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

Answer:

A)3.8196 * $10^{9}$ Newtons        B) 2.153 * $10^{9}$ Newtons

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 =    $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 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