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The distributed force was never turned off at the end of the beam. |
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Load distribution:
:<math>w=-3\text{ N}\langle x-0 \rangle^{-1}\ +\ 6\text{ Nm}^{-1}\langle x-2\text{ m} \rangle^0\ -\ 9\text{ N}\langle x-4\text{ m}\rangle^{-1}\ -\ 6\text{ Nm}^{-1}\langle x-2\text{ m} \rangle^0\</math>
Shear force:
:<math>S=\int w\, dx</math>
:<math>S=-3\text{ N}\langle x-0\rangle^0\ +\ 6\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^1\ -\ 9\text{ N}\langle x-4\text{ m}\rangle^0\ -\ 6\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^1\,</math>
Bending moment:
:<math>M = -\int S\, dx</math>
:<math>M=3\text{ N}\langle x-0\rangle^1\ -\ 3\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^2\ +\ 9\text{ N}\langle x-4\text{ m} \rangle^1\ +\ 3\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^2\,</math>
Slope:
:<math>u'=\frac{1}{EI}\int M\, dx</math>
:Because the slope is not zero at <var>x</var> = 0, a constant of integration, <var>c</var>, is added
:<math>u'=\frac{1}{EI}\left(\frac{3}{2}\text{ N}\langle x-0\rangle^2\ -\ 1\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^3\ +\ \frac{9}{2}\text{ N}\langle x-4\text{ m}\rangle^2\ +\ c\right)\ +\ 1\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^3\,</math>
Deflection:
:<math>u=\int u'\, dx</math>
:<math>u=\frac{1}{EI}\left(\frac{1}{2}\text{ N}\langle x-0\rangle^3\ -\ \frac{1}{4}\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^4\ +\ \frac{3}{2}\text{ N}\langle x-4\text{ m}\rangle^3\ +\ cx\right)\ +\ \frac{1}{4}\text{ Nm}^{-1}\langle x-2\text{ m}\rangle^4\,</math>
The boundary condition <var>u</var> = 0 at <var>x</var> = 4 m allows us to solve for <var>c</var> = −7 Nm<sup>2</sup>
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