three-moment equation

Problem 835 | Reactions of Continuous Beams

Problem 835
Refer to Problem 827 for which M2 = -1215 lb·ft and M3 = -661 lb·ft.
 

835-shear-diagram.gif

 

Problem 833 | Reactions of Continuous Beams

Problem 833
Refer to Problem 825 for which M2 = -980 lb·ft and M3 = -1082 lb·ft.
 

833-shear-diagram.gif

Problem 822 | Continuous Beam by Three-Moment Equation

Problem 822
Solve Prob. 821 if the concentrated load is replaced by a uniformly distributed load of intensity wo over the middle span.
 

822-beta-alpha-span-uniform-load.gif

 

Answers:
$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\beta}{4(\alpha + 1)(1 + \beta) - 1}$

$M_3 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\alpha}{4(1 + \alpha)(1 + \beta) - 1}$
 

Problem 821 | Continuous Beam by Three-Moment Equation

Problem 821
See Fig. P-821.
 

821-alpha-beta-continuous-beam.gif

 

$M_2 = -\dfrac{3PL}{8} \cdot \dfrac{1 + 2\beta}{4(1 + \alpha)(1 + \beta) - 1}$           answer

$M_3 = -\dfrac{3PL}{8} \cdot \dfrac{1 + 2\alpha}{4(1 + \alpha)(1 + \beta) - 1}$           answer
 

Problem 820 | Continuous Beam by Three-Moment Equation

Problem 820
Solve Prob. 819 if the concentrated load is replaced by a uniformly distributed load of intensity wo over the first span.
 

820-continuous-beam-uniform-load.gif

 

Problem 819 | Continuous Beam by Three-Moment Equation

Problem 819
Find the moment under the supports R2 and R3 for the beam shown in Fig. P-819.
 

819-length-of-spans-continuous-beams.gif

 

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