Project 1

1.

AIM: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.

1. 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.
2. For the same beam, consider load patterning to determine maximum positive and negative moments in each span.

SOLUTION:

Given

• thickness of the floor=120mm or 0.12m
• concrete floor unit weight=25KN/m^3
• floor finish=3KN/m^3
• Live load=5KN/m^3
• E=10Gpa or 10*10^6=10^7 KN/m^2
• I=bd^3/12
• = 0.5*0.75^3/12
• =0.017578125m^4
• EI=175781.25KNm^2

Determine the loading on the beam

load from the slab

For only dead load	For dead load+ live load
Unit weight of concrete=25KN/m^3 Thickness =0.12m =25*0.12=3KN/m^2 total dead load from slab= 3+3KN/m^2(floor finish) =6KN/m^2 since it is a one way slab = 6*Ly(distance bet. beams in m) =6*3=18KN/m D.L of beam = 25KN/^3 *(0.5m*0.75m) =9.375KN/m Total D.L of beam and slab= 18KN/m+9.375KN/m =27.375KN/m	Total D.L of slab= 6KN/m^2 Live load on slab= 5KN/m^2 Total load(DL+LL)from slab=11KN/^2 load on beam =11KN/m^2* Ly =11*3=33KN/m D.L of beam =9.375KN/m Total DL+LL on beam= 33KN/m+9.375KN/m =42.375KN/m

QUESTION 2:

To determine maximum +ve and -ve moment in a span we have to put the live load according to the condition

1. to get maximum +ve moment in each span

• We take BC span to calculate maximum moment
• we Draw theILD for maximum moment in span BC

• To get the maximum +ve moment, we have to load the span we want maximum and the alternate span
• Upto 2 alternate span is enough as more span consideration may cause error in calculation.
• The given below is the loading according to the above condition.

• since Live load considers the load patterning on the beams therefore the above drawing

STEP1- FIXED END MOMENT

• FEM'AB=-PL^2/12
• =-27.375*9^2/12
• =-184.78KNm //FEM'CD=-184.78KNm
• FEM'BA=+PL^2/12
• =+27.375*9^2/12
• =+184.78KNm //FEM'DC=+184.78KNm
• FEM'BC=-PL^2/12
• =-42.375*9^2/12
• =-286.03KNm //FEM'DE-286.03KNm
• FEM'CB=+PL^2/12
• =+42.375*9^2/12
• =+286.03KNm //FEM'ED=+286.03KNm

STEP2- SLOPE DEFLECTION EQUATION

• M'AB= FEM'AB+4EIθA/L+2EIθB/L
• =-184.78+(4/9)*175781.25*θA+(2/9)*175781.25*θB
• =78125θA+39062.5θB-184.78
• M'BA= FEM'BA+4EIθB/L+2EIθA/L
• =+184.78+(4/9)*175781.25*θB+(2/9)*175781.25*θA
• =+39062.5θA+78125θB+184.78
• M'BC= FEM'BC+4EIθB/L+2EIθC/L
• =-286.03+(4/9)*175781.25*θB+(2/9)*175781.25*θC
• =+78125θB+39062.5θC-286.03
• M'CB= FEM'CB+4EIθC/L+2EIθB/L
• =+286.03+(4/9)*175781.25*θC+(2/9)*175781.25*θB
• =78125θC+39062.5θB+286.03
• M'CD= FEM'CD+4EIθC/L+2EIθD/L
• =-184.78+(4/9)*175781.25*θC+(2/9)*175781.25*θD
• =78125θC+39062.5θD-184.78
• M'DC= FEM'DC+4EIθD/L+2EIθC/L
• =+184.78+(4/9)*175781.25*θD+(2/9)*175781.25*θC
• =78125θD+39062.5θC+184.78
• M'DE= FEM'DC+4EIθD/L+2EIθC/L
• =-286.03+(4/9)*175781.25*θD+(2/9)*175781.25*θE
• =78125θD+39062.5θE-286.03
• M'ED= FEM'ED+4EIθE/L+2EIθD/L
• =+286.03+(4/9)*175781.25*θD+(2/9)*175781.25*θE
• =78125θE+39062.5θD+286.03

STEP3- Joint Equilibrium Equation

• ∑M'A
• =78125θA+39062.5θB=184.78 ------------------(i)
• ∑M'B=M'BA+M'BC
• =(39062.5θA+78125θB+184.78)+(78125θB+39062.5θC-286.03)
• =39062.5θA+156250θB+39062.5θC=101.25 -----------------(ii)
• ∑M'C=M'CB+M'CD
• =(78125θC+39062.5θB+286.03)+(78125θC+39062.5θD-184.78)
• =39062.5θB+156250θC+39062.5θD=-101.25 ------------------(iii)
• ∑M'D=M'DC+M'DE
• =(78125θD+39062.5θC+184.78)+(78125θD+39062.5θE-286.03)
• =39062.5θC+156250θD+39062.5θE=101.25 ------------------(iv)
• ∑M'E
• =78125θE+39062.5θD=-286.03 --------------------(v)

STEP4- FINDING UNKNOWNS

//from the elemination method we find the values of the equations

• θA=2.14*10^-3 rad
• θB=0.46*10^-3 rad
• θC=-1.29*10^-3 rad
• θD=2.15*10^-3 rad
• θE=-4.74*10^-3 rad

On putting the values of Theta on the equations we find that

• M'AB=0 0.37KNm
• M'BA=304.70KNm
• M'BC=-304.70KNm
• M'CB=203.22KNm
• M'CD=-203.22KNm
• M'DC=302.35KNm
• M'DE=-302.35KNm
• M'ED=0

STEP5-SHEAR FORCE AND BMD

• span AB
• ∑M'B=Ra*9-27.375*9*4.5+304.70
• =Ra=89.33KN
• ∑V=0
• =Ra+Rb=27.375*9
• =Rb=157.045KN
• span BC
• ∑M'C=Rb1*9-42.375*9*4.5-304.70+203.22
• =Rb1=201.96KN
• ∑V=0
• =Rb1+Rc=42.375*9
• =Rc=179.41KN
• span CD
• ∑M'D=Rc1*9-27.375*9*4.5+302.35-203.22
• =Rc1=112.17KN
• ∑V=0
• =Rc1+Rd=27.375*9
• =Rc=134.20KN
• span DE
• ∑M'E=Rd1*9-42.375*9*4.5-302.35
• =Rd1=224.28KN
• ∑V=0
• =Rc1+Rd=27.375*9
• =Rc=157.09KN

Finding Maximum +ve BM in BC span

• from above shear force calculation we already know that
• Rb=201.96KN
• Now we find the point where SF is 0,zero.since BM is max where SF is zero.
• we take X distance from the Support B.
• we take the moment at that distance X
• =201.96*X-42.375*X^2/2 -----(i)
• On differentiation once w.r.t X we find that
• =201.96-42.375*X=0
• =>X=4.76m //from the B support
• On putting the value of X on equation (i) we get the max BMD
• =201.96*4.76-42.375*4.76^2/2

RESULT:

• Max B.M'BC=481.27KNm

//Similarly the BM max will be same for the other spans with the respective loading pattern

b.TO FIND MAXIMUM NEGATIVE MOMENT IN A SPAN

• Below shown is the ILD for maximum negative moment

• below shown is the maximum negative moment loading pattern.
• the moment will be calculated at the BC span though, since it gets the Hogging moment there.

• since Live load considers the load patterning on the beams therefore the above drawing

STEP1- FIXED END MOMENT

• FEM'AB=-PL^2/12
• =-42.375*9^2/12
• =-286.03KNm //FEM'CD=-286.03KNm
• FEM'BA=+PL^2/12
• =+42.375*9^2/12
• =+286.03KNm //FEM'DC=+286.03KNm
• FEM'BC=-PL^2/12
• =-27.375*9^2/12
• =-184.78KNm //FEM'DE-184.78KNm
• FEM'CB=+PL^2/12
• =+27.375*9^2/12
• =+184.78KNm //FEM'ED=+184.78KNm

STEP2- SLOPE DEFLECTION EQUATION

• M'AB= FEM'AB+4EIθA/L+2EIθB/L
• =-286.03+(4/9)*175781.25*θA+(2/9)*175781.25*θB
• =78125θA+39062.5θB-286.03
• M'BA= FEM'BA+4EIθB/L+2EIθA/L
• =+286.03+(4/9)*175781.25*θB+(2/9)*175781.25*θA
• =+39062.5θA+78125θB+286.03
• M'BC= FEM'BC+4EIθB/L+2EIθC/L
• =-184.78+(4/9)*175781.25*θB+(2/9)*175781.25*θC
• =+78125θB+39062.5θC-184.78
• M'CB= FEM'CB+4EIθC/L+2EIθB/L
• =+184.78+(4/9)*175781.25*θC+(2/9)*175781.25*θB
• =78125θC+39062.5θB+184.78
• M'CD= FEM'CD+4EIθC/L+2EIθD/L
• =-286.03+(4/9)*175781.25*θC+(2/9)*175781.25*θD
• =78125θC+39062.5θD-286.03
• M'DC= FEM'DC+4EIθD/L+2EIθC/L
• =+286.03+(4/9)*175781.25*θD+(2/9)*175781.25*θC
• =78125θD+39062.5θC+286.03
• M'DE= FEM'DC+4EIθD/L+2EIθE/L
• =-184.78+(4/9)*175781.25*θD+(2/9)*175781.25*θE
• =78125θD+39062.5θE-184.78
• M'ED= FEM'ED+4EIθE/L+2EIθD/L
• =+184.78+(4/9)*175781.25*θD+(2/9)*175781.25*θE
• =78125θE+39062.5θD+184.78

STEP3- Joint Equilibrium Equation

• ∑M'A
• =78125θA+39062.5θB=286.03 ------------------(i)
• ∑M'B=M'BA+M'BC
• =(39062.5θA+78125θB+286.03)+(78125θB+39062.5θC-184.78)
• =39062.5θA+156250θB+39062.5θC=-101.25 -----------------(ii)
• ∑M'C=M'CB+M'CD
• =(78125θC+39062.5θB+184.78)+(78125θC+39062.5θD-286.03)
• =39062.5θB+156250θC+39062.5θD=101.25 ------------------(iii)
• ∑M'D=M'DC+M'DE
• =(78125θD+39062.5θC+286.03)+(78125θD+39062.5θE-184.78)
• =39062.5θC+156250θD+39062.5θE=-101.25 ------------------(iv)
• ∑M'E
• =78125θE+39062.5θD=-184.78 --------------------(v)

STEP3- FINDING UNKNOWNS

• from the above equations we find the unknowns by elimination method
• we get the following Theta values
• θA= 4.7*10^-3 rad
• θB= -2.1*10^-3 rad
• θC= 1.3*10^-3 rad
• θD= -0.44*10^-3 rad
• θE= -2.14*10^-3 rad

By putting the value of theta on the given equation on step2 we find the moment

• M'AB=0KNm
• M'BA=305.56KNm
• M'BC=-305.56KNm
• M'CB=204.31KNm
• M'CD=-204.65KNm
• M'DC=302.75KNm
• M'DE=-302.75KNm
• M'ED=0KNm

STEP4-MAXIMUM -ve BM AT BC SPAN

• first we find the shear force on the support
• Let ∑M'C=Rb*9-305.56+204.65-27.375*9*4.5
• =>Rb=134.39KN
• to find the max -ve mom. we have to fing th point weher SF=0
• SInce BM is max at point where SF=0
• we take moment at X distance from B
• Now => Rb*X-27.375*X*X/2=0 -----(i)
• On differentiatig the above equation we get
• =>134.39-27.375*X=0
• =>X=5.09m //SF is zero at 5.09m from the support B
• Put the value on eq(i) to get max BM at the span

RESULT:

• MAX -ve BM => 134.39*5.09-27.375*5.09^2/2
• =>329.42KNm