Project 1
Que.
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.
Solution:-
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
