Answer:42.57x 10^7 N/mm2
Explanation:
The normal stress is derived in a body that is under tension and is calculated by taking the ratio of axial force applied on the body to the cross-sectional area of that same body.
Therefore,
The dimensions are 1/16 x 1/4
Let the force along AC be Fac then stress at AC will be
Phi (N) = Fac / A
Fac Cos 30 degrees x (.175m + .075m) x 960 = equilibrium of moment about point B.
Therefore
Phi N = 665.127(N) / (1/16x 1/4)
Therefore normal stress at the centre of link AC = 42.57 x 10^3 N/mm2
