Question 11.8: A block of mass m moving with a velocity v0 hits squarely th...
A block of mass m moving with a velocity v_0 hits squarely the prismatic member AB at its midpoint C (Fig. 11.31). Determine (a) the equivalent static load P_m, (b) the maximum stress s_m in the member, and (c) the maximum deflection x_m at point C.
(a) Equivalent Static Load. The maximum strain energy of the member is equal to the kinetic energy of the block before impact. We have
U_{m}=\frac{1}{2} m v_{0}^{2} (11.50)
On the other hand, expressing U_m as the work of the equivalent horizontal static load as it is slowly applied at the midpoint C of the member, we write
U_{m}=\frac{1}{2} P_{m} x_{m} (11.51)
where x_m is the deflection of C corresponding to the static load P_m. From the table of Beam Deflections and Slopes of Appendix D, we find that
x_{m}=\frac{P_{m} L^{3}}{48 E I} (11.52)
| Beam Deflections and Slopes | ||||
| Beam and Loading | Elastic Curve | Maximum Deflection | Slope at End | Equation of Elastic Curve |
 |
 |
-\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) |
 |
 |
-\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) |
 |
 |
-\frac{M L^{2}}{2 E I} | -\frac{M L}{E I} | y=-\frac{M}{2 E I} x^{2} |
 |
 |
-\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) |
 |
 |
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} |
 |
 |
-\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) |
 |
 |
\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) |
Substituting for x_m from (11.52) into (11.51), we write
U_{m}=\frac{1}{2} \frac{P_{m}^{2} L^{3}}{48 E I}Solving for P_m and recalling Eq. (11.50), we find that the static load equivalent to the given impact loading is
P_{m}=\sqrt{\frac{96 U_{m} E I}{L^{3}}}=\sqrt{\frac{48 m v_{0}^{2} E I}{L^{3}}} (11.53)
(b) Maximum Stress. Drawing the free-body diagram of the member (Fig. 11.32), we find that the maximum value of the bending moment occurs at C and is M_{\max }=P_{m} L / 4. The maximum stress, therefore, occurs in a transverse section through C and is equal to
s _{m}=\frac{M_{\max } c}{I}=\frac{P_{m} L c}{4 I}Substituting for P_m from (11.53), we write
s_{m}=\sqrt{\frac{3 m v_{0}^{2} E I}{L(I / c)^{2}}}(c) Maximum Deflection. Substituting into Eq. (11.52) the expression obtained for P_m in (11.53), we have
x_{m}=\frac{L^{3}}{48 E I} \sqrt{\frac{48 m v_{0}^{2} E I}{L^{3}}}=\sqrt{\frac{m v_{0}^{2} L^{3}}{48 E I}}