three-moment equation factors

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

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
See Fig. P-821.
 

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

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

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

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

Problem 818
In Problem 817, determine the changed value of the applied couple that will cause M₂ to become zero.
 

Problem 817
Find the moment under the support R₂ of the continuous beam shown in Fig. P-817.
 

Problem 816
Determine the lengths of the overhangs in Fig. P-816 so that the moments over the supports will be equal
 

Problem 815
Find the moment at R₂ and R₃ of the continuous beam shown in Fig. P-815.
 

Problem 814
Find the moment at R₂ of the continuous beam shown in Fig. P-814.
 

Problem 813
Determine the moment over the support R₂ of the beam shown in Fig. P-813.