Question

A horizontal rigid bar ABC is pinned at end A and supported by two cables at points B and C. A vertical load P 5 10 kN acts at end C of the bar. The two cables are made of steel with a modulus elasticity E 5 200 GPa and have the same cross-sectional area. Calculate the minimum cross-sectional area of each cable if the yield stress of the cable is 400 MPa and the factor of safety is 2.0. Consider load P only; ignore the weight of bar ABC and the cables.

Answer 1

Answer:

vertical load = 10 kN

Modulus of elasticity = 200GPa

Yield stress on the cable = 400 MPa

Safety factor = 2.0

Explanation:

Data

let L = $\sqrt{(1.5)^{2} + (1.5)^{2} }$

        = 3.35 m

substituting 1.5 m for h and 3 m for the tern (a + b)

$\theta$ = tan⁻¹($\frac{1.5}{1.5}$)

  = 45⁰

substituting 1.5 for h and 3 m for (a+ b) yields:

$\theta$₂ = tan⁻¹ ($\frac{1}{3}$)

   =25.56⁰

checking all the forces, they add up to zero. This means that the system is balanced and there is no resultant force.