The block D of mass m is released from rest and falls a distance h before it strikes the

Question

The block [latex]D[/latex] of mass [latex]m[/latex] is released from rest and falls a distance [latex]h[/latex] before it strikes the midpoint [latex]C[/latex] of the aluminum beam [latex]A B[/latex]. Using [latex]E=73 \mathrm{GPa}[/latex], determine [latex](a)[/latex] the maximum deflection of point [latex]C,(b)[/latex] the maximum stress that occurs in the beam.

Accepted Answer

Principle of Work and Energy. Since the block is released from rest, we note that in position 1 both the kinetic energy and the strain energy are zero. In position 2 , where the maximum deflection [latex]y_{m}[/latex] occurs, the kinetic energy is again zero. Referring to the table of Beam Deflections and Slopes of Appendix D, we find the expression for [latex]y_{m}[/latex] shown. The strain energy of the beam in position 2 is

[latex]U_{2}=\frac{1}{2} P_{m} y_{m}=\frac{1}{2} \frac{48 E I}{L^{3}} y_{m}^{2} \quad U_{2}=\frac{24 E I}{L^{3}} y_{m}^{2}[/latex]

We observe that the work done by the weight [latex]\mathbf{W}[/latex] of the block is [latex]W\left(h+y_{m} ight)[/latex]. Equating the strain energy of the beam to the work done by [latex]\mathbf{W}[/latex], we have

[latex]\frac{24 E I}{L^{3}} y_{m}^{2}=W\left(h+y_{m} ight)[/latex] (1)

a. Maximum Deflection of Point [latex]C[/latex]. From the given data we have

[latex] \begin{gathered} E I=\left(73 imes 10^{9} \mathrm{~Pa} ight) \frac{1}{12}(0.04 \mathrm{~m})^{4}=15.573 imes 10^{3} \mathrm{~N} \cdot \mathrm{m}^{2} \ L=1 \mathrm{~m} \quad h=0.040 \mathrm{~m} \quad W=m g=(80 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight)=784.8 \mathrm{~N} \end{gathered} [/latex]

Substituting into Eq. (1), we obtain and solve the quadratic equation

[latex]\left(373.8 imes 10^{3} ight) y_{m}^{2}-784.8 y_{m}-31.39=0 \quad y_{m}=10.27 \mathrm{~mm}[/latex]

b. Maximum Stress. The value of [latex]P_{m}[/latex] is

[latex]P_{m}=\frac{48 E I}{L^{3}} y_{m}=\frac{48\left(15.573 imes 10^{3} \mathrm{~N} \cdot \mathrm{m} ight)}{(1 \mathrm{~m})^{3}}(0.01027 \mathrm{~m}) \quad P_{m}=7677 \mathrm{~N}[/latex]

Recalling that [latex]\sigma_{m}=M_{\max } c / I[/latex] and [latex]M_{\max }=\frac{1}{4} P_{m} L[/latex], we write

[latex]\sigma_{m}=\frac{\left(\frac{1}{4} P_{m} L ight) c}{I}=\frac{\frac{1}{4}(7677 \mathrm{~N})(1 \mathrm{~m})(0.020 \mathrm{~m})}{\frac{1}{12}(0.040 \mathrm{~m})^{4}} \quad \sigma_{m}=179.9 \mathrm{MPa}[/latex]

An approximation for the work done by the weight of the block can be obtained by omitting [latex]y_{m}[/latex] from the expression for the work and from the right-hand member of Eq. (1), as was done in Example 11.07. If this approximation is used here, we find [latex]y_{m}=9.16 \mathrm{~mm}[/latex]; the error is [latex]10.8 \%[/latex]. However, if an 8-kg block is dropped from a height of [latex]400 \mathrm{~mm}[/latex], producing the same value of Wh, omitting [latex]y_{m}[/latex] from the right-hand member of Eq. (1) results in an error of only [latex]1.2 \%[/latex]. A further discussion of this approximation is given in Prob. 11.70.

Question 11.SP.3: The block D of mass m is released from rest and falls a dist...

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