Temperature changes problems
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1) The rod consists of two portions made of different materials joined together. When the temperature increases by 50°C, a compressive force of 142.55 kN is induced in the rod. 2) A 0.5 mm gap exists in an aluminum bar at 24°C. The temperature at which the stress in the bar will be -75 MPa is 94.2°C. The length of the aluminum bar at this temperature is 0.45027m.
Temperature changes problems
- 1. 1- A rod consistingof twocylindrical portionsABandBC isrestrainedatboth ends. PortionABis made of steel (Es = 200 GPa, αs = 11.7 * 10-6 /o C) andportionBC is made of brass(Eb = 105 GPa, αb = 20.9 * 10-6 /o C). Knowingthatthe rod isinitially unstressed,determine the compressive force inducedinABCwhenthere isa temperature rise of 50o C. Solution: By dividing the rod to two parts 1) BC and 2) AB Δ1 = 𝛼 Δ𝑇 𝐿 = 20.9 ∗ 50 ∗ 0.3 ∗ 10−6 = 3.135∗ 10−4 𝑚 Δ2 = 11.7 ∗ 0.25∗ 50 ∗ 10−6 = 1.4625∗ 10−4 𝑚 Δ1 + Δ2 = 4.5975∗ 10−4 𝑚 𝐹 𝐿1 𝐴1 𝐸1 + 𝐹 𝐿2 𝐴2 𝐸2 = 4.5975∗ 10−4 Then, F = 142.55 KN 2- Knowingthata 0.5 mm gap existswhenthe temperature is24o C,determine(a) the temperature atwhichthe normal stressinthe aluminumbarwill be equal to -75 MPa, (b) the correspondingexactlengthof the aluminumbar.
- 2. Solution: In the first case: Δ1 = 𝛼1 Δ𝑇 𝐿1 Δ2 = 𝛼2 Δ𝑇 𝐿2 Δ = Δ1 + Δ2 In the second case: Δ1 = 𝐹 ( 𝐿1+ Δ1) 𝐴1 𝐸1 Δ2 = 𝐹 ( 𝐿2 + Δ2) 𝐴2 𝐸2 Δ = Δ1 + Δ2 Δ − Δ = 0.5 × 10−3 𝑚 Δ = 𝛼1 Δ𝑇 𝐿1 + 𝛼2 Δ𝑇 𝐿2 = ( 𝛼1 𝐿1 + 𝛼2 𝐿2)Δ𝑇 Δ = 𝐹 ( 𝐿1 + Δ1) 𝐴1 𝐸1 + 𝐹 ( 𝐿2 + Δ2) 𝐴2 𝐸2 = 𝐹 ( ( 𝐿1 + Δ1) 𝐴1 𝐸1 + ( 𝐿2 + Δ2) 𝐴2 𝐸2 ) = 𝐹 ( ( 𝐿1 + 𝛼1 Δ𝑇 𝐿1) 𝐴1 𝐸1 + ( 𝐿2 + 𝛼2 Δ𝑇 𝐿2) 𝐴2 𝐸2 ) ( 𝛼1 𝐿1 + 𝛼2 𝐿2)Δ𝑇 − 𝐹 ( ( 𝐿1 + 𝛼1 Δ𝑇 𝐿1) 𝐴1 𝐸1 + ( 𝐿2 + 𝛼2 Δ𝑇 𝐿2) 𝐴2 𝐸2 ) = 0.5 × 10−3 𝑚 ∴ Δ𝑇 = 𝐹 ( ( 𝐿1 ) 𝐴1 𝐸1 + ( 𝐿2) 𝐴2 𝐸2 )+ 0.5 × 10−3 ( 𝛼1 𝐿1 + 𝛼2 𝐿2)− 𝐹 ( ( 𝛼1 𝐿1) 𝐴1 𝐸1 + ( 𝛼2 𝐿2) 𝐴2 𝐸2 ) L1 = 0.45 m , L2 = 0.35 m , A1 = 1.8 * 10-3 m2 , A2 = 1.5 * 10-3 m2 , 𝛼1 = 23.2∗ 10−6 𝐶−1 , 𝛼1 = 21.6 ∗ 10−6 𝐶−1
- 3. E1 = 73 * 109 Pa , E2 = 105 * 109 Pa , F = 75 * 106 * 1.8 * 10-3 = 135 * 103 N Then, Δ𝑇 = 70,1957C T = 70,1957 + 24 = 94.2 C The length of Aluminum rod = 𝛼1 Δ𝑇 𝐿1 − 𝐹 ( 𝐿1) 𝐴1 𝐸1 + 450∗ 10−3 = 0.45027 m