Project 1
Sol.
LyLy/LxLx=9/3 =3 which is greater than 2 so it is a one way slab.
Thickness of slab= 0.120m
Self weight of the slab= 25x0.120 =3 KN/m^2
Live load= 5 KN/m^2
Floor Finish= 3 KN/m^2
Total Load= 3+5+3= 11 KN/m^2
Load from Slab to Beam= 11x9x1.5x2
= 297 KN
Converting it to UDL load= 297/9 = 33 KN/m
Self Weight of the beam= 25x0.5x0.75
= 9.375 KN/m
Total Load on the beam= 33+9.375
= 42.375 KN/m
Calculation of fixed end moments,
MFABMFAB= −wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFBAMFBA= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFBCMFBC= −wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFCBMFCB= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFCDMFCD=−wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFDCMFDC= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFDEMFDE= −wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFEDMFED= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
Calculation of support end moments,
MABMAB= MFABMFAB+4EITa4EITa/L/L+ 2EITb2EITb/L/L
0= -286.031+ 4EITa4EITa/9/9+ 2EITb2EITb/9/9
286.031- 2EITb2EITb/9/9= 4EITa4EITa/9/9
2EITa2EITa/9/9= 143.0155 - EITbEITb/9/9
MBAMBA= MFBAMFBA+ 4EITB4EITB/L/L+ 2EITA2EITA/L/L
MBAMBA= 286.031+ 4EITB4EITB/9/9+ 143.0155 - EITbEITb/9/9
MBAMBA= 429.0465+ 3EITB3EITB/9/9
MBCMBC= MFBCMFBC+ 4EITB4EITB/L/L+ 2EITC2EITC/L/L
MBCMBC= -286.031+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9
MCBMCB= MFCBMFCB+ 4EITC4EITC/L/L+ 2EITB2EITB/L/L
MCBMCB= 286.031+ 4EITC4EITC/9/9+ 2EITB2EITB/9/9
MCDMCD= MFCDMFCD+ 4EITC4EITC/L/L+ 2EITD2EITD/L/L
MCDMCD= -286.031+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9
MDCMDC= MFDCMFDC+ 4EITD4EITD/L/L+ 2EITC2EITC/L/L
MDCMDC= 286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9
MDEMDE= MFDEMFDE+ 4EITD4EITD/L/L+ 2EITE2EITE/L/L
MDEMDE= -286.031+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9
MEDMED= MFEDMFED+ 4EITE4EITE/L/L+ 2EITD2EITD/L/L
0= 286.031+ 4EITE4EITE/9/9+ 2EITD2EITD/9/9
2EITE2EITE/9/9= -143.0155`- EITDEITD/9/9
At Joint B
MBAMBA+ MBCMBC= 0
429.0465+ 3EITB3EITB/9/9-286.031+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9=0
7EITB7EITB/9/9+2EITC2EITC/9/9= -143.0155
At Joint C
MCBMCB+ MCDMCD= 0
286.031+ 4EITC4EITC/9/9+ 2EITB2EITB/9/9-286.031+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9=0
2EITB2EITB/9/9+ 8EITC8EITC/9/9+ 2EITD2EITD/9/9=0
At Joint D
MDCMDC+ MDEMDE= 0
286.01+4EITd4EITd/9/9+2EITc2EITc/9-286.031+4EITd4EITd/9/9-143.0155-EITdEITd/9/9=0
2EITC2EITC/9/9+7EITD7EITD/9/9= 143.0155
Solving the above Three equations we get,
EITDEITD= 183.8770
EITCEITC= 0
EITBEITB= -183.8770
EITEEITE= -735.5058
EITAEITA= 735.5058
Substituting the above values we get:
MABMAB=0
MBAMBA= 367.754 KN-m
MBCMBC= -367.754 KN-m
MCBMCB= 245.169 KN-m
MCDMCD= -245.169 KN-m
MDCMDC= 367.757 KN-m
MDEMDE= -367.757 KN-m
MEDMED= 0
2. With Load Patterning
Thickness of slab= 0.120m
Self weight of the slab= 25x0.120 =3 KN/m^2
Live load= 5 KN/m^2
Floor Finish= 3 KN/m^2
Total Load= 3+5+3= 11 KN/m^2
Load from Slab to Beam= 11x9x1.5x2
= 297 KN
Converting it to UDL load= 297/9 = 33 KN/m
Self Wight of the beam= 25x0.5x0.75
= 9.375 KN/m
Total Load on the beam= 33+9.375
= 42.375 KN/m
Calculation for Spans without Live Load
Thickness of slab= 0.120m
Self weight of the slab= 25x0.120 =3 KN/m^2
Floor Finish= 3 KN/m^2
Total Load= 3+3= 6 KN/m^2
Load from Slab to Beam= 6x9x1.5x2
= 162 KN
Converting it to UDL load= 162/9 = 18 KN/m
Self Weight of the beam= 25x0.5x0.75
= 9.375 KN/m
Total Load on the beam= 18+9.375
= 27.375 KN/m
Calculation For Maximum Bending Moment at D
We calculate the Fixed End Moments for New Loading
MFABMFAB= −wL^2/12 = −42.375⋅9^2/12= -286.031 KN-m
MFBAMFBA= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFBCMFBC= −wL^2/12= −27.375⋅9^2/12= -184.78 KN-m
MFCBMFCB= wL^2/12= 27.375⋅9^2/12= 184.78 KN-m
MFCDMFCD=−wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFDCMFDC= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFDEMFDE= −wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFEDMFED= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
Calculation For the Support End Moments
MABMAB= MFABMFAB+4EITa4EITa/L/L+ 2EITb2EITb/L/L
0= -286.031+ 4EITa4EITa/9/9+ 2EITb2EITb/9/9
286.031- 2EITb2EITb/9/9= 4EITa4EITa/9/9
2EITa2EITa/9/9= 143.0155 - EITbEITb/9/9
MBAMBA= MFBAMFBA+ 4EITB4EITB/L/L+ 2EITA2EITA/L/L
MBAMBA= 286.031+ 4EITB4EITB/9/9+ 143.0155 - EITbEITb/9/9
MBAMBA= 429.0465+ 3EITB3EITB/9/9
MBCMBC= MFBCMFBC+ 4EITB4EITB/L/L+ 2EITC2EITC/L/L
MBCMBC= -184.78+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9
MCBMCB= MFCBMFCB+ 4EITC4EITC/L/L+ 2EITB2EITB/L/L
MCBMCB= 184.78+ 4EITC4EITC/9/9+ 2EITB2EITB/9/9
MCDMCD= MFCDMFCD+ 4EITC4EITC/L/L+ 2EITD2EITD/L/L
MCDMCD= -286.031+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9
MDCMDC= MFDCMFDC+ 4EITD4EITD/L/L+ 2EITC2EITC/L/L
MDCMDC= 286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9
MDEMDE= MFDEMFDE+ 4EITD4EITD/L/L+ 2EITE2EITE/L/L
MDEMDE= -286.031+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9
MEDMED= MFEDMFED+ 4EITE4EITE/L/L+ 2EITD2EITD/L/L
0= 286.031+ 4EITE4EITE/9/9+ 2EITD2EITD/9/9
2EITE2EITE/9/9= -143.0155−-EITundersetD/9`
At Joint B
MBAMBA+ MBCMBC= 0
429.0465+ 3EITB3EITB/9/9-184.78+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9=0
7EITB7EITB/9/9+ 2EITC2EITC/9/9= -244.266
At Joint C
MCBMCB+ MCDMCD= 0
184.78+ 4EITC4EITC/9/9+ 2EITB2EITB/9/9-286.031+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9=0
2EITB2EITB/9/9+ 8EITC8EITC/9/9+2EITD2EITD/9/9= 101.251
At Joint D
MDCMDC+ MDEMDE= 0
286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9-286.031+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9=0
286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9-286.031+ 4EITD4EITD/9/9-143.0155−-EITundersetD/9`=0
2EITC2EITC/9/9+ 7EITD7EITD/9/9= 143.0155
Solving the above three Joint Equations:
EITBEITB= -362.873
EITCEITC= 170.861
EITDEITD= 135.058
Now Substituting the above values we get
MDCMDC= 384.025 KN-m
MDEMDE= -384.025 KN-m
Calculation Max Bending Moment at B
Since the Loading and structure is symmetrical so we can say that
MBAMBA= 384.025 KN-m
MBCMBC=-384.025 KN-m
Calculation for Maximum Bending Moment At C
We calculate the Fixed End Moments for New Loading,
MFABMFAB= −wL^2/12= −27.375⋅9^2/12= -184.781 KN-m`
MFBAMFBA= wL^2/12= 42.375⋅9^2/12= 184.781 KN-m
MFBCMFBC= −wL^2/12= −42.375⋅9^2/12=-286.031 KN-m
MFCBMFCB= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFCDMFCD=−wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFDCMFDC= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFDEMFDE= −wL^2/12= −27.375⋅9^2/12=-184.781 KN-m
MFEDMFED= wL^2/12= 27.375⋅9^2/12= 184.781 KN-m
Calculation For Support End Moments:
MABMAB= MFABMFAB+4EITa4EITa/L/L+ 2EITb2EITb/L/L
0= -184.781+ 4EITa4EITa/9/9+ 2EITb2EITb/9/9
2EITa2EITa/9/9= 92.3905-EITbEITb/9/9
MBAMBA= MFBAMFBA+ 4EITB4EITB/L/L+ 2EITA2EITA/L/L
MBAMBA= 184.781+ 4EITB4EITB/9/9+ 92.3905-EITbEITb/9/9
MBAMBA= 277.172+3EITB3EITB/9/9
MBCMBC= MFBCMFBC+ 4EITB4EITB/L/L+ 2EITC2EITC/L/L
MBCMBC=-286.031+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9
MCBMCB= MFCBMFCB+ 4EITC4EITC/L/L+ 2EITB2EITB/L/L
MCBMCB= 286.031+4EITC4EITC/9/9+ 2EITB2EITB/9/9
MCDMCD= MFCDMFCD+ 4EITC4EITC/L/L+ 2EITD2EITD/L/L
MCDMCD= -286.031+4EITC4EITC/9/9+ 2EITD2EITD/9/9
MDCMDC= MFDCMFDC+ 4EITD4EITD/L/L+ 2EITC2EITC/L/L
MDCMDC= 286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9
MDEMDE= MFDEMFDE+ 4EITD4EITD/L/L+ 2EITE2EITE/L/L
MDEMDE= -184.781+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9
MEDMED= MFEDMFED+ 4EITE4EITE/L/L+ 2EITD2EITD/L/L
0= 184.781+ 4EITE4EITE/9/9+ 2EITD2EITD/9/9
2EITE2EITE/9/9= -92.3905`- EITDEITD/9/9=0
At Joint B
MBAMBA+ MBCMBC= 0
277.172+3EITB3EITB/9/9+ -286.031+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9=0
7EITB7EITB/9/9+ 2EITC2EITC/9/9= 8.859
At Joint C
MCBMCB+ MCDMCD= 0
286.031+4EITC4EITC/9/9+ 2EITB2EITB/9/9-286.031+4EITC4EITC/9/9+ 2EITD2EITD/9/9=0
2EITB2EITB/9/9+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9=0
At Joint D
MDCMDC+ MDEMDE= 0
286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9-184.781+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9=0
286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9-184.781+ 4EITD4EITD/9/9-92.3905`- EITDEITD/9/9=0
2EITC2EITC/9/9+ 7EITD7EITD/9/9= -8.86
Solving the above Joint Equations:
EITBEITB= 11.244
EITCEITC= 0.5109
EITDEITD=- 13.288
Substituting the above values we get
MCBMCB= 288.757 KN-m
MDEMDE= -288.757 KN-m
Calculation For Bending Moment at Mid Span of Beam
Calculation For Fixed End Moments
MFABMFAB= −wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFBAMFBA= wL^2/12 = 42.375⋅9^2/12= 286.031 KN-m
MFBCMFBC= −wL^2/12= −27.375⋅9^2/12= -184.78 KN-m
MFCBMFCB= wL^2/12= 27.375⋅9^2/12= 184.78 KN-m
MFCDMFCD=−wL^2/12= −42.375⋅9^2/12= -286.031 KN-m
MFDCMFDC= wL^2/12= 42.375⋅9^2/12= 286.031 KN-m
MFDEMFDE= −wL^2/12= −42.375⋅9^2/12= -184.781 KN-m
MFEDMFED= wL^2/12= 42.375⋅9^2/12= 184.781 KN-m
Calculation For Support End Moments
MABMAB= MFABMFAB+4EITa4EITa/L/L+ 2EITb2EITb/L/L
0= -286.031+ 4EITa4EITa/9/9+ 2EITb2EITb/9/9
286.031- 2EITb2EITb/9/9= 4EITa4EITa/9/9
2EITa2EITa/9/9= 143.0155 - EITbEITb/9/9
MBAMBA= MFBAMFBA+ 4EITB4EITB/L/L+ 2EITA2EITA/L/L
MBAMBA= 286.031+ 4EITB4EITB/9/9+ 143.0155 - EITbEITb/9/9
MBAMBA= 429.0465+ 3EITB3EITB/9/9
MBCMBC= MFBCMFBC+ 4EITB4EITB/L/L+ 2EITC2EITC/L/L
MBCMBC= -184.78+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9
MCBMCB= MFCBMFCB+ 4EITC4EITC/L/L+ 2EITB2EITB/L/L
MCBMCB= 184.78+ 4EITC4EITC/9/9+ 2EITB2EITB/9/9
MCDMCD= MFCDMFCD+ 4EITC4EITC/L/L+ 2EITD2EITD/L/L
MCDMCD= -286.031+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9
MDCMDC= MFDCMFDC+ 4EITD4EITD/L/L+ 2EITC2EITC/L/L
MDCMDC= 286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9
MDEMDE= MFDEMFDE+ 4EITD4EITD/L/L+ 2EITE2EITE/L/L
MDEMDE= -184.781+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9
MEDMED= MFEDMFED+ 4EITE4EITE/L/L+ 2EITD2EITD/L/L
0= 184.781+ 4EITE4EITE/9/9+ 2EITD2EITD/9/9
2EITE2EITE/9/9= -92.3905`- EITDEITD/9/9
At Joint B
MBAMBA+ MBCMBC= 0
429.0465+ 3EITB3EITB/9/9-184.78+ 4EITB4EITB/9/9+ 2EITC2EITC/9/9=0
7EITB7EITB/9/9+ 2EITC2EITC/9/9= -244.266
At Joint C
MCBMCB+ MCDMCD= 0
184.78+ 4EITC4EITC/9/9+ 2EITB2EITB/9/9-286.031+ 4EITC4EITC/9/9+ 2EITD2EITD/9/9=0
2EITB2EITB/9/9+ 8EITC8EITC/9/9+2EITD2EITD/9/9= 101.251
At Joint D
MDCMDC+ MDEMDE= 0
286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9-184.781+ 4EITD4EITD/9/9+ 2EITE2EITE/9/9=0
286.031+ 4EITD4EITD/9/9+ 2EITC2EITC/9/9-184.781+ 4EITD4EITD/9/9-92.3905 - EITDEITD/9=0/9=0
2EITC2EITC/9/9+ 7EITD7EITD/9/9= -8.8595
Solving the above Joint Equations
EITBEITB= -377.273
EITCEITC= 221.175
EITDEITD= -51.802
Calculation For Reaction at RARA
MBAMBA= 429.0465+ 3EITB3EITB/9/9
= 429.0465- 125.757
= 303.288 KN-m
Now taking Span AB
Taking Moment About Point B
RARAx9+303.288-42.375x9x4.5=0
RARA= 156.988 KN
Calculation for the Maximum Positive Bending Moment For Every Span
So the maximum Bending moment occurs where shear force is zero.
RARA- w⋅Xw⋅X=0
156.988-42.375⋅X⋅X=0
156.988=42.375⋅X⋅X
X=3.7m
M3.7mM3.7mfrom A
=RARAx(3.7)-42.375x3.7x(3.7)/2
= 156.988x(3.7)-42.375x3.7x(3.7)/2
= 290.798 KN-m
