Project 1
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.
(Answer):
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 = 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 :
Mac = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mca = wL^2/12 = 42.375*81/12 = +286.03Knm
Mcd = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mdc = wL^2/12 = 42.375*81/12 = +286.03Knm
Med = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mde = wL^2/12 = 42.375*81/12 = +286.03Knm
Meb = -wL^2/12 = -42.375*81/12 = -286.03Knm
Mbe = wL^2/12 = 42.375*81/12 = +286.03Knm
moments at each point:
Ma = Mac + 4EITa/L + 2EITc/L
Ma = -286.03 + 78125Ta + 39062.5Tc ------------eq.1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.03 + 78125Tc + 39062.5Ta ------------eq.2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.03 + 78125Tc + 39062.5Td ------------eq.3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.03 + 78125Td + 39062.5Tc ------------eq.4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -412.59 + 78125Td + 39062.5Te ------------eq.5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.03 + 78125Te + 39062.5Td ------------eq.6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.03 + 78125Te + 39062.5Tb ------------eq.7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.03 + 78125Tc + 39062.5Ta ------------eq.8
from eq1.
78125Ta = 286.03 - 39062.5Tc ---------eq.9
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 ------------eq.10
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 ------------eq.11
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 ----------eq.12
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 = -0.001046 -----------eq.13
similarly substitute the above value in all the equations and derive the values of unknowns.
Tb = -0.00418
Te = 0.001046
Td = -1.84*10^-6
Ta = 0.00418
substitute the above unknown values in moment equations and get the moment values.
Ma = -0.4
Mc,ca = 367.7Knm
Mc,cd = -367.7Knm
Md,dc = 245.17Knm
Md,de = -245.17KNm
Me,ed = 367.7Knm
Me,eb = -367.7KNm
Mb,be = 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
2nd part of question
let us findout the maximum bending moment at C by loading span
AC ,CD and EB.
i.e 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).
we get the fixed end moments as follows.
Mac = -wL^2/12=-286.01kNm
Mca = 286.01Knm
Mcd = -286.01kNm
Mdc = 286.01kNm
Meb = -286.01kNm
Mbe = 286.01knm
Mde = -121.5kNm
Med = 121.5kNm
Ma = Mac+4EITa/L + 2EITc/L
Ma = -286.01+0.4EITa+0.2EITc ------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.01 + 0.4EITc + 0.2EITa -------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.01 + 0.4EITc + 0.2EITd ------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.01 + 0.4EITd + 0.2EITc -------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -121.5 + 0.4EITd + 0.2EITe ------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 121.5 + 0.4EITe + 0.2EITd -------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.01 + 0.4EITe + 2EITb --------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.01 + 0.4EITb + 0.2EITe ------- 8
from equation 1,
0.4EITa = 286.01 - 0.2EITc
0.2EITa = 143 - 0.1EITc --------- 9
substitute the above value in summation of eq.2 and 3
143 - 0.1EITc + 0.8EITc + 0.2EITd = 0
0.7EITc + 0.2EITd + 143 = 0
0.7EITc = -143 - 0.2EITd
0.2EITc = -40.85 - 0.0571EITd ------- 10
substitute the above value in summation of eq.4 and 5
Md = 0
-40.85 - 0.0571EITd + 0.8EITd + 0.2EITe + 164.51 = 0
0.74EITd + 0.2EITe + 123.66 = 0 ------- 11
0.2EITe = -0.74EITd - 123.66 * 2
0.4EITe = -1.48EITd - 247.32 * 2
0.8EITe = -2.96EITd - 494.64
substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 0.8EITe - 164.51 = 0
0.2EITb + 0.2EITd - 2.96EITd - 494.64 - 164.51 = 0
0.2EITb - 2.76EITd - 659.15 * 2
0.4EITb = 5.52EITd + 1318.3 -------- 12
substitute the above value in summation of moment at B,
Mb = 0
5.52EITd + 1318.3 + 286.01 - 0.74EITd - 123.66 = 0
4.78EITd = -1480.65
EITd = -309.75
substitute the value of EITd in other equations and derive unknowns as follows.
EITb = 978.8
EITc = -115.81
EITa = 772.93
EIT = 572.35
substitute the values of these unknowns in all the moment equations and derive
the moment values.
Ma = 0
Mc,ca = 394.27 kNM
Mc,cd = -394.87 kNM
Md,dc = 198.948 kNM
Md,de = -137.34 kNM
Me,ed = 288.49 kNM
Me,Eb = -288.49 kNM
Mbe = 8.96 kNM
similarly findout the maximum moment at E,
let us findout the maximum bending moment at C by loading span
DE,AC and EB.
i.e 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).
we get the fixed end moments as follows.
Mac = -wL^2/12=-286.01kNm
Mca = 286.01Knm
Mcd = -121.5kNm
Mdc = 121.5kNm
Mde = -286.01kNm
Med = 286.01knm
Meb = -286.01kNm
Mbe = 286.01kNm
Ma = Mac+4EITa/L + 2EITc/L
Ma = -286.01+0.4EITa+0.2EITc ------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.01 + 0.4EITc + 0.2EITa -------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -121.5 + 0.4EITc + 0.2EITd ------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 121.5 + 0.4EITd + 0.2EITc -------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -286.01 + 0.4EITd + 0.2EITe ------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.01 + 0.4EITe + 0.2EITd -------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.01 + 0.4EITe + 0.2EITb --------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.01 + 0.4EITb + 0.2EITe ------- 8
from equation 1,
0.4EITa = 286.01 - 0.2EITc
0.2EITa = 143 - 0.1EITc --------- 9
substitute the above value in summation of eq.2 and 3
143 - 0.1EITc + 0.8EITc + 0.2EITd = 0
0.7EITc + 0.2EITd + 143 = 0
0.7EITc = -143 - 0.2EITd
0.2EITc = -40.85 - 0.0571EITd ------- 10
substitute the above value in summation of eq.4 and 5
Md = 0
-40.85 - 0.0571EITd + 0.8EITd + 0.2EITe - 164.51 = 0
0.74EITd + 0.2EITe -205.36 = 0 ------- 11
0.2EITe = 205.36 - 0.74EITd
0.4EITe = -1.48EITd - 247.32 * 2
0.8EITe = 821.44 - 2.96EITd ------------ 12
substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 821.44 - 2.96EITd = 0
0.2EITb - 2.76EITd +821.44 = 0
0.2EITb = 2.76EITd - 821.44
0.4EITb = 5.52EITd - 1642.88 -------- 13
substitute the above value in summation of moment at B,
Mb = 0
5.52EITd -1642.88 + 286.01 +205.36 - 0.74EITd = 0
4.78EITd = 1151.51
EITd = 240.9
substitute the value of EITd in other equations and derive unknowns as follows.
EITb = -782.78
EITc = -273.11
EITa = 851.58
EITe = 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
let us findout the maximum moment at D,
let us findout the maximum bending moment at D by loading span
CD and DE.
i.e ,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).
we get the fixed end moments as follows.
Mac = -wL^2/12=-121.5kNm
Mca = 121.5Knm
Mcd = -286.01kNm
Mdc = 286.01kNm
Meb = -121.5kNm
Mbe = 121.5knm
Mde = -286.01kNm
Med = 286.01kNm
Ma = Mac+4EITa/L + 2EITc/L
Ma = -121.5+0.4EITa+0.2EITc ------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 121.5 + 0.4EITc + 0.2EITa -------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.01 + 0.4EITc + 0.2EITd ------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.01 + 0.4EITd + 0.2EITc -------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -286.01 + 0.4EITd + 0.2EITe ------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.01 + 0.4EITe + 0.2EITd -------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -121.5 + 0.4EITe + 2EITb --------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.01 + 0.4EITb + 0.2EITe ------- 8
from equation 1,
0.4EITa = 121.5 - 0.2EITc
0.2EITa = 60.75 - 0.1EITc --------- 9
substitute the above value in summation of eq.2 and 3
60.75 - 0.1EITc + 0.8EITc + 0.2EITd - 164.5= 0
0.7EITc + 0.2EITd - 103.75 = 0
0.2EITd = 103.75 - 0.7EITc
0.4EITd = 207.5 - 1.4EITc ------- 10
substitute the above value in summation of eq.4 and 5
Md = 0
415 - 2.8EITc + 0.2EITc + 0.2EITe = 0
0.2EITe - 2.6EITc + 415 = 0 ------- 11
0.2EITe = 2.6EITc - 415 ----------12
0.8EITe = 10.4EITc - 1660 ------------13
substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 10.4EITc - 1494.5 = 0
0.2EITb =- 0.2EITd - 10.4EITc +1495.5
0.4EITb = 2990.4 -0.4EITd - 20.8EITc -------- 14
substitute the above value in summation of moment at B,
Mb = 0
2990.4 - 0.4EITd - 20.8EITc +2.6EITc -415 +121.5 = 0
2990.4+ 207.5 - 1.4EITc -20.8EITc + 2.6EITc -415+121.5 = 0
19.6EITc = 2904.4
EITc = 148.18
substitute the value of EITd in other equations and derive unknowns as follows.
EITd = 0.12
EITe = -140.16
EITa = 229.6
EITb = -233.67
substitute the values of these unknowns in all the moment equations and derive
the moment values.
Ma = 0
Mc,ca = 226.69 kNM
Mc,cd = -226.71 kNM
Md,dc = 315.69 kNM
Md,de = -313.9 kNM
Me,ed = 229.97 kNM
Me,Eb = -229.97 kNM
Mbe = 0 kNM