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

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 y_{m} occurs, the kinetic energy is again zero. Referring to the table of Beam Deflections and Slopes of Appendix D, we find the expression for y_{m} shown. The strain energy of the beam in position 2 is

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}

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

\frac{24 E I}{L^{3}} y_{m}^{2}=W\left(h+y_{m}\right)  (1)

a. Maximum Deflection of Point C. From the given data we have

\begin{gathered} E I=\left(73 \times 10^{9} \mathrm{~Pa}\right) \frac{1}{12}(0.04 \mathrm{~m})^{4}=15.573 \times 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}\right)=784.8 \mathrm{~N} \end{gathered}

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

\left(373.8 \times 10^{3}\right) y_{m}^{2}-784.8 y_{m}-31.39=0 \quad y_{m}=10.27 \mathrm{~mm}

b. Maximum Stress. The value of P_{m} is

P_{m}=\frac{48 E I}{L^{3}} y_{m}=\frac{48\left(15.573 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}\right)}{(1 \mathrm{~m})^{3}}(0.01027 \mathrm{~m}) \quad P_{m}=7677 \mathrm{~N}

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

\sigma_{m}=\frac{\left(\frac{1}{4} P_{m} L\right) 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}

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