Problem 6.7. Deflection of a cantilever

Problem a)

End deflection, e [mm]=

Problem b)

End rotation, φ [rad]=

Problem a)

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

Check moment diagramsHide moment diagrams

Step 2.  Apply Castigliano’s II theorem. Calculate the deflection performing the integration visually.

Check deflectionHide deflection

See Hint in Problem 6.1.

See Footnote j in Section 6.4.

$$\begin{gathered} \mathit{e}=\frac{1}{EI}{\int }_L{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}Ah=\frac{1}{EI}\left(\frac{p{L}^2}{6}L\frac{1}{4}\right)L\frac{4}{5}=\frac{p{L}^4}{30EI}= \\ =\frac{8000\times 5^4}{30\times 4.2\times {10}^6}=0.0397 \mathrm{m}\mathbf{=}\mathbf{39}\mathbf{.}\mathbf{7}\mathbf{ }\mathbf{mm} \end{gathered}$$

where

$A=\frac{M_{A}L}{4}$

Problem b)

Step 1.  Draw moment diagrams from the linearly distributed load, and from a unit moment placed at the free end of the cantilever.

Check moment diagramsHide moment diagrams

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

Check rotationHide rotation

See Footnote j in Section 6.4.

$\mathit{\phi }=\frac{1}{EI}{\int }_L{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}Ah=\frac{1}{EI}\left(\frac{p{L}^2}{6}L\frac{1}{4}\right)\times 1=\frac{p{L}^3}{24EI}=\frac{8000\times 5^3}{24\times 4.2\times {10}^6}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00992}\mathbf{ }\mathbf{rad}$

Problem a)

Moment diagrams from the linearly distributed load, and from a unit force placed at the free end of the beam are given in the Figure:

Applying Castigliano’s II theorem the deflection is calculated by performing the integration visually:

See Footnote j in Section 6.4.

$$\begin{gathered} \mathit{e}=\frac{1}{EI}{\int }_L{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}Ah=\frac{1}{EI}\left(\frac{p{L}^2}{6}L\frac{1}{4}\right)L\frac{4}{5}=\frac{p{L}^4}{30EI}= \\ =\frac{8000\times 5^4}{30\times 4.2\times {10}^6}=0.0397 \mathrm{m}\mathbf{=}\mathbf{39}\mathbf{.}\mathbf{7}\mathbf{ }\mathbf{mm} \end{gathered}$$

where

$A=\frac{M_{A}L}{4}$

Problem b)

Moment diagrams from the linearly distributed load, and from a unit moment placed at the free end of the beam are given in the Figure:

Applying Castigliano’s II theorem the rotation is calculated by performing the integration visually (see Hint in Problem 6.1.):

See Footnote j in Section 6.4.

$\mathit{\phi }=\frac{1}{EI}{\int }_L{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}Ah=\frac{1}{EI}\left(\frac{p{L}^2}{6}L\frac{1}{4}\right)\times 1=\frac{p{L}^3}{24EI}=\frac{8000\times 5^3}{24\times 4.2\times {10}^6}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00992}\mathbf{ }\mathbf{rad}$