Determine the vertical displacement of end B of the cantilevered 6061-T6 aluminum alloy rectangular beam. Consider both shearing and bending strain energy

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Internal Loadings. Referring to the FBD of beam's right cut segment, Fig. a,

[latex]+\uparrow \Sigma F_{y}=0 ; \quad V-150\left(10^{3} ight)=0 \quad V=150\left(10^{3} ight) N \\curvearrowleft +\Sigma M_{0}=0 ; \quad -M-150\left(10^{3} ight) x=0 \quad M=-150\left(10^{3} ight) x[/latex]

Shearing Strain Energy. For the rectangular beam, the form factor is [latex]f_{s}=\frac{6}{5}[/latex].

[latex]\left(U_{i} ight)_{v}=\int_{0}^{L} \frac{f_{s} V^{2} d x}{2 G A}=\int_{0}^{1 m } \frac{\frac{6}{5}\left[150\left(10^{3} ight) ight]^{2} d x}{2\left[26\left(10^{9} ight) ight][0.1(0.3)]}=17.31 J[/latex]

Bending Strain Energy. [latex]I=\frac{1}{12}(0.1)\left(0.3^{3} ight)=0.225\left(10^{-3} ight) m ^{4}[/latex]. We obtain

[latex]\left(U_{i} ight)_{b}=\int_{0}^{L} \frac{M^{2} d x}{2 E I}=\int_{0}^{1 m } \frac{\left[-150\left(10^{3} ight) x ight]^{2} d x}{2\left[68.9\left(10^{9} ight) ight]\left[0.225\left(10^{-3} ight) ight]} \=725.689 \int_{0}^{1 m } x^{2} d x \=\left.725.689\left(\frac{x^{3}}{3} ight) ight|_{0} ^{1 m } \=241.90 J[/latex]

Thus, the strain energy stored in the beam is

[latex]U_{i} =\left(U_{i} ight)_{v}+\left(U_{i} ight)_{b} \=17.31+241.90 \=259.20 J[/latex]

External Work. The work done by the external force P=150 kN is

[latex]U_{e}=\frac{1}{2} P \Delta=\frac{1}{2}\left[150\left(10^{3} ight) ight] \Delta_{B}=75\left(10^{3} ight) \Delta_{B}[/latex]

Conservation of Energy.

[latex]U_{e} =U_{i} \75\left(10^{3} ight) \Delta_{B} =259.20 \\Delta_{B} =3.456\left(10^{-3} ight) m =3.46 mm[/latex]

Question 14.30: Determine the vertical displacement of end B of the cantilev...

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