Handbook of Civil Engineering Calculations

3. Determine the compressive stress
   Use the relation s = EkLIL, where the symbols are as given earlier. Thus, s =
   30(10^6 )(0.0823)/[18(12)] 11,430 lb/in^2 (78,809.9 kPa).

THERMAL EFFECTS IN COMPOSITE

MEMBER HAVING ELEMENTS IN PARALLEL

A^1 Xi-Ui (12.7-mm) diameter Copperweld bar consists of a steel core^3 /s in (9.53 mm) in di-
ameter and a copper skin %6 in (1.6 mm) thick. What is the elongation of a 1-ft (0.3-m)
length of this bar, and what is the internal force between the steel and copper arising from
a temperature rise of 8O^0 F (44.4^0 C)? Use the following values for thermal expansion co-
efficients: cs = 6.5 x 10"^6 and cc = 9.0 x 1O-6, where the subscripts s and c refer to steel
and copper, respectively. Also, EC=15* 106 lb/in^2 (1.03 x 108 kPa).

Calculation Procedure:

1. Determine the cross-sectional areas of the metals
   The total area A = 0.1963 in^2 (1.266 cm^2 ). The area of the steel A 5 = 0.1105 in^2 (0.712
   cm^2 ). By difference, the area of the copper A 0 = 0.0858 in^2 (0.553 cm^2 ).


2. Determine the coefficient of expansion of the
   composite member
   Weight the coefficients of expansion of the two members according to their respective
   AE values. Thus

Af 3 (relative) = 0.1105 x 30 x IQ^6 = 3315

Afc (relative) = 0.0858 x 15 x IQ

6

= 1287

Total 4602

Then the coefficient of thermal expansion of the composite member is c = (3315c 5 +
1287cc)/4602 = 7.2 x 10-V^0 F (1.30 x 10-^5 /°C).

3. Determine the thermal expansion of the 1-ft (0.3-m) section
   Using the relation AL - cLNo, we get M = 7.2(10^6 )(12)(80) = 0.00691 in (0.17551 mm).


4. Determine the expansion of the first material without restraint
   Using the same relation as in step 3 for copper without restraint yields ALC = 9.0(1O-6) x
   (12)(80) = 0.00864 in (0.219456 mm).


5. Compute the restraint of the first material
   The copper is restrained to the amount computed in step 3. Thus, the restraint exerted by
   the steel is Mcs = 0.00864 - 0.00691 - 0.00173 in (0.043942 mm).


6. Compute the restraining force exerted by the second material
   Use the relation P = (A 0 E 1 AL 0 ^IL 9 where the symbols are as given before: P =
   [1,287,000(0.00173)]/12 = 185 Ib (822.9 N).


7. Verify the results obtained
   Repeat steps 4, 5, and 6 with the two materials interchanged. So AI 5 = 6.5(10~^6 )(12)(80)
   = 0.00624 in (0.15849 mm); Msc = 0.00691 - 0.00624 = 0.00067 in (0.01701 mm). Then
   P = 3,315,000(0.00067)/12 - 185 Ib (822.9 N), as before.