Posted On :

The method solves for the joint mo ments in continuous beams and rigid frames by successive approxi mation

**MOMENT DISTRIBUTION METHOD **

**INTRODUCTION
AND BASIC PRINCIPLES**

**Introduction**

(Method
developed by Prof. Hardy Cross in 1932)

The method solves for the joint mo ments in continuous beams and rigid frames by successive approxi mation

**Statement
of Basic Principles**

Consider
the continuous beam ABC D, subjected to the given loads,

as shown
in Figure below. Assume that only rotation of joints occur

at B, C
and D, and that no support d isplacements occur at B, C and

D. Due to the applied loads in spans AB, BC and
CD, rotations occur at B, C and D

In order to solve the problem in a su ccessively approximating
manner,

it can be visualized to be made up of a continued two-stage
problems

viz., that of locking and releasing the joints in a continuous
sequence.

**The joints B, C and D are locked
in position before any load is applied on the b eam ABCD; then given loads are
applied on the bea m. Since the joints of beam ABCD are locke d in position,
beams AB, BC and CD acts as ind ividual and separate fixed beams, subjected to
the applied loads; these loads develop fixed ennd moments.**

In beam AB

Fixed end moment at A = -wl^{2}/12 = - (15)(8)(8)/12 =
- 80 kN.m

Fixed end moment at B = +wl^{2}/12 = +(15)(8)(8)/12 =
+ 80 kN.m

In beam BC

Fixed end moment at B = - (Pab^{2})/l^{2} = -
(150)(3)(3)^{2}/6^{2}

= -112.5
kN.m

Fixed end moment at C = + (Pab^{2})/l^{2} = +
(150)(3)(3)^{2}/6^{2}

= + 112.5

In beam AB

Fixed end moment at C = -wl^{2}/12 = - (10)(8)(8)/12 =
- 53.33 kN.m

Fixed end moment at D = +wl^{2}/12 = +(10)(8)(8)/12
= + 53.33kN.m

Since the
joints B, C and D were fix ed artificially (to compute the the fixed-end
moments), now the joints B, C and D are released and allowed to rotate. Due to
the joint release, the joints rotate
maintaining the continuous nature of the beam. Due to the joint release, the
fixed en d moments on either side of joints B, C and D act in the opposite
direction now, and cause a net un balanced moment to occur at the joint.

These unbalanced moments act at th e joints and modify the
joint moments at B, C a nd D, according to their relative stiffnesses at the
respective joints. The joint moments are distributed t o either side of the
joint B, C or D, according to their reelative stiffnesses. These distributed
moments al so modify the moments at the opposite side of the beam span, viz.,
at joint A in span AB, at joints B and C in span BC and at joints C and D in
span CD. This modification is dependent on the carry-over factor (which is
equal to 0.5 in this case);

**The carry-over moment becomes the unbalanced
moment at the joints to whic h they are carried over. Steps 3 and 4 are
repeated t ill the carry-over or distributed moment beco mes small**.

**Sum up
all the moments at each of the joint **to obtain the joint moments.

**SOME BASIC
DEFINITIONS**

In order to understand the five steps mentioned in section
7.3, some words need to be defined and relevant derivations made.

**1Stiffness
and Carry-over Factors**

Stiffness
= Resistance offered by m ember to a unit displacement or rotation at a point,
for given support constraint conditions

A clockwise moment M_{A} is applied at A to produce a
+ve bending in beam AB. Fin d q_{A} and M_{B}.

**Using
method of consistent defor mations**

**Considering moment M _{B,}**

**M _{B}
+ M_{A} + R_{A}L = 0**

**\****M _{B} = M_{A}/2= (1/2)M_{A}**

**Carry - over Factor = 1/2**

**2 Distribution Factor**

Distribution
factor is the ratio according to which an externally applied unbalanced moment
M at a joint is apportioned to the various m embers mating at the joint

**M = M _{BA}
+ M_{BC} + M_{BD}**

**Modified Stiffness Factor**

The
stiffness factor changes when t he far end of the beam is simply-supported.

As per
earlier equations for deforma tion, given in Mechanics of Solids text-books.

**Solve the previously given proble m by the moment
distribution method**

**Fixed end
moments**

**Stiffness
Factors (Unmodified Stifffness**

**Distribution
Factors**

**Computation
of Shear Forces**

Tags : Civil - Structural Analysis - Moment Distribution Method

Last 30 days 282 views
Related words : ### What is Structural Analysis - Moment Distribution Method with Solved Problems Define Structural Analysis - Moment Distribution Method with Solved Problems Definition of Structural Analysis - Moment Distribution Method with Solved Problems where how
meaning of Structural Analysis - Moment Distribution Method with Solved Problems
lecturing notes for Structural Analysis - Moment Distribution Method with Solved Problems lecture notes question and answer for Structural Analysis - Moment Distribution Method with Solved Problems answer
Structural Analysis - Moment Distribution Method with Solved Problems study material Structural Analysis - Moment Distribution Method with Solved Problems assignment Structural Analysis - Moment Distribution Method with Solved Problems reference description of Structural Analysis - Moment Distribution Method with Solved Problems
explanation of Structural Analysis - Moment Distribution Method with Solved Problems brief detail of Structural Analysis - Moment Distribution Method with Solved Problems easy explanation solution Structural Analysis - Moment Distribution Method with Solved Problems wiki
Structural Analysis - Moment Distribution Method with Solved Problems wikipedia how why is who were when is when did where did Structural Analysis - Moment Distribution Method with Solved Problems list of Structural Analysis - Moment Distribution Method with Solved Problems school assignment college assignment
Structural Analysis - Moment Distribution Method with Solved Problems college notes school notes kids with diagram or figure or image difference between Structural Analysis - Moment Distribution Method with Solved Problems www.readorrefer.in - Read Or Refer

Recent New Topics :