A block of mass m moving with a velocity v0 hits squarely the prismatic member AB at its midpoint C (Fig. 11.31). Determine (a) the equivalent static load Pm, (b) the maximum stress sm in the member, and (c) the maximum deflection xm at point C.

Question

A block of mass m moving with a velocity [latex]v_0[/latex] hits squarely the prismatic member AB at its midpoint C (Fig. 11.31). Determine (a) the equivalent static load [latex]P_m[/latex], (b) the maximum stress [latex]s_m[/latex] in the member, and (c) the maximum deflection [latex]x_m[/latex] at point C.

Accepted Answer

(a) Equivalent Static Load. The maximum strain energy of the member is equal to the kinetic energy of the block before impact. We have

[latex]U_{m}=\frac{1}{2} m v_{0}^{2}[/latex] (11.50)

On the other hand, expressing [latex]U_m[/latex] as the work of the equivalent horizontal static load as it is slowly applied at the midpoint C of the member, we write

[latex]U_{m}=\frac{1}{2} P_{m} x_{m}[/latex] (11.51)

where [latex]x_m[/latex] is the deflection of C corresponding to the static load [latex]P_m[/latex]. From the table of Beam Deflections and Slopes of Appendix D, we find that

[latex]x_{m}=\frac{P_{m} L^{3}}{48 E I}[/latex] (11.52)

Beam Deflections and Slopes
Beam and Loading	Elastic Curve	Maximum Deflection	Slope at End	Equation of Elastic Curve
		[latex]-\frac{P L^{3}}{3 E I}[/latex]	[latex]-\frac{P L^{2}}{2 E I}[/latex]	[latex]y=\frac{P}{6 EI}\left(x^{3}-3 L x^{2}\right)[/latex]
		[latex]-\frac{w L^{4}}{8 E I}[/latex]	[latex]-\frac{w L^{3}}{6 E I}[/latex]	[latex]y=-\frac{w}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}\right)[/latex]
		[latex]-\frac{M L^{2}}{2 E I}[/latex]	[latex]-\frac{M L}{E I}[/latex]	[latex]y=-\frac{M}{2 E I} x^{2}[/latex]
		[latex]-\frac{P L^{3}}{48 E I}[/latex]	[latex]\pm \frac{P L^{2}}{16 E I}[/latex]	For [latex]x \leq \frac{1}{2} L[/latex] : [latex]y=\frac{P}{48 E I}\left(4 x^{3}-3 L^{2} x\right)[/latex]
		For a > b: [latex]-\frac{P b\left(L^{2}-b^{2}\right)^{3 / 2}}{9 \sqrt{3} E I L}[/latex] at [latex]x_{m}=\sqrt{\frac{L^{2}-b^{2}}{3}}[/latex]	[latex]u _{A}=-\frac{P b\left(L^{2}-b^{2}\right)}{6 E I L}[/latex] [latex]u_{B}=+\frac{P a\left(L^{2}-a^{2}\right)}{6 E I L}[/latex]	For x < a: [latex]y=\frac{P b}{6 E I L}\left[x^{3}-\left(L^{2}-b^{2}\right) x\right][/latex] For x = a: [latex]y=-\frac{P a^{2} b^{2}}{3 E I L}[/latex]
		[latex]-\frac{5 w L^{4}}{384 E I}[/latex]	[latex]\pm \frac{w L^{3}}{24 E I}[/latex]	[latex]y=-\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right)[/latex]
		[latex]\frac{M L^{2}}{9 \sqrt{3} E I}[/latex]	[latex]u _{A}=+\frac{M L}{6 E I}[/latex] [latex]u _{B}=-\frac{M L}{3 E I}[/latex]	[latex]y=-\frac{M}{6 E I L}\left(x^{3}-L^{2} x\right)[/latex]

Substituting for [latex]x_m[/latex] from (11.52) into (11.51), we write

[latex]U_{m}=\frac{1}{2} \frac{P_{m}^{2} L^{3}}{48 E I}[/latex]

Solving for [latex]P_m[/latex] and recalling Eq. (11.50), we find that the static load equivalent to the given impact loading is

[latex]P_{m}=\sqrt{\frac{96 U_{m} E I}{L^{3}}}=\sqrt{\frac{48 m v_{0}^{2} E I}{L^{3}}}[/latex] (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 [latex]M_{\max }=P_{m} L / 4[/latex]. The maximum stress, therefore, occurs in a transverse section through C and is equal to

[latex]s _{m}=\frac{M_{\max } c}{I}=\frac{P_{m} L c}{4 I}[/latex]

Substituting for [latex]P_m[/latex] from (11.53), we write

[latex]s_{m}=\sqrt{\frac{3 m v_{0}^{2} E I}{L(I / c)^{2}}}[/latex]

(c) Maximum Deflection. Substituting into Eq. (11.52) the expression obtained for [latex]P_m[/latex] in (11.53), we have

[latex]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}}[/latex]

Question 11.8: A block of mass m moving with a velocity v0 hits squarely th...

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