A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel (Es = 200 GPa, αs = 11.7 * 10-6/oC) and portion BC is made of brass (Eb = 105 GPa, αb = 20.9 * 10-6/oC). Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 50oC.
Solution:
By dividing the rod to two parts 1) BC and 2) AB
Then, F = 142.55 KN 2.
Knowing that a 0.5 mm gap exists when the temperature is 24oC, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -75 MPa, (b) the corresponding exact length of the aluminum bar.
Solution:
In the first
case:
In the second case: (
)
(
)
( ) (
)
(
)
( (
( (
(
(
) )
)
(
(
(
)
( L1 = 0.45 m
)
,
(
)
(
)
( ) )
)
(
L2 = 0.35 m
(
,
(
) ,
(
)
E2 = 105 * 109 Pa
C
)
A1 = 1.8 * 10-3 m2 ,
F = 75 * 106 * 1.8 * 10-3 = 135 * 103 N Then,
)
)
A2 = 1.5 * 10-3 m2 , E1 = 73 * 109 Pa
)
,
,
)
T = 70,1957 + 24 = 94.2 C The length of Aluminum rod = ( ) = 0.45027 m
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