Project 1
soln.
self weight of slab = 0.12*25 = 3kn/m
self weight of beam = 0.5*0.75*25 = 9.375Kn/m
floor finish = 3kn/m
live load = 5kn/m
Total weight = 11kn/m
load on beam = total weight * tributary width
let us give the names to all the beams in order as B1,B2,B3.....etc.
load on B1 = W * tributary width+self weight of beam
= 11* 0.5(3+3)+9.375
= 42.375 Kn/m
load on B2 = W * tributary width
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
load on B3 = W * tributary width
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
load on B4 = W * tributary width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
load on B5 = W * tributary width
= 11 * 0.5(3+3) +9.375
= 42.375Kn/m
load on B6 = W * tributary width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
load on B7 = W * tributary 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
2nd part of question
let us findout the maximum bending moment at C by loading span
AC ,CD and EB.
i.e AC,CD and EB will be subjected to a load of 42.375kn/m and DE will be loaded with
UDL of 18kn/m(excluding live load).
we get the fixed end moments as follows.
Mac = -wL^2/12=-286.01kNm
Mca = 286.01Knm
Mcd = -286.01kNm
Mdc = 286.01kNm
Meb = -286.01kNm
Mbe = 286.01knm
Mde = -121.5kNm
Med = 121.5kNm
Ma = Mac+4EITa/L + 2EITc/L
Ma = -286.01+0.4EITa+0.2EITc ------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.01 + 0.4EITc + 0.2EITa -------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.01 + 0.4EITc + 0.2EITd ------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.01 + 0.4EITd + 0.2EITc -------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -121.5 + 0.4EITd + 0.2EITe ------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 121.5 + 0.4EITe + 0.2EITd -------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.01 + 0.4EITe + 2EITb --------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.01 + 0.4EITb + 0.2EITe ------- 8
from equation 1,
0.4EITa = 286.01 - 0.2EITc
0.2EITa = 143 - 0.1EITc --------- 9
substitute the above value in summation of eq.2 and 3
143 - 0.1EITc + 0.8EITc + 0.2EITd = 0
0.7EITc + 0.2EITd + 143 = 0
0.7EITc = -143 - 0.2EITd
0.2EITc = -40.85 - 0.0571EITd ------- 10
substitute the above value in summation of eq.4 and 5
Md = 0
-40.85 - 0.0571EITd + 0.8EITd + 0.2EITe + 164.51 = 0
0.74EITd + 0.2EITe + 123.66 = 0 ------- 11
0.2EITe = -0.74EITd - 123.66 * 2
0.4EITe = -1.48EITd - 247.32 * 2
0.8EITe = -2.96EITd - 494.64
substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 0.8EITe - 164.51 = 0
0.2EITb + 0.2EITd - 2.96EITd - 494.64 - 164.51 = 0
0.2EITb - 2.76EITd - 659.15 * 2
0.4EITb = 5.52EITd + 1318.3 -------- 12
substitute the above value in summation of moment at B,
Mb = 0
5.52EITd + 1318.3 + 286.01 - 0.74EITd - 123.66 = 0
4.78EITd = -1480.65
EITd = -309.75
substitute the value of EITd in other equations and derive unknowns as follows.
EITb = 978.8
EITc = -115.81
EITa = 772.93
EIT = 572.35
substitute the values of these unknowns in all the moment equations and derive
the moment values.
Ma = 0
Mc,ca = 394.27 kNM
Mc,cd = -394.87 kNM
Md,dc = 198.948 kNM
Md,de = -137.34 kNM
Me,ed = 288.49 kNM
Me,Eb = -288.49 kNM
Mbe = 8.96 kNM
similarly findout the maximum moment at E,
let us findout the maximum bending moment at C by loading span
DE,AC and EB.
i.e AC,DE and EB will be subjected to a load of 42.375kn/m and CD will be loaded with
UDL of 18kn/m(excluding live load).
we get the fixed end moments as follows.
Mac = -wL^2/12=-286.01kNm
Mca = 286.01Knm
Mcd = -121.5kNm
Mdc = 121.5kNm
Mde = -286.01kNm
Med = 286.01knm
Meb = -286.01kNm
Mbe = 286.01kNm
Ma = Mac+4EITa/L + 2EITc/L
Ma = -286.01+0.4EITa+0.2EITc ------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.01 + 0.4EITc + 0.2EITa -------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -121.5 + 0.4EITc + 0.2EITd ------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 121.5 + 0.4EITd + 0.2EITc -------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -286.01 + 0.4EITd + 0.2EITe ------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.01 + 0.4EITe + 0.2EITd -------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.01 + 0.4EITe + 0.2EITb --------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.01 + 0.4EITb + 0.2EITe ------- 8
from equation 1,
0.4EITa = 286.01 - 0.2EITc
0.2EITa = 143 - 0.1EITc --------- 9
substitute the above value in summation of eq.2 and 3
143 - 0.1EITc + 0.8EITc + 0.2EITd = 0
0.7EITc + 0.2EITd + 143 = 0
0.7EITc = -143 - 0.2EITd
0.2EITc = -40.85 - 0.0571EITd ------- 10
substitute the above value in summation of eq.4 and 5
Md = 0
-40.85 - 0.0571EITd + 0.8EITd + 0.2EITe - 164.51 = 0
0.74EITd + 0.2EITe -205.36 = 0 ------- 11
0.2EITe = 205.36 - 0.74EITd
0.4EITe = -1.48EITd - 247.32 * 2
0.8EITe = 821.44 - 2.96EITd ------------ 12
substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 821.44 - 2.96EITd = 0
0.2EITb - 2.76EITd +821.44 = 0
0.2EITb = 2.76EITd - 821.44
0.4EITb = 5.52EITd - 1642.88 -------- 13
substitute the above value in summation of moment at B,
Mb = 0
5.52EITd -1642.88 + 286.01 +205.36 - 0.74EITd = 0
4.78EITd = 1151.51
EITd = 240.9
substitute the value of EITd in other equations and derive unknowns as follows.
EITb = -782.78
EITc = -273.11
EITa = 851.58
EITe = 135.51
substitute the values of these unknowns in all the moment equations and derive
the moment values.
Ma = 0
Mc,ca = 347.08 kNM
Mc,cd = -347.87 kNM
Md,dc = 163.23 kNM
Md,de = -162.54 kNM
Me,ed = 388.49 kNM
Me,Eb = -388.49 kNM
Mbe = 0 kNM
let us findout the maximum moment at D,
let us findout the maximum bending moment at D by loading span
CD and DE.
i.e ,CD and DE will be subjected to a load of 42.375kn/m and AC, AB will be loaded with
UDL of 18kn/m(excluding live load).
we get the fixed end moments as follows.
Mac = -wL^2/12=-121.5kNm
Mca = 121.5Knm
Mcd = -286.01kNm
Mdc = 286.01kNm
Meb = -121.5kNm
Mbe = 121.5knm
Mde = -286.01kNm
Med = 286.01kNm
Ma = Mac+4EITa/L + 2EITc/L
Ma = -121.5+0.4EITa+0.2EITc ------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 121.5 + 0.4EITc + 0.2EITa -------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.01 + 0.4EITc + 0.2EITd ------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.01 + 0.4EITd + 0.2EITc -------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -286.01 + 0.4EITd + 0.2EITe ------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.01 + 0.4EITe + 0.2EITd -------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -121.5 + 0.4EITe + 2EITb --------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.01 + 0.4EITb + 0.2EITe ------- 8
from equation 1,
0.4EITa = 121.5 - 0.2EITc
0.2EITa = 60.75 - 0.1EITc --------- 9
substitute the above value in summation of eq.2 and 3
60.75 - 0.1EITc + 0.8EITc + 0.2EITd - 164.5= 0
0.7EITc + 0.2EITd - 103.75 = 0
0.2EITd = 103.75 - 0.7EITc
0.4EITd = 207.5 - 1.4EITc ------- 10
substitute the above value in summation of eq.4 and 5
Md = 0
415 - 2.8EITc + 0.2EITc + 0.2EITe = 0
0.2EITe - 2.6EITc + 415 = 0 ------- 11
0.2EITe = 2.6EITc - 415 ----------12
0.8EITe = 10.4EITc - 1660 ------------13
substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 10.4EITc - 1494.5 = 0
0.2EITb =- 0.2EITd - 10.4EITc +1495.5
0.4EITb = 2990.4 -0.4EITd - 20.8EITc -------- 14
substitute the above value in summation of moment at B,
Mb = 0
2990.4 - 0.4EITd - 20.8EITc +2.6EITc -415 +121.5 = 0
2990.4+ 207.5 - 1.4EITc -20.8EITc + 2.6EITc -415+121.5 = 0
19.6EITc = 2904.4
EITc = 148.18
substitute the value of EITd in other equations and derive unknowns as follows.
EITd = 0.12
EITe = -140.16
EITa = 229.6
EITb = -233.67
substitute the values of these unknowns in all the moment equations and derive
the moment values.
Ma = 0
Mc,ca = 226.69 kNM
Mc,cd = -226.71 kNM
Md,dc = 315.69 kNM
Md,de = -313.9 kNM
Me,ed = 229.97 kNM
Me,Eb = -229.97 kNM
Mbe = 0 kNM
