ALPHA COLLEGE OF ENGINEERING AND TECHNOLOGY STRUCTURAL ANALYSIS-1 TOPIC – INTRODUCTION OF MOMENT DISTRIBUTION METHOD PREPARED BY, KHUNT AKSHAY (1305106026) LAKHANI HANNY(1305106027) LEUVA NISARG (1305106028)

MOMENT DISTRIBUTION METHOD Introduction Important terms Sign convention Steps for moment distribution method Portal frames with side sway or no side sway

INTRODUCTION Professor Hardy Cross of University of Illinois of U.S.A. invented this method in 1930. It can be used to analyze all types of statically indeterminate beams or rigid frames. It consist of solving the linear simultaneous equations that were obtain in the slope deflection method by successive.

INTRODUCTION This method is very simple and gives the results with sufficient accuracy. In this method, the numerical solution is performed in a tabular form. It is basically an iterative process. In this method the joints are released one by one in succession.

INTRODUCTION At each released joint the unbalanced moments are distributed to all the ends of the members meeting at that joint. A certain fraction of these distributed moments are carried over to the far end of members. This completes one cycle of operation. Increased no of cycles would result in more accuracy.

IMPORTANT TERMS STIFFNESS: Resistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions. MB MA A B A A RA RB L E, I – Member properties

Stiffness = moment rotation k = M θ but, θ = 1

When far end fixed : The moment required to rotate the near end of a prismatic beam through a until angle, without translation, the far end being fixed, is given by M = k = 4EI/L MB MA A B A A RA RB L E, I – Member properties

When far end is simply supported: The moment required to rotate the near end of a prismatic beam through a unit angle, without translation, the far end being simple, is given by M = K = MA A A B L RA RB

2. CARRY OVER MOMENT : A beam with the far end fixed , if M moment is applied at the near end , M moment is produced at the far end. This M moment is called carry over moment. For abeam with far end fixed M’ = 2EI L And, M = 4EI/L

When far end simple And, M’ = 0 M = moment applied at near end M’ = carry over moment M =

3. Carry over factor (C.O.F) : c.o. factor = moment produce at far end moment applied at near end = M’ M When far end of beam is fixed, c.o. factor = 2EI/L 4EI/L = ½ = 0.5

DISTIBUTION FACTOR : The factor by which the applied moment is distributed to the member is known as the distribution factor(D.F). M B A MBC C MBA MBD D

Distribution factor=k = stiffness of a member Σk sum of stiffness total stiffness = Σk i.e., M = MBA + MBC + MBD

SIGN CONVENTIONS Clockwise moments that act on the member are considered positive, whereas anticlockwise moments considered negative. Clockwise rotation is considered positive and anticlockwise rotation considered negative. When beam rotates in clockwise direction then the settlement will be taken as positive and beam rotates in anticlockwise direction then settlement will be taken as negative.

STEPS FOR MOMENT DISTRIBUTION METHOD Find out fixed end moments for each member. Find out distribution factor. Prepare moment distribution table, in which the vertical lines represent the joints. At each joint , find out unbalanced moment and distribute them. Then they are balanced. The distribution moments are carried over to the other end of the member with same sign.

The cycle may be repeated depending on the accuracy desired. The final may be found out by summing all the moments. At , the intermediate joints , the sum of moments must be zero. At internal joint, sum of distribution factor is 1.

PORTAL FRAMES WITH SIDE SWAY

FRAME WITH SIDE SWAY

FRAME WITH NO SIDE SWAY

Properly restrained in the horizontal direction. No restrained, but the frame is symmetry in respect to both loading and geometry.

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