Project 1
AIM:-
Given below is a floor plan of a building. The red lines running East-west are beams supported by the girders spanning North-South between columns. The floor system is 120 mm thick normal weight concrete floor. The floor finish and mech loads on the floor is 3 kPa. The live load on the floor is 5 kPa. All beams have cross section 500 mmx750 mm. The beams are equally spaced. The rectangles in cyan color are columns. Modulus of elasticity of concrete is 10 GPa.
- Determine the loading on the beam indicated, analyze the beams and draw bending moment and shear force diagram. Use slope deflection method. Do not consider any load patterning. Also, do not consider any girder deflections for this analysis purpose.
- For the same beam, consider load patterning to determine maximum positive and negative moments in each span.
PROJECT 1:-
Given Values:-
- Determine the loading on the beam indicated, analyze the beams and draw bending moment and shear force diagram. Use slope deflection method. Do not consider any load patterning. Also, do not consider any girder deflections for this analysis purpose.
Concrete floor Thickness = 120mm = 0.12m
Self Weight of Floor = 0.12X25 = 3KPa
Floor finish and mech loads = 3KPa
Live load = 5 KPa
Total load = 11KPa
Self Weight of Beam = 0.5X 0.75X25 = 9.375KN/m
Load on the indicated beam = (11X1.5)X2 = 33KN/m
Adding Self Weight of beam = 33 + 9.375 = 42.375KN/m
- Analyzing without cnsidering load patterning
- Solving By Slope deflection method
Step1:-
Fixed End Momrnt :-
MFab = `-(frac{WXL^2}{12})`
MFab = `-(frac{42.375X9^2}{12})` = -286.03KN/m
MFba = 286.03KN/m
MFbc = - 286.03KN/m
MFcb = 286.03KN/m
MFcd = -286.03KN/m
MFdc = 286.03KN/m
MFde = -286.03KN/m
MFed = 286.03KN/m
Step2:-
Find out Unknowns:-
`thetaa`
`thetab`
`thetac`
`thetad`
`thetae`
Given valuese:-
E = 10GPa = 10X 10^6 KPa
I = bd^3/12
I = (0.5X0.75^3)/12 = 0.0176m^4
Where
EI = 1.76X10^5KN/m
Step3:-
Find out End Moments:-
Spane AB:-
Mab = MFab +`(frac{2EI}{L})` (2`thetaa`+`thetab`)
Mab = -286.03 +`(frac{2X1.76X10^5}{9})` (2`thetaa`+`thetab`)
Mab = -286.03 + 78222.2`thetaa` + 39111.1`thetab`.............................1
Mba = MFba +`(frac{2EI}{L})` (2`thetab`+`thetaa`)
Mba = 286.03 + 78222.2`thetab`+39111.1`thetaa`................................2
Mbc = MFbc +`(frac{2EI}{L})` (2`thetab`+`thetac`)
Mbc = -286.03+ 78222.2`thetab`+ 39111.1`thetac`................................3
Mcb = MFcb +`(frac{2EI}{L})` (2`thetac`+`thetab`)
Mcb = 286.03 + 78222.2`thetac`+ 39111.1`thetab`................................4
Mcd = MFcd +`(frac{2EI}{L})` (2`thetac`+`thetad`)
Mcd = -286.03 + 78222.2`thetac`+ 39111.1`thetad`................................5
Mdc = MFdc +`(frac{2EI}{L})` (2`thetad`+`thetac`)
Mdc =286.03 + 78222.2`thetad`+ 39111.1`thetac`...................................6
Mde = MFde +`(frac{2EI}{L})` (2`thetad`+`thetae`)
Mde = -286.03 + 78222.2`thetad`+ 39111.1`thetae`................................7
Med= MFed +`(frac{2EI}{L})` (2`thetae`+`thetad`)
Med= 286.03 + 78222.2`thetae`+ 39111.1`thetad`.................................8
Step4:-
Equlibrium equations:-
Mab = 0
-286.03 + 78222.2`thetaa` + 39111.1`thetab`= 0
- 78222.2`thetaa` + 39111.1`thetab` = 286.03
Mba + Mbc = 0
(286.03 + 78222.2`thetab`+39111.1`thetaa`)+ (-286.03+ 78222.2`thetab`+ 39111.1`thetac`)= 0
- 39111.1`thetaa`+ 1.56X10^5`thetab`+ 39111.1`thetac` = 0
Mcb + Mcd = 0
(286.03 + 78222.2`thetac`+ 39111.1`thetab`)+(-286.03 + 78222.2`thetac`+ 39111.1`thetad`)= 0
- 39111.1`thetab`+ 1.56X10^5`thetac`+ 39111.1`thetad` = 0
Mdc + Mde = 0
(286.03 + 78222.2`thetad`+ 39111.1`thetac`) + (-286.03 + 78222.2`thetad`+ 39111.1`thetae`) = 0
- 39111.1`thetac`+ 1.56X10^5`thetad`+39111.1`thetae` =0
Med = 0
286.03 + 78222.2`thetae`+ 39111.1`thetad`
- 286.03 + 78222.2`thetae`+ 39111.1`thetad` = -286.03
Solving The unknowns :-
`thetaa` = 0.00418
`thetab` = - 0.00105
`thetac` = 0
`thetad` = 0.00105
`thetae` = -0.00418
On Substituting :-
Mab = 0
Mba = 367.38KNm
Mbc = -367.38Knm
Mcb = 244.96KNm
Mcd = -244.96KNm
Mdc = 367.38KNm
Mde = -367.38KNm
Med = 0
- 2. For the same beam, consider load patterning to determine maximum positive and negative moments in each span.
Concrete floor Thickness = 120mm = 0.12m
Self Weight of Floor = 0.12X25 = 3KPa
Floor finish and mech loads = 3KPa
Total load = 6KPa
Self Weight of Beam = 0.5X 0.75X25 = 9.375KN/m
Load on the indicated beam = (6X1.5)X2 = 18KN/m
Adding Self Weight of beam = 18+ 9.375 = 27.375KN/m
Live load on spane AB and CD = (5X1.5X2) = 15KN/m
Total Load on AB and CD = 15 + 27.375 = 42.375KN/m
- Solving By Slope deflection method
Step1:-
Fixed End Momrnt :-
MFab = `-(frac{WXL^2}{12})`
MFab = `-(frac{42.375X81}{12})` = -286.03KNm
MFba = 286.03
MFbc = `-(frac{27.375X81}{12})` = -184.78KNm
MFcb = 184.78KNm
MFcd = -286.03KNm
Mdc = 286.03KNm
MFde = -184.78KNm
Step2:-
Find out Unknowns:-
`thetaa`
`thetab`
`thetac`
`thetad`
`thetae`
Given valuese:-
E = 10GPa = 10X 10^6 KPa
I = bd^3/12
I = (0.5X0.75^3)/12 = 0.0176m^4
Where
EI = 1.76X10^5KN/m
Step3:-
Find out End Moments:-
Spane AB:-
Mab = MFab +`(frac{2EI}{L})` (2`thetaa`+`thetab`)
Mab = -286.03 + 78222.2`thetaa` + 39111.1`thetab`.............................1
Mba = MFba +`(frac{2EI}{L})` (2`thetab`+`thetaa`)
Mba = 286.03 + 78222.2`thetab`+39111.1`thetaa`................................2
Mbc = MFbc +`(frac{2EI}{L})` (2`thetab`+`thetac`)
Mbc = -184.78 + 78222.2`thetab`+ 39111.1`thetac`................................3
Mcb = MFcb +`(frac{2EI}{L})` (2`thetac`+`thetab`)
Mcb = 184.78 + 78222.2`thetac`+ 39111.1`thetab`................................4
Mcd = MFcd +`(frac{2EI}{L})` (2`thetac`+`thetad`)
Mcd = -286.03 + 78222.2`thetac`+ 39111.1`thetad`................................5
Mdc = MFdc +`(frac{2EI}{L})` (2`thetad`+`thetac`)
Mdc =286.03 + 78222.2`thetad`+ 39111.1`thetac`...................................6
Mde = MFde +`(frac{2EI}{L})` (2`thetad`+`thetae`)
Mde = -184.78 + 78222.2`thetad`+ 39111.1`thetae`................................7
Med= MFed +`(frac{2EI}{L})` (2`thetae`+`thetad`)
Med= 184.78 + 78222.2`thetae`+ 39111.1`thetad`.................................8
Step4:-
Equlibrium equations:-
Mab = 0
-286.03 + 78222.2`thetaa` + 39111.1`thetab`= 0
- 78222.2`thetaa` + 39111.1`thetab` = 286.03
Mba + Mbc = 0
(286.03 + 78222.2`thetab`+39111.1`thetaa`)+ (-184.78 + 78222.2`thetab`+ 39111.1`thetac`)= 0
- 39111.1`thetaa`+ 1.56X10^5`thetab`+ 39111.1`thetac` = -101.25
Mcb + Mcd = 0
(184.78 + 78222.2`thetac`+ 39111.1`thetab`)+(-286.03 + 78222.2`thetac`+ 39111.1`thetad`)= 0
- 39111.1`thetab`+ 1.56X10^5`thetac`+ 39111.1`thetad` = 101.25
Mdc + Mde = 0
(286.03 + 78222.2`thetad`+ 39111.1`thetac`) + (-184.78 + 78222.2`thetad`+ 39111.1`thetae`) = 0
- 39111.1`thetac`+ 1.56X10^5`thetad`+39111.1`thetae` = -101.25
Med = 0
184.78 + 78222.2`thetae`+ 39111.1`thetad`
- 78222.2`thetae`+ 39111.1`thetad` = -184.78
Solving The unknowns :-
`thetaa` = 0.00474
`thetab` = - 0.00216
`thetac` = 0.00130
`thetad` = -0.00044
`thetae` = -0.00214
On Substituting :-
Mab = 0
Mba = 302.46KNm
Mbc = -302.46Knm
Mcb = 201.99KNm
Mcd = -201.99KNm
Mdc = 302.46KNm
Mde = -302.46KNm
Med = 0
Concrete floor Thickness = 120mm = 0.12m
Self Weight of Floor = 0.12X25 = 3KPa
Floor finish and mech loads = 3KPa
Total load = 6KPa
Self Weight of Beam = 0.5X 0.75X25 = 9.375KN/m
Load on the indicated beam = (6X1.5)X2 = 18KN/m
Adding Self Weight of beam = 18+ 9.375 = 27.375KN/m
Live load on spane AB and CD = (5X1.5X2) = 15KN/m
Total Load on AB and CD = 15 + 27.375 = 42.375KN/m
- Solving By Slope deflection method
Step1:-
Fixed End Momrnt :-
MFab = `-(frac{WXL^2}{12})`
MFab = `-(frac{27.375X81}{12})` = -184.78KNm
MFba = 184.78KNm
MFbc = `-(frac{27.375X81}{12})` = -286.03KNm
MFcb = 286.03KNm
MFcd = -184.78KNm
Mdc = 184.78KNm
MFde = -286.03KNm
Step2:-
Find out Unknowns:-
`thetaa`
`thetab`
`thetac`
`thetad`
`thetae`
Given valuese:-
E = 10GPa = 10X 10^6 KPa
I = bd^3/12
I = (0.5X0.75^3)/12 = 0.0176m^4
Where
EI = 1.76X10^5KN/m
Step3:-
Find out End Moments:-
Spane AB:-
Mab = MFab +`(frac{2EI}{L})` (2`thetaa`+`thetab`)
Mab = -184.78+ 78222.2`thetaa` + 39111.1`thetab`.............................1
Mba = MFba +`(frac{2EI}{L})` (2`thetab`+`thetaa`)
Mba = 184.78 + 78222.2`thetab`+39111.1`thetaa`................................2
Mbc = MFbc +`(frac{2EI}{L})` (2`thetab`+`thetac`)
Mbc = -286.03 + 78222.2`thetab`+ 39111.1`thetac`................................3
Mcb = MFcb +`(frac{2EI}{L})` (2`thetac`+`thetab`)
Mcb = 286.03 + 78222.2`thetac`+ 39111.1`thetab`................................4
Mcd = MFcd +`(frac{2EI}{L})` (2`thetac`+`thetad`)
Mcd = -184.78 + 78222.2`thetac`+ 39111.1`thetad`................................5
Mdc = MFdc +`(frac{2EI}{L})` (2`thetad`+`thetac`)
Mdc =184.78 + 78222.2`thetad`+ 39111.1`thetac`...................................6
Mde = MFde +`(frac{2EI}{L})` (2`thetad`+`thetae`)
Mde = -286.03 + 78222.2`thetad`+ 39111.1`thetae`................................7
Med= MFed +`(frac{2EI}{L})` (2`thetae`+`thetad`)
Med= 286.03 + 78222.2`thetae`+ 39111.1`thetad`.................................8
Step4:-
Equlibrium equations:-
Mab = 0
-184.78 + 78222.2`thetaa` + 39111.1`thetab`= 0
- 78222.2`thetaa` + 39111.1`thetab` = 184.78
Mba + Mbc = 0
(184.78 + 78222.2`thetab`+39111.1`thetaa`)+ (-286.03 + 78222.2`thetab`+ 39111.1`thetac`)= 0
- 39111.1`thetaa`+ 1.56X10^5`thetab`+ 39111.1`thetac` = 101.25
Mcb + Mcd = 0
(184.78 + 78222.2`thetad`+ 39111.1`thetac`)+(-184.78 + 78222.2`thetac`+ 39111.1`thetad`)= 0
- 39111.1`thetab`+ 1.56X10^5`thetac`+ 39111.1`thetad` = -101.25
Mdc + Mde = 0
(286.03 + 78222.2`thetad`+ 39111.1`thetac`) + (-286.03 + 78222.2`thetad`+ 39111.1`thetae`) = 0
- 39111.1`thetac`+ 1.56X10^5`thetad`+39111.1`thetae` = 0
Med = 0
286.03 + 78222.2`thetae`+ 39111.1`thetad`
- 78222.2`thetae`+ 39111.1`thetad` = -286.03
Solving The unknowns :-
`thetaa` = 0.00217
`thetab` = 0.00038
`thetac` = -0.00108
`thetad` = 0.00136
`thetae` = -0.00434
On Substituting :-
Mab = 0
Mba = 299.38KNm
Mbc = -299.38Knm
Mcb = 216.42KNm
Mcd = -216.42KNm
Mdc = 344.35KNm
Mde = -344.35KNm
Med = 0