Project 1
1)
Aim -To determine the loading on the beam indicated and analyze the beams and draw bending moment and shear force diagram.
Given Data -
i) The floor system is 120 mm thick normal weight concrete floor
ii) The floor finish and mech loads on the floor is 3 kPa = 3kN/m^2
iii) The live load on the floor is 5 kPa = 5kN/m^2
iv) All beams have cross section 500 mmx750 mm.
So, I =bd^3 /12 = 0.5*0.75 ^3 /12 =0.0176 m^4
v) Modulus of elasticity of concrete is 10 GPa. = 1*10^6 KPa
Procedure -
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
Total weight = self wt + live load + floor finish
= 3 + 5 + 3 = 11kN/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
= 11* 0.5(3+3)+9.375
= 42.375 Kn/m
ii) Load on B2 = W * tributary width + self wt of beam
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
iii) Load on B3 = W * tributary width + self wt of beam
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
iv) Load on B4 = W * tributary width + self wt of beam
= 11 * 0.5(3+3)+9.375
= 42.75 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 AC -
Mac = -wL^2/12 = -42.375*9*9/12 = -286.03Knm
Mca = wL^2/12 = 42.375*9*9/12 = +286.03Knm
For Span CD -
Mcd = -wL^2/12 = -42.375*9*9/12 = -286.03Knm
Mdc = wL^2/12 = 42.375*81/12 = +286.03Knm
For Span DE -
Mde = -wL^2/12 = -42.375*81/12 = -286.03Knm
Med = wL^2/12 = 42.375*81/12 = +286.03Knm
For Span EB -
Meb = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mbe = wL^2/12 = 42.375*81/12 = +286.03Knm
Step 2) End Moment Equation -
For Span AC -
MA = MFab + 4EI/L * Ta + 2EI/L * Tc
= -286.03 + 78125 Ta + 39062.5 Tc
MC = MFca + 4EI*Tc/L + 2EI*Ta/L
= 286.03 + 78125Tc + 39062.5Ta
For Span CD -
MC = MFcd + 4EI*Tc/L + 2EI*Td/L
= -286.03 + 78125Tc + 39062.5Td
MD = MFdc + 4EI*Td/L + 2EI*Tc/L
= 286.03 + 78125*Td + 39062.5*Tc
For Span DE -
MD = MFde + 4EI*Td/L + 2EI*Te/L
= -286.03 + 78125Td + 39062.5Te
ME = MFed + 4EI*Te/L + 2EI*Td/L
= 286.03 + 78125Te + 39062.5Td
FOR Span EB -
ME = MFeb + 4EI*Te/L + 2EI*Tb/L
= -286.03 + 78125Te + 39062.5Tb
MB = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.03 + 78125Tb + 39062.5Te
Step 3 ) Using Moment Equilibrium eqn
i) Ma = 0
-286.03 + 78125Ta + 39062.5Tc = 0 ............................................................(1)
Considering the equilibrium at C, Mc = 0
ii) Mc = Mcd + Mca
78125Tc + 78125Tc + 39062.5Ta + 39062.5Ta + 39062.5Td = 0
156250Tc + 78125Ta + 39062.5Td = 0................................................................(2)
Considering the equilibrium at D, Md = 0
iii) Md = Mdc + Mde
286.03 + 78125*Td + 39062.5*Tc - 286.03 + 78125Td + 39062.5Te = 0
156250Td + 39062.5Tc + 39062.5Te = 0 ........................................................................(3)
Considering equilibrium at E, Me = 0
iii) Me = Med + Meb
= 286.03 + 78125Te + 39062.5Td - 286.03 + 78125Te + 39062.5Tb = 0
= 156250Te + 39062.5Td + 39062.5Tb = 0 .....................................................................(4)
Considering the equilibrium at B, Mbe = 0
iv) Mb = 0
286.03 + 78125Tb + 39062.5Te = 0 ................................................................................(5)
Solving above equation
we get ,
Tc = 1.046 x 10^-3 rad
Tb =4.18*10^-3 rad
Te = 1.046*10^-3 rad
Td = -1.84*10^-6 rad
Ta = 4.0*10^-3 rad
Step 4) Substitute the above unknown values in moment equations and get the moment values.
i) MA = MFac + 4EI*Ta/L + 2EI*Tc/L
= -286.03 + 78125Ta + 39062.5Tc
MA = -0.00
ii) MC = MFca + 4EI*Tc/L + 2EI*Ta/L
= 286.03 + 78125Tc + 39062.5Ta
= 367.7Knm
iii) MD = MFdc + 4EI*Td/L + 2EI*Tc/L
= 286.03 + 78125Td + 39062.5Tc
MD = 245.17Knm
iv) ME = MFed + 4EI*Te/L + 2EI*Td/L
= 286.03 + 78125Te + 39062.5Td
ME = 367.7Knm
v) MB = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.03 + 78125Tc + 39062.5Ta
MB = 0 KNm
Step 5) To determine the shear forces , consider each span individually,
i) Consider span AC,taking moment about A,
Hc * 9 + 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
ii) 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
iii) 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
iv) Consider span EB,taking moment about E,
9Hb - 367.7 - 42.375*9*4.5 = 0
Hb = - 231.535Kn
He = 149.84Kn
SFD
BMD
2)
Aim - To determine the load patterning to determine maximum positive and negative moments in each span.
Procedure -
i) Maximum bending moment at C by loading span
ii) 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)
Step 1) - FIXED END MOMENTS -
FOR AC -
MFac = -wL^2/12=-286.01kNm
MFca = wL^2/12=286.01kNm
FOR CD -
MFcd = -wL^2/12=-286.01kNm
MFdc = wL^2/12=286.01kNm
FOR DE -
MFeb = -wL^2/12=-286.01kNm
MFbe = wL^2/12=286.01kNm
FOR EB -
MFde = -121.5kNm
MFed = 121.5kNm
Step 2) End Moment Equation
FOR SPAN AC:
Mac = MFac+4EI*Ta/L + 2EI*Tc/L
= -286.01+0.4EI*Ta+0.2EI*Tc
Mca = MFca + 4EI*Tc/L + 2EI*Ta/L
= 286.01 + 0.4EI*Tc + 0.2EI*Ta
FOR SPAN CD:
Mcd = MFcd + 4EI*Tc/L + 2EI*Td/L
= -286.01 + 0.4EI*Tc + 0.2EI*Td
Mdc = MFdc + 4EI*Td/L + 2EI*Tc/L
= 286.01 + 0.4EI*Td + 0.2EI*Tc
FOR SPAN DE:
Mde = MFde + 4EITd/L + 2EITe/L
= -121.5 + 0.4EITd + 0.2EITe
Med = MFed + 4EITe/L + 2EITd/L
= 121.5 + 0.4EITe + 0.2EITd
FOR SPAN EB:
Meb = MFeb + 4EI*Te/L + 2EI*Tb/L
= -286.01 + 0.4EI*Te + 2EI*Tb
Mbe = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.01 + 0.4EI*Tb + 0.2EI*Te
from equation 1,
0.4EI*Ta = 286.01 - 0.2EI*Tc
0.2EI*Ta = 143 - 0.1EI*Tc --------- 1
Substitute the above value:
143 - 0.1EI*Tc + 0.8EI*Tc + 0.2EI*Td = 0
0.7EI*Tc + 0.2EI*Td + 143 = 0
0.7EI*Tc = -143 - 0.2EI*Td
0.2EI*Tc = -40.85 - 0.0571EI*Td ------- 2
Substitute the above value :
Md = 0
-40.85 - 0.0571EI*Td + 0.8EI*Td + 0.2EI*Te + 164.51 = 0
0.74EI*Td + 0.2EI*Te + 123.66 = 0 ------- 11
0.2EI*Te = -0.74EI*Td - 123.66 * 2
0.4EI*Te = -1.48EI*Td - 247.32 * 2
0.8EI*Te = -2.96EI*Td - 494.64
Substitute the above value in summation of E,
Me = 0
0.2EI*Tb + 0.2EI*Td + 0.8EI*Te - 164.51 = 0
0.2EI*Tb + 0.2EI*Td - 2.96EI*Td - 494.64 - 164.51 = 0
0.2EI*Tb - 2.76EI*Td - 659.15 * 2
0.4EI*Tb = 5.52EI*Td + 1318.3 -------- 12
Substitute the above value in summation of moment at B,
Mb = 0
5.52EI*Td + 1318.3 + 286.01 - 0.74EI*Td - 123.66 = 0
4.78EI*Td = -1480.65
EI*Td = -309.75
Step 3) Substitute the value of EI*Td in other equations and derive unknowns as follows.
EI*Tb = 978.8
EI*Tc = -115.81
EI*Ta = 772.93
EI*T = 572.35
Step 4) Substitute the values of these unknowns in all the moment equations and derive the moment values.
Ma = 0
Mca = 394.27 kNM
Mcd = -394.87 kNM
Mdc = 198.948 kNM
Mde = -137.34 kNM
Med = 288.49 kNM
MEb = -288.49 kNM
Mbe = 8.96 kNM
B) Similarly find out the maximum moment at E,
i) Let us findout the maximum bending moment at E by loading span
ii) 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).
Step 1) Fixed End Moments -
FOR AC -
MFac = -wL^2/12=-286.01kNm
MFca = 286.01Knm
FOR CD -
MFcd = -121.5kNm
MFdc = 121.5kNm
FOR DE -
MFde = -286.01kNm
MFed = 286.01knm
FOR EB -
MFeb = -286.01kNm
MFbe = 286.01kNm
Step 2) END MOMENTS Equation -
Mac = MFac+4EI*Ta/L + 2EI*Tc/L
= -286.01+0.4EI*Ta+0.2EI*Tc
Mca = MFca + 4EI*Tc/L + 2EI*Ta/L
= 286.01 + 0.4EI*Tc + 0.2EI*Ta
Mcd = MFcd + 4EI*Tc/L + 2EI**Td/L
= -121.5 + 0.4EI*Tc + 0.2EITd
Mdc = MFdc + 4EI*Td/L + 2EI*Tc/L
= 121.5 + 0.4EI*Td + 0.2EI*Tc
Mde = MFde + 4EI*Td/L + 2EI*Te/L
= -286.01 + 0.4EI*Td + 0.2EI*Te
Med = MFed + 4EI*Te/L + 2EI*Td/L
= 286.01 + 0.4EI*Te + 0.2EI*Td
Meb = MFeb + 4EI*Te/L + 2EI*Tb/L
= -286.01 + 0.4EI*Te + 0.2EI*Tb
Mbe = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.01 + 0.4EI*Tb + 0.2EI*Te
Step 3) Using Moment Equilibrium Equation -
MA = 0
0.4EI*Ta = 286.01 - 0.2EI*Tc
0.2EI*Ta = 143 - 0.1EI*Tc --------- 1
substitute the above value in summation
143 - 0.1EI*Tc + 0.8EI*Tc + 0.2EI*Td = 0
0.7EI*Tc + 0.2EI*Td + 143 = 0
0.7EI*Tc = -143 - 0.2EI*Td
0.2EI*Tc = -40.85 - 0.0571EI*Td ------- 2
substitute the above value in summation
Md = 0
-40.85 - 0.0571EI*Td + 0.8EI*Td + 0.2EI*Te - 164.51 = 0
0.74EITd + 0.2EI*Te -205.36 = 0 ------- 11
0.2EI**Te = 205.36 - 0.74EI*Td
0.4EI*Te = -1.48EI*Td - 247.32 * 2
0.8EI*Te = 821.44 - 2.96EI*Td ------------ 12
substitute the above value in summation of E,
Me = 0
0.2EI*Tb + 0.2EI*Td + 821.44 - 2.96EI*Td = 0
0.2EI*Tb - 2.76EI*Td +821.44 = 0
0.2EI*Tb = 2.76EI*Td - 821.44
0.4EI*Tb = 5.52EI*Td - 1642.88 -------- 13
substitute the above value in summation of moment at B,
Mb = 0
5.52EI*Td -1642.88 + 286.01 +205.36 - 0.74EI*Td = 0
4.78EI*Td = 1151.51
EI*Td = 240.9
substitute the value of EI*Td in other equations and derive unknowns as follows.
EI*Tb = -782.78
EI*Tc = -273.11
EI*Ta = 851.58
EI*Te = 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
C) Let us find out the maximum moment at D,
i) Let us find out the maximum bending moment at D by loading span
ii) 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).
Step 1) FIXED END MOMENTS:
MFac = -wL^2/12=-121.5kNm
MFca = 121.5Knm
MFcd = -286.01kNm
MFdc = 286.01kNm
MFeb = -121.5kNm
MFbe = 121.5knm
MFde = -286.01kNm
MFed = 286.01kNm
Step 2) END MOMENT Equation -
Ma = MFac+4EI*Ta/L + 2EI*Tc/L
= -121.5+0.4EI*Ta+0.2EI*Tc
Mca = MFca + 4EI*Tc/L + 2EITa/L
= 121.5 + 0.4EI*Tc + 0.2EI*Ta
Mcd = MFcd + 4EI*Tc/L + 2EI*Td/L
= -286.01 + 0.4EI*Tc + 0.2EI*Td
Mdc = MFdc + 4EI*Td/L + 2EI*Tc/L
= 286.01 + 0.4EI*Td + 0.2EI*Tc
Mde = MFde + 4EITd/L + 2EITe/L
= -286.01 + 0.4EITd + 0.2EITe
Med = MFed + 4EI*Te/L + 2EI*Td/L
= 286.01 + 0.4EI*Te + 0.2EI*Td
Meb = MFeb + 4EI*Te/L + 2EI*Tb/L
= -121.5 + 0.4EI*Te + 2EI*Tb
Mbe = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.01 + 0.4EI*Tb + 0.2EI*Te
Step 3) Using Moment Equilibrium equation -
Ma = 0
0.4EI*Ta = 121.5 - 0.2EI*Tc
0.2EI*Ta = 60.75 - 0.1EI*Tc --------- 1
Substitute the above value in summation
60.75 - 0.1EI*Tc + 0.8EI*Tc + 0.2EI*Td - 164.5= 0
0.7EI*Tc + 0.2EI*Td - 103.75 = 0
0.2EI*Td = 103.75 - 0.7EI*Tc
0.4EI*Td = 207.5 - 1.4EI*Tc ------- 2
Substitute the above value in summation
Md = 0
415 - 2.8EI*Tc + 0.2EI*Tc + 0.2EI*Te = 0
0.2EI*Te - 2.6EI*Tc + 415 = 0
0.2EI*Te = 2.6EI*Tc - 415
0.8EI*Te = 10.4EI*Tc - 1660
substitute the above value in summation of E,
Me = 0
0.2EI*Tb + 0.2EI*Td + 10.4EI*Tc - 1494.5 = 0
0.2EI*Tb =- 0.2EI*Td - 10.4EI*Tc +1495.5
0.4EI*Tb = 2990.4 -0.4EI*Td - 20.8EI*Tc
substitute the above value in summation of moment at B,
Mb = 0
2990.4 - 0.4EI*Td - 20.8EI*Tc +2.6EI*Tc -415 +121.5 = 0
2990.4+ 207.5 - 1.4EI*Tc -20.8EI*Tc + 2.6EI*Tc -415+121.5 = 0
19.6EI*Tc = 2904.4
EI*Tc = 148.18
substitute the value of EI*Td in other equations and derive unknowns as follows.
EI*Td = 0.12
EI*Te = -140.16
EI*Ta = 229.6
EI*Tb = -233.67
substitute the values of these unknowns in all the moment equations and derive the moment values.
Ma = 0
Mca = 226.69 kNM
Mcd = -226.71 kNM
Mdc = 315.69 kNM
Mde = -313.9 kNM
Med = 229.97 kNM
MEb = -229.97 kNM
Mbe = 0 kNM
