Question

A horizontal aluminum rod 4.8 cm in diameter projects 5.6 cm from a wall. A 1200 kg object is suspended from the end of the rod. The shear modulus of aluminum is 3.0 ? 1010 N/m2. Neglect the rod's mass.(a) Find the shear stress on the rod. (N/m2)(b) Find the vertical deflection of the end of the rod. (m)

Answer 1

(a) Shear stress will be 4780487.80
`N/m^2`

(b)
`\Delta l=7.648* 10^(-6)m`

Step-by-step explanation:

We have given length of the aluminium rod l = 4.8 cm = 0.048 m

Diameter
`d=5.6cm,\ so\ r=(5.6)/(2)=2.8cm=0.028m`

Area
`A=\pi r^2=3.14* 0.028^2=0.00246m^2`

Shear modulus
`G=3* 10^(10)N/m^2`

Mass m = 1200 kg

So weight
`w=1200* 9.8=11760N`

(a) So hear stress P
`=(F)/(A)=(11760)/(0.00246)=4780487.80N/m^2`

(b) Now shear modulus is
`G=(stress)/(strain)`

So
`3* 10^(10)=(4780487.80)/(strain)`

`strain=1.593* 10^(-4)`

Now strain is given by

`strain=(\Deltal)/(l)`

So
`1.593* 10^(-4)=(\Delta l)/(0.048)`

`\Delta l=7.648* 10^(-6)m`