chap11

# chap11 - PROBLEM 11.1 KNOWN A steel rod(circular...

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PROBLEM 11.1 KNOWN: A steel rod (circular cross-section) having: 6 2 m 10 in 0.25 in 30 10 lb/in 40 lb m L D E m = = = × = FIND: The change in length of the rod, L δ , ASSUMPTIONS: Rod is elastically deformed, such that m E σ ε = SOLUTION: The force resulting from 40 lb m in standard gravity is ( 29 ( 29 2 m m 2 40 lb 32.174 ft/sec 40 lb ft lb 32.174 lb sec N c ma F g = = = The resulting uniaxial stress is N a c F A σ = where ( 29 2 2 2 2 0.25 0.049 in 4 40 lb 814.9 lb/in 0.049 in c a A π σ = = = = and 2 5 6 2 814.9 lb/in 2.716 10 30 10 lb/in a a m E σ ε - = = = × × The change in length is then ( 29 ( 29 5 2.716 10 10 in 0.00027 in a L L δ ε - = = × = 10 in D=0.25 in F N Cross Section, A c

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PROBLEM 11.2 KNOWN: A steel rod (circular cross-section) having: 10 0.3 m 5 mm 20 10 Pa 50 kg m L D E m = = = × = FIND: The change in length of the rod, L δ . ASSUMPTIONS: Rod is elastically deformed, such that m E σ ε = SOLUTION: The force resulting from 50 kg in standard gravity is ( 29 ( 29 2 2 50 kg 9.8 m/s kg m 1.0 s N N c ma F g = = 490 N N F = The resulting uniaxial stress is N a c F A σ = where ( 29 2 3 5 2 6 -5 2 5 10 1.96 10 m 4 490 N 25 10 Pa 1.96 10 m c a A π σ - - = × = × = = × × and 6 6 10 25 10 Pa 125 10 20 10 Pa a a m E σ ε - × = = = × × The change in length is then ( 29 ( 29 6 6 0.3 125 10 37.5 10 m a L L δ ε - - = = × = × 0.3 m D=5 mm F N Cross Section, A c
PROBLEM 11.3 KNOWN: An electrical coil with 20,000 0.051 in 2.0 in N D r = = = FIND: The resistance, R . SOLUTION: We know e c L R A ρ = where 6 1.673 10 cm e ρ - = × for copper, and ( 29 2 2 5 2 0.051 1.42 10 ft 4 4 12 c A D π π - = = = × The length is then found as ( 29 ( 29 2 2 2 20,000 20,944 ft 12 L r N π π = = = which yields a resistance of ( 29 ( 29 ( 29 6 5 2 1.673 10 -cm 0.0328 ft/cm 20,944 ft 1.42 10 ft 81 R R - - × = × =

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PROBLEM 11.4 KNOWN: Aluminum having a volume of 3.14159 x 10 -5 m 3 , and a resistivity, 8 2.66 10 m e ρ - = × . FIND: Resistance, R , of 2 mm and 1 mm diameter wires having the same total volume. SOLUTION: Volume for a cylindrical wire is ( 29 2 4 V D L π = yielding L 2mm = 10 m and L 1mm = 40 m The resistance values are then calculated as ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 8 2 2 3 8 1 2 3 4 2.66 10 -m 10 m 0.085 2 10 m 4 2.66 10 -m 40 m 1.355 1 10 m 4 e mm c c mm mm L R A D A R R ρ π π π - - - - = = × = = × × = = ×
PROBLEM 11.5 KNOWN: A nickel conductor with ρ e = 6.8 x 10 -8 m A c = 5 x 2 mm (rectangular) L = 5 m FIND: a) R - the total resistance b) The diameter of a 5 m long copper wire having a circular cross-section to yield the same resistance. SOLUTION: The resistance is found from 2 10 mm 5 m e c c L R A L A ρ = = = which in this case yields ( 29 ( 29 8 6 2 6.8 10 m 5 m 0.034 10 10 m R - - × = = × For the copper, 8 1.7 10 m e ρ - = × ( 29 ( 29 8 6 2 1.7 10 m 5 m 0.034 2.5 10 m 1.8 mm c c A A D - - × = = × =

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PROBLEM 11.6 KNOWN: A Wheatstone bridge with all resistances initially equal to 100 . The maximum power through R 1 is 0.25 W. FIN D: Maximum applied voltage Bridge sensitivity ASSUMPTIONS: Infinite meter resistance SOLUTION: From the circuit shown below ( 29 1 1 2 2 1 2 1 2 i i i R i R E i i i i R R E + = = = + = But we know power, P , is given by 2 P i R = , and 0.25 W 0.05 A 100 i = = At node A 1 3 2 0.1 A i i i i i i i = + = = ( 29 ( 29 where is the equivalent bridge resistance so 0.1 A 100 10 V i i B B i E i R R E = = Ω =
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