Question

Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the force in each rod if the temperature is increased by 30° C. E copper = 120GPa, αcopper = 16.9x10-6/ ° C.

Answer 1

A)3.8196 *
`10^(9)` Newtons B) 2.153 *
`10^(9)` Newtons

Step-by-step explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π
`R^(2)`.

Area Expansivity β =
`(Change in Area)/(Original Area * Temperature Rise)`

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm =
`(0.2m)/(2)` = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm =
`(0.15)/(2)` = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π *
`0.1^(2)` = 0.0314
`m^(2)`

Area of Rod BC = π *
`0.075^(2)` = 0.0177
`m^(2)`.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9*
`10^(-6)` * 0.0314 * 30 = 3.183 *
`10^(-5)`
`m^(2)`.

For Rod BC we have = 2 * 16.9*
`10^(-6)` * 0.0177 * 30 = 1.794∈-5
`m^(2)`.

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 *
`10^(-5)` gives 3.8196 *
`10^(9)` Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 *
`10^(-5)` gives 2.153 *
`10^(9)` Newtons