fixed-end moment

Problem 883
Compute the moments over the supports of the beam shown in Fig. P-853.
 

Moment distribution is based on the method of successive approximation developed by Hardy Cross (1885–1959) in his stay at the University of Illinois at Urbana-Champaign (UIUC). This method is applicable to all types of rigid frame analysis.
 

Problem 852
Find the moments over the supports for the continuous beam in Figure P-852. Use the results of Problems 850 and 851.
 

Answers
$M_1 = -146.43 ~ \text{N}\cdot\text{m}$

$M_2 = -307.14 ~ \text{N}\cdot\text{m}$

$M_3 = -521.43 ~ \text{N}\cdot\text{m}$
 

Problem 851
Replace the distributed load in Problem 850 by a concentrated load P at the midspan and solve for the moment over the supports.
 

Problem 850
Determine the moment over the supports for the beam loaded as shown in Fig. P-850.
 

Problem 849
Find the moments over the support for the beam shown in Fig. P-849.
 

Problem 848
Determine the support moments and reactions for the beam shown in Fig. P-848.
 

Summary for the value of end moments and deflection of perfectly restrained beam carrying various loadings. Note that for values of EIy, y is positive downward.
 

Case 1: Concentrated load anywhere on the span of fully restrained beam

End moments
$M_A = -\dfrac{Pab^2}{L^2}$

$M_B = -\dfrac{Pa^2b}{L^2}$
 

Value of EIy
$\text{Midspan } EI\,y = \dfrac{Pb^2}{48}(3L - 4b)$

Note: only for b > a

Problem 737
In the perfectly restrained beam shown in Fig. P-737, support B has settled a distance Δ below support A. Show that M_B = -M_A = 6EIΔ/L².