Determine the required stiffness k so that the uniform 7 kg bar AC is in equilibrium when θ=30o.

Due to the collar guide at B the spring remains vertical and is unstretched when θ = 0o.Use principle of virtual work.

#### Solution

Solution:

Given : : Mass of bar AC = 7 kg

θ = 30°

To find : Required stiffness k

Solution:**Weight of rod = 7g N**Assume rod AC have a small virtual angular displacement 𝛿θ in anti-clockwise direction

Reaction forces H

_{A}and V

_{A}do not do any virtual work

Un-stretched length of the spring = BD

Extension of the spring (x) = CD = 2sin θ

Assume F

_{S}be the spring force at end C of the rod

**FS = Kx = 2Ksin θ**

Assume A to be the origin and AD be the X-axis of the system

Active force | Co-ordinate of the point of action along the force | Virtual Displacement |

W=7g | -sin θ | 𝛿yM=-cos θ 𝛿 θ |

FS=2Ksin θ | -2sin θ | 𝛿yC’=-2cos θ 𝛿 θ |

APPLYING PRINCIPLE OF VIRTUAL WORK

𝛿U = 0

-W X 𝛿YM + FS X 𝛿YC’ + 50 X 𝛿 θ = 0

2Ksin θ x (-2cos θ 𝛿 θ) = 7g x (-cos θ 𝛿 θ) - 50 x 𝛿 θ

Substituting the value of θ and solving**K=63.2025 NmThe required stiffness K for bar AC to remain in equilibrium is 63.2025 Nm**