fully restrained beam

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

000-fully-restrained-beam-point-load.gifEnd 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 738 | Fully restrained beam with moment load

Problem 738
A perfectly restrained beam is loaded by a couple M applied where shown in Fig. P-738. Determine the end moments.



Solution 738

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 MB = -MA = 6EIΔ/L2.



Problem 736 | Shear and moment diagrams of fully restrained beam under triangular load

Problem 736
Determine the end shears and end moments for the restrained beam shown in Fig. P-736 and sketch the shear and moment diagrams.



Problem 730 | Uniform loads at each end of fully restrained beam

Problem 703
Determine the end moment and maximum deflection for a perfectly restrained beam loaded as shown in Fig. P-730.


Problem 729 | Uniform load over the center part of fixed-ended beam

Problem 729
For the restrained beam shown in Fig. P-729, compute the end moment and maximum EIδ.



Problem 727 | Fully restrained beam with uniform load over the entire span

Problem 727
Repeat Problem 726 assuming that the concentrated load is replaced by a uniformly distributed load of intensity wo over the entire length.

Problem 726 | Fully restrained beam with concentrated load at midspan

Problem 726
A beam of length L, perfectly restrained at both ends, supports a concentrated load P at midspan. Determine the end moment and maximum deflection.

Application of Area-Moment Method to Restrained Beams

See deflection of beam by moment-area method for details.

Rotation of beam from A to B

$\theta_{AB} = \dfrac{1}{EI}(\text{Area}_{AB})$


Deviation of B from a tangent line through A

$t_{B/A} = \dfrac{1}{EI} (Area_{AB}) \, \bar{X}_B$



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