Respuesta :
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
What is the magnitude of the force?
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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