Question 11.CA.8: A block of mass m moving with a velocity v0 hits the prismat......
A block of mass m moving with a velocity v_0 hits the prismatic member AB squarely at its midpoint C (Fig. 11.27a). Determine (a) the equivalent static load P_m , (b) the maximum stress σ_m in the member, (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.
U_m=\frac{1}{2} m ν_0^2 (1)
On the other hand, U_m can be given as the work of the equivalent horizontal static load as it is slowly applied at the midpoint C
U_m=\frac{1}{2} P_m x_m (2)
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 F,
x_m=\frac{P_m L^3}{48 E I} (3)
Substituting for x_m from Eq. (3) into Eq. (2),
U_m=\frac{1}{2} \frac{P_m^2 L^3}{48 E I}
Solving for P_m and recalling Eq. (1), 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 ν_0^2 E I}{L^3}} (4)
b. Maximum Stress. Drawing the free-body diagram of the member (Fig. 11.27b), the maximum value of the bending moment occurs at C and is M_{\max }=P_m L / 4 . The maximum stress occurs in a transverse section through C and is equal to
\sigma_m=\frac{M_{\max } c}{I}=\frac{P_m L c}{4 I}
Substituting for P_m from Eq. (4),
\sigma_m=\sqrt{\frac{3 m ν_0^2 E I}{L(I / c)^2}}
c. Maximum Deflection. Substituting into Eq. (3) the expression obtained for P_m in Eq. (4):
x_m=\frac{L^3}{48 E I} \sqrt{\frac{48 m ν_0^2 E I}{L^3}}=\sqrt{\frac{m ν_0^2 L^3}{48 E I}}
