Skip to main content

SUBSTITUTE FRAME METHOD

Analysis of multi-storey building frames involves lot of complications and tedious calculations by using conventional methods. To carry out exact analysis is a time consuming task. Substitute frame method for analysis of multistory frame can be handy in approximate and quick analysis. This method has been applied only for vertical loading conditions. 

Substitute frame method 

Substitute frame method assumes that the moments in the beams of any floor are influenced by loading on that floor alone. The influence of loading on the lower or upper floors is ignored altogether.
The process involves the division of multi-storied structure into smaller frames. These sub frames are known as Equivalent frames or Substitute frames.
The substitute frames are formed by the beams at the floor level under consideration, together with the columns above and below with their far ends fixed. 

The sub frames are usually analyzed by the moment distribution method, using only Two cycle of distribution. It is only necessary to consider the loads on the two nearest spans on each side of the point .The distributed B.M are not carried over far ends of the columns in this process.
The moments in the columns are computed at each floor level independently and retained at that floor irrespective of further analysis.

Steps for the Analysis 

(1) Select a substitute frame, by taking-floor beam with columns of lower and upper storeys fixed at far ends. 
(2) Cross sectional dimensions of beams and columns may be chosen such that moment of inertia of beam is 1.5 to 2 times that of a column and find distribution factors at a joint considering stiffnesses of beams and columns. 
(3) Calculate the dead load and live load on beam. Live load should be placed in such a way that it causes worst effect at the section considered. (Live loads are placed alternate and adjacent spans as shown below should be adopted)
a) For maximum positive moments at mid point of span : Live loads are on alternate spans 

b) For maximum negative moments at the point of support : Live loads are on adjacent spans.

c) For maximum moments in columns : Live loads are on alternate spans.


(4) Find the initial fixed end moments and analyse this frame by moment distribution method. 
(5) Finally draw shear and moment diagram indicating values at critical section. 
Note: 
1.The Height of all columns should be same in a particular storey.
2. Sway of substitute frame is ignored even during unsymmetrical loading.

Comments

Post a Comment

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           ...

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 co...

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 e...