Page 54 - Handbook of Civil Engineering Calculations, Second Edition

STATICS, STRESS AND STRAIN, AND FLEXURAL ANALYSIS 1.37

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

THERMAL EFFECTS IN COMPOSITE
MEMBER HAVING ELEMENTS IN PARALLEL

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A /2-in. (12.7-mm) diameter Copperweld bar consists of a steel core /8 in. (9.53 mm) in
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diameter and a copper skin /16 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 80°F (44.4°C)? Use the following values for thermal expansion co-
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efficients: c s 6.5 10 6 and c c 9.0 10 , where the subscripts s and c refer to
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steel and copper, respectively. Also, E c 15 10 lb/sq.in. (1.03 10 kPa).
Calculation Procedure:
1. Determine the cross-sectional areas of the metals
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The total area A 0.1963 sq.in. (1.266 cm ). The area of the steel A s 0.1105 sq.in.
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(0.712 cm ). By difference, the area of the copper A c 0.0858 sq.in. (0.553 cm ).
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
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A s E s (relative) 0.1105 30 10
3315
A c E c (relative) 0.0858 15 10
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1287
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Total 4602
Then the coefficient of thermal expansion of the composite member is c (3315c s
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1287c c )/4602 7.2 10 /°F (1.30 10 /°C).
3. Determine the thermal expansion of the 1-ft (0.3-m) section
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Using the relation L cL T, we get L 7.2(10 )(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 L c 9.0(10 )
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(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 L cs 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 c E c L cs )/L, where the symbols are as given before: P
[1,287,000(0.00173)]/12 185 lb (822.9 N).
7. Verify the results obtained
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Repeat steps 4, 5, and 6 with the two materials interchanged. So L s 6.5(10 )(12)(80)
0.00624 in. (0.15849 mm); L sc 0.00691 0.00624 0.00067 in. (0.01701 mm).
Then P 3,315,000(0.00067)/12 185 lb (822.9 N), as before.