# fixed-end moment

## Problem 883 | Continuous Beam by Moment Distribution Method

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## The Moment Distribution Method

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.

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## Problem 852 | Continuous Beams with Fixed Ends

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## Problem 851 | Continuous Beams with Fixed Ends

**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.

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## Problem 850 | Continuous Beams with Fixed Ends

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## Problem 849 | Continuous Beams with Fixed Ends

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## Problem 848 | Continuous Beams with Fixed Ends

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## Fixed-end moments of fully restrained beam

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**

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

$M_B = -\dfrac{Pa^2b}{L^2}$

Value of EIy

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

Note: only for b > a

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## Problem 738 | Fully restrained beam with moment load

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## Problem 737 | Fully restrained beam with one support settling

**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/L^{2}.

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