Problem 6.8. Rotation from linearly distributed load

End rotation, φ_A [rad]=

Step 1.  Draw moment diagrams from the linearly distributed load, and from a unit force placed at left end of the beam.

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Step 2.  Apply Castigliano’s II theorem. Calculate the rotation.

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See Footnote j in Section 6.4.

$$\begin{gathered} {\mathit{\phi }}_{\mathbf{A}}=\frac{1}{EI}{\int }_L{M}_p{M}_1\mathrm{d}x=\frac{p_0}{EI}{\int }_L\left(\frac{Lx}{6}–\frac{x^3}{6L}\right)\frac{x}{L}=\frac{p_0}{EI}{\left[\frac{L{x}^3}{18L}–\frac{x^5}{30L^2}\right]}_0^L=\frac{p{L}^3}{45EI}= \\ =\frac{8000\times 5^3}{45\times 4.2\times {10}^6}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00529}\mathbf{ }\mathbf{rad} \end{gathered}$$

First moment diagrams are drawn from the linearly distributed load, and from a unit force placed at left end of the beam.

To calculate the rotation Castigliano’s II theorem is applied.

See Footnote j in Section 6.4.

$$\begin{gathered} {\mathit{\phi }}_{\mathbf{A}}=\frac{1}{EI}{\int }_L{M}_p{M}_1\mathrm{d}x=\frac{p_0}{EI}{\int }_L\left(\frac{Lx}{6}–\frac{x^3}{6L}\right)\frac{x}{L}=\frac{p_0}{EI}{\left[\frac{L{x}^3}{18L}–\frac{x^5}{30L^2}\right]}_0^L=\frac{p{L}^3}{45EI}= \\ =\frac{8000\times 5^3}{45\times 4.2\times {10}^6}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00529}\mathbf{ }\mathbf{rad} \end{gathered}$$