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		<title>en&gt;Marcos: correctly placed bracket</title>
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		<updated>2012-07-09T08:08:05Z</updated>

		<summary type="html">&lt;p&gt;correctly placed bracket&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;The &amp;#039;&amp;#039;&amp;#039;slope deflection method&amp;#039;&amp;#039;&amp;#039; is a [[structural analysis]] method for [[beam (structure)|beams]] and [[Framing (construction)|frames]] introduced in 1915 by George A. Maney.&amp;lt;ref name=&amp;quot;history1&amp;quot;&amp;gt;{{Cite journal|first=George A.|last=Maney|year=1915|title=Studies in Engineering|publisher=University of Minnesota|location=Minneapolis|postscript=&amp;lt;!--None--&amp;gt;}}&amp;lt;/ref&amp;gt; The slope deflection method was widely used for more than a decade until the [[moment distribution method]] was developed.&lt;br /&gt;
&lt;br /&gt;
== Introduction ==&lt;br /&gt;
By forming &amp;#039;&amp;#039;&amp;#039;slope deflection equations&amp;#039;&amp;#039;&amp;#039; and applying joint and shear equilibrium conditions, the rotation angles (or the slope angles) are calculated. Substutituting them back into the slope deflection equations, member end moments are readily determined.&lt;br /&gt;
&lt;br /&gt;
== Slope deflection equations ==&lt;br /&gt;
&lt;br /&gt;
The slope deflection equations can also be written using the stiffness factor &amp;lt;math&amp;gt;K=\frac{I_{ab}}{L_{ab}}&amp;lt;/math&amp;gt; and the chord rotation &amp;lt;math&amp;gt;\psi =\frac{ \Delta}{L_{ab}}&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
=== Derivation of slope deflection equations ===&lt;br /&gt;
When a [[Beam (structure)#Types of beams|simple beam]] of length &amp;lt;math&amp;gt;L_{ab}&amp;lt;/math&amp;gt; and flexural rigidity &amp;lt;math&amp;gt;E_{ab} I_{ab}&amp;lt;/math&amp;gt; is loaded at each end with counterclockwise moments &amp;lt;math&amp;gt;M_{ab}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;M_{ba}&amp;lt;/math&amp;gt;, member end rotations occur in the same direction. These rotation angles can be calculated using the [[unit dummy force method]] or [[Darcy&amp;#039;s Law]].&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;\theta_a - \frac{\Delta}{L_{ab}}= \frac{L_{ab}}{3E_{ab} I_{ab}} M_{ab} - \frac{L_{ab}}{6E_{ab} I_{ab}} M_{ba}&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;\theta_b - \frac{\Delta}{L_{ab}}= - \frac{L_{ab}}{6E_{ab} I_{ab}} M_{ab} + \frac{L_{ab}}{3E_{ab} I_{ab}} M_{ba}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Rearranging these equations, the slope deflection equations are derived.&lt;br /&gt;
&lt;br /&gt;
== Equilibrium conditions ==&lt;br /&gt;
&lt;br /&gt;
=== Joint equilibrium ===&lt;br /&gt;
Joint equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments i.e. be in equilibrium. Therefore,&lt;br /&gt;
:&amp;lt;math&amp;gt;\Sigma \left( M^{f} + M_{member} \right) = \Sigma M_{joint}&amp;lt;/math&amp;gt;&lt;br /&gt;
Here, &amp;lt;math&amp;gt;M_{member}&amp;lt;/math&amp;gt; are the member end moments, &amp;lt;math&amp;gt;M^{f}&amp;lt;/math&amp;gt; are the [[fixed end moments]], and &amp;lt;math&amp;gt;M_{joint}&amp;lt;/math&amp;gt; are the external moments directly applied at the joint.&lt;br /&gt;
&lt;br /&gt;
=== Shear equilibrium ===&lt;br /&gt;
When there are chord rotations in a frame, additional equilibrium conditions, namely the shear equilibrium conditions need to be taken into account.&lt;br /&gt;
&lt;br /&gt;
== Example ==&lt;br /&gt;
[[Image:MomentDistributionMethod.jpg|thumb|434px|right|Example]]&lt;br /&gt;
The statically indeterminate beam shown in the figure is to be analysed.&lt;br /&gt;
*Members AB, BC, CD have the same length &amp;lt;math&amp;gt; L = 10 \ m &amp;lt;/math&amp;gt;.&lt;br /&gt;
*Flexural rigidities are EI, 2EI, EI respectively.&lt;br /&gt;
*Concentrated load of magnitude &amp;lt;math&amp;gt; P = 10 \ kN &amp;lt;/math&amp;gt; acts at a distance &amp;lt;math&amp;gt; a = 3 \ m &amp;lt;/math&amp;gt; from the support A.&lt;br /&gt;
*Uniform load of intensity &amp;lt;math&amp;gt; q = 1 \ kN/m&amp;lt;/math&amp;gt; acts on BC.&lt;br /&gt;
*Member CD is loaded at its midspan with a concentrated load of magnitude &amp;lt;math&amp;gt; P = 10 \ kN &amp;lt;/math&amp;gt;.&lt;br /&gt;
In the following calculations, clockwise moments and rotations are positive.&lt;br /&gt;
&lt;br /&gt;
=== Degrees of freedom ===&lt;br /&gt;
Rotation angles &amp;lt;math&amp;gt;\theta_A&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;\theta_B&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;\theta_C&amp;lt;/math&amp;gt;, of joints A, B, C, respectively are taken as the unknowns. There are no chord rotations due to other causes including support settlement.&lt;br /&gt;
&lt;br /&gt;
=== Fixed end moments ===&lt;br /&gt;
Fixed end moments are:&lt;br /&gt;
:&amp;lt;math&amp;gt;M _{AB} ^f = - \frac{P a b^2 }{L ^2} = - \frac{10 \times 3 \times 7^2}{10^2} = -14.7 \mathrm{\,kN \,m}&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M _{BA} ^f = \frac{P a^2 b}{L^2} = \frac{10 \times 3^2 \times 7}{10^2} = 6.3 \mathrm{\,kN \,m}&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M _{BC} ^f = - \frac{qL^2}{12} = - \frac{1 \times 10^2}{12} = - 8.333 \mathrm{\,kN \,m}&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M _{CB} ^f = \frac{qL^2}{12} = \frac{1 \times 10^2}{12} = 8.333 \mathrm{\,kN \,m}&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M _{CD} ^f = - \frac{PL}{8} = - \frac{10 \times 10}{8} = -12.5 \mathrm{\,kN \,m}&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M _{DC} ^f = \frac{PL}{8} = \frac{10 \times 10}{8} = 12.5 \mathrm{\,kN \,m}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Slope deflection equations ===&lt;br /&gt;
The slope deflection equations are constructed as follows:&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{AB} = \frac{EI}{L} \left( 4 \theta_A + 2 \theta_B \right) = 0.4EI \theta_A + 0.2EI \theta_B&amp;lt;/math&amp;gt; &lt;br /&gt;
:&amp;lt;math&amp;gt;M_{BA} = \frac{EI}{L} \left( 2 \theta_A + 4 \theta_B \right) = 0.2EI \theta_A + 0.4EI \theta_B&amp;lt;/math&amp;gt; &lt;br /&gt;
:&amp;lt;math&amp;gt;M_{BC} = \frac{2EI}{L} \left( 4 \theta_B + 2 \theta_C \right) = 0.8EI \theta_B + 0.4EI \theta_C&amp;lt;/math&amp;gt; &lt;br /&gt;
:&amp;lt;math&amp;gt;M_{CB} = \frac{2EI}{L} \left( 2 \theta_B + 4 \theta_C \right) = 0.4EI \theta_B + 0.8EI \theta_C&amp;lt;/math&amp;gt; &lt;br /&gt;
:&amp;lt;math&amp;gt;M_{CD} = \frac{EI}{L} \left( 4 \theta_C \right) = 0.4EI \theta_C &amp;lt;/math&amp;gt; &lt;br /&gt;
:&amp;lt;math&amp;gt;M_{DC} = \frac{EI}{L}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Joint equilibrium equations ===&lt;br /&gt;
Joints A, B, C should suffice the equilibrium condition. Therefore&lt;br /&gt;
:&amp;lt;math&amp;gt;\Sigma M_A = M_{AB} + M_{AB}^f = 0.4EI \theta_A + 0.2EI \theta_B  - 14.7 = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;\Sigma M_B = M_{BA} + M_{BA}^f + M_{BC} + M_{BC}^f = 0.2EI \theta_A + 1.2EI \theta_B + 0.4EI \theta_C - 2.033 = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;\Sigma M_C = M_{CB} + M_{CB}^f + M_{CD} + M_{CD}^f = 0.4EI \theta_B + 1.2EI \theta_C  - 4.167 = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Rotation angles ===&lt;br /&gt;
The rotation angles are calculated from simultaneous equations above.&lt;br /&gt;
:&amp;lt;math&amp;gt;\theta_A = \frac{40.219}{EI} &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;\theta_B = \frac{-6.937}{EI} &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;\theta_C = \frac{5.785}{EI} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Member end moments ===&lt;br /&gt;
Substitution of these values back into the slope deflection equations yields the member end moments (in kNm):&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{AB} = 0.4 \times 40.219 + 0.2 \times \left( -6.937 \right) - 14.7 = 0 &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{BA} = 0.2 \times 40.219 + 0.4 \times \left( -6.937 \right) + 6.3 = 11.57 &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{BC} = 0.8 \times \left( -6.937 \right) + 0.4 \times 5.785 - 8.333 = -11.57 &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{CB} = 0.4 \times \left( -6.937 \right) + 0.8 \times 5.785 + 8.333 = 10.19 &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{CD} = 0.4 \times 5.785 - 12.5 = -10.19 &amp;lt;/math&amp;gt;&lt;br /&gt;
:&amp;lt;math&amp;gt;M_{DC} = 0.2 \times 5.785 + 12.5 = 13.66 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
*[[Beam theory]]&lt;br /&gt;
&lt;br /&gt;
== Notes ==&lt;br /&gt;
{{reflist|1}}&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
*{{cite book|last=Norris|first=Charles Head|coauthors=John Benson Wilbur, Senol Utku|title=Elementary Structural Analysis|edition=3rd|year=1976|publisher=McGraw-Hill|isbn=0-07-047256-4|pages=313–326}}&lt;br /&gt;
*{{cite book|last=McCormac|first=Jack C.|coauthors=James K. Nelson, Jr.|title=Structural Analysis: A Classical and Matrix Approach|edition=2nd |year=1997|publisher=Addison-Wesley|isbn=0-673-99753-7|pages=430–451}}&lt;br /&gt;
*{{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=357–389}}&lt;br /&gt;
&lt;br /&gt;
[[Category:Structural analysis]]&lt;/div&gt;</summary>
		<author><name>en&gt;Marcos</name></author>
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