Question

The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Answer 1

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

$\theta$ = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑$f_{y}$ = 0, ∑M = 0

If ∑$M_{A}$ = 0

⇒$F_{BC}$×0.9 - P × 0.6 = 0

⇒$F_{BC}$×3 - P × 2 = 0

⇒$F_{BC}$ = $\frac{2P}{3}$

If ∑$M_{B}$ = 0

⇒$F_{AD}$×0.9 = P × 0.3

⇒$F_{AD}$ ×3 = P

⇒$F_{AD}$ = $\frac{P}{3}$

Now,

Area, A = $\frac{\pi }{4} X (0.012)^{2}$ = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , $\delta$ = $\frac{P l}{A E}$

Now,

$\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 9.1626 × 10⁻⁹ P

$\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 1.83253 × 10⁻⁸ P

Given that,

$\theta$ = 0.015°

⇒$\theta$ = 2.618 × 10⁻⁴ rad

So,

$\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}$

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N