Ian: New page: The diagram shows the forces acting on the beam Image:14.5a.gif Taking moments about the left hand support: $\begin{align} & 1\textrm{.}2\times {{R}_{B}}+0\textrm{.}4\time...$

Ian — Thu, 02 Sep 2010 13:50:02 GMT

New page: The diagram shows the forces acting on the beam Image:14.5a.gif Taking moments about the left hand support: <math>\begin{align} & 1\textrm{.}2\times {{R}_{B}}+0\textrm{.}4\time...

New page

The diagram shows the forces acting on the beam

[[Image:14.5a.gif]]

Taking moments about the left hand support:

<math>\begin{align}
& 1\textrm{.}2\times {{R}_{B}}+0\textrm{.}4\times 11\textrm{.}76=0\textrm{.}6\times 58\textrm{.}8 \\
& {{R}_{B}}=\frac{0.6\times 58\textrm{.}8-0\textrm{.}4\times 11\textrm{.}76}{1\textrm{.}2}=25\textrm{.}48\text{ N} \\
\end{align}</math>

Taking moments about the right hand support:

<math>\begin{align}
& 1\textrm{.}2\times {{R}_{A}}=1\textrm{.}6\times 11\textrm{.}76+0\textrm{.}8\times 58\textrm{.}8 \\
& {{R}_{A}}=\frac{1\textrm{.}6\times 11\textrm{.}76+0\textrm{.}6\times 58\textrm{.}8}{1\textrm{.}2}=45\textrm{.}08\text{ N} \\
\end{align}</math>

Or

<math>\begin{align}
& {{R}_{A}}+25\textrm{.}48=11\textrm{.}76+58\textrm{.}8 \\
& {{R}_{A}}=11\textrm{.}76+58\textrm{.}8-25\textrm{.}48=45\textrm{.}08\text{ N} \\
\end{align}</math>

Solution 14.5a - Revision history

Ian: New page: The diagram shows the forces acting on the beam Image:14.5a.gif Taking moments about the left hand support: \begin{align} & 1\textrm{.}2\times {{R}_{B}}+0\textrm{.}4\time...

Ian: New page: The diagram shows the forces acting on the beam Image:14.5a.gif Taking moments about the left hand support: $\begin{align} & 1\textrm{.}2\times {{R}_{B}}+0\textrm{.}4\time...$