Let's break down the problem and solve for the **maximum stress** and **maximum deflection** in the rod. ### **Problem Analysis** - **Rod:** Vertical, cross-sectional area \(A\), length \(L\) - **Collar:** Attached to the bottom, inextensible mass \(M\) - **Mass dropped:** From height \(h\) above the collar - **Neglect:** Energy and other losses - **Assume:** Mass is an annular ring (acts instantly on the collar) - **Find:** Maximum stress and maximum deflection due to impact --- ## **Step 1: Energy Consideration** When the mass \(M\) is dropped from height \(h\), it gains kinetic energy as it falls and imparts this energy to the rod upon impact. **Total Potential Energy Lost by Mass:** \[ E = Mgh \] When the mass hits the collar, it causes the rod to extend by a maximum deflection \(\delta_{max}\), so the total energy absorbed by the rod is: \[ Mgh = \text{Work done in stretching the rod} \] The work done in stretching the rod (if the maximum force is \(F_{max}\)) is: \[ \text{Work} = \frac{1}{2} F_{max} \delta_{max} \] --- ## **Step 2: Static Deflection** If the mass \(M\) is gently placed, the static deflection (\(\delta_{st}\)) is: \[ \delta_{st} = \frac{MgL}{AE} \] where: - \(L\) = Length of the rod - \(A\) = Cross-sectional area - \(E\) = Young's modulus --- ## **Step 3: Maximum Deflection from Impact** From energy balance, for impact loading: \[ Mgh = \frac{1}{2} F_{max} \delta_{max} \] But \(F_{max}\) is related to \(\delta_{max}\) by: \[ F_{max} = \frac{AE}{L} \delta_{max} \] So, \[ Mgh = \frac{1}{2} \left(\frac{AE}{L} \delta_{max}\right) \delta_{max} \] \[ Mgh = \frac{AE}{2L} \delta_{max}^2 \] \[ \delta_{max} = \sqrt{\frac{2MghL}{AE}} \] But, including the static deflection due to the weight \(Mg\) as well, the total energy is: \[ Mgh + Mg\delta_{max} = \frac{1}{2} \frac{AE}{L} \delta_{max}^2 \] So, \[ Mgh + Mg\delta_{max} = \frac{AE}{2L} \delta_{max}^2 \] \[ \frac{AE}{2L} \delta_{max}^2 - Mg\delta_{max} - Mgh = 0 \] This is a quadratic equation in \(\delta_{max}\): \[ \delta_{max}^2 - \frac{2MgL}{AE} \delta_{max} - \frac{2MghL}{AE} = 0 \] Let \(a = 1\), \(b = -\frac{2MgL}{AE}\), \(c = -\frac{2MghL}{AE}\): \[ \delta_{max} = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \] Plug in the values: \[ \delta_{max} = \frac{\frac{2MgL}{AE} + \sqrt{\left(\frac{2MgL}{AE}\right)^2 + \frac{8MghL}{AE}}}{2} \] \[ \delta_{max} = \frac{MgL}{AE} + \frac{1}{2}\sqrt{\left(\frac{2MgL}{AE}\right)^2 + \frac{8MghL}{AE}} \] Or, more simply: \[ \delta_{max} = \frac{MgL}{AE} \left[1 + \sqrt{1 + \frac{2hAE}{MgL}}\,\right] \] --- ## **Step 4: Maximum Stress in the Rod** The maximum force in the rod at maximum deflection: \[ F_{max} = \frac{AE}{L} \delta_{max} \] So, \[ \sigma_{max} = \frac{F_{max}}{A} = \frac{E}{L} \delta_{max} \] --- ## **Step 5: Special Case (\(h = 0\))** If \(h = 0\), mass is applied gently (no drop, only the static case): \[ \delta_{max} = \frac{MgL}{AE} \] \[ \sigma_{max} = \frac{Mg}{A} \] --- ## **Summary of Answers** ### **(a) Maximum deflection:** \[ \boxed{ \delta_{max} = \frac{MgL}{AE} \left[ 1 + \sqrt{1 + \frac{2hAE}{MgL}}\,\right] } \] ### **(b) Maximum stress:** \[ \boxed{ \sigma_{max} = \frac{E}{L} \delta_{max} } \] ### **(c) For \(h = 0\) (static case):** \[ \boxed{ \sigma_{max} = \frac{Mg}{A} } \] --- ### **Let me know if you need a numerical example or further clarification!**