Project 1
PROJECT - 1
Given below is a floor plan of building. The red lines running East-west are beams supported by the girders spanning North-South between columns. The floor system is 12 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 mm x 750 mm. The beams are equally spaced. The rectangles in cyan colour are columns. Modulus of elasticity of concrete is 10 GPa.
- Determine the loading on the beam indicated, analyse 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 = 3 KN/m
Self-weight of beam = 0.5*0.75*25 = 9.375 KN/m
Floor finish = 3 KN/m
Live load = 5 KN/m
Total weight = 11 KN/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
= 42.375 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
FIXED END MOMENTS:
Mac = -wl^2/12 = -42.375*81/12 = -286.03 KNm
Mca = wl^2/12 = 42.375*81/12 = +286.03 KNm
Mcd = -wl^2/12 = -42.375*81/12 = -286.03 KNm
Mdc = wl^2/12 = 42.375*81/12 = +286.03 KNm
Med = -wl^2/12 = -42.375*81/12 = -286.03 KNm
Mde = wl^2/12 = 42.375*81/12 = +286.03 KNm
Meb = -wl^2/12 = -42.375*81/12 = -286.03 KNm
Mbe = wl^2/12 = 42.375*81/12 = +286.03 KNm
MOMENTS AT EACH POINT:
Ma = Mac + 4EITa/L + 2EITc/L
Ma = -286.03 + 78125Ta + 39062.5Tc ---------- 1
Mc,ca = Mca + 4EITc/L + 2EITa/L
Mc,ca = 286.03 + 78125Tc + 39062.5Ta ---------- 2
Mc,cd = Mcd + 4EITc/L + 2EITd/L
Mc,cd = -286.03 + 78125Tc + 39062.5Td ---------- 3
Md,dc = Mdc + 4EITd/L + 2EITc/L
Md,dc = 286.03 + 78125Td + 39062.5Te ---------- 4
Md,de = Mde + 4EITd/L + 2EITe/L
Md,de = -412.59 + 78125Td + 39062.5Te ---------- 5
Me,ed = Med + 4EITe/L + 2EITd/L
Me,ed = 286.03 + 78125Te + 39062.5Td ---------- 6
Me,eb = Meb + 4EITe/L + 2EITb/L
Me,eb = -286.03 + 78125Te + 39062.5Tb ---------- 7
Mb,be = Mbe + 4EITb/L + 2EITe/L
Mb,be = 286.03 + 78125Tc + 39062.5Ta ---------- 8
From 1,
78125Ta = 286.03 – 39062.5Tc ---------- 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
136718.75Td = -143.01 – 136718.75Tc ---------- 10
Considering the equilibrium at D, Md = 0,
156250Td + 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 ---------- 11
Considering the equilibrium at E, Me = 0,
156250Te + 39062.5Tc + 39062.5Tb = 0
2031250Tc + 2288.24 – 143.014 - 136718.75Tc + 39062.5Tb = 0
1894531.25Tc + 39062.5Tb = - 2145.25 ---------- 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 ---------- 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 = 367.7 KNm
Mc,cd = -367.7 KNm
Md,dc = 245.17 KNm
Md,de = -245.17 KNm
Me,ed = 367.7 KNm
Me,eb = -367.7 KNm
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 = 231.53 KN
Consider span CD,
Taking moment about c,
- 367.59 + 245.78 + Hd*9 – 42.375*9*4.5 = 0
Hd = 204.28 KN
Hc + Hd = 381.375
Hc = 177.09 KN
Consider span DE,
Taking moment about D,
- 225.65 + 367.7 + He*9 – 42.375*9*4.5 = 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 = 149.84 KN
QUESTION – 2:
Let us findout the maximum bending moment at C by loading span AC, CD and EB.
AC, CD and EB will be subjected to a load of 42.375 KN/m and DE will be loaded with UDL of 18 KN/m (excluding live load)
We get the fixed end moments as follows.
Mac = -wl^2/12 = -286.01 KNm
Mca = 286.01 KNm
Mcd = -286.01 KNm
Mdc = 286.01 KNm
Med = -286.01 KNm
Mbe = 286.01 KNm
Mde = - 121.5 KNm
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 = 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 = Med + 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 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 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
0.8EITe = -2.96EITd – 494.64
Substitute the above value in summation of E,
Me = 0
0.2EITb + 0.2EITd + 0.8EITe -164.51 = 0
0.4EITb = 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
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.
AC, DE and EB will be subjected to a load of 42.375 KN/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 = -286.01 KNm
Mca = 286.01 KNm
Mcd = -121.5 KNm
Mdc = 121.5 KNm
Mde = -286.01 KNm
Med = 286.01 KNm
Meb = -286.01 KNm
Mbe = 286.01 KNm
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 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 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 = -0.74EITd + 205.36
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
0.4EITb = 5.52EITd – 1642.88 ---------- 13
Substitute the above value in summation of moment at B,
Mb = 0
5.52EITd – 1642.88 + 286.01 – 0.74EITd + 205.36 = 0
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 bending moment at D by loading span CD and DE
CD and DE will be subjected to a load of 42.375 KN/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 = -121.5 KNm
Mca = 121.5 KNm
Mcd = -286.01 KNm
Mdc = 286.01 KNm
Mde = -286.01 KNm
Med = 286.01 KNm
Meb = -121.5 KNm
Mbe = 121.5 KNm
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 + 0.2EITb ---------- 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 2 and 3,
60.75 – 0.1EITc + 0.8EITc + 0.2EITd – 164.5 = 0
0.7EITc + 0.2EITd – 103.75 = 0
0.4EITc = 207.5 – 1.4Tc ---------- 10
Substitute the above value in summation of 4 and 5,
Md = 0
415 – 2.8EITc + 0.2EITc + 0.2EITe = 0
02EITe – 2.6EITc + 415 = 0 ---------- 11
0.2EITe = - 415 + 2.6EITc ---------- 12
0.8EITe = 10.4EITc – 1660 ---------- 13
Substitute the above value in summation of moment at E,
Me = 0
0.2EITb – 1494.5 + 0.2EITd + 10.4EITc = 0
0.4EITb = 2990.4 – 0.4EITd – 20.8EITc ---------- 14
Substitute the above value in summation of moment at B,
Mb = 0
2990.4 - 0.4EITb – 20.81EITc + 2.6EITc – 415 + 121.5 = 0
EITc = 148.18
Substitute the value of EITd in other equations and derive unknowns as follows:
EITb = -233.67
EITd = 0.12
EITa = 229.6
EITe = -140.16
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
