Project 1
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.
AIM:
To determine the loading on the beam indicated and analyze the beams and draw bending moment and shear force diagram.
GIVEN DATA:
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.
So, I =bd^3 /12 = 0.5*0.75 ^3 /12 =0.0176 m^4 = 1.76*10^-14 m^2
Modulus of elasticity of concrete is 10 GPa. = 1*10^6 KPa
PROCEDURE:
Self weight of slab = 0.12*25 = 3kn/m
Self weight of beam = 0.5*0.75*25 = 9.375Kn/m
Floor finish = 3kn/m
Live load = 5kn/m
Total weight = self wt + live load + floor finish
= 3+5+3 = 11kN/m
Load on beam = total weight * tributary width
Let us give the names to all the beams in order as B1,B2,B3.....etc.
Load on B1 = W * tributary width+ self weight of beam
= 11* 0.5(3+3)+9.375
= 42.375 Kn/m
Load on B2 = W * tributary width
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
Load on B3 = W * tributary width
= 11* 0.5(3+3)+9.375
= 42.375Kn/m
Load on B4 = W * tributary width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
Load on B5 = W * tributary width
= 11 * 0.5(3+3) +9.375
= 42.375Kn/m
Load on B6 = W * tributary width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
Load on B7 = W * tributary width
= 11 * 0.5(3+3)+9.375
= 42.375Kn/m
FIXED END MOMENTS:
FOR SPAN AC:
Mac = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mca = wL^2/12 = 42.375*81/12 = +286.03Knm
FOR SPAN CD:
Mcd = -wL^2/12 = -42.375*81/12 = -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
MOMENTS AT EACH POINT:
FOR SPAN AC:
Mac = MFac + 4EI*Ta/L + 2EI*Tc/L
= -286.03 + 78125Ta + 39062.5Tc
Mca = MFca + 4EI*Tc/L + 2EI*Ta/L
= 286.03 + 78125Tc + 39062.5Ta
FOR SPAN CD:
Mcd = MFcd + 4EI*Tc/L + 2EI*Td/L
= -286.03 + 78125Tc + 39062.5Td
Mdc = MFdc + 4EI*Td/L + 2EI*Tc/L
= 286.03 + 78125*Td + 39062.5*Tc
FOR SPAN DE:
Mde = MFde + 4EI*Td/L + 2EI*Te/L
= -412.59 + 78125Td + 39062.5Te
Med = MFed + 4EI*Te/L + 2EI*Td/L
= 286.03 + 78125Te + 39062.5Td
FOR SPAN EB:
Meb = MFeb + 4EI*Te/L + 2EI*Tb/L
= -286.03 + 78125Te + 39062.5Tb
Mbe = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.03 + 78125Tc + 39062.5Ta
FROM EQUILIBRIUM EQUN:
Considering the equilibrium at A, Ma = 0
78125Ta = 286.03 - 39062.5Tc
Considering the equilibrium at C, Mc = 0
Mc = Mcd + Mca
78125Tc + 78125Tc + 39062.5Ta + 39062.5Ta + 39062.5Td = 0
156250Tc + 143.015 - 19531.25tc + 39062.5Td = 0
136718.75Tc + 39062.5Td = -143.015
39062.5Td = -143.015 - 136718.75Tc
Considering the equilibrium at D, Md = 0
156250Td + 39062.5Tc + 39062.5Te = 0
-572.06 - 546875Tc + 39062.5Tc + 39062.5Te = 0
-572.06 - 546875Tc + 39062.5Tc + 39062.5Te = 0
-572.06 - 507812.5Tc = -39062.5Te
39062.5Te = 507812.5Tc + 572.06
Considering equilibrium at E, Me = 0
156250Te + 39062.5Td + 39062.5Tb = 0
2031250Tc + 2288.24 - 143.014 - 136718.75Tc + 39062.5Tb = 0
1894531.25Tc + 39062.5Tb = -2145.25
Considering the equilibrium at B, Mbe = 0
78125Tb + 39062.5Te + 286.03 = 0
-4290.452 - 3789062.5Tc + 507812.5Tc + 572.06 + 286.03 = 0
-3281250Tc = 3432.36
Tc = 1.046*10^-3 rad
Similarly substitute the above value in all the equations and derive the values of unknowns.
Tb =4.18*10^-3 rad
Te = 1.046*10^-3 rad
Td = -1.84*10^-6 rad
Ta = 4.0*10^-3 rad
Substitute the above unknown values in moment equations and get the moment values.
FOR SPAN AC:
Mac = MFac + 4EI*Ta/L + 2EI*Tc/L
= -286.03 + 78125Ta + 39062.5Tc
Ma = - 0.4knm
Mca = MFca + 4EI*Tc/L + 2EI*Ta/L
= 286.03 + 78125Tc + 39062.5Ta
Mca = 367.7Knm
FOR SPAN CD:
Mcd = MFcd + 4EI*Tc/L + 2EI*Td/L
= -286.03 + 78125Tc + 39062.5Td
Mcd = -367.7Knm
Mdc = MFdc + 4EI*Td/L + 2EI*Tc/L
= 286.03 + 78125Td + 39062.5Tc
Mdc = 245.17Knm
FOR SPAN DE:
Mde = MFde + 4EI*Td/L + 2EI*Te/L
= -412.59 + 78125Td + 39062.5Te
Mde = -245.17KNm
Med = MFed + 4EI*Te/L + 2EI*Td/L
= 286.03 + 78125Te + 39062.5Td
Med = 367.7Knm
FOR SPAN EB:
Meb = MFeb + 4EI*Te/L + 2EI*Tb/L
= -286.03 + 78125Te + 39062.5Tb
Meb = -367.7KNm
Mbe = MFbe + 4EI*Tb/L + 2EI*Te/L
= 286.03 + 78125Tc + 39062.5Ta
Mbe = 0 KNm
To determine the shear forces , consider each span individually,
consider span AC,taking moment about A,
Hc * 9 - 0.4 + 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
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
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
Consider span EB,taking moment about E,
9Hb - 367.7 - 42.375*9*4.5 = 0
Hb = - 231.535Kn
He = 149.84Kn
2.For the same beam, consider load patterning to determine maximum positive and negative moments in each span.
AIM:
To determine the load patterning to determine maximum positive and negative moments in each span.
PROCEDURE:
Maximum bending moment at C by loading span
AC ,CD and EB.
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).
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
END MOMENTS:
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
Td = -309.75/EI rad
Substitute the value of EI*Td in other equations and derive unknowns as follows.
Tb = 978.8/EI rad
Tc = -115.81/EI rad
Ta = 772.93/EI rad
Te = 572.35/EI rad
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
Similarly find out the maximum moment at E,
NOw, let us findout the maximum bending moment at C by loading span DE,AC and EB.
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).
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
END MOMENTS:
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
from equation 1,
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
Td = 240.9/EI rad
substitute the value of EI*Td in other equations and derive unknowns as follows.
Tb = -782.78/EI rad
Tc = -273.11/EI rad
Ta = 851.58 /EI rad
Te = 135.51/EI rad
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
LET US FINDOUT THE MAXIMUM MOMENT AT D,
let us find out the maximum bending moment at D by loading span
CD and DE.
CD and DE will be subjected to a load of 42.375kn/m and AC, EB will be loaded with UDL of 18kn/m(excluding live load).
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
END MOMENTS:
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
from equation 1,
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
Tc = 148.18/EI rad
substitute the value of EI*Td in other equations and derive unknowns as follows.
Td = 0.12/EI rad
Te = -140.16/EI rad
Ta = 229.6/EI rad
Tb = -233.67/EI rad
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