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Structural Analysis-II
CEE-317
(2.0 credits)
Lecture: 3
Bijit Kumar Banik
Assistant Professor, CEE, SUST
Room No.: 115 (“C...
From Previous Classes
Stiffness: The end moment required to produce a unit rotation
at one end of the member while the oth...
From Previous Classes
Joint Stiffness Factor: This is the case when there is more
then one member in a joint.
KA = ∑K = KA...
From Previous Classes
Distribution Factor (DF): Fraction of the total resisting moment
supplied by the member is called di...
From Previous Classes
Distribution Factor for Fixed End = 0
Distribution Factor for Pin/Hinge/Roller End = 1
Important:
A
...
Procedure
1. Identify joints & calculate stiffness factor for each span
2. Determine DF for each span
3. Calculate fixed e...
Fixed End Moment
Ex-4.3: Beam Problem (Shoraf sir’s book, pp-11)
EI EI
1 k/ft 20 k
60 ft 20 ft 20 ft
A B C
Find out,
MA, MBA, MBC & MC
Soln...
Ex-4.3: Beam Problem (Shoraf sir’s book, pp-11)
BA C
0.4 0.60 0
-300 +300 -100 +100
-80 -120CO CO-40 -60
-340 +220 -220 +4...
Cee 317 (3) (structural analysis)
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Cee 317 (3) (structural analysis)

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Cee 317 (3) (structural analysis)

  1. 1. Structural Analysis-II CEE-317 (2.0 credits) Lecture: 3 Bijit Kumar Banik Assistant Professor, CEE, SUST Room No.: 115 (“C” building) bijit_sustbd@yahoo.com Department of Civil and Environmental Engineering
  2. 2. From Previous Classes Stiffness: The end moment required to produce a unit rotation at one end of the member while the other end is fixed. Carryover Stiffness K = 4EI/L is also called Member stiffness factor
  3. 3. From Previous Classes Joint Stiffness Factor: This is the case when there is more then one member in a joint. KA = ∑K = KAB + KAC + KAD AD B C KAD KAB KAC
  4. 4. From Previous Classes Distribution Factor (DF): Fraction of the total resisting moment supplied by the member is called distribution factor. K K DF i ∑ = AD B C KAD = 1000 KAB = 4000 KAC = 5000 M = 2000 k-ft4.0 10000 4000 ==ABDF 5.0 10000 5000 ==ACDF 1.0 10000 1000 ==ADDF ftkM AB −== 8002000*4.0 ftkM AC −== 10002000*5.0 ftkM AD −== 2002000*1.0
  5. 5. From Previous Classes Distribution Factor for Fixed End = 0 Distribution Factor for Pin/Hinge/Roller End = 1 Important: A D B C 0.1 0.4 0.5 0 1
  6. 6. Procedure 1. Identify joints & calculate stiffness factor for each span 2. Determine DF for each span 3. Calculate fixed end moment. [Lock joints] 4. Determine the moment that is needed to put each joint in equilibrium 5. Release or unlock the joints and distribute the counter balancing moment into the connecting span at each joint 6. Carry these moment in each span over to its other end by multiplying by multiplying the COF 1/2 7. Repeat locking & unlocking the joints until a small correction value obtained In the case of more than one locking unlocking process, stop With no carry over (i.e. stop after distribution) 8. Before final value stop with double line 9. Sum up the moments to get final value
  7. 7. Fixed End Moment
  8. 8. Ex-4.3: Beam Problem (Shoraf sir’s book, pp-11) EI EI 1 k/ft 20 k 60 ft 20 ft 20 ft A B C Find out, MA, MBA, MBC & MC Soln EI EI 1 k/ft 20 k 60 ft 20 ft 20 ft A B C Lock (say) ftk WL −== 300 12 60*1 12 22 ftk PL −== 100 8 40*20 8 300 k-ft 300 k-ft 100 k-ft 100 k-ft B 200 k-ft A C B 80 k-ft A C 120 k-ft Corrected moment applied at joint B Moment at B is distributed
  9. 9. Ex-4.3: Beam Problem (Shoraf sir’s book, pp-11) BA C 0.4 0.60 0 -300 +300 -100 +100 -80 -120CO CO-40 -60 -340 +220 -220 +40 40 4EI KBC = 60 4EI KBA = 24 4EI K =∑ 6.0= ∑ = K K D BC BC 4.0= ∑ = K K D BA BA EI EI 1 k/ft 20 k 60 ft 20 ft 20 ft A B C H:W:

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