Project 1
Given data:
Slab thickness = 120mm
floor finish = 3 kN/m2
Live load = 5 kN/m2
Beam size = 500 x 750 mm
Therefore I = 17.5 x 10^-3 m4
E = 10x10^6 kN/m2
Solution :
Self weight of the slab = 0.12 x 25
= 3 kN/m2
Total UDL acting on an beam = Tributory area of the slab to that particular beam x load on slab + self weight of the beam
= (1.5+1.5) x (3+3+5) + (0.75x0.5x25)
= 42.375 kN/m
Fixed end moments for span AB
Mfab = -wl^2/12
= -42.375 x9x9/12
= -286 kNm
Mfba = wl^2/12
= 42.375 x9x9/12
= 286 kNm
Similarly
Mfbc = -286 kNm
Mfcb = 286 kNm
Mfcd = -286 kNm
Mfdc = 286 kNm
Mfde = -286 kNm
Mfed = 286 kNm
By using Slope deflection method
Mab = Mfab + 2EI/L(2θa+θb)
Mab = -286+0.44EIθa+0.22EIθb
Mba = Mfba + 2EI/L(θa+2θb)
Mba = 286+0.22EIθa+0.44EIθb
By repeating the same for all spans
Mbc = -286+0.44EIθb+0.22EIθc
Mcb = 286+0.22EIθb+0.44EIθc
Mcd = -286+0.44EIθc+0.22EIθd
Mdc = 286+0.22EIc+0.44EIθd
Mde = -286+0.44EIθd+0.22EIθe
Med = 286+0.22EIθd+0.44EIθe
By applying the end conditions
Mab =0
Mba+Mbc =0
Mcb+Mcd =0
Mdc+Mde =0
Med = 0 We will get five equation
77000θa + 38500θb = 286 --------------------------------------(1)
38500θa +154000θb + 38500θc = 0 ---------------------------(2)
38500θb + 154000θc + 38500θd = 0 --------------------------(3)
38500θc + 154000θd + 38500θe = 0 --------------------------(4)
38500θd + 77000θe = -286 -------------------------------------(5)
By solving these 5 eqn
θa = 4240 rad
θb = -1060 rad
θc = 0
θd = 1060 rad
θe = -4240 rad
By substituting θ values in previos equations
Mab = 0
Mba = 368 kNm
Mbc = -368 kNm
Mcb = 245.2 kNm
Mcd = -245.2 kNm
Mdc = 368 kNm
Mde = -368 kNm
Med = 0
Application of moment equilibrium for span 1
42.375x 9x4.5 -Vb1(9) + 368 = 0
Vb1 = 231.6 kN
From vertical force equilibrium
Ra + Vb1 -42.375(9) = 0
Ra = 150 kN
similarly for span 2,3&4
Vc1 = 177 kN
Vb2 = 204.375 kN
Vd1 = 204.33 kN
Vc2 = 177 kN
Re = 150 kN
Vd2 = 231.6 kN
Case (ii)a
In load patterning case first consider the span AB alone loaded with the live load of 5 kN/m2, and the remaining spans only carry a self weight and floor finish load.
Therefore load on span AB = 42.375 kN/m
load on other spans = 27.375 kN/m
For this condition, the fixed end moments are
Mfab =-286 kNm
Mfba = 286 kNm
Mfbc = -185 kNm
Mfcb = 185 kNm
Mfcd = -185 kNm
Mfdc = 185 kNm
Mfde = -185 kNm
Mfed = 185 kNm
And the rotation values are
θa = 4670 rad
θb = -1900 rad
θc = 330 rad
θd = 590 rad
θe = -2700 rad
And the corresponding end moments are
Mab = 0
Mba = 319. kNm
Mbc = -319 kNm
Mcb = 137 kNm
Mcd = -137 kNm
Mdc = 243 kNm
Mde = -243 kNm
Med = 0
And the reactions for this particular patterning
Vb1 = 226 kN
Ra = 155 kN
Vc1 = 103 kN
Vb2 = 143 kN
Vd1 = 135 kN
Vc2 = 111 kN
Re = 96 kN
Vd2 = 150 kN
Case (ii)b
In load patterning case first consider the span BC alone loaded with the live load of 5 kN/m2, and the remaining spans only carry a self weight and floor finish load.
Therefore load on span BC = 42.375 kN/m
load on other spans = 27.375 kN/m
For this condition, the fixed end moments are
Mfab =-185 kNm
Mfba = 185 kNm
Mfbc = -286 kNm
Mfcb = 286 kNm
Mfcd = -185 kNm
Mfdc = 185 kNm
Mfde = -185 kNm
Mfed = 185 kNm
And the rotation values are
θa = 2230 rad
θb = 340 rad
θc = -980 rad
θd = 970 rad
θe = -2890 rad
And the corresponding end moments are
Mab = 0
Mba = 297 kNm
Mbc = -297 kNm
Mcb = 223 kNm
Mcd = -223 kNm
Mdc = 222 kNm
Mde = -222 kNm
Med = 0
And the reactions for this particular patterning
Vb1 = 156 kN
Ra = 90 kN
Vc1 = 182.5 kN
Vb2 = 199 kN
Vd1 = 123 kN
Vc2 = 123 kN
Re = 148 kN
Vd2 = 98.5 kN
Conclusion:
From the above calculations it is evident that the end support and the support next to it carries a maximum reaction when the live load is acting on the end span alone.
And the support next to end support ,carries a maximum moment when the load is acting on end span alone.
When we apply live load on any one or both of the mid spans, the middle support carries a reaction and bending moment.