Project 1
A. LOAD CALCULATION
0.120 m x 25 KN/m^3 = 3Kpa , FF & mech =3 Kpa , live load = 5 Kpa
Total load = 11 Kpa
Load on the beam = (11x 1.5 ) + 11x 1.5 = 33 KN /m^2
self weight of beam = 0.5x0.75 x25 = 9.375 KN/m
total load on beam = 42.375 KN/m
FEM
Mf AB = -WL^2/12 = - 286 KNm
Mf BA = WL^2/12 = 286 KNm
Mf BC =-WL^2/12 = - 286 KNm
Mf CB = WL^2/12 = 286 KNm
Mf CD =-WL^2/12 = - 286 KNm
Mf DC =WL^2/12 = 286 KNm
Mf DE= -WL^2/12 = - 286 KNm
Mf ED = WL^2/12 = 286 KNm
M AB = Mf AB + 2EI/L (2 `theta`A + `theta`B) = -286 +80000`theta`A + 40000`theta`B
M BA = Mf BA+ 2EI/L (2 `theta`B + `theta`A) = 286 +80000`theta`B + 40000`theta`A
M BC = Mf BC + 2EI/L (2 `theta`B+ `theta`C) = -286 +80000`theta`B + 40000`theta`C
M CB = Mf CB+ 2EI/L (2 `theta`C+ `theta`B) = 286 +80000`theta`C + 40000`theta`B
M CD = Mf CD + 2EI/L (2 `theta`C+ `theta`D) = -286 +80000`theta`C + 40000`theta`D
M DC = Mf DC+ 2EI/L (2 `theta`D + `theta`C) = 286 +80000`theta`D + 40000`theta`C
M DE = Mf DE+ 2EI/L (2 `theta`D + `theta`E) = -286 +80000`theta`D+ 40000`theta`E
M ED = Mf ED + 2EI/L (2 `theta`E + `theta`D) = 286 +80000`theta`E + 40000`theta`D
Equlibrium condition
M AB =0 , MBA + MBC =0 , MCB +MCD=0 , MDC+ MDE =0 , M ED =0
-286 +80000`theta`A + 40000`theta`B =0
160000 `theta`B +40000 `theta`C + 40000`theta`A =0
160000 `theta`C+40000 `theta`B + 40000`theta`D =0
160000 `theta`D+40000 `theta`C + 40000`theta`E=0
286 +80000`theta`E + 40000`theta`D=0
By solving above equation
`theta`A = 0.004 Radian , `theta`B = -0.001 Radian , `theta`C = 0.00 Radian, `theta`D = 0.001 Radian ,
`theta`E = -0.004 Radian
M AB = -6 KNm , M BA = 366 KNM, M BC = -366 KNM , M, CB = 246 KNM , M CD = -246 KNM , M DC = 366 KNM ,
M DE= -366KNM , M ED = 6 KNM
From span AB
R A = 150 KN , RB = 231.3 KN
From span BC
R B = 163.3 KN , RC = 218 KN
From span CD
R C= 218 KN , RD = 163.3 KN
From span DE
R D = 231.4 KN , RE = 150 KN
Load patterning
Fixed end moments
Mf AB = -WL^2/12 = - 286 KNm
Mf BA = WL^2/12 = 286 KNm
Mf BC =-WL^2/12 = -184.74 KNm
Mf CB = WL^2/12 = 184.74 KNm
Mf CD =-WL^2/12 = - 286 KNm
Mf DC =WL^2/12 = 286 KNm
Mf DE= -WL^2/12 = - 184.74 KNm
Mf ED = WL^2/12 = 184.74 KNm
M AB = Mf AB + 2EI/L (2 `theta`A + `theta`B) = -286 +80000`theta`A + 40000`theta`B
M BA = Mf BA+ 2EI/L (2 `theta`B + `theta`A) = 286 +80000`theta`B + 40000`theta`A
M BC = Mf BC + 2EI/L (2 `theta`B+ `theta`C) = -184.74 +80000`theta`B + 40000`theta`C
M CB = Mf CB+ 2EI/L (2 `theta`C+ `theta`B) = 184.74 +80000`theta`C + 40000`theta`B
M CD = Mf CD + 2EI/L (2 `theta`C+ `theta`D) = -286 +80000`theta`C + 40000`theta`D
M DC = Mf DC+ 2EI/L (2 `theta`D + `theta`C) = 286 +80000`theta`D + 40000`theta`C
M DE = Mf DE+ 2EI/L (2 `theta`D + `theta`E) = -184.74 +80000`theta`D+ 40000`theta`E
M ED = Mf ED + 2EI/L (2 `theta`E + `theta`D) = 184.74 +80000`theta`E + 40000`theta`D
Equlibrium condition
M AB =0 , MBA + MBC =0 , MCB +MCD=0 , MDC+ MDE =0 , M ED =0
-286 +80000`theta`A + 40000`theta`B =0
101.26 +160000 `theta`B + 40000 `theta`A + 40000`theta`C=0
-101.26 + 160000`theta`C + 40000`theta`B + 40000`theta`D =0
101.26 + 160000`theta`D + 40000`theta`C + 40000`theta`E =0
184.74 +80000`theta`E + 40000`theta`D=0
By solving above equation
`theta`A = 0.004 , `theta`B = -0.002 , `theta`C =0.001 , `theta`D = -0.000 , `theta`E = - 0.002
M AB= -46 KNM , MBA = 286 KNM , M BC = -304.74 KNM , M CB = 184.74 KNM , MCD = -206 KNM , M DC = 326 KNM
M DE = -264.74 KNM , M ED = 24.5 KNM
From span AB
R A = 158.8 KN , RB = 222.5 KN
From span BC
R B = 102.6 KN , RC = 143.7 KN
From span CD
R C= 154.4 KN , RD = 226.3 KN
From span DE
R D = 120 KN , RE = 126.3 KN
Fixed end moments
Mf AB = -WL^2/12 = - 184.74 KNm
Mf BA = WL^2/12 = 184.74KNm
Mf BC =-WL^2/12 = - 286 KNm
Mf CB = WL^2/12 = 286 KNm
Mf CD =-WL^2/12 = -184.74 KNm
Mf DC =WL^2/12 = 184.74 KNm
Mf DE= -WL^2/12 = - 286 KNm
Mf ED = WL^2/12 = 286 KNm
M AB = Mf AB + 2EI/L (2 `theta`A + `theta`B) = -184.74 +80000`theta`A + 40000`theta`B
M BA = Mf BA+ 2EI/L (2 `theta`B + `theta`A) = 184.74 +80000`theta`B + 40000`theta`A
M BC = Mf BC + 2EI/L (2 `theta`B+ `theta`C) = - 286+80000`theta`B + 40000`theta`C
M CB = Mf CB+ 2EI/L (2 `theta`C+ `theta`B) = 286 +80000`theta`C + 40000`theta`B
M CD = Mf CD + 2EI/L (2 `theta`C+ `theta`D) = -184.74+80000`theta`C + 40000`theta`D
M DC = Mf DC+ 2EI/L (2 `theta`D + `theta`C) = 184.74 +80000`theta`D + 40000`theta`C
M DE = Mf DE+ 2EI/L (2 `theta`D + `theta`E) = -286 +80000`theta`D+ 40000`theta`E
M ED = Mf ED + 2EI/L (2 `theta`E + `theta`D) = 286+80000`theta`E + 40000`theta`D
Equlibrium condition
M AB =0 , MBA + MBC =0 , MCB +MCD=0 , MDC+ MDE =0 , M ED =0
-184.74 +80000`theta`A + 40000`theta`B =0
-101.26 + 160000 `theta`B + 40000`theta`A + 40000`theta`C =0
101.26 + 160000 `theta`C + 40000`theta`B+ 40000`theta`D=0
-101.26 + 160000 `theta`D+ 40000`theta`E+ 40000`theta`C=0
286+80000`theta`E + 40000`theta`D=0
By solving above equation
`theta`A = 0.002 , `theta`B = 0.0002 , `theta`C =-0.0006 , `theta`D = -0.0001, `theta`E = - 0.003
M AB= -16.75 KNM , MBA = 280.75 KNM , M BC = -294 KNM , M CB = 246 KNM , MCD = -236.74 KNM , M DC = 152.74 KNM
M DE = -174 KNM , M ED = 522 KNM
From span AB
R A = 91.9 KN , RB = 154.4 KN
From span BC
R B = 218 KN , RC = 163.7 KN
From span CD
R C= 106.1 KN , RD = 140.2 KN
From span DE
R D = 132.6 KN , RE = 248.7 KN