Question

A horizontal aluminium rod of diameter 4.8 cm projects 5.3 cm from a wall. A 1200 kg object is suspended from the end of the rod. The shear modulus of aluminium is 3.0xx10^(10)N//m^(2). Nelecting the mass of te rod find a shearing stress on the rod and b the vertical deflection of the end of the rod.

Answer 1

Answer:

Stress in the rod is given as

$Stress = 6.5 \times 10^6 N/m^2$

Deflection of the end of the rod is

$\Delta L = 1.15 \times 10^{-3} cm$

Explanation:

Part a)

Shearing stress of the rod is given as

$Stress = \frac{F}{A}$

now we know that

$Area = \pi r^2$

$Area = \pi (2.4\times 10^{-2})^2$

$Area = 1.81 \times 10^[-3} m^2$

now we have

$Stress = \frac{1200 \times 9.81}{1.81 \times 10^{-3}}$

$Stress = 6.5 \times 10^6 N/m^2$

Part b)

Now by the definition of elasticity we know that

$\frac{Stress}{Strain}$ = modulus of elasticity

so we have

$3\times 10^{10} = \frac{6.5 \times 10^6}{strain}$

now we have

$strain = 2.17 \times 10^{-4}$

now we have

$\Delta L = 2.17 \times 10^{-4} (5.3)$

$\Delta L = 1.15 \times 10^{-3} cm$

#Learn

Topic : Elasticity of the rod

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