Project 1
- SW of slab = .120 m * 25 = 3Kn/m^2
Live load + Dead load = 8 KN/m^2
Total load = 11 KN/m^2
uniform Load on the beam = 11 KN/m^2 * (3/2) * 2 = 33 KN/m
Self weight of beam = 25 KN/m^3 * 0.5 m * 0.75m = 9.375 KN/m
Total load on beam = 33 + 9.375 = 42.375 KN/m
Fixed end moments
MFab = -WL^2/12 = -286.031 KNm
MFba = WL^2/12 = 286.031 KNm
MFbc = -286.031 KNm
MFcb = 286.031 KNm
MFcd = -286.031 KNm
MFdc = 286.031 KNm
MFde = -286.031 KNm
MFed = 286.031 KNm
End moments
Mab = MFab + 4EIQa/L + 2EIQb/L
Mab =-286.031+0.44EIQa + 0.22EIQb
Mba = 286.031 + 0.44EIQb + 0.22EIQa
Mbc = -286.031 + 0.44EIQb + 0.22EIQc
Mcb = 286.031 + 0.44EIQc + 0.22EIQb
Mcd = -286.031 + 0.44EIQc + 0.22EIQd
Mdc = 286.031 + 0.44EIQd + 0.22EIQc
Mde = -286.031 + 0.44EIQd + 0.22EIQe
Med = 286.031 + 0.44EIQe + 0.22EIQd
Equilibrium Equations :
Mba + Mbc = 0
0 = 0.88EIQb + 0.22EIQa + 0.22EIQc ————- (i)
Mcb + Mcd = 0
0 = 0.88EIQd + 0.22EIQb + 0.22EIQc ————- (ii)
Mdc + Mde = 0
0 = 0.88EIQd + 0.22EIQc + 0.22EIQe ————— (iii)
Mab = 0
EIQa = 650.0681818 - 0.5EIQb ———— (iv)
Med = 0
EIQe = -650.0681818 - 0.5EIQd ———- (v)
By solving above equations:
EIQc = 0
EIQd = 185.7337662
EIQb = -185.7337662
EIQe = -742.9350649
EIQa = 742.9350649
Using the above rotational reactions:
We find the moments:
Mab = 0
Mba = 367.753 KNm
Mbc = -367.753 KNm
Mcb = 245.168 KNm
Mcd = -245.168 KNm
Mdc = 367.753 KNm
Mde = -367.753 KNm
Med = 0
Shear force of span AB
Summation of Mb = 0,
9Va - 42.375(9)(9/2) + 367.753 = 0
Va = 149.826
Va + Vb - 43.375 * 9 = 0
Vb +149.826 - 381.375 = 0
Vb = 231.549 Kn/m^2
Shear force of span BC
Summation of Mc = 0
9Vb - 42.375(9)(9/2) + 245.169 - 367.753 = 0
Vb = 204.308 KN/m^2
Summation for vertical forces = 0
Vb +vc - 381.375 = 0
Vc = 177.067 KN/m^2
Shear force of span CD
Summation of Md = 0
9Vc - 42.375(9)(9/2) + 367.753 - 245.169 = 0
Vc = 177.067 KN/m^2
Summation for vertical forces = 0
Vd = 204.308 KN/m^2
Shear force of span DE
Summation of Me = 0
9Vd - 1716.1875 - 367.753 = 0
Vd = 231.549 KN/m^2
Summation for vertical forces = 0
Ve = 149.826 KN/m^2
(ii) Load Patterning
With Live Load:
SW of slab = .120 m * 25 = 3Kn/m^2
Live load + Dead load = 8 KN/m^2
Total load = 11 KN/m^2
uniform Load on the beam = 11 KN/m^2 * (3/2) * 2 = 33 KN/m
Self weight of beam = 25 KN/m^3 * 0.5 m * 0.75m = 9.375 KN/m
Total load on beam = 33 + 9.375 = 42.375 KN/m
Without Live Load:
SW of slab = .120 m * 25 = 3Kn/m^2
Dead load = 3 KN/m^2
Total load = 6 KN/m^2
uniform Load on the beam = 6 KN/m^2 * (3/2) * 2 = 18 KN/m
Self weight of beam = 25 KN/m^3 * 0.5 m * 0.75m = 9.375 KN/m
Total load on beam = 18 + 9.375 = 27.375 KN/m
To find max negative bending moment at D
Fixed end moments:
Mfab = -286.031 KNm
Mfba = 286.031 KNm
Mfbc = -184.781 KNm
Mfcb = 184.781 KNm
Mfcd = -286.031 KNm
Mfdc = 286.031 KNm
Mfde = -286.031 KNm
Mfed = 286.031 KNm
End Moments:
Mab = -286.031 + 4EIQa/9 + 2EIQb/9
Mba = 286.031 + 4EIQb/9 + 2EIQa/9
Mbc = -184.78 + 4EIQb/9 + 2EIQc/9
Mcb = 184.78 + 4EIQc/9 + 2EIQb/9
Mcd = -286.031 + 4EIQc/9 + 2EIQd/9
Mdc = 286.031 + 4EIQd/9 + 2EIQc/9
Mde = -286.031 + 4EIQd/9 + 2EIQe/9
Med = 286.031 + 4EIQe/9 + 2EIQd/9
Solving the equations above we get the rotational reactions in each Point
EIQc = 170.839 KN/m^2
EIQb = -362.744
EIQd = 135.188
EIQa = -462.198
EIQe = -711.162
Substituting the rotational reactions in the moment equations to find the moment
Mab = 0
Mba = 308.035
Mbc = -308.035
Mcb = 180.099
Mcd = -180. 061
Mdc = 384.079
Mde = -384.015
Med = 0
Moment at D = 384.079 KNm
Max negative Bending moment at B
Fixed end moments:
Mfab = -286.031 KNm
Mfba = 286.031 KNm
Mfbc = -286.031 KNm
Mfcb = 286.031 KNm
Mfcd = -184.781 KNm
Mfdc = 184.781 KNm
Mfde = -286.031 KNm
Mfed = 286.031 KNm
End moments:
Mab = 0
Mba = 384.079 KNm
Mbc = -384.079 KNm
Mcb = 180.061 KNm
Mcd = -180.061 KNm
Mdc = 308.035 KNm
Mde = -308.035 KNm
Med = 0
Moment at b = 384.079 KNm
Bending moment at C:
Fixed end moments:
Mfab = -184.781
Mfba = 184.781
Mfbc = -286.031
Mfcb= 286.031
Mfcd = -286.031
Mfdc = 286.031
Mde = -184.781
Med = 184.781
End moments:
Mab = -184.781 + 4EIQa/9 + 2EIQb/9
Mba = 277.1715 + 3EIQb/9
Mbc = -286.031 + 4EIQb/9 + 2EIQc/9
Mcb = 286.031 + 4EIQc/9 + 2EIQb/9
Mcd = -286.031 + 4EIQc/9 + 2EIQd/9
Mdc = 286.031 + 4EIQd/9 + 2EIQc/9
Mde = -184.781 + 4EIQd/9 + 2EIQe/9
Med = 184.781 + 4EIQe/9 + 2EIQd/9
Equilibrium equations:
Mba +Mbc = 0
-8.8595 + 7EIQb/9 + 2EIQc/9 = 0
Mcb + Mcd = 0
8EIQc/9 + 2EIQb/9 + 2EIQd/9 = 0
Mdc + Mde = 0
8.8595 + 7EIQd/9 + 2EIQc/9=0
Solving the above equations we get the rotational values of each point
EIQa = 410.06175 KNm
EIQb = 11.391 KNm
EIQc = 0 KNm
EIQd = -11.391 KNm
EIQe = -410.06175 KNm
Moment Values after substituting the rotational reactions in the moment equations
Mab = 0
Mba = 280.9685 KNm
Mbc = -280.9685 KNm
Mcb = 288.5623 KNm
Mcd = -288.5623 KNm
Mdc = 280.9683 KNm
Mde = -280.9683 KNm
Med = 0
Moment at c = 288.5623 KNm
BENDING MOMENT AT MID SPAN OF BEAM
ILD of Bending moment at midspan AB (Sagging)
Fixed End Moments:
Mfab = -286.031
Mfba = 286.031
Mfbc = -184.78
Mfcb = 184.78
Mfcd = -286.031
Mfdc = 286.031
Mfde = -184.78
Mfed = 184.78
Support End Moments:
Mab = -286.031 + 4EIQa/9 + 2EIQb/9
Mba = 286.031 + 4EIQb/9 + 2EIQa/9
Mbc = -184.78 + 4EIQb/9 + 2EIQc/9
Mcb = 184.78 + 4EIQc/9 + 2EIQb/9
Mcd = -286.031 + 4EIQc/9 + 2EIQd/9
Mdc = 286.031 + 4EIQd/9 + 2EIQc/9
Mde = -184.78 + 4EIQd/9 + 2EIQe/9
Med = 184.78 + 4EIQe/9 + 2EIQd/9
The rotational values:
EIQd = -51.802
EIQc = 221.175
EIQb = -377.273
EIQa= 832.206
EIQe = -389.854
Moment Values:
Mab = 0
Mba = 303.2887778 KNm
Mbc = -303.2887778 KNm
Mcb = 199.2425556 KNm
Mcd = -199.2425556 KNm
Mdc = 312.1578889 KNm
Mde = -312.1578889 KNm
Med = 0
To find Moment at midspan AB:
Summation of Mba = 0
Mba - 42.375 * 9 * 4.5 +Ra9 = 0
Ra = 156.988 KN
M at midspan AB
Ra4.5 -42.375 * 4.5 * 2.25 = M (Midspan AB)
M ( Midspan Ab) = 277.399 KNm
At other midspan also The sagging moment will be = 277.399KNm
Maximum positive bending moment for every span
Max bending moment occurs when the shear force=0
Ra - Wx = 0
156.988 - 42.375x = 0
X = 3.7m
Moment at 3.7m from A=
=Ra(3.7) - 42.375(3.7*3.7/2)
=156.988(3.7) - (42.375)(3.7*3.7/2)
= 290.798 KNm
Similarly for every span Max bending moment = 290.798 KNm