Project 1
Question:-
Given below is a floor plan of a building. The red lines running East-west are beams supported by the girders spanning North-South between columns. The floor system is 120 mm thick normal weight concrete floor. The floor finish and mech loads on the floor is 3 kPa. The live load on the floor is 5 kPa. All beams have cross section 500 mmx750 mm. The beams are equally spaced. The rectangles in cyan color are columns. Modulus of elasticity of concrete is 10 GPa.
- Determine the loading on the beam indicated, analyze the beams and draw bending moment and shear force diagram. Use slope deflection method. Do not consider any load patterning. Also, do not consider any girder deflections for this analysis purpose.
- For the same beam, consider load patterning to determine maximum positive and negative moments in each span.
Ans=
Aim:- To analysis the beam using slope-deflection method without load patterning
Case-1
i) Self weight of slab = 0.12*25 = 3kn/m
ii) Self weight of beam = 0.5*0.75*25 = 9.375Kn/m
iii) Floor finish = 3kn/m
iv) Live load = 5kn/m
E= 10 Gpa = 10 X 10^6 KN/m2 = 10^7 KN/m2
I= bd3/12 = (0.5 X 0.75^3)/12 = 0.0175
EI= (10^7) X 0.0175 = 175000
Total weight = self wt + live load + floor finish
= 3 + 5 + 3 = 11kN/m
Factored Total Load (W)= 1.5 X 11= 16.5 KN/m
Therefore , Load on beam ,
i) Load on beam = Total weight * Tributary width
So , Let us give the names to all the beams in order as B1,B2,B3,B4 starting from left side
i) Load on B1 = W * tributary width+ self weight of beam
= 16.5* 0.5(3+3)+9.375
= 58.875 Kn/m
ii) Load on B2 = W * tributary width + self wt of beam
= 16.5* 0.5(3+3)+9.375
= 58.875 Kn/m
iii) Load on B3 = W * tributary width + self wt of beam
= 16.5* 0.5(3+3)+9.375
= 58.875 Kn/m
iv) Load on B4 = W * tributary width + self wt of beam
= 16.5 * 0.5(3+3)+9.375
= 58.875 kN/m
The figure is as shown below for 1st Case - Considered a Continuous beam with simply supported
Step 1 ) Fixed End Moments
For Span AB -
Mab = -wL^2/12 = -58.875*9*9/12 = -397.4 Knm
Mba = wL^2/12 = 58.875*9*9/12 = +397.4 Knm
For Span BC -
Mbc = -wL^2/12 = -58.875*9*9/12 = -397.4 Knm
Mcb = wL^2/12 = 58.875*81/12 = +397.4 Knm
For Span CD -
Mcd = -wL^2/12 = -58.875*81/12 = -397.4 Knm
Mde = wL^2/12 = 58.875*81/12 = +397.4 Knm
For Span DE -
Mde = -wL^2/12 = -58.875*81/12 = -397.4 Knm
Med = wL^2/12 = 58.875*81/12 = +397.4 Knm
Step 2) End Moment Equation -
For Span AB -
MAB = MFab + 4EI/L * Ta + 2EI/L * Tb
= -397.4 + [(4 X 175000)/9] Ta + [(2 X 175000)/9] Tb
= -397.4 + 77778 Ta + 38889 Tb
MBA = MFba + 4EI*Tb/L + 2EI*Ta/L
= 397.4 + [(4 X 175000)/9] Tb + [(2 X 175000)/9] Ta
= 397.4 + 77778 Tb + 38889 Ta
For Span BC -
MBC = MFbc + 4EI*Tb/L + 2EI*Tc/L
= -397.4 + [(4 X 175000)/9] Tb + [(2 X 175000)/9] Tc
= -397.4 + 77778 Tb + 38889 Tc
MCB = MFcb + 4EI*Tc/L + 2EI*Tb/L
= 397.4 + [(4 X 175000)/9] Tc + [(2 X 175000)/9] Tb
= 397.4 + 77778 Tc + 38889 Tb
For Span CD -
MCD = MFcd + 4EI*Tc/L + 2EI*Td/L
= -397.4 + [(4 X 175000)/9] Tc + [(2 X 175000)/9] Td
= -397.4 + 77778 Tc + 38889 Td
MDC = MFdc + 4EI*Td/L + 2EI*Tc/L
= 397.4 + [(4 X 175000)/9] Td + [(2 X 175000)/9] Tc
= 397.4 + 77778 Td + 38889 Tc
FOR Span DE -
MDE = MFde + 4EI*Td/L + 2EI*Te/L
= -397.4 + [(4 X 175000)/9] Td + [(2 X 175000)/9] Te
= -397.4 + 77778 Td + 38889 Te
MED = MFed + 4EI*Te/L + 2EI*Td/L
= 397.4 + [(4 X 175000)/9] Te + [(2 X 175000)/9] Td
= 397.4 + 77778 Te + 38889 Td
Step 3 ) Using Moment Equilibrium eqn
i) Mab = 0
-397.4 + 77778 Ta + 38889 Tb = 0
77778*Ta + 38889*Tb = 397.4 ............................................................(1)
Considering the equilibrium at B, Mb = 0
ii) Mb = Mba + Mbc
-397.4 + 77778 Tb + 38889 Ta + 397.4 + 77778 Tb + 38889 Tc = 0
77778Tb + 77778Tb + 38889 Ta + 38889 Tc = 0
155556Tb + 38889Ta + 38889Tc = 0................................................................(2)
Considering the equilibrium at C, Mc = 0
iii) Mc = Mcb + Mcd
397.4 + 77778 Tc + 38889 Tb - 397.4 + 77778 Tc + 38889 Td = 0
77778*Tc + 38889*Tb + 77778*Tc + 38889*Td = 0
155556*Tc + 38889*Tb + 38889*Td = 0 ........................................................................(3)
Considering equilibrium at D, Md = 0
iii) Md = Mdc + Mde
397.4 + 77778 Td + 38889 Tc - 397.4 + 77778 Td + 38889 Te = 0
77778*Td + 38889*Tc + 77778*Td + 38889*Te = 0
= 155556*Td + 38889*Tc + 38889*Te = 0 .....................................................................(4)
Considering the equilibrium at E, Med = 0
iv) Me = 0
397.4 + 77778*Te + 38889*Td = 0
77778*Te + 38889*Td = -397.4 ................................................................................(5)
Solving above equation
we get ,
Ta = 0.00584
Tb = -0.00146
Tc = 0.00
Td = -0.00146
Te = -0.00584
Step 4) Substitute the above unknown values in moment equations and get the moment values.
i) MAB = MFab + 4EI*Ta/L + 2EI*Tb/L
= -397.4 + 77778 Ta + 38889 Tb
= -397.4 + (77778 X 0.00584)+ (38889 X -0.00146)
= -397.4 + 453.44 - 56.73
= 0 KN-m
MAB = 0.00
ii) MBA = MFba + 4EI*Tb/L + 2EI*Ta/L
= 397.4 + 77778 Tb + 38889 Ta
= 397.4 + [77778 X (-0.00146)] + (38889 X 0.00584)
= 510.64 KN-m
iii) MBC = MFbc + 4EI*Tb/L + 2EI*Tc/L
= -397.4 + 77778 Tb + 38889 Tc
= -397.4 + [77778 X (-0.00146)] + [38889 X (0.00)]
= -397.4 - 113.56
= - 510.96 KN-m
iii) MCB = MFcb + 4EI*Tc/L + 2EI*Tb/L
= 397.4 + 77778 Tc + 38889 Tb
= 397.4 + [77778 X (0.00)] + [38889 X (-0.00146)]
= 340.63 KN-m
MCB = 340.63 Kn-m
iv) MCD = MFcd + 4EI*Tc/L + 2EI*Td/L
= -397.4 + 77778 Tc + 38889 Td
= -397.4 + [ 77778 X (0.00)] + [ 38889 X (0.00146)]
= - 340.63 KN-m
iv) MDC = MFdc + 4EI*Td/L + 2EI*Tc/L
= 397.4 + 77778 Td + 38889 Tc
= 397.4 + [77778 X (0.00146) + [38889 X (0.00)]
= 510.96 KN-m
MDC = 367.7Knm
iv) MDE = MFde + 4EI*Td + 2EI*Te/L
= -397.4 + 77778 Td + 38889 Te
= -397.4 + [77778 X (0.00146)] + [38889 X (-0.00584)]
= -397.4 + 113.56 - 227.12
= - 510.96 KN-m
v) MED = MFed + 4EI*Te/L + 2EI*Td/L
= 397.4 + 77778 Te + 38889 Td
= 397.4 + [77778 X (-0.00584)] + [38889 X (0.00146)]
= 397.4 - 454.22 + 56.78
= 0 KN-m
MED = 0 KNm
Step 5) To determine the shear forces , consider each span individually,
i) Consider span AB,taking moment about A,
We know, Va + Vb1 = (58.875X9) =529.875
Now I will take the moment about A,
(-Vb1 * 9) + 58.875*9*4.5 + Mab + Mba = 0
-9Vb1 + 2384.43 + 0 + 510.64 = 0
-9Vb1 = - 2895.08
Vb1 = 321.68 KN
We Know,
Va + Vb1 = 529.875
Va + 321.68 = 529.875
Va = 208.2 KN
ii) Consider span BC,taking moment about B,
We Know,
Vb2 + Vc1 = 58.875 X 9 = 529.875
Taking Moment about B,
(- Vc1 X 9) + 58.875*9*4.5 + MBC + MCB = 0
-9Vc1 +2384.43 -510.96 + 340.63 = 0
Vc1 = 246.02 KN
We Know,
Vb2 + Vc1 = 529.875
Vb2 = 283.87 KN
iii) Consider span CD,taking moment about C,
We Know,
Vc2 + Vd1 = (58.875 X 9) = 529.875
Taking moment about C we get,
(- Vd1 X 9) + 58.875*9*4.5 + MCD + MDC = 0
-9Vd1 + 2384.44 - 340.63 + 510.96 = 0
Vd1 = 283.86 KN
We Know,
Vc2 + Vd1 = 529.875
Vc2 = 529.875 - 283.86 = 246.02 KN
Vc2= 246.02 KN
iv) Consider span DE,taking moment about D,
We Know,
Vd2 + Ve = (58.875 X 9) = 529.875
Taking moment about D we get,
(- Ve X 9) + 58.875*9*4.5 + MDE + MED = 0
-9Ve + 2384.44 - 510.96 + 0 = 0
Ve = 208.16 KN
We Know,
Vd2 + Ve = 529.875
Vd2 = 529.875 - 208.16 = 321.71 KN
Vd2= 321.71 KN
Summarization of all vertical reaction Forces:-
Va = 208.2 KN
Vb = Vb1 + Vb2 = 321.68 + 283.87 = 605.55 KN
Vc = Vc1 + Vc2 = 246.02 + 246.02 = 492.04 KN
Vd = Vd1 + Vd2 = 283.86 + 321.71 = 605.57 KN
Ve = 208.2 KN
Now I Will Draw the Shear Force Diagram:-
SFD
BMD
If I want to calculate Bending moment at every location then it will be like this
Case-2
Aim:- To analysis the beam using slope-deflection method with load patterning
For the same beam, consider load patterning to determine maximum positive and negative moments in each span:-
As per the given question I will consider DL + FF for first and third span and I will consider DL+FF+LL for second and fourth span.
i) Self weight of slab = 0.12*25 = 3kn/m
ii) Self weight of beam = 0.5*0.75*25 = 9.375Kn/m
iii) Floor finish = 3kn/m
iv) Live load = 5kn/m
E= 10 Gpa = 10 X 10^6 KN/m2 = 10^7 KN/m2
I= bd3/12 = (0.5 X 0.75^3)/12 = 0.0175
EI= (10^7) X 0.0175 = 175000
Load combination for 1st & 3rd span(B1&B3) = self weight + Floor finish = 3 + 3 = 6 KN/m
So factored load for 1st & 3rd span (W1) = (6 X 1.5) = 9 KN/m
Load combination for 2nd & 4th span (B2&B4) = self weight + live load + floor finish
= 3 + 5 + 3 = 11kN/m
So, Factored load for 2nd & 4th span (W2)= 1.5 X 11= 16.5 KN/m
Therefore , Load on beam ,
i) Load on beam = Total weight * Tributary width
So , Let us give the names to all the beams in order as B1,B2,B3,B4 starting from left side
i) Load on B1 = W1 * tributary width+ self weight of beam
= 9* 0.5(3+3)+9.375
= 36.375 Kn/m
ii) Load on B2 = W2 * tributary width + self wt of beam
= 16.5* 0.5(3+3)+9.375
= 58.875 Kn/m
iii) Load on B3 = W1 * tributary width + self wt of beam
= 9* 0.5(3+3)+9.375
= 36.375 Kn/m
iv) Load on B4 = W2 * tributary width + self wt of beam
= 16.5 * 0.5(3+3)+9.375
= 58.875 kN/m
The figure is as shown below for 1st Case - Considered a Continuous beam with simply supported
Step 1 ) Fixed End Moments
For Span AB -
Mab = -wL^2/12 = -36.375*9*9/12 = -245.53 Knm
Mba = wL^2/12 = 36.375*9*9/12 = +245.53 Knm
For Span BC -
Mbc = -wL^2/12 = -58.875*9*9/12 = -397.4 Knm
Mcb = wL^2/12 = 58.875*81/12 = +397.4 Knm
For Span CD -
Mcd = -wL^2/12 = -36.375*9*9/12 = -245.53 Knm
Mde = wL^2/12 = 36.375*9*9/12 = +245.53 Knm
For Span DE -
Mde = -wL^2/12 = -58.875*81/12 = -397.4 Knm
Med = wL^2/12 = 58.875*81/12 = +397.4 Knm
Step 2) End Moment Equation -
For Span AB -
MAB = MFab + 4EI/L * Ta + 2EI/L * Tb
= -245.53 + [(4 X 175000)/9] Ta + [(2 X 175000)/9] Tb
= -245.53 + 77778 Ta + 38889 Tb
MBA = MFba + 4EI*Tb/L + 2EI*Ta/L
= 245.53 + [(4 X 175000)/9] Tb + [(2 X 175000)/9] Ta
= 245.53 + 77778 Tb + 38889 Ta
For Span BC -
MBC = MFbc + 4EI*Tb/L + 2EI*Tc/L
= -397.4 + [(4 X 175000)/9] Tb + [(2 X 175000)/9] Tc
= -397.4 + 77778 Tb + 38889 Tc
MCB = MFcb + 4EI*Tc/L + 2EI*Tb/L
= 397.4 + [(4 X 175000)/9] Tc + [(2 X 175000)/9] Tb
= 397.4 + 77778 Tc + 38889 Tb
For Span CD -
MCD = MFcd + 4EI*Tc/L + 2EI*Td/L
= -245.53 + [(4 X 175000)/9] Tc + [(2 X 175000)/9] Td
= -245.53 + 77778 Tc + 38889 Td
MDC = MFdc + 4EI*Td/L + 2EI*Tc/L
= 245.53 + [(4 X 175000)/9] Td + [(2 X 175000)/9] Tc
= 245.53 + 77778 Td + 38889 Tc
FOR Span DE -
MDE = MFde + 4EI*Td/L + 2EI*Te/L
= -397.4 + [(4 X 175000)/9] Td + [(2 X 175000)/9] Te
= -397.4 + 77778 Td + 38889 Te
MED = MFed + 4EI*Te/L + 2EI*Td/L
= 397.4 + [(4 X 175000)/9] Te + [(2 X 175000)/9] Td
= 397.4 + 77778 Te + 38889 Td
Step 3 ) Using Moment Equilibrium eqn
i) Mab = 0
-245.53 + 77778 Ta + 38889 Tb = 0
77778*Ta + 38889*Tb = 245.53 ............................................................(1)
Considering the equilibrium at B, Mb = 0
ii) Mb = Mba + Mbc
245.53 + 77778 Tb + 38889 Ta - 397.4 + 77778 Tb + 38889 Tc = 0
77778Tb + 77778Tb + 38889 Ta + 38889 Tc = 151.87
155556Tb + 38889Ta + 38889Tc = 151.87 ................................................................(2)
Considering the equilibrium at C, Mc = 0
iii) Mc = Mcb + Mcd
397.4 + 77778 Tc + 38889 Tb - 245.53 + 77778 Tc + 38889 Td = 0
77778*Tc + 38889*Tb + 77778*Tc + 38889*Td = - 151.87
155556*Tc + 38889*Tb + 38889*Td = - 151.87 ........................................................................(3)
Considering equilibrium at D, Md = 0
iii) Md = Mdc + Mde
245.53 + 77778 Td + 38889 Tc - 397.4 + 77778 Td + 38889 Te = 0
77778*Td + 38889*Tc + 77778*Td + 38889*Te = 151.87
155556*Td + 38889*Tc + 38889*Te = 151.87 .....................................................................(4)
Considering the equilibrium at E, Med = 0
iv) Me = 0
397.4 + 77778*Te + 38889*Td = 0
77778*Te + 38889*Td = -397.4 ................................................................................(5)
Solving above equation
we get ,
Ta = 0.00277
Tb = 0.00077
Tc = -0.00195
Td = 0.00313
Te = -0.00332
Step 4) Substitute the above unknown values in moment equations and get the moment values.
i) MAB = MFab + 4EI*Ta/L + 2EI*Tb/L
= -245.53 + 77778 Ta + 38889 Tb
= -245.53 + (77778 X 0.00277)+ (38889 X 0.00077)
= -245.53 + 215.44 + 29.95
= 0 KN-m
MAB = 0.00
ii) MBA = MFba + 4EI*Tb/L + 2EI*Ta/L
= 245.53 + 77778 Tb + 38889 Ta
= 245.53 + [77778 X (0.00077)] + (38889 X 0.00277)
= 413.14 KN-m
iii) MBC = MFbc + 4EI*Tb/L + 2EI*Tc/L
= -397.4 + 77778 Tb + 38889 Tc
= -397.4 + [77778 X (0.00077)] + [38889 X (-0.00195)]
= -397.4 + 59.88 - 75.83
= - 413.14 KN-m
iii) MCB = MFcb + 4EI*Tc/L + 2EI*Tb/L
= 397.4 + 77778 Tc + 38889 Tb
= 397.4 + [77778 X (-0.00195)] + [38889 X (0.00077)]
= 397.4 - 151.66 + 29.95
= 275.69 KN-m
MCB = 275.69 Kn-m
iv) MCD = MFcd + 4EI*Tc/L + 2EI*Td/L
= -245.53 + 77778 Tc + 38889 Td
= -245.53 + [ 77778 X (-0.00195)] + [ 38889 X (0.00313)]
= - 245.53 - 151.66 + 121.73
= - 275.69 KN-m
iv) MDC = MFdc + 4EI*Td/L + 2EI*Tc/L
= 245.53 + 77778 Td + 38889 Tc
= 245.53 + [77778 X (0.00313) + [38889 X (-0.00195)]
= 245.53 + 113.56 - 75.84
= 283.25 KN-m
MDC = 283.25 Knm
iv) MDE = MFde + 4EI*Td + 2EI*Te/L
= -397.4 + 77778 Td + 38889 Te
= -397.4 + [77778 X (0.00313)] + [38889 X (-0.00332)]
= -397.4 + 243.44 - 129.2
= - 283.25 KN-m
v) MED = MFed + 4EI*Te/L + 2EI*Td/L
= 397.4 + 77778 Te + 38889 Td
= 397.4 + [77778 X (-0.00332)] + [38889 X (-0.00313)]
= 397.4 - 258.22 - 121.72
= 0 KN-m
MED = 0 KNm
Step 5) To determine the shear forces , consider each span individually,
i) Consider span AB,taking moment about A,
We know, Va + Vb1 = (36.375X9) = 327.375
Now I will take the moment about A,
(-Vb1 * 9) + 36.375*9*4.5 + Mab + Mba = 0
-9Vb1 + 1473.18 + 0 + 413.14 = 0
-9Vb1 = - 1886.32
Vb1 = 209.6 KN
We Know,
Va + Vb1 = 327.375
Va + 209.6 = 327.375
Va = 117.775 KN
ii) Consider span BC,taking moment about B,
We Know,
Vb2 + Vc1 = 58.875 X 9 = 529.875
Taking Moment about B,
(- Vc1 X 9) + 58.875*9*4.5 + MBC + MCB = 0
-9Vc1 +2384.43 - 413.14 + 275.69 = 0
Vc1 = 249.67 KN
We Know,
Vb2 + Vc1 = 529.875
Vb2 = 280.205 KN
iii) Consider span CD,taking moment about C,
We Know,
Vc2 + Vd1 = (36.375X9) = 327.375
Taking moment about C we get,
(- Vd1 X 9) + 36.375*9*4.5 + MCD + MDC = 0
-9Vd1 + 1473.18 - 275.69 + 283.25 = 0
Vd1 = 164.53 KN
We Know,
Vc2 + Vd1 = 327.375
Vc2 = 327.375 - 164.53 = 162.845 KN
Vc2= 162.845 KN
iv) Consider span DE,taking moment about D,
We Know,
Vd2 + Ve = (58.875 X 9) = 529.875
Taking moment about D we get,
(- Ve X 9) + 58.875*9*4.5 + MDE + MED = 0
-9Ve + 2384.44 - 283.25 + 0 = 0
Ve = 233.46 KN
We Know,
Vd2 + Ve = 529.875
Vd2 = 529.875 - 233.46 = 296.41 KN
Vd2= 296.41 KN
Summarization of all vertical reaction Forces:-
Va = 117.775 KN
Vb = Vb1 + Vb2 = 209.6 + 280.205 = 489.805 KN
Vc = Vc1 + Vc2 = 249.67 + 162.845 = 412.515 KN
Vd = Vd1 + Vd2 = 164.53 + 296.41 = 460.94 KN
Ve = 233.46 KN
Now I Will Draw the Shear Force Diagram:-
SFD
BMD
