Project 1
`Lundersetx/Lundersety` = 9/3 = 3>2 therefore is a ONE WAY SLAB
1. WITHOUT LOAD PATTERNING
STEP 1
LOADS FROM SLAB
1. SW of slab
t = 120mm = 0.120m
= unit weight of cncrete x 0.120m
= 25kN/m3 x 0.120m
- SW = 3kN/m2
- LL = 5kPa = 5kN/m2
- SDL = 3kPa = 3kN/m2
- TL = SW+ DL + LL = 3 + 5 + 3 = 11 kN/m2
STEP 2
LOAD FROM SLAB TO BEAM
= 11 `(kN)/m^2` x (9x1.5) `m^2`
= 11x 13.5
= 148.5kN
= 148.5kN/ 9m
= 16.5 `(kN)/m`
since load from slab gets transmitted from both sides
= 16.5 x 2
= 33 kN/m
STEP 3
SW OF BEAM
= unit wt of concrete x area of c/s of beam
= 25 `(kN)/m^3` x 0.5m x 0.75m
= 9.375kN/m
STEP 4
TOTAL LOAD ON BEAM
= 33 + 9.375
= 42.375kN/m
STEP 5
SOLVING BEAM USING SLOPE DEFLECTION METHOD
TO FIND FIXED END MOMENTS
`Munderset(FAB)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FBA)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FBC)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FCB)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FCD)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FDC)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FDE)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FED)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
STEP 5
SUPPORT END MOMENT
`Munderset(AB) = Munderset(FAB) + 4EIthetaundersetA/L + 2EIthetaundersetB/L`
`Munderset(AB) = -286.031 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`0 = -286.031 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`286.031 - 2EIthetaundersetB/9 = 4EIthetaundersetA/9`
`2EIthetaundersetA/9 = 143.0155 - EIthetaundersetB/9`
`Munderset(BA) = Munderset(FBA) + 4EIthetaundersetB/L + 2EIthetaundersetA/L`
`Munderset(BA) = 286.031 + 4EIthetaundersetB/9 + 143.0155 - EIthetaundersetB/9`
`Munderset(BA) = 429.0465 + 3EIthetaundersetB/9`
`Munderset(BC) = Munderset(FBC) + 4EIthetaundersetB/L + 2EIthetaundersetC/L`
`Munderset(BC) = -286.031 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`Munderset(CB) = Munderset(FCB) + 4EIthetaundersetC/L + 2EIthetaundersetB/L`
`Munderset(CB) = 286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9`
`Munderset(CD) = Munderset(FCD) + 4EIthetaundersetC/L + 2EIthetaundersetD/L`
`Munderset(CD) = -286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`Munderset(DC) = Munderset(FDC) + 4EIthetaundersetD/L + 2EIthetaundersetC/L`
`Munderset(DC) = 286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`
`Munderset(DE) = Munderset(FDE) + 4EIthetaundersetD/L + 2EIthetaundersetE/L`
`Munderset(DE) = -286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetE/9`
`Munderset(ED) = Munderset(FED) + 4EIthetaundersetE/L + 2EIthetaundersetD/L`
`0 = 286.031 + 4EIthetaundersetE/9 + 2EIthetaundersetD/9`
`4EIthetaundersetE/9 = -286.031 - 2EIthetaundersetD/9`
`2EIthetaundersetE/9 = -143.0155 - EIthetaundersetD/9`
`sum MundersetB` =0,
`Munderset (BA) + Munderset(BC)` = 0
`429.0465 + 3EIthetaundersetB/9`` -286.031 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`143.0155+ 7EIthetaundersetB/9 + 2EIthetaundersetC/9` =0
`7EIthetaundersetB/9 + 2EIthetaundersetC/9` = -143.0155 ... (1)
`sum MundersetC` =0,
`Munderset (CB) + Munderset(CD)` = 0
`286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9``-286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`2EIthetaundersetB/9 + 8EIthetaundersetC/9 + 2EIthetaundersetD/9` =0 ... (2)
`sum MundersetD` =0,
`Munderset (DC) + Munderset(DE)` = 0
`286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`` -286.031 + 4EIthetaundersetD/9 -143.0155 - EIthetaundersetD/9 `
`-143.0155+ 7EIthetaundersetD/9 + 2EIthetaundersetC/9` =0
`2EIthetaundersetC/9 + 7EIthetaundersetD/9` = 143.0155 ... (3)
`[[7/9,2/9,0,|,EIthetab],[2/9,8/9,2/9,|,EIthetac],[0,2/9,7/9,|,EIthetad]]` = `((-143.0155),(0),(143.0155))`
R2`->`R1 - 3.5 R2
`[[7/9,2/9,0,|,EIthetab],[0,-26/9,-7/9,|,EIthetac],[0,2/9,7/9,|,EIthetad]]` = `((-143.0155),(-143.0155),(143.0155))`
R3`->`13 R3 + R2
`[[7/9,2/9,0,|,EIthetab],[0,-26/9,-7/9,|,EIthetac],[0, 0 ,84/9,|,EIthetad]]` = `((-143.0155),(-143.0155),(1716.186))`
`EIthetaunderset(D)`= 183.8770
`EIthetaunderset(C)`= 0
`EIthetaunderset(B)`= -183.8770
`EIthetaunderset(E)`= -735.5083
`EIthetaunderset(A)`= 735.5083
Substituting the above values we get
`Munderset(AB) = 0`
`Munderset(BA) = 367.754 kNm`
`Munderset(BC) = -367.754 kNm`
`Munderset(CB) = 245.169 kNm`
`Munderset(CD) = -245.169 kNm`
`Munderset(DC) = 367.757 kNm`
`Munderset(DE) = -367.757 kNm`
`Munderset(ED) = 0`
2. WITH LOAD PATTERNING
LL always varies therefore is never 100% full in a building , so based on its occupancy using Muller Breslau's principle its location are determined to find max positive and negative bending moment
STEP 1
LOADS FROM SLAB
1. SW of slab
t = 120mm = 0.120m
= unit weight of cncrete x 0.120m
= 25kN/m3 x 0.120m
- SW = 3kN/m2
- LL = 5kPa = 5kN/m2
- SDL = 3kPa = 3kN/m2
- TL = SW+ DL + LL = 3 + 5 + 3 = 11 kN/m2
STEP 2
LOAD FROM SLAB TO BEAM
= 11 `(kN)/m^2` x (9x1.5) `m^2`
= 11x 13.5
= 148.5kN
= 148.5kN/ 9m
= 16.5 `(kN)/m`
since load from slab gets transmitted from both sides
= 16.5 x 2
= 33 kN/m
STEP 3
SW OF BEAM
= unit wt of concrete x area of c/s of beam
= 25 `(kN)/m^3` x 0.5m x 0.75m
= 9.375kN/m
STEP 4
TOTAL LOAD ON BEAM
= 33 + 9.375
= 42.375kN/m
PLACES WITHOUT LL
STEP 1
LOADS FROM SLAB
1. SW of slab
t = 120mm = 0.120m
= unit weight of cncrete x 0.120m
= 25kN/m3 x 0.120m
- SW = 3kN/m2
- SDL = 3kPa = 3kN/m2
- TL = SW+ DL = 3 + 3 = 6 kN/m2
STEP 2
LOAD FROM SLAB TO BEAM
= 6 `(kN)/m^2` x (9x1.5) `m^2`
= 6x 13.5
= 81kN
= 81kN/ 9m
= 9`(kN)/m`
since load from slab gets transmitted from both sides
= 9 x 2
= 18 kN/m
STEP 3
SW OF BEAM
= unit wt of concrete x area of c/s of beam
= 25 `(kN)/m^3` x 0.5m x 0.75m
= 9.375kN/m
STEP 4
TOTAL LOAD ON BEAM
= 18 + 9.375
= 27.375kN/m
(a) TO FIND MAX BENDING MOMENT AT D
STEP 5
SOLVING BEAM USING SLOPE DEFLECTION METHOD
TO FIND FIXED END MOMENTS
`Munderset(FAB)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FBA)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FBC)` = `-(wl^2)/12` = `-( 27.375 xx 9^2)/12` = -184.78kNm
`Munderset(FCB)` = `(wl^2)/12` = `( 27.375 xx 9^2)/12` = 184.78kNm
`Munderset(FCD)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FDC)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FDE)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FED)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
STEP 5
SUPPORT END MOMENT
`Munderset(AB) = Munderset(FAB) + 4EIthetaundersetA/L + 2EIthetaundersetB/L`
`Munderset(AB) = -286.031 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`0 = -286.031 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`286.031 - 2EIthetaundersetB/9 = 4EIthetaundersetA/9`
`2EIthetaundersetA/9 = 143.0155 - EIthetaundersetB/9`
`Munderset(BA) = Munderset(FBA) + 4EIthetaundersetB/L + 2EIthetaundersetA/L`
`Munderset(BA) = 286.031 + 4EIthetaundersetB/9 + 143.0155 - EIthetaundersetB/9`
`Munderset(BA) = 429.0465 + 3EIthetaundersetB/9`
`Munderset(BC) = Munderset(FBC) + 4EIthetaundersetB/L + 2EIthetaundersetC/L`
`Munderset(BC) = -184.78 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`Munderset(CB) = Munderset(FCB) + 4EIthetaundersetC/L + 2EIthetaundersetB/L`
`Munderset(CB) = 184.78 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9`
`Munderset(CD) = Munderset(FCD) + 4EIthetaundersetC/L + 2EIthetaundersetD/L`
`Munderset(CD) = -286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`Munderset(DC) = Munderset(FDC) + 4EIthetaundersetD/L + 2EIthetaundersetC/L`
`Munderset(DC) = 286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`
`Munderset(DE) = Munderset(FDE) + 4EIthetaundersetD/L + 2EIthetaundersetE/L`
`Munderset(DE) = -286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetE/9`
`Munderset(ED) = Munderset(FED) + 4EIthetaundersetE/L + 2EIthetaundersetD/L`
`0 = 286.031 + 4EIthetaundersetE/9 + 2EIthetaundersetD/9`
`4EIthetaundersetE/9 = -286.031 - 2EIthetaundersetD/9`
`2EIthetaundersetE/9 = -143.0155 - EIthetaundersetD/9`
`sum MundersetB` =0,
`Munderset (BA) + Munderset(BC)` = 0
`429.0465 + 3EIthetaundersetB/9`` -184.78 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`244.266+ 7EIthetaundersetB/9 + 2EIthetaundersetC/9` =0
`7EIthetaundersetB/9 + 2EIthetaundersetC/9` = -244.266 ... (1)
`sum MundersetC` =0,
`Munderset (CB) + Munderset(CD)` = 0
`-286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9``184.78 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`2EIthetaundersetB/9 + 8EIthetaundersetC/9 + 2EIthetaundersetD/9` = 101.251 ... (2)
`sum MundersetD` =0,
`Munderset (DC) + Munderset(DE)` = 0
`286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`` -429.0465+ 4EIthetaundersetD/9 -143.0155 - EIthetaundersetD/9 `
`-143.0155+ 7EIthetaundersetD/9 + 2EIthetaundersetC/9` =0
`2EIthetaundersetC/9 + 7EIthetaundersetD/9` = 143.0155 ... (3)
`[[7/9,2/9,0],[2/9,8/9,2/9],[0,2/9,7/9]]` `((EIthetaa),(EIthetab),(EIthetac))` =`((-244.266),(101.251),(143.0155))``
R2`->`3.5 R2 - R1
`[[7/9,2/9,0],[0,26/9,7/9],[0,2/9,7/9]]` `((EIthetaa),(EIthetab),(EIthetac))` = `((-244.266),(598.645),(143.015))`
R3`->`13 R3 - R2
`[[7/9,2/9,0],[0,26/9,7/9],[0, 0 ,84/9]]` `((EIthetaa),(EIthetab),(EIthetac))` = `((-244.266),(598.645),(1260.55))`
`EIthetaunderset(D)`= 135.058
`EIthetaunderset(C)`= 170.861
`EIthetaunderset(B)`= -362.873
Substituting the above values we get
`Munderset(DC) = 384.025 kNm`
`Munderset(DE) = -384.025 kNm`
(b) TO FIND BENDING MOMENT AT B
Following similar procedure
Since structure and loads are symmetrical
`Munderset(BA) = 384.025 kNm`
`Munderset(BC) = -384.025 kNm`
(c) TO FIND MAX BENDING MOMENT AT C
STEP 5
SOLVING BEAM USING SLOPE DEFLECTION METHOD
TO FIND FIXED END MOMENTS
`Munderset(FAB)` = `-(wl^2)/12` = `-( 27.375 xx 9^2)/12` = - 184.781kNm
`Munderset(FBA)` = `(wl^2)/12` = `( 27.375 xx 9^2)/12` = 184.781kNm
`Munderset(FBC)` = `-(wl^2)/12` = `-( 18.375 xx 9^2)/12` = -124.03kNm
`Munderset(FCB)` = `(wl^2)/12` = `( 18.375 xx 9^2)/12` = 124.03kNm
`Munderset(FCD)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FDC)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FDE)` = `-(wl^2)/12` = `-( 27.375 xx 9^2)/12` = - 184.781kNm
`Munderset(FED)` = `(wl^2)/12` = `( 27.375 xx 9^2)/12` = 184.781kNm
STEP 5
SUPPORT END MOMENT
`Munderset(AB) = Munderset(FAB) + 4EIthetaundersetA/L + 2EIthetaundersetB/L`
`Munderset(AB) = -184.781 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`0 = -184.781 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`184.781 - 2EIthetaundersetB/9 = 4EIthetaundersetA/9`
`2EIthetaundersetA/9 = 92.3905 - EIthetaundersetB/9`
`Munderset(BA) = Munderset(FBA) + 4EIthetaundersetB/L + 2EIthetaundersetA/L`
`Munderset(BA) = 184.781 + 4EIthetaundersetB/9 + 92.3905 - EIthetaundersetB/9`
`Munderset(BA) = 277.172 + 3EIthetaundersetB/9`
`Munderset(BC) = Munderset(FBC) + 4EIthetaundersetB/L + 2EIthetaundersetC/L`
`Munderset(BC) = -286.031 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`Munderset(CB) = Munderset(FCB) + 4EIthetaundersetC/L + 2EIthetaundersetB/L`
`Munderset(CB) = 286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9`
`Munderset(CD) = Munderset(FCD) + 4EIthetaundersetC/L + 2EIthetaundersetD/L`
`Munderset(CD) = -286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`Munderset(DC) = Munderset(FDC) + 4EIthetaundersetD/L + 2EIthetaundersetC/L`
`Munderset(DC) = 286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`
`Munderset(DE) = Munderset(FDE) + 4EIthetaundersetD/L + 2EIthetaundersetE/L`
`Munderset(DE) = -184.781 + 4EIthetaundersetD/9 + 2EIthetaundersetE/9`
`Munderset(ED) = Munderset(FED) + 4EIthetaundersetE/L + 2EIthetaundersetD/L`
`0 = 184.781 + 4EIthetaundersetE/9 + 2EIthetaundersetD/9`
`4EIthetaundersetE/9 = -184.781 - 2EIthetaundersetD/9`
`2EIthetaundersetE/9 = -92.3905 - EIthetaundersetD/9`
`sum MundersetB` =0,
`Munderset (BA) + Munderset(BC)` = 0
`7EIthetaundersetB/9 + 2EIthetaundersetC/9` = 8.859 ... (1)
`sum MundersetC` =0,
`Munderset (CB) + Munderset(CD)` = 0
`2EIthetaundersetB/9 + 8EIthetaundersetC/9 + 2EIthetaundersetD/9` = 0 ... (2)
`sum MundersetD` =0,
`Munderset (DC) + Munderset(DE)` = 0
`2EIthetaundersetC/9 + 7EIthetaundersetD/9` = -8.86 ... (3)
`[[7/9,2/9,0],[2/9,8/9,2/9],[0,2/9,7/9]]` `((EIthetaa),(EIthetab),(EIthetac))` =`((8.859),(0),(-8.859))`
R2`->`3.5 R2 - R1
`[[7/9,2/9,0],[0,26/9,7/9],[0,2/9,7/9]]` `((EIthetaa),(EIthetab),(EIthetac))` = `((8.859),(-8.859),(-8.859))`
R3`->`13 R3 - R2
`[[7/9,2/9,0],[0,26/9,7/9],[0, 0 ,84/9]]` `((EIthetaa),(EIthetab),(EIthetac))` = `((8.859),(-8.859),(-124.026))`
`EIthetaunderset(D)`= -13.288
`EIthetaunderset(C)`= 0.5109
`EIthetaunderset(B)`= 11.244
Substituting the above values we get
`Munderset(CB) = 288.757 kNm`
`Munderset(CD) = -288.757 kNm`
(d) TO FIND BM AT MID SPAN OF BEAM
STEP 5
SOLVING BEAM USING SLOPE DEFLECTION METHOD
TO FIND FIXED END MOMENTS
`Munderset(FAB)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FBA)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FBC)` = `-(wl^2)/12` = `-( 27.375 xx 9^2)/12` = -184.78kNm
`Munderset(FCB)` = `(wl^2)/12` = `( 27.375 xx 9^2)/12` = 184.78kNm
`Munderset(FCD)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 286.031kNm
`Munderset(FDC)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 286.031kNm
`Munderset(FDE)` = `-(wl^2)/12` = `-( 42.375 xx 9^2)/12` = - 184.781kNm
`Munderset(FED)` = `(wl^2)/12` = `( 42.375 xx 9^2)/12` = 184.781kNm
STEP 5
SUPPORT END MOMENT
`Munderset(AB) = Munderset(FAB) + 4EIthetaundersetA/L + 2EIthetaundersetB/L`
`Munderset(AB) = -286.031 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`0 = -286.031 + 4EIthetaundersetA/9 + 2EIthetaundersetB/9`
`286.031 - 2EIthetaundersetB/9 = 4EIthetaundersetA/9`
`2EIthetaundersetA/9 = 143.0155 - EIthetaundersetB/9`
`Munderset(BA) = Munderset(FBA) + 4EIthetaundersetB/L + 2EIthetaundersetA/L`
`Munderset(BA) = 286.031 + 4EIthetaundersetB/9 + 143.0155 - EIthetaundersetB/9`
`Munderset(BA) = 429.0465 + 3EIthetaundersetB/9`
`Munderset(BC) = Munderset(FBC) + 4EIthetaundersetB/L + 2EIthetaundersetC/L`
`Munderset(BC) = -184.78 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`Munderset(CB) = Munderset(FCB) + 4EIthetaundersetC/L + 2EIthetaundersetB/L`
`Munderset(CB) = 184.78 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9`
`Munderset(CD) = Munderset(FCD) + 4EIthetaundersetC/L + 2EIthetaundersetD/L`
`Munderset(CD) = -286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`Munderset(DC) = Munderset(FDC) + 4EIthetaundersetD/L + 2EIthetaundersetC/L`
`Munderset(DC) = 286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`
`Munderset(DE) = Munderset(FDE) + 4EIthetaundersetD/L + 2EIthetaundersetE/L`
`Munderset(DE) = -184.781 + 4EIthetaundersetD/9 + 2EIthetaundersetE/9`
`Munderset(ED) = Munderset(FED) + 4EIthetaundersetE/L + 2EIthetaundersetD/L`
`0 = 184.781 + 4EIthetaundersetE/9 + 2EIthetaundersetD/9`
`4EIthetaundersetE/9 = -184.781 - 2EIthetaundersetD/9`
`2EIthetaundersetE/9 = -92.3905 - EIthetaundersetD/9`
`sum MundersetB` =0,
`Munderset (BA) + Munderset(BC)` = 0
`429.0465 + 3EIthetaundersetB/9`` -184.78 + 4EIthetaundersetB/9 + 2EIthetaundersetC/9`
`244.266+ 7EIthetaundersetB/9 + 2EIthetaundersetC/9` =0
`7EIthetaundersetB/9 + 2EIthetaundersetC/9` = -244.266 ... (1)
`sum MundersetC` =0,
`Munderset (CB) + Munderset(CD)` = 0
`-286.031 + 4EIthetaundersetC/9 + 2EIthetaundersetB/9``184.78 + 4EIthetaundersetC/9 + 2EIthetaundersetD/9`
`2EIthetaundersetB/9 + 8EIthetaundersetC/9 + 2EIthetaundersetD/9` = 101.251 ... (2)
`sum MundersetD` =0,
`Munderset (DC) + Munderset(DE)` = 0
`286.031 + 4EIthetaundersetD/9 + 2EIthetaundersetC/9`` -182.781+ 4EIthetaundersetD/9 -143.0155 - EIthetaundersetD/9 `
`8.8595+ 7EIthetaundersetD/9 + 2EIthetaundersetC/9` =0
`2EIthetaundersetC/9 + 7EIthetaundersetD/9` = -8.8595 ... (3)
`[[7/9,2/9,0],[2/9,8/9,2/9],[0,2/9,7/9]]` `((EIthetaa),(EIthetab),(EIthetac))` =`((-244.266),(101.251),(-8.8595))``
R2`->`3.5 R2 - R1
`[[7/9,2/9,0],[0,26/9,7/9],[0,2/9,7/9]]` `((EIthetaa),(EIthetab),(EIthetac))` = `((-244.266),(598.645),(-8.8595))`
R3`->`13 R3 - R2
`[[7/9,2/9,0],[0,26/9,7/9],[0, 0 ,84/9]]` `((EIthetaa),(EIthetab),(EIthetac))` = `((-244.266),(598.645),(-483.456))`
`EIthetaunderset(D)`= -51.802
`EIthetaunderset(C)`= 221.175
`EIthetaunderset(B)`= -377.273
`sumMunderset(BA)` =0,
`Munderset(BA)` - 42.375x9x9/2 + `RundersetA`x9 =0
Ra= 156.988 kN
`Munderset(@mid AB)`
= Ra(4.5) - 42.375 (4.5x4.5/2)
`Munderset(@mid AB) = 277.399 kNm`
Similarly at other mid spans
the mid span sagging moment at each span = 277.399kNm
(e) TO FIND MAX POSITIVE BENDING MOMENT FOR EVERY SPAN
The max. Bending moment occurs where the shear force becomes zero.
Ra - w x X = 0
156.988 - 42.375(x) = 0
156.988 = 42.375x
x=3.7m
`Munderset(3.5m A)` from A
= Ra(3.7) - 42.375(3.7x3.7/2)
= 156.988(3.7) - (42.375)(3.7x3.7/2)
= 290.798kNm
SIMILARLY FOR EVERY SPAN
= 290.798kNm
(since structure and loads are symmetrical and load patterning to find max positive or sagging moment is placing udl on alternate beams.)
1. WITHOUT LOAD PATTERNING
MAXIMUM NEGATIVE MOMENTS
Mmax @ B = 367.754kNm (hogging)
Mmax @ C = 245.169kNm (hogging)
Mmax @ D = 367.754kNm (hogging)
2. WITH LOAD PATTERNING
MAXIMUM NEGATIVE MOMENTS
Mmax @ B = 384.025kNm (hogging)
Mmax @ C = 288.757kNm (hogging)
Mmax @ D = 384.025kNm (hogging)
MAXIMUM POSITIVE MOMENTS
Mmax @ for everyspan = 290.798kNm (sagging)