Question

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C. E copper = 120GPa, αcopper = 16.9x10-6/ ° C.

Answer 1

`T&=(\sigma_(AB)+\sigma_(BC))/(2\alpha E)`

Step-by-step explanation:

The given data :-

• Diameter of rod AB ( d₁ ) = 200 mm.

• Diameter of rod BC ( d₂ ) = 150 mm.

• The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

• The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
• Consider the required temperature increase to close the gap at C = T °C
• Consider the change in length of the rod = бL

Solution :-

`\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_(A)&=\text{reaction\:force\:at\:A}\\R_(C)&=\text{reaction\:force\:at\:C}\\\sigma_(AB)&=\text{axial\:stress\:at\:A}\\\sigma_(BC)&=\text{axial\:stress\:at\:B}\\\sigma_(AB)&=(R)/(A_(A))\\&=(R_(A))/(A_(A))\\\sigma_(BC)&=(R_(B))/(A_(B))\\&=(R)/(A_(B))\\(\sigma_(AB))/(\sigma_(BC))&=(A_(B))/(A_(B))\\&=((\pi)/(4)\cdot 150^(2))/((\pi)/(4)\cdot 200^(2))\\&=(9)/(16)\end{aligned}`

`\begin{aligned}\delta L&= (\delta L _(thermal) +\delta L_(axial))_(AB) + ( \delta L _(thermal) +\delta L_(axial))_(BC)\\0& = (\delta L _(thermal) +\delta L_(axial))_(AB) + ( \delta L _(thermal) +\delta L_(axial))_(BC)\\&=\left[\alpha\:T\:L+\left((-RL)/(AE)\right)\right]_(AB)+\left[\alpha\:T\:L+\left((-RL)/(AE)\right)\right]_(BC)\\&=2\:\alpha\:T\:L-(L)/(E)(\sigma_(AB)+\sigma_(BC))\\T&=(\sigma_(AB)+\sigma_(BC))/(2\alpha E)\end{aligned}`