Skip to main content

Moment Distribution method

The Moment Distribution method is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross,  in 1930 .The moment distribution method falls into the category of displacement method of structural analysis.
The method only accounts for flexural effects and ignores axial and shear effects.
In the moment distribution method, every joint of the structure to be analysed is fixed so as to develop the fixed-end moments. Then each fixed joint is sequentially released and the fixed-end moments (which by the time of release are not in equilibrium) are distributed to adjacent members until equilibrium is achieved. 
The moment distribution method in mathematical terms can be demonstrated as the process of solving a set of simultaneous equations by means of iteration.

Distribution factors

When a joint is released and begins to rotate under the unbalanced moment, resisting forces develop at each member framed together at the joint. Although the total resistance is equal to the unbalanced moment, the magnitudes of resisting forces developed at each member differ by the members' flexural stiffness. Distribution factors can be defined as the proportions of the unbalanced moments carried by each of the members. 
In mathematical terms, distribution factor of member k framed at joint j is given as:
D_{jk} = \frac{\frac{E_k I_k}{L_k}}{\sum_{i=1}^{i=n} \frac{E_i I_i}{L_i}}
where n is the number of members framed at the joint.
The sum of all distribution factors at joint will always equal to 'one' 

Carryover factors

When a joint is released, balancing moment occurs to counterbalance the unbalanced moment which is initially the same as the fixed-end moment. This balancing moment is then carried over to the member's other end. The ratio of the carried-over moment at the other end to the fixed-end moment of the initial end is the carryover factor. It is equal to 0.5
Method of Analysis

  •  Calculate stiffness factors for each member 
  •  Calculate distribution factors at both ends of  each member
  •  Determine carryover factors at both ends of each member  
  • Assume all joints are fixed and calculate fixed-end moments for each member 
  • Balance pinned (to zero) and cantilevered ends and distribute half the moment to the opposite end.
  • Distribute the unbalanced moments at all other joints to each adjacent member based on the distribution factor.
  • Carryover the distributed moments to the opposite ends of the each member using the carryover factors.
  • Iterate steps 6 and 7 until moment imbalance at each joint approaches zero 

Labels:

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           ...
5 comments
Read more

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...
1 comment
Read more

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...
3 comments
Read more