A 5 kg mass is suspended at free end of overhanging beam, having a pin support and roller support as shown in figure. E = 200 GPa, Area moment of inertia = 10^–8 m⁴, The natural frequency in rad/s is

The correct answer is 10.

1. Extracting the Given Parameters from the Problem and the Diagram:
By analyzing the provided image, we can observe the configuration of the overhanging beam:
- The left end of the beam is attached to a pin support.
- A roller support is positioned at a distance of 1 m from the pin support. Let this span length be L=1 m.
- The beam extends beyond the roller support as an overhang of length a=2 m to a free end.
- A mass of m=5 kg is suspended at this free end.
- The Young's modulus of the beam material is E=200 GPa=200×109 N/m2.
- The area moment of inertia of the cross-section is I=10-8 m4.

2. Calculating the Flexural Rigidity (EI):
The flexural rigidity of the beam is calculated as:
EI=(200×109)×10-8=2000 N·m2

3. Determining the Deflection at the Free End:
To find the equivalent stiffness of the beam at the location of the mass, we apply a virtual vertical load P at the free end. The total downward deflection δC at the free end consists of two parts:
1. Deflection due to the bending of the overhanging portion acting like a cantilever beam: Pa33EI
2. Deflection due to the rotation of the beam at the roller support: θ·a=Pa2L3EI
Adding these two components gives the standard formula for deflection at the free end of an overhanging beam:

δC=Pa2(L+a)3EI

Substituting the values of L=1 m, a=2 m, and EI=2000 N·m2:

δC=P·22·(1+2)3·2000=12P6000=P500

4. Finding the Equivalent Stiffness (k):
The equivalent spring stiffness k at the free end is:

k=PδC=500 N/m

5. Calculating the Natural Frequency (ωn):
The natural frequency of a single degree of freedom system is given by:

ωn=km

Substituting k=500 N/m and m=5 kg:

ωn=5005=100=10 rad/s