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{{Short description|Structural analysis technique for statically indeterminate structures}}
{{Distinguish|
The '''moment distribution method''' is a [[structural analysis]] method for [[statically indeterminate]] [[Beam (structure)|beam]]s and [[Framing (construction)|frames]] developed by [[Hardy Cross]]. It was published in 1930 in an [[American Society of Civil Engineers|ASCE]] journal.<ref name="asce1">{{Cite news|first=Hardy|last=Cross|year=1930|title=Analysis of Continuous Frames by Distributing Fixed-End Moments|periodical=Proceedings of the American Society of Civil Engineers|publisher=ASCE|pages=919–928
== Introduction ==
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=== Fixed end moments ===
[[Fixed end moments]] are the moments produced at member ends by external loads.
===
The [[
=== Distribution factors ===
When a joint is being 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'
:<math>D_{jk} = \frac{\frac{E_k I_k}{L_k}}{\sum_{i=1}^{i=n} \frac{E_i I_i}{L_i}}</math>
where n is the number of members framed at the joint.
=== Carryover factors ===
When a joint is released, balancing moment occurs to counterbalance the unbalanced moment.
==== Determination of carryover factors ====
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Once a sign convention has been chosen, it has to be maintained for the whole structure. The traditional engineer's sign convention is not used in the calculations of the moment distribution method although the results can be expressed in the conventional way. In the BMD case, the left side moment is clockwise direction and other is anticlockwise direction so the bending is positive and is called sagging.
=== Framed
Framed
== Example ==
[[Image:MomentDistributionMethod.jpg|thumb|434px|right|Example]]
The statically indeterminate beam shown in the figure is to be analysed.
The beam is considered to be three separate members, AB, BC, and CD, connected by fixed end (moment resisting) joints at B and C.
*Members AB, BC, CD have the same [[Span (architecture)|span]] <math> L = 10 \ m </math>.
*Flexural rigidities are EI, 2EI, EI respectively.
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*Uniform load of intensity <math> q = 1 \ kN/m</math> acts on BC.
*Member CD is loaded at its midspan with a concentrated load of magnitude <math> P = 10 \ kN </math>.
In the following calculations,
=== Fixed end moments ===
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:<math>M _{DC} ^f =\frac{PL}{8} =\frac{10 \times 10}{8} = + 12.500 \ kN\cdot m</math>
===
The
:<math>D_{BA} = \frac{\frac{3EI}{L}}{\frac{3EI}{L}+\frac{4\times 2EI}{L}} = \frac{\frac{3}{10}}{\frac{3}{10}+\frac{8}{10}} = \frac{3}{11} = 0.(27)</math>
:<math>D_{BC} = \frac{\frac{4\times 2EI}{L}}{\frac{3EI}{L}+\frac{4\times 2EI}{L}} = \frac{\frac{8}{10}}{\frac{3}{10}+\frac{8}{10}} = \frac{8}{11} = 0.(72)</math>
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Numbers <span style="background-color:#F8F8F8; border-style:solid; border-width:1px; border-color:#AAAAAA;">in grey</span> are balanced moments; arrows (<span style="border-style:solid; border-width:1px; border-color:#AAAAAA;"> → / ← </span>) represent the carry-over of moment from one end to the other end of a member.
* Step 1: As joint A is released, balancing moment of magnitude equal to the fixed end moment <math>M_{AB}^{f} = 14.700 \mathrm{\,kN \,m}</math> develops and is carried-over from joint A to joint B. * Step 2: The unbalanced moment at joint B now is the summation of the fixed end moments <math>M_{BA}^{f}</math>, <math>M_{BC}^{f}</math> and the carry-over moment from joint A. This unbalanced moment is distributed to members BA and BC in accordance with the distribution factors <math>D_{BA} = 0.2727</math> and <math>D_{BC} = 0.7273</math>. Step 2 ends with carry-over of balanced moment <math>M_{BC}=3.867 \mathrm{\,kN \,m}</math> to joint C. Joint A is a roller support which has no rotational restraint, so moment carryover from joint B to joint A is zero. * Step 3: The unbalanced moment at joint C now is the summation of the fixed end moments <math>M_{CB}^{f}</math>, <math>M_{CD}^{f}</math> and the carryover moment from joint B. As in the previous step, this unbalanced moment is distributed to each member and then carried over to joint D and back to joint B. Joint D is a fixed support and carried-over moments to this joint will not be distributed nor be carried over to joint C. * Step 4: Joint B still has balanced moment which was carried over from joint C in step 3. Joint B is released once again to induce moment distribution and to achieve equilibrium. * Steps 5 - 10: Joints are released and fixed again until every joint has unbalanced moments of size zero or neglectably small in required precision. Arithmetically summing all moments in each respective columns gives the final moment values. === Result ===
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=== Result via displacements method ===
As the Hardy Cross method provides only approximate results, with a margin of error inversely proportionate to the number of iterations, it is important{{citation needed|date=September 2012}} to have an idea of how accurate this method might be. With this in mind, here is the result obtained by using an exact method: the [[
For this, the displacements method equation assumes the following form:
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== References ==
*{{cite book|last=Błaszkowiak|first=Stanisław|author2=Zbigniew Kączkowski|title=Iterative Methods in Structural Analysis|year=1966|publisher=Pergamon Press, Państwowe Wydawnictwo Naukowe}}
*{{cite book|last=Norris|first=Charles Head|author2=John Benson Wilbur
*{{cite book
*{{cite book|last=Yang|first=Chang-hyeon|title=Structural Analysis|url=http://www.cmgbook.co.kr/category/sub_detail.html?no=1017|edition=4th|date=2001-01-10|publisher=Cheong Moon Gak Publishers|language=Korean|___location=Seoul|isbn=89-7088-709-1|pages=391–422|access-date=2007-08-31|archive-url=https://web.archive.org/web/20071008135424/http://www.cmgbook.co.kr/category/sub_detail.html?no=1017|archive-date=2007-10-08|url-status=dead}}
*{{cite
[[Category:Structural analysis]]
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