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

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PROBLEM 11.1 KNOWN: A steel rod (circular cross-section) having: 62 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 ( )( ) 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 = where () 2 2 2 2 0.25 0.049 in 4 40 lb 814.9 lb/in 0.049 in c a A π and 2 5 814.9 lb/in 2.716 10 a a m E = × × The change in length is then ( ) 5 2.716 10 10 in 0.00027 in a LL δε =⋅ = × = 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 () ( ) 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 = where 2 35 2 6 -5 2 51 0 1 . 9 61 0 m 4 490 N 25 10 Pa 1.96 10 m c a A π −− × × and 6 6 10 125 10 a a m E × × The change in length is then 66 0.3 125 10 37.5 10 m a LL δε =⋅ = × = × 0.3 m D=5 mm F N Cross Section, A c
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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 for copper, and () 2 2 52 0.051 1.42 10 ft 44 1 2 c AD ππ ⎛⎞ == = × ⎜⎟ ⎝⎠ The length is then found as 2 2 2 20,000 20,944 ft 12 Lr N = which yields a resistance of ( )( ) 6 -cm 0.0328 ft/cm 20,944 ft 81 R R ×Ω = × =Ω
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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 =×Ω . FIND: Resistance, R , of 2 mm and 1 mm diameter wires having the same total volume. SOLUTION: Volume for a cylindrical wire is () 2 4 VD L π = yielding L 2mm = 10 m and L 1mm = 40 m The resistance values are then calculated as 2 2 8 2 2 3 8 1 2 3 4 2.66 10 -m 10 m 0.085 21 0 m 4 2.66 10 -m 40 m 1.355 11 4 e mm c c mm mm L RA D A R R == ×Ω × ×
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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 RA L A ρ == = which in this case yields () ( ) 8 62 6.8 10 m 5 m 0.034 10 10 m R ×Ω × For the copper, 8 1.7 10 m e ( ) 8 m 5 m 0.034 2.5 10 m 1.8 mm c c A AD = =
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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. FIND: Maximum applied voltage Bridge sensitivity ASSUMPTIONS: Infinite meter resistance SOLUTION: From the circuit shown below () 11 2 2 12 i i iR iR E iii iR R E += == But we know power, P , is given by 2 PiR = , and 0.25 W 0.05 A 100 i At node A 13 2 0.1 A ii iii ii =+= = ( ) where is the equivalent bridge resistance so 0.1 A 100 10 V B B i Ei R R E = =Ω = The bridge sensitivity is defined as 0 1 B E K R δ = and for a bridge with all resistances initially equal and assuming R R ± ( ) 0 10 V 100 2.5 V/ 4 400 i EE R RR ≈=
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PROBLEM 11.7 KNOWN: A strain gauge with R 1 = 120 , GF
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chap11 - PROBLEM 11.1 KNOWN A steel rod(circular...

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