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Loads on Pin-ended Beams

When the ends of beam are not fully fixed, some adjustment has to be made to the forces and moments that are applied at the nodes and the pinned end cannot sustain any moment. The simplest case is when both ends are pinned so

m1=m2=0m_{1} = m_{2} = 0

The forces and moments are then modified to maintain equilibrium as follows

f1→f1±(m1+m2)l,m1→0f_{1} \rightarrow f_{1} \pm \frac{\left( m_{1} + m_{2} \right)}{l},\qquad m_{1} \rightarrow 0
f2→f2∓(m1+m2)l,m2→0f_{2} \rightarrow f_{2} \mp \frac{\left( m_{1} + m_{2} \right)}{l},\qquad m_{2} \rightarrow 0

for y/z.

When the element is pinned at one end only the corrections depend on the material properties in the general case

pin at end 1pin at end 2
f1→f1±(64+α)m1lf_{1} \rightarrow f_{1} \pm \left( \frac{6}{4 + \alpha} \right)\frac{m_{1}}{l}f1→f1∓(64+α)m2lf_{1} \rightarrow f_{1} \mp \left( \frac{6}{4 + \alpha} \right)\frac{m_{2}}{l}
f2→f2∓(64+α)m1lf_{2} \rightarrow f_{2} \mp \left( \frac{6}{4 + \alpha} \right)\frac{m_{1}}{l}f2→f2±(64+α)m2lf_{2} \rightarrow f_{2} \pm \left( \frac{6}{4 + \alpha} \right)\frac{m_{2}}{l}
m1→0m_{1} \rightarrow 0m1→m1−(2−α4+α)m2m_{1} \rightarrow m_{1} - \left( \frac{2 - \alpha}{4 + \alpha} \right)m_{2}
m2→m2−(2−α4+α)m1m_{2} \rightarrow m_{2} - \left( \frac{2 - \alpha}{4 + \alpha} \right)m_{1}m2→0m_{2} \rightarrow 0

where

α=12EIGAslAs=kA\alpha = 12\frac{EI}{GA_{s}l}\qquad A_{s} = kA

For a simple beam this reduces to

pin at end 1pin at end 2
f1→f1±3m12lf_{1} \rightarrow f_{1} \pm \frac{3m_{1}}{2l}f1→f1∓3m22lf_{1} \rightarrow f_{1} \mp \frac{3m_{2}}{2l}
f2→f2∓3m12lf_{2} \rightarrow f_{2} \mp \frac{3m_{1}}{2l}f2→f2±3m22lf_{2} \rightarrow f_{2} \pm \frac{3m_{2}}{2l}
m1→0m_{1} \rightarrow 0m1→m1−m22m_{1} \rightarrow m_{1} - \frac{m_{2}}{2}
m2→m2−m12m_{2} \rightarrow m_{2} - \frac{m_{1}}{2}m2→0m_{2} \rightarrow 0