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Valentin [98]
4 years ago
6

Both portions of the rod ABC are made of an aluminum for whichE = 70 GPa. Knowing that the magnitude of P is 4 kN, determine(a)

the value of Q so that the deflection at A is zero, (b) the correspondingdeflection of B.0.4 m0.5 m
Engineering
1 answer:
san4es73 [151]4 years ago
4 0

Explanation:

ΔL_{BC} = ΔL_{AB}

\frac{(Q - 4000)(0.5)}{3.14* 0.03 *0.03 *70*10^{9} }     (1)

= \frac{4000*0.4}{3.14*0.01*0.01*70*10^{9} }

Q = 32,800 N

now put this value in equation 1.

Deflection of B = \frac{(32800-4000)(0.5)}{3.14*0.03*0.03*70*10^{9} }

                       = 0.0728 mm

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A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
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Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

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