Problem 885 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 08:56 pm

Problem 885
Solve for the support moments in Problem 825 if the ends are perfectly fixed instead of simply supported.
 

825-continuous-beam.gif

 

Problem 884 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 08:51 pm

Problem 884
Compute the moments over the supports of the beam shown in Fig. P-856.
 

856-overhanging-fixed-beam.gif

 

Problem 882 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 08:26 pm

Problem 882
Compute the moments over the supports of the beam shown in Fig. P-849.
 

849-fixed-continuous.gif

 

Problem 881 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 08:21 pm

Problem 881
Compute the moments over the supports of the beam shown in Fig. P-846.
 

846-continuous-beam-with-one-end-fixed.gif

 

Problem 880 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 06:46 pm

Problem 880
Compute the moments over the supports of the beam shown in Fig. P-845.
 

845-continuous-beam-overhang-fixed.gif

 

Problem 879 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 06:35 pm

Problem 879
Using moment-distribution method, solve for the moments over supports R2 and R3 of the continuous beam in Figure P-827.
 

827-continuous-beam.gif

 

Problem 878 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 06:26 pm

Problem 878
Using moment-distribution method, solve for the moments over supports R2 and R3 of the continuous beam in Figure P-826.
 

826-continuous-beam.gif

 

Problem 877 | Continuous Beam by Moment Distribution Method Jhun Vert Sat, 04/25/2020 - 04:08 pm

Problem 877
By means of moment-distribution method, solve the moment at R2 and R3 of the continuous beam shown in Fig. P-815.
 

815-continuous-beam-triangular-concentrated-loads.gif

 

Problem 872 | Continuous Beam with Spring End-Support Jhun Vert Sat, 04/25/2020 - 04:01 pm

Problem 872
Repeat Problem 871 assuming the loadings on the spans are interchanged.
 

Solution 872
$E = 1.5 \times 10^6 ~ \text{psi} = \left( 1.5 \times 10^6 \dfrac{\text{lb}}{\text{in.}^2} \right)\left( \dfrac{12 ~ \text{in.}}{1 ~ \text{ft.}} \right)^2 = 216 \times 10^6 ~ \text{psf}$