Question

A horizontal aluminum rod 3.3 cm in diameter projects 7.0 cm from a wall. A 1000 kg object is suspended from the end of the rod. The shear modulus of aluminum is 3.0·1010 N/m2. Neglecting the rod's mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.

Answer 1

Answer:

a) τ= 2.59 MPa

b)y=28.4 x 10⁻⁵ cm

Explanation:

Given that

L= 3.3 cm

d= 7 cm

m= 1000 kg

G= 3 x 10¹⁰ Pa

Force due to load 1000 kg

P = m g= 1000 x 10 N

P = 10000 N

Area A

$A=\dfrac{\pi}{4}d^2$

$A=\dfrac{\pi}{4}\times 70^2\ mm^2$

A=3848.45 mm²

The shear stress τ

$\tau=\dfrac{P}{A}$

$\tau=\dfrac{10000}{3848.45}\ MPa$

τ= 2.59 MPa

Lets take vertical deflection is y

We know that

τ = G Φ

For small deflection

$\phi=\dfrac{y}{L}$

So

τ L = G y

$y=\dfrac{\tau L}{G}$

$y=\dfrac{2.59\times 3.3}{3\times 10000}$

y=0.000284 cm

y=28.4 x 10⁻⁵ cm

Answer 2

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