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: Find the support reactions.
The beam has a pin at A (x = 0 m), a roller at B (x = 1 m), and the mass hangs at the free end C (x = 3 m). For a downward force \(P\) at C, take moments about A: \(R_B(1) = P(3)\), so \(R_B = 3P\) upward. From \(R_A + R_B = P\), we get \(R_A = -2P\), meaning the pin actually pulls down with magnitude \(2P\).
Step 2: Write the bending moment in each span.
For \(0 \le x \le 1\), only \(R_A\) acts to the left of the cut, so \(M(x) = R_A x = -2Px\). For \(1 \le x \le 3\), both reactions act, so \(M(x) = R_A x + R_B(x-1) = P(x-3)\). Check: at \(x=1\) both give \(M=-2P\), and at the free end \(x=3\), \(M=0\), which is correct.
Step 3: Get the tip deflection with the unit load method.
Let \(m(x) = M(x)/P\) be the moment from a unit load at C. Then \(\delta_C = \dfrac{P}{EI}\int_0^3 m(x)^2\,dx\). Splitting the beam, \(\int_0^1 (2x)^2\,dx = \dfrac{4}{3}\) and \(\int_1^3 (x-3)^2\,dx = \dfrac{8}{3}\), which add up to \(4\). So \(\delta_C = \dfrac{4P}{EI}\), giving an effective spring stiffness \(k = P/\delta_C = EI/4\) felt by the hanging mass.
Step 4: Compute k and the natural frequency.
\(EI = (200\times10^{9})(10^{-8}) = 2000\) N.m\(^2\), so \(k = 2000/4 = 500\) N/m. With \(m = 5\) kg, \(\omega_n = \sqrt{k/m} = \sqrt{500/5} = \sqrt{100} = 10\) rad/s. Options with a \(\sqrt{3}\) in them come from carrying the reaction or span numbers through the calculation without simplifying at the end.
Final Answer:
The overhang acts like a spring of stiffness 500 N/m under the 5 kg tip mass.
\[ \boxed{\omega_n = 10\ \text{rad/s}} \]
