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.

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