Restrained Beams
In addition to the equations of static equilibrium, relations from the geometry of elastic curve are essential to the study of indeterminate beams. Such relations can be obtained from the study of deflection and rotation of beam. This section will focus on two types of indeterminate beams; the propped beams and the fully restrained beams.
 

Propped beam and fully restrained beam

 

A propped beam is fixed at one end and propped either at the other end or at any other point along its span. If the simple support is removed, propped beam will become cantilever beam. Fully restrained beam is fixed at both ends as shown in the figure above.
 

Deflection and Rotation of Propped Beam
Unless otherwise specified, the boundary conditions of propped beams are as follows.

  • Deflection at both ends is zero.
  • Rotation at fixed support is zero.

 

Deflection and Rotation of Fully Restrained Beam
Unless otherwise specified, the boundary conditions of propped beams are as follows.

  • Deflection at both ends is zero.
  • Rotation at both ends is zero.

 

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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