Skip to main content

Analysis of Frames - Kani's Method

Kani's method was introduced by Gasper Kani in 1940’s. It involves distributing the unknown fixed end moments of structural members to adjacent joints, in order to satisfy the conditions of continuity of slopes and displacements.
Kani's method is also known as Rotation contribution method.

Advantages of Kani's Method :

1. Hardy Cross method distributed only the unbalanced moments at joints, whereas Kani’s method distributes the total joint moment at any stage of iteration.
2. The more significant feature of Kani’s method is that the process is self corrective. Any error at any stage of iteration is corrected in subsequent steps.
Framed structures are rarely symmetric and subjected to side sway, hence Kani’s method is best and much simpler than other methods like moment distribution method and slope displacement method.

Procedure 

1. Rotation stiffness at each end of all members of a structure is determined depending upon the end conditions.
The stiffness factors 
a) If Both ends are fixed
Kab= Kba = I/L
b) If Near end fixed and far end simply supported
Kab = ¾ I/L and Kba = 0
2. The Rotational factors are computed for all the members at each joint it is given by
rab= - 0.5 (Kab/ Sum of Kab)
Note: 
The sum of rotational factors at a joint is 0.5 
3.Fixed end moments (FEM) including transitional moments, moment releases and carry over moments are computed for members. 
The sum of the FEM at a joint ,and rotational factors are entered in the square drawn at the joint .
3. Iterations can be commenced at any joint however the iterations commence from the left end of the structure generally given by the equation
mab = rab [(Mfi + msi) +sum of mba)]
4. Initially the rotational components ? Mji (sum of the rotational moments at the far ends of the joint) can be assumed to be zero. Further iterations take into account the rotational moments of the previous joints.
5. Rotational moments are computed at each joint successively till all the joints are processed. This process completes one cycle of iteration.
6. Steps 4 and 5 are repeated till the difference in the values of rotation moments from successive cycles is neglected.
7. Final moments in the members at each joint are computed from the rotational members of the final iterations step.
Mab = (Mfab + msab) + 2 mab + mba
The lateral translation of joints (side sway) is taken into consideration by including column shear in the iterative procedure.
8. Displacement factors are calculated for each storey given by
Uij = -1.5 (Kij/ sum of Kij)

Comments

Popular posts from this blog

Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E)   It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E =  s/e Modulus of Rigidity (G)  It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G =    Ƭ / ø   Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube  PQRS is  subjected to a shearing force F.  Let  Ƭ     be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ   Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point   S moved to S' and point R moved to R' as shown in fig.  The shear strain = The angle of distortion  ø                          ø = RR'/ RQ ---(1) Shear strain = Shear stress /Rigidity modulus                

PORTAL METHOD and CANTILEVER METHOD

The behavior of a structure subjected to horizontal forces depends on its height to width ratio. The deformation in low-rise structures, where the height is smaller than its width, is characterized predominantly by shear deformations. In high rise building, where height is several times greater than its lateral dimensions, is dominated by bending action. To analyze the structures subjected to horizontal loading we have two methods. Portal method  and Cantilever method 1. PORTAL METHOD The portal method is an approximate analysis used for analysing building frames subjected to lateral loads such as Wind loads/ seismic forces.  Since shear deformations are dominant in low rise structures, the method makes simplifying assumptions regarding horizontal shear in columns.  Each bay of a structure is treated as a portal frame, and horizontal force is distributed equally among them. Assumptions in portal method   1. The points of inflection are located at the mid-height of each column above th

Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is  τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.  Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us consider on