Project 1

PROJECT 1 : MASTERING SHEAR FORCE AND BENDING MOMENT DIAGRAM

Aim:

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.

Solution for question No.1:

Slab's own weight = 0.12*25 = 3 kN/m

Floor finish = 3kn/m; self weight of beam = 0.5*0.75*25 = 9.375KN/m

Live load is equivalent to 5kN/m.

Weight total: 11 kg/m

Total weight x tributary width equals load on beam.
Let's call each beam by its name, starting with B1, then B2, then B3, etc.

Load on each Beams:

load on B1 = W * tributary width + beam's own weight = 11 * 0.5(3+3) + 9.375 = 42.375 Kn/m

load on B2 = W * tributary width = 11 * 0.5(3+3) + 9.375 = 42.375 Kn/m

Load on B3 = W * tributary width = 11 * 0.5(3+3) + 9.375 = 42.375 kN/m

Load on B4 = W * tributary width = 11 * 0.5(3+3) + 9.375 = 42.375 kN/m

Load on B5 = W * tributary width = 11 * 0.5(3+3) + 9.375 = 42.375 Kn/m

load on B6 = W * tributary width = 11 * 0.5(3+3)+9.375 = 42.375 kN/m

Load on B7 = W * tributary width = 11 * 0.5(3+3) + 9.375 = 42.375 kN/m

Calculation of 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 = -286.03 KNm
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 = -286.03 kNm
Mbe = wL^2/12 = 42.375*81/12 = +286.03Knm

Moment at each span:

Ma = Mac + 4EITa/L + 2EITc/L
Ma = 39062.5Tc + 78125Ta + -286.03 ------------eq.1

Mc,ca = 4EITc/L + 2EITa/L + Mca
Mc,ca = 78125Tc + 39062.5Ta + 286.03 ------------eq.2

Mc,cd = Mc + 4EITc/L + 2EITd/L
Mc,cd = -286.03 + 78125Tc + 39062.5Td ------------eq.3

Md,dc = Md + 4EITd/L + 2EITc/L
Md,dc = 78125Td + 39062.5Tc + 286.03 ------------eq.4

Md,de = -412.59 + 78125Td + 39062.5Te
Md,de = Mde + 4EITd/L + 2EITe/L ------------eq. 5

Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 78125Te + 39062.5Td + 286.03 ------------eq.6

Meb,eb = Me + 4EITe/L + 2EITb/L
Me,eb = -286.03 + 78125 Te + 39062.5 Tb ------------eq.7

Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.03 + 78125Tc + 39062.5Ta ------------eq.8

Now, from equation 1
78125Ta = 286.03 - 39062.5Tc ---------eq.9

Taking into account the equilibrium at point C:

Mc = 0
Mc = Mcd + Mca

156250Tc + 143.015 - 19531.25tc + 39062.5Td = 0
78125Tc + 78125Tc + 39062.5Ta + 39062.5Ta + 39062.5Td = 0
39062.5Td + 136718.75Tc = -143.015

39062.5Td = 136718.75Tc - 143.015 ------------eq.10

Taking into account the D equilibrium, Md = 0.

39062.5Te + 39062.5Tc + 156250Td = 0.
546875 Tc + 39062.5 Tc + 39062.5 Te - 572.06 = 0
546875 Tc + 39062.5 Tc + 39062.5 Te - 572.06 = 0
-572.06 - 507812.5Tc = -39062.5Te

39062.5Te = 507812.5Tc + 572.06 ------------eq.11

Taking into account equilibrium at E, Me = 0

39062.5Td + 39062.5Tb + 156250Te = 0

2031250Tc + 2288.24 - 143.014 - 136718.75Tc + 39062.5Tb = 0

39062.5Tb + 1894531.25Tc = -2145.25 ----------eq.12

Taking into account equilibrium at B, Mbe = 0

78125Tb + 286.03 + 39062.5Te = 0

389062.5Tc + 507812.5Tc + 572.06 + 286.03 + -4290.452 = 0

-3281250Tc = 3432.36

Tc = -0.001046 ————eq. 13

In a similar manner, replace the above value in each equation to determine the values of the unknowns.
Te = 0.001046
Tb = -0.00418

Td = -1.84*10^-6.
Ta = 0.00418.

To obtain the moment values, solve the moment equations with the above unknown values substituted:
Ma = -0.4.

Mc,ca = 367.7 kNm
Mc,cd = -367.7kNm

Md,dc = 245.17 kNm
Md,de = -245.17 KNm.

Me,ed = 367.7 kNm
Me,eb = -367.7 KNm.

Mb,be = 0 KNm

In order to ascertain the shear forces, evaluate each span separately. For example, evaluate span AC,
taking moment about A,
Then, calculate
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 = 231.53 KN

Consider span CD, taking moment about c,
42.375*9*4.5 - -367.59 + 245.78 + Hd * 9 = 0

Hd = 204.28KN

Hc+Hd = 381.375

Hc = 177.09KN

Consider span DE, Taking moment about D,
-225.65 - 42.375*9*4.5 - 367.7 + He*9 = 0

He = 174.9 KN
Hd = 206.45 KN


Consider span EB, Taking moment about E,
9Hb - 367.7 - 42.375*9*4.5 = 0

Hb = 231.535 KN.
He = 149.84KN

Bending Moment and Shear Force Diagram, below:

Solution for question No.2:

Let us use the loading spans AC, CD, and EB to get the maximum bending moment at C.
i.e., a load of 42.375 kn/m will be applied to AC, CD, and EB, and a load of 18 kN/m will be applied to DE (excluding live load).
The fixed end moments are obtained as follows.

Mac = -wL^2/12 = 286.01kNm
Mca = 286.01 kNm.

Mcd = -286.01kNm.
Mdc = 286.01 kNm.

Meb = -286.01kNm.
Mbe = 286.01 kNm.

Mde = -121.5kNm.
Med = 121.5 kNm.

Ma = Mac + 4EITa/L + 2EITc/L
Ma = -286.01 +0.4EITa + 0.2EITc ------- 1

Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 26.01 + 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

Med + 4EITe/L + 2EITd/L = Me,ed
Me,ed = 0.2EITd + 0.4EITe + 121.5 -------- 6

Me + 4EITe/L + 2EITb/L = Meb,eb
Me,eb = -4EITe + 2EITb + -286.01 --------- 7

Mbe + 4EITb/L + 2EITe/L = Mb,be.
Mb,be = 0.4EITb + 0.2EITe + 286.01 ------- 8

0.4EITa = 286.01 - 0.2EITc
0.2EITa = 143 - 0.1EITc --------- 9

from equation replace the value above in the sum of equations 2 and 3

-0.1EITc + 0.8EITc + 0.2EITd + 143 = 0
0.7EITc = -143 - 0.2EITd
0.2EITc = -40.85 - 0.0571EITd ------- 10

substitute the value above in the sum of equations 4 and 5.
Md = 0.
-0.0571EITd + 0.8EITd + 0.2EITe + 164.51 -40.85 = 0
0.2EITe + 123.66 + 0.74EITd = 0 ------- 11

-0.74EITd - 123.66 * 2 = 0.2EITe
0.4EITe = - 247.32 * 2 - -1.48EITd
0.8EITe = - 494.64 -2.96EITd

substitute the above-mentioned value in the total of E.
Me = 0

- 164.51 + 0.2EITb + 0.2EITd + 0.8EITe = 0

2.96EITd - 494.64 - 164.51 - 0.2EITb + 0.2EITd = 0

2.76EITd - 659.15 * 2 + 0.2EITb

0.4EITb = 5.52EITd +1318.3 -------- e.q12


substitute the value above in the summation of the the moment at B.
Mb = 0.

0.74EITd - 123.66 + 5.52EITd + 1318.3 + 286.01 = 0

4.78EITd = -1480.65

EITd = -309.75.

Substitute the value of EITd in other equations, then use the following method to derive the unknowns.

EITb = 978.8.

EITc = -115.81.

EITa = 772.93.

EIT = 572.35.

For calculating the moment values, substitute the values of these unknowns in each moment equation.
Ma = 0

Mc,ca = 394.27 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.