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

 

Case 2: Concentrated load on the midspan of fully restrained beam

000-fully-restrained-beam-point-load-midspan.gifEnd moments
$M_A = M_B = -\dfrac{PL}{8}$
 

Value of EIy
$\text{Maximum } EI\,y = \dfrac{PL^3}{192}$

 

Case 3: Uniformly distributed load over the entire span of fully restrained beam

000-fully-restrained-beam-uniform-load.gifEnd moments
$M_A = M_B = -\dfrac{w_oL^2}{12} = -\dfrac{WL}{12}$
 

Value of EIy
$\text{Maximum } EI\,y = \dfrac{w_oL^4}{384} = \dfrac{WL^3}{384}$

 
Case 4: Uniformly distributed load over half the span of fully restrained beam

000-fully-restrained-beam-uniform-load-half-span.gifEnd moments
$M_A = -\dfrac{5w_oL^2}{192} = -\dfrac{5WL}{96}$

$M_B = -\dfrac{11w_oL^2}{192} = -\dfrac{11WL}{96}$
 

Value of EIy
$\text{Midspan } EI\,y = \dfrac{w_oL^4}{384} = \dfrac{WL^3}{384}$

 

Case 5: Triangular load over the entire span of fully restrained beam

000-fully-restrained-beam-triangular-load-increasing.gifEnd moments
$M_A = -\dfrac{w_oL^2}{30} = -\dfrac{WL}{15}$

$M_B = -\dfrac{w_oL^2}{20} = -\dfrac{WL}{10}$
 

Value of EIy
$\text{Midspan } EI\,y = \dfrac{w_oL^4}{768} = \dfrac{WL^3}{384}$

 

Case 6: Isosceles triangle loadings over the entire span of fully restrained beam

000-fully-restrained-beam-triangular-load-symmetrical.gifEnd moments
$M_A = M_B = -\dfrac{5w_oL^2}{96} = -\dfrac{5WL}{48}$
 

Value of EIy
$\text{Maximum } EI\,y = \dfrac{7w_oL^4}{3840} = \dfrac{7WL^3}{1920}$

 

Case 7: Moment load anywhere on the span of fully restrained beam

000-fully-restrained-beam-moment-load.gifEnd moments
$M_A = \dfrac{Mb}{L}\left( \dfrac{3a}{L} - 1 \right)$

$M_B = -\dfrac{Ma}{L}\left( \dfrac{3b}{L} - 1 \right)$

 

Case 8: Fully restrained beam with one support settling

000-fully-restrained-beam-settling-support.gifEnd moments
$M_A = -\dfrac{6EI \Delta}{L^2}$

$M_B = \dfrac{6EI \Delta}{L^2}$

 

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