Project 1
SOLUTION-
self weight of beam = 0.5*0.75*25 = 9.375Kn/m
self weight of slab = 0.12*25 = 3kn/m
live load = 5kn/m
floor finish = 3kn/m
Total weight = 11kn/m
load on beam = total weight * width
let us give the names to all the beams in order as B1,B2,B3.....etc.
load on B1 = W * width+self weight of beam
= 11* 0.5(3+3)+9.375
= 42.375 Kn/m
load on B2 = W * width
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
load on B3 = W * width
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
load on B4 = W * width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
load on B5 = W * width
= 11 * 0.5(3+3) +9.375
= 42.375Kn/m
load on B6 = W * width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
load on B7 = W * width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
FIXED END MOMENTS :
Mac = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mca = wL^2/12 = 42.375*81/12 = +286.03Knm
Mcd = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mdc = wL^2/12 = 42.375*81/12 = +286.03Knm
Med = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mde = wL^2/12 = 42.375*81/12 = +286.03Knm
Meb = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mbe = wL^2/12 = 42.375*81/12 = +286.03Knm
moments at each point:
Ma = Mac + 4EITa/L + 2EITc/L
Ma = -286.03 + 78125Ta + 39062.5Tc ------------eq.1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.03 + 78125Tc + 39062.5Ta ------------eq.2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.03 + 78125Tc + 39062.5Td ------------eq.3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.03 + 78125Td + 39062.5Tc ------------eq.4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -412.59 + 78125Td + 39062.5Te ------------eq.5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.03 + 78125Te + 39062.5Td ------------eq.6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.03 + 78125Te + 39062.5Tb ------------eq.7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.03 + 78125Tc + 39062.5Ta ------------eq.8
from eq1.
78125Ta = 286.03 - 39062.5Tc ---------eq.9
considering the equilibrium at C,
Mc = 0
Mc = Mcd + Mca
78125Tc + 78125Tc + 39062.5Ta + 39062.5Ta + 39062.5Td = 0
156250Tc + 143.015 - 19531.25tc + 39062.5Td = 0
136718.75Tc + 39062.5Td = -143.015
39062.5Td = -143.015 - 136718.75Tc ------------eq.10
considering the equilibrium at D, Md = 0
156250Td + 39062.5Tc + 39062.5Te = 0
-572.06 - 546875Tc + 39062.5Tc + 39062.5Te = 0
-572.06 - 546875Tc + 39062.5Tc + 39062.5Te = 0
-572.06 - 507812.5Tc = -39062.5Te
39062.5Te = 507812.5Tc + 572.06 ------------eq.11
considering equilibrium at E, Me = 0
156250Te + 39062.5Td + 39062.5Tb = 0
2031250Tc + 2288.24 - 143.014 - 136718.75Tc + 39062.5Tb = 0
1894531.25Tc + 39062.5Tb = -2145.25 ----------eq.12
considering the equilibrium at B, Mbe = 0
78125Tb + 39062.5Te + 286.03 = 0
-4290.452 - 3789062.5Tc + 507812.5Tc + 572.06 + 286.03 = 0
-3281250Tc = 3432.36
Tc = -0.001046 -----------eq.13
similarly substitute the above value in all the equations and derive the values of unknowns.
Tb = -0.00418
Te = 0.001046
Td = -1.84*10^-6
Ta = 0.00418
substitute the above unknown values in moment equations and get the moment values.
Ma = -0.4
Mc,ca = 367.7Knm
Mc,cd = -367.7Knm
Md,dc = 245.17Knm
Md,de = -245.17KNm
Me,ed = 367.7Knm
Me,eb = -367.7KNm
Mb,be = 0 Knm
To determine the shear forces , consider each span individually,
consider span AC,
taking moment about A,
Hc * 9 - 0.4 + 367.59 - 42.375*9*4.5 = 0
Hc = 149.84 KN
Ha + Hc = 42.375*9
Ha + 149.84 = 381.375
Ha = 231.53Kn
consider span CD,
taking moment about c,
-367.59 + 245.78 + Hd * 9 - 42.375*9*4.5 = 0
Hd = 204.28Kn
Hc+Hd = 381.375
Hc = 177.09Kn
consider span DE,
taking moment about D,
-225.65 + 367.7 + He*9 - 42.375*9*4.5 = 0
He = 174.9Kn
Hd=206.45Kn
consider span EB,
taking moment about E,
9Hb - 367.7 - 42.375*9*4.5 = 0
Hb = 231.535Kn
He = 149.84Kn
BM = wl^2/8 = 42.375 x 81 / 8
= 429.04 kNm
LIve load will be excluded for 1 span:
The maximum BM @ E by loading span AB,BC,CD. shall be subjected to a load of 42.375KN/m
And DE will be loaded with udl of 27.35KN/m (excluding L.L)
FEM:
M AC = -WL^2 / 12 = -286.03 kNm
M CA = -WL^2 / 12 = 286.03 kNm
M CD = -286.03 kNm
M DC = 286.03 kNm
M DE = -286.03 kNm
M ED = 286.03 kNm
M EB = - 184.62 kNm
M BE = 184.62 kNm
Equilibrium equations ;
MAC=0
M DC + M DE = 0
M BE = 0
M CA + M CD = 0
M ED + M EB = 0
- 80000 θA + 40000θC = 286.03
By solving the equations
θA = 7.1 x 10 ^ -3
θC = - 7.1 x 10 ^ -3
θD = 2.1 x 10 ^ -2
θE = -7.7 x 10 ^ -2
θB = 3.7 x 10 ^ -2
calculating the moments
M AC = 0
M CA = 1678
M CD = -1678
M DC = 1678
M DE = -1678
M ED = 7455
M EB = -7455
M BE = 0
SPAN AC
Moment about AC
RA x 9 – 42.375 x 9 x 4.5 -1678 =0
RA = 281.79 kN
RA + RC1 = 381.375
RC1 = 381.375 -281.79
RC1 = 99.58 kN
SPAN CD
RC2 x 9 -42.375 x 9 x 4.5/2 +1678 = 0
RC2 = -91 kN
RC2 + RD1 = 381.375
RD1 = 381.375 + 91
RD1 = 472 kN
SPAN DE
RD2 x9 -42.395 x 9 x 4.5/2 +7455 = 0
= RD2 = -733 kN
RD2 + RE1 = 0
-733 + RE1 = 381.375
RD1 = 1114.39 kN
SPAN EB
RE2 x 9 -27.35 x 9 x 4.5/2 +7455 = 0
RE2 = 766 KN
RE2 + RB1 = 27.35 x 9
RB = 519.8 KN
