A 5 kg mass is suspended at the free end of an overhanging massless beam, having a pin support and a roller support, as shown in the figure below. Young's modulus of the material of the beam is 200 GPa and area moment of inertia of the beam is \(10^{-8}\) m\(^4\). The natural frequency of the beam in rad/s is
Show Hint
Find the tip stiffness of the overhanging beam using the bending moment diagram, then apply \(\omega_n=\sqrt{k/m}\).
Step 1: Set up the moment equations for direct integration.
With the pin at A (x = 0), roller at B (x = 1 m), and load $P$ at the tip C (x = 3 m), the reactions work out to $R_A = -2P$ and $R_B = 3P$ (found from moment and force balance). The bending moment is $M_1(x) = -2Px$ for $0 \le x \le 1$ and $M_2(x) = P(x-3)$ for $1 \le x \le 3$.
Step 2: Integrate twice in each span.
Using $EIy'' = M(x)$: in span 1, $EIy_1' = -Px^2 + C_1$ and $EIy_1 = -\dfrac{Px^3}{3} + C_1x + C_2$. In span 2, $EIy_2' = \dfrac{P(x-3)^2}{2} + C_3$ and $EIy_2 = \dfrac{P(x-3)^3}{6} + C_3x + C_4$.
Step 3: Apply the support and matching conditions.
No deflection at A gives $C_2 = 0$. No deflection at B ($x=1$) gives $C_1 = P/3$. Matching slope and deflection at $x = 1$ between the two spans gives $C_3 = -8P/3$ and $C_4 = 4P$. These four constants pin down the full deflected shape of the beam.
Step 4: Read off the tip deflection and close the problem.
At the free end $x = 3$: $EIy_2(3) = 0 + 3(-8P/3) + 4P = -4P$, so the tip deflects by $\delta_C = 4P/EI$ (the sign only shows direction). The stiffness felt by the 5 kg mass is $k = P/\delta_C = EI/4 = (200\times10^9)(10^{-8})/4 = 500$ N/m, and $\omega_n = \sqrt{k/m} = \sqrt{500/5} = 10$ rad/s, matching the unit load result.
Final Answer:
Both approaches agree that the beam behaves like a 500 N/m spring under the tip mass.
\[ \boxed{\omega_n = 10\ \text{rad/s}} \]
