M^Fbc`+ `(2EI)/L`(2θb + θc)
= - 286.03 + (2x1.75x`10^5`)/9 (2θb + θc)
= - 286.03 +7.77x`10^4`θb + 3.88x`10^4`θc-----(3)
Mcb = `M^Fcb`+ `(2EI)/L`(θb + 2θc)
= 286.03 + (2x1.75x`10^5`)/9 (θb + 2θc)
= 286.03 +3.88x`10^4`θb+ 7.77x`10^4`θc-----(4)
Mcd = `M^Fcd`+ `(2EI)/L`(2θc + θd)
= - 184.78 + (2x1.75x`10^5`)/9 (2θc + θd)
= - 184.78 +7.77x`10^4`θc + 3.88x`10^4`θd-----(5)
Mdc = `M^Fdc `(2EI)/L`(θc+ 2θd)
= 184.78 + (2x1.75x`10^5`)/9 (θc + 2θd)
= 184.78 +3.88x`10^4`θc+ 7.77x`10^4`θd-----(6)
Mde = `M^Fde`+ `(2EI)/L`(2θd + θe)
= - 286.03 + (2x1.75x`10^5`)/9 (2θd + θe)
= - 286.03 +7.77x`10^4`θd+ 3.88x`10^4`θe-----(7)
Med = `M^Fed`+ `(2EI)/L`(θd + 2θe)
= 286.03 + (2x1.75x`10^5`)/9 (θd + 2θe)
= 286.03 +3.88x`10^4`θd + 7.77x`10^4`θe-----(8)
Step-4: Equilibrium Equation
Mab = 0
Mba + Mbc = 0
Mcb + Mcd = 0
Mdc + Mde = 0
Med = 0
Finding Equation Mab = 0
7.77x`10^4`θa + 3.88x`10^4`θb = 184.78-----(1)
Finding Equation Mba + Mbc = 0
3.88x`10^4`θa + 154700 θb + 3.88x`10^4`θc =101.25-----(2)
Finding Equation Mcb + Mcd = 0
3.88x`10^4`θb +154700 θc + 3.88x`10^4`θd =-101.25-----(3)
Finding Equation Mdc + Mde = 0
3.88x`10^4`θc +154700 θd + 3.88x`10^4`θe =101.25-----(4)
Finding Equation Med = 0
3.88x`10^4`θd + 7.77x`10^4`θe = -286.03-----(5)
By substituting the above equations 1, 2, 3, 4 we get.
θa = 0.00216
θb = 0.00044
θc = -0.00131
θd = 0.00218
θe = - 0.00477
By substituting the θa, θb, θc,θd & θe in End Moments equations, we get.
Mab = - 184.78 +7.77x`10^4`θa + 3.88x`10^4`θb
= -184.78 + (77700 x 0.00216) + (38800 x 0.00044)
= - 184.78 + 184.8
= 0
Mba = 184.78 +3.88x`10^4`θa + 7.77x`10^4`θb
= 184.78 +(38800 x 0.00216) + (77700 x 0.00044)
= 184.78 + 117.99
= 302.77 KNm
Mbc = - 286.03 +7.77x`10^4`θb + 3.88x`10^4`θc
= -286.03 + (77700 x 0.00044) + (38800 x -0.00131)
= - 286.03 - 16.64
= -302.67 KNm
Mcb = 286.03 +3.88x`10^4`θb + 7.77x`10^4`θc
= 286.03 + (38800 x 0.00044) + (77700 x -0.00131)
= 286.03 - 84.71
= 201.32 KNm
Mcd = - 184.78 +7.77x`10^4`θc + 3.88x`10^4`θd
= -184.78 + (77700 x -0.00131) + (38800 x 0.00218)
= - 184.78 - 17.20
= - 201.98 KNm
Mdc = 184.78 +3.88x`10^4`θc + 7.77x`10^4`θd
= 184.78 + (38800 x -0.00131) + (77700 x 0.00218)
= 184.78 + 118.55
= 303.3 KNm
Mde = - 286.03 +7.77x`10^4`θd + 3.88x`10^4`θe
= -286.03 + (77700 x 0.00218) + (38800 x -0.00477)
= - 286.03 - 15.69
= -301.72 KNm
Med = 286.03 +3.88x`10^4`θd + 7.77x`10^4`θe
= 286.03 + (38800 x 0.00218) + (77700 x -0.00477)
= 286.03 - 286.04
= 0
Span @ AB:
From moment equilibrium @ B
`sumMb=0`
Ra x 9 + 302.77-(27.375 x 9 x 4.5) = 0
Ra =89.47 KN
`sumV=0`
Ra + Rb1 = 381.37 KN
Rb1 = 381.37 - Ra(89.47)
Rb1 = 291.9 KN
Span @ BC:
From moment equilibrium @ C
`sumMc=0`
Rb2 x 9 -302.67+201.32 -( 42.375 x 9 x 4.5) = 0
Rb2 =201.92 KN
`sumV=0`
Rb2 + Rc1 = 381.37 KN
Rc1 = 381.37 - Rb2(201.92)
Rc1 = 179.45 KN
Span @ CD:
From moment equilibrium @ D
`sumMd=0`
Rc2 x 9 -201.32+303.3 -( 27.375 x 9 x 4.5) = 0
Rc2 =111.78 KN
`sumV=0`
Rc2 + Rd1 = 381.37 KN
Rd1 = 381.37 - Rc2(111.78)
Rd1 = 269.59 KN
Span @ DE:
From moment equilibrium @ E
`sumMe=0`
Rd2 x 9 -301.72 -( 42.375 x 9 x 4.5) = 0
Rd2 =224.2 KN
`sumV=0`
Rd2 + Re = 381.37 KN
Re = 381.37 - Rd2(224.2)
Re = 157.17 KN
SHEAR FORCE AND BENDING MOMENT DIAGRAM
RESULT:
Consedering the beam, Analysis id done by using slope deflection method.
Considering the load patterning, The Maximum positive and negative moments in each span is done successfully
Bending moment & Shear Force Diagram is shown successfully.