Question 11.SP.3: The block D of mass m is released from rest and falls a dist......
The block D of mass m is released from rest and falls a distance h before it strikes the midpoint C of the aluminum beam AB. Using E = 73 GPa, determine (a) the maximum deflection of point C, (b) the maximum stress in the beam.
STRATEGY: Calculate the strain energy of the beam in terms of the deflection and equate this to the work done by the block. This then can be used with the data to solve part a. Using the relation between the applied load and deflection (Appendix F), obtain the equivalent static load and use this to get the normal stress due to bending.
MODELING:
Principle of Work and Energy. The block is released from rest (Fig. 1, position 1). Note that in this position both the kinetic and strain energy are zero. In position 2 (Fig. 1), where the maximum deflection y_m occurs, the kinetic energy is also zero. Use to the table of Beam Deflections and Slopes in Appendix F to find the expression for y_m shown in Fig. 2. 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
The work done by the weight W of the block is W\left(h+y_m\right) . Equating the strain energy of the beam to the work done by W gives
\frac{24 E I}{L^3} y_m^2=W\left(h+y_m\right) (1)
ANALYSIS:
a. Maximum Deflection of Point C. From the given data,
\begin{gathered} E I=\left(73 \times 10^9 Pa \right) \frac{1}{12}(0.04 m )^4=15.573 \times 10^3 N \cdot m ^2 \\ L=1 m \quad h=0.040 m \quad W=m g=(80 kg )\left(9.81 m / s ^2\right)=784.8 N \end{gathered}
Substituting W into Eq. (1), we obtain a quadratic equation that can be solved for the deflection:
\left(373.8 \times 10^3\right) y_m^2-784.8 y_m-31.39=0 \quad y_m=10.27 mm
b. Maximum Stress. The value of P_m (Fig. 2) is
P_m=\frac{48 E I}{L^3} y_m=\frac{48\left(15.573 \times 10^3 N \cdot m \right)}{(1 m )^3}(0.01027 m ) \quad P_m=7677 N
Recalling that \sigma_m=M_{\max } c / I \text { and } M_{\max }=\frac{1}{4} P_m L , write
\sigma_m=\frac{\left(\frac{1}{4} P_m L\right) c}{I}=\frac{\frac{1}{4}(7677 N )(1 m )(0.020 m )}{\frac{1}{12}(0.040 m )^4}
\sigma_m=179.9 MPa
REFLECT and THINK: An approximation for the work done by the weight of the block is obtained by omitting y_m from the expression for the work and from the right-hand member of Eq. (1), as was done in Concept Application 11.7.
If this approximation is used here, y_m = 9.16 mm, and the error is 10.8%.
However, if an 8-kg block is dropped from a height of 400 mm (producing the same value for Wh), omitting y_m from the right-hand member of Eq. (1) results in an error of only 1.2%
