THEORY OF INDETERMINATE STRUCTURES

CHAPTER SIX 6. KANIS METHOD OR ROTATION CONTRIBUTION METHOD OF FRAME ANALYSIS This method may be considered as a further simplification of moment distribution method wherein the problems involving sway were attempted in a tabular form thrice (for double story frames) and two shear co-efficients had to be determined which when inserted in end moments gave us the final end moments. All this effort can be cut short very considerably by using this method. → Frame analysis is carried out by solving the slope − deflection equations by successive approximations. Useful in case of side sway as well. → Operation is simple, as it is carried out in a specific direction. If some error is committed, it will be eliminated in subsequent cycles if the restraining moments and distribution factors have been determined correctly. Please note that the method does not give realistic results in cases of columns of unequal heights within a storey and for pin ended columns both of these cases are in fact extremely rare even in actual practice. Even codes suggest that RC columns framing into footings or members above may be considered more or less as fixed for analysis and design purposes. Case 1. No side sway and therefore no translation of joints derivation. Consider a typical member AB loaded as shown below: Tangent at B Mab A Tangent at A P1 a Elastic Curve L P2 b Mba B

A GENERAL BEAM ELEMENT UNDER END MOMENTS AND LOADS General Slope deflection equations are. 2EI Mab = MFab + ( − 2θa − θb ) L 2EI Mba = MFba + ( − θa −2θb ) L equation (1) can be re-written as Mab = MFab + 2 M′ab + M′ba and

→ (1) → (2) → (3)

where MFab = fixed end moment at A due to applied loads. EI M′ab = rotation contribution of near end A of member AB = − (2θa) L =− → (4) where k1=

2EI θa = − 2E k1 θa L / M ba = rotation contribution of for end B of member AB. 2 EI θb So M/ba = − = − 2Ek1 θb L

I1 L1

→ (5)

KANIS METHOD OF FRAME ANALYSIS

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Now consider a generalized joint A in a frame where members AB, AC, AD.........meet. It carries a moment M. B k1 M A k2 k3 D C

E

k3

For equilibrium of joint A, ∑Ma = 0 or or Mab + Mac + Mad + Mae..................= 0 Putting these end moments in form of eqn. (3)

∑MF (ab, ac, ad) + 2 ∑M′ (ab, ac, ad ) + ∑M′ (ba, ca, da) = 0 Let ∑MF (ab, ac, ad) = MFa (net FEM at A) So MFa + 2 ∑M′ (ab, ac, ad) + ∑M′ (ba, ca, da) = 0 → (6)

From (6), ∑M′ (ab, ac, ad) = −

1 [(MFa + ∑M′ (ba, ca, da)] → (7) 2

From (4), ∑M′ (ab, ac, ad) = − 2Ek1 θa − 2 Ek2 θa − 2 Ek3 θa + ............... = − 2 Eθa ( k1 + k2 + k3) = − 2 Eθa (∑k), ( sum of the member stiffnesses framing in at joint A) or θa = − ∑M′ (ab, ac, ad) 2E (∑k) → (8)

From (4), M′ab = − 2 Ek1 θa. Put θa from (8), we have M′ab = − 2E k1 −

∑M′ (ab, ac, ad) k1 = [ ∑M′ (ab, ac, ad)] 2E (∑k) ∑k

From (7), Put ∑M′ (ab, ac, ad) k1 1 So M′ab = − (MFa + ∑M′ (ba, ca, da)) ∑k 2

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THEORY OF INDETERMINATE STRUCTURES

or

M′ab = −

1 k1 [ MFa + ∑M′ (ba, ca, da)] 2 ∑k 1 k2 [ MFa + ∑M′ (ba, ca, da)] 2 ∑k

on similar lines

M′ac = −

and

M/ad = −

1 k3 [ MFa + ∑M′ (ba, ca, da)] 2 ∑k sum of the rotations contributions of far ends of members meeting at A.

rotation contribution of near end of member ad.

Sum of rotation factors at near end of members ab, ac, ad is − 1 k1 1 k2 1 k3 1 k1 + k2 + k3 + ......... − − =− 2 ∑k 2 ∑k 2 ∑k 2 ∑k =− 1 , sum of rotation factors of different members meeting at a 2 joint is equal to – 1 2

[

]

Therefore, if net fixed end moment at any joint along with sum of the far end contribution of members meeting at that joint are known then near end moment contribution can be determined. If far end contributions are approximate, near end contributions will also be approximate. When Far end contributions are not known (as in the first cycle), they can be assumed to be zero. 6.1. RULES FOR CALCULATING ROTATION CONTRIBUTIONS :__...