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KIC000114

KIC000114 - B The OM is the moment that would need to be...

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Example 1 Solve using moment distribution, EI is constant. ~A r B C~ l1f1iT I, 5 ft ,I. 5 ft .1. 20 ft ,I 1. First, let's set up a table to keep track of things: Joint A B C Member AB BA BC CB Stiffness,k DF FEM UnbalancedMoments (UM) Distribute UM's Carry over 2. At each joint that is free to rotate, compute the stiffness (k) and Distribution Factors (OFs). 4El k oA = -- 10 k _ 4El oc- 20 Calculate the OFs 4El 10 -'-~'-"----'--) = 0.667 ( 4EI + 4EI 10 20 4EI _ 20 _ OF oc - ( ) - 0.333 4EI 4El --+-- 20 10 DFco=O 3. Lock the structure by applying a "clamp" at each joint that is free to rotate, creating a temporary fixed condition. Calculate the respective FEM that result on each member. (Remember the sign convention) ew moments that act on the end of the member = pas. CCW moments that act on the end of the member = neg.

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Tilam p " I c~ Go to the tables for the FEM values (Remember, you are to memorize two cases) C ~~A _f'------------.:B ~) I. U2 .1. U2 .1 C~I-"-B -=c~) I. L .1 FEM AB = - PL = -50" 8 FEMBc=O FEMc.=O 4. Calculate the unbalanced moment (OM) at the first joint to be unlocked, (Joint
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Unformatted text preview: B). The OM is the moment that would need to be applied to the joint to bring the sum of all moments to zero. In this case +50 + UM B = 0 + UM. = -50 ft.k 5. Unlock the joint and distribute the UM's according to the Distribution Factors already calculated. 6. Carry over moment Carry over the appropriate amount of distributed end moment to the far, fixed ends of the beam, i.e., the "carry over" moment. The "carry over" moment has the same direction as the distributed moment, (remember the derivation). 7. If the structure had more than one joint that rotated, relockjoint B and proceed to the next rotating joint and unlock it., repeat steps 4 through 6. Continue to repeat steps 4-6 until UM's are satisfactorily small. 8. Sum the moments at each joint to complete the moment distribution. Note: do not include UM's in the sum!...
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