Question 11.S-P.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 that occurs in the beam.
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} U_{2}=\frac{24 E I}{L^{3}} y_{m}^{2}
| Beam Deflections and Slopes | ||||
| Beam and Loading | Elastic Curve | Maximum Deflection | Slope at End | Equation of Elastic Curve |
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-\frac{P L^{3}}{3 E I} | -\frac{P L^{2}}{2 E I} | y=\frac{P}{6 EI}\left(x^{3}-3 L x^{2}\right) |
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-\frac{w L^{4}}{8 E I} | -\frac{w L^{3}}{6 E I} | y=-\frac{w}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}\right) |
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-\frac{M L^{2}}{2 E I} | -\frac{M L}{E I} | y=-\frac{M}{2 E I} x^{2} |
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-\frac{P L^{3}}{48 E I} | \pm \frac{P L^{2}}{16 E I} | For x \leq \frac{1}{2} L :
y=\frac{P}{48 E I}\left(4 x^{3}-3 L^{2} x\right) |
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For a > b:
-\frac{P b\left(L^{2}-b^{2}\right)^{3 / 2}}{9 \sqrt{3} E I L} at x_{m}=\sqrt{\frac{L^{2}-b^{2}}{3}} |
u _{A}=-\frac{P b\left(L^{2}-b^{2}\right)}{6 E I L}
u_{B}=+\frac{P a\left(L^{2}-a^{2}\right)}{6 E I L} |
For x < a:
y=\frac{P b}{6 E I L}\left[x^{3}-\left(L^{2}-b^{2}\right) x\right] For x = a: y=-\frac{P a^{2} b^{2}}{3 E I L} |
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-\frac{5 w L^{4}}{384 E I} | \pm \frac{w L^{3}}{24 E I} | y=-\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right) |
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\frac{M L^{2}}{9 \sqrt{3} E I} | u _{A}=+\frac{M L}{6 E I}
u _{B}=-\frac{M L}{3 E I} |
y=-\frac{M}{6 E I L}\left(x^{3}-L^{2} x\right) |
We observe that 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, 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
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 h = 0.040 m W = mg = (80 kg) (9.81 m/s²) = 784.8 N
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 y_{m}=10.27 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} N \cdot m \right)}{(1 m )^{3}}(0.01027 m ) P_{m}=7677 N
Recalling that s _{m}=M_{\max } c / I and M_{\max }=\frac{1}{4} P_{m} L, we write
s _{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}} s _{m}=179.9 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 mm; the error is 10.8%. However, if an 8-kg block is dropped from a height of 400 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.
