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

a) τ= 2.59 MPa

b)y=28.4 x 10⁻⁵ cm

Step-by-step 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=(\pi)/(4)d^2`

`A=(\pi)/(4)* 70^2\ mm^2`

A=3848.45 mm²

The shear stress τ

`\tau=(P)/(A)`

`\tau=(10000)/(3848.45)\ MPa`

τ= 2.59 MPa

Lets take vertical deflection is y

We know that

τ = G Φ

For small deflection

`\phi=(y)/(L)`

So

τ L = G y

`y=(\tau L)/(G)`

`y=(2.59* 3.3)/(3* 10000)`

y=0.000284 cm

y=28.4 x 10⁻⁵ cm