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:

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

[tex]\theta[/tex] = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0

If ∑[tex]M_{A}[/tex] = 0

⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0

⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0

⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]

If ∑[tex]M_{B}[/tex] = 0

⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3

⇒[tex]F_{AD}[/tex] ×3 = P

⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]

Now,

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

We know that,

Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]

Now,

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

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

Given that,

[tex]\theta[/tex] = 0.015°

⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad

So,

[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]

⇒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

The magnitude of Force P is; P = 25715.15 N

If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;

Taking moments about point A, we have;

(F_bc * 0.9) - P(0.6) = 0

F_bc = ²/₃P

Taking moments about B gives;

P(0.3) - F_ad * 0.9 = 0

F_ad = ¹/₃P

Normal stress for BC is;

σ_bc = F_bc/A_bc

σ_bc = (²/₃P)/(π * 0.006²)

σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²

σ_ad = (¹/₃P)/(π * 0.006²)

σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²

We know that;

Elongation is; ΔL = PL/AE = (P/A) * (L/E)

Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa

Thus;

ΔL_bc =  (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_bc = 1.83253P × 10⁻⁸

Likewise;

ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_ad = 9.1626P × 10⁻⁹ m

Converting the beam tilt angle from degrees to radians gives;

θ = 0.015° = 0.00026179939 rads

Using small angle analysis, we can say that;

θ = (ΔL_bc - ΔL_ad)/36

θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36

Solving gives P = 25715.15 N

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