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Building Design and Construction Handbook - Merritt - Ventech!

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STRUCTURAL THEORY 5.89<br />

� �<br />

4EI 1 6EI d<br />

F<br />

ML � �L � �R � M L � (5.135)<br />

L 2 L L<br />

4EI 1 6EI d<br />

F<br />

MR � ��R � �L� � M R � (5.136)<br />

L 2 L L<br />

where E � modulus of elasticity<br />

I � moment of inertia of the cross section<br />

Note that if end L moves downward with respect to R, the sign for d in the<br />

preceding equations is changed.<br />

If the end moments M L <strong>and</strong> M R are known <strong>and</strong> the end rotations are to be<br />

determined, Eqs. (5.131) to (5.134) can be solved for � L <strong>and</strong> � R or derived by<br />

superposition of simple-beam components, as is done in Art. 5.11.4. For beams<br />

with moment of inertia varying along the span:<br />

d<br />

F F<br />

�L � (ML � M L) �L � (MR � M R)<br />

� � (5.137)<br />

L<br />

d<br />

F F<br />

�R � (MR � M R) �R � (ML � M L)<br />

� � (5.138)<br />

L<br />

where � is given by Eq. (5.105) <strong>and</strong> � by Eq. (5.106). For beams with constant<br />

moment of inertia:<br />

L L d<br />

F F<br />

�L � (ML � M L) � (MR � M R)<br />

�<br />

3EI 6EI L<br />

(5.139)<br />

L L d<br />

F F<br />

�R � (MR � M R) � (ML � M L)<br />

�<br />

3EI 6EI L<br />

(5.140)<br />

The slope-deflection equations can be used to determine end moments <strong>and</strong> rotations<br />

of the spans of continuous beams by writing compatibility <strong>and</strong> equilibrium<br />

equations for the conditions at each support. For example, the sum of the moments<br />

at each support must be zero. Also, because of continuity, the member must rotate<br />

through the same angle on both sides of every support. Hence, M L for one span,<br />

given by Eq. (5.133) or (5.135), must be equal to �M R for the adjoining span,<br />

given by Eq. (5.134) or (5.136), <strong>and</strong> the end rotation � at that support must be the<br />

same on both sides of the equation. One such equation with the end rotations at<br />

the supports as the unknowns can be written for each support. With the end rotations<br />

determined by simultaneous solution of the equations, the end moments can be<br />

computed from the slope-deflection equations <strong>and</strong> the continuous beam can now<br />

be treated as statically determinate.<br />

See also Arts. 5.11.9 <strong>and</strong> 5.13.2.<br />

(C. H. Norris et al., ‘‘Elementary Structural Analysis,’’ 4th ed., McGraw-Hill<br />

Book Company, New York.)<br />

5.11.6 Moment Distribution<br />

The frame in Fig. 5.74 consists of four prismatic members rigidly connected together<br />

at O at fixed at ends A, B, C, <strong>and</strong> D. If an external moment U is applied at

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