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.

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.

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