THERMAL EFFECTS IN COMPOSITE
MEMBER HAVING ELEMENTS IN SERIES
The aluminum and steel bars in Fig. 23 have cross-sectional areas of 1.2 and 1.0 in
2
(7.7
and 6.5 cm
2
), respectively. The member is restrained against lateral deflection. A temper-
ature rise of 10O
0
F (55
0
C) causes the length of the member to increase to 42.016 in
(106.720 cm). Determine the stress and deformation of each bar. For aluminum, £ = 10 x
10
6
c = 13.0 x 10-
6
; for steel, c = 6.5 x IQ-
6
.
Calculation Procedure:
- Express the deformation of each bar resulting from the
temperature change and the compressive force
The temperature rise causes the bar to expand,
whereas the compressive force resists this expan-
sion. Thus, the net expansion is the difference be-
tween these two changes, or ALa = cLAT -
PLf(AE), where the subscript a refers to the alu-
minum bar; the other symbols are the same as
given earlier. Substituting gives AZa = 13.0 x
10^(24)(100)-P(24)/[l.2(10 x 106 )] = (31,200-
2P)IO"^6 , Eq. a. Likewise, for steel: Ms = 6.5 x
10-
6
(18)(100)-P(18)/[1.0(30x 10
6
)] = (11,700-
0.6P) IO-^6 , Eq. b. - Sum the results in step 1 to obtain
the total deformation of the member
Set the result equal to 0.016 in (0.4064 mm); FIGURE 23
solve for P. Or, AL = (42,900 - 2.6P)IO"^6 = 0.016
in (0.4064 mm); P = (42,900 - 16,000)72.6 =
10,350 Ib (46,037 N). - Determine the stresses and deformation
Substitute the computed value of P in the stress equation 5- = PIA. For aluminum sa =
10,350/1.2 = 8630 lb/in^2 (59,503.9 kPa). Then Ma = (31,200 - 2 x 10,35O)IQ-^6 = 0.0105
in (0.2667 mm). Likewise, for steel ss = 10,350/1.0 = 10,350 lb/in^2 (71,363.3 kPa); and
M 8 = (11,700 - 0.6 x 10,350)10~^6 = 0.0055 in (0.1397 mm).
SHRINK-FIT STRESS AND RADIAL
PRESSURE
An open steel cylinder having an internal diameter of 4 ft (1.2 m) and a wall thickness of
5
/i6 in (7.9 mm) is to be heated to fit over an iron casting. The internal diameter of the
cylinder before heating is Vn in (0.8 mm) less than that of the casting. How much must
the temperature of the cylinder be increased to provide a clearance of VM in (0.8 mm) all
around between the cylinder and casting? If the casting is considered rigid, what stress
will exist in the cylinder after it cools, and what radial pressure will it then exert on the
casting?