Ejercicios sobre Resistencia de Materiales

[pic 1]

l = No pude encontrar la longitud de este alambre

Ƿ = 1.7X10^-6 Ω.inc = 4.318x10^-6 Ω.cm

A = (0.16256 cm)^2(3.1416) = 0.083018953cm^2

Ƿ = RA/l

R = Ƿl/A

[pic 2]

€ = 0.0001 inc/inc
F = 1500 lb
E = 30x10^6 Lb/inc^2

E = [pic 3]/€
[pic 4] = (30x10^6 Lb/inc^2)/(0.0001 inc/inc)
[pic 5] = 3000 Lb/inc^2

A = F/[pic 6]
A = (1500 lb)/(3000 Lb/inc^2)
A = 0.5 inc^2

[pic 7]

E = 16x10^6 Lb/inc^2
[pic 8]= 30,000 lb/inc^2
€ = (30,000 lb/inc^2)/(16x10^6 Lb/inc^2)
€ = 1.875x10^-3
Δl = (1.875x10^-3)(96 inc)
Δl = 0.18 inc

 [pic 9]

NDB = ____________P____________
                 (ΠD/2)(D-√D²-d²)

NDB = __________________________3000 kg________________________
                 {(Π)(10 mm)/2}{10 mm-√[(10 mm)²-(4 mm)²]}

NDB = 228.7671434 kg/mm^2

NDB = __________________________3000 kg________________________
                 {(Π)(10 mm)/2}{10 mm-√[(10 mm)²-(3.2 mm)²]}

NDB = 364.2122596 kg/mm^2

[pic 10]

l = 45 cm
[pic 11] = 60 kg/cm^2
E = 7x10^10 N/m^2

[pic 12]

α = 9x10^-6 °C
Lo= 1.7        
Tf = 45°C      
To = 80 °C  
[pic 13] = (1.7 m)(9x10^-6/°C)(45-80)°F = -5.34x10^-4 m

[pic 14]

ΔV = 0.27m^3(39 x 10-6/°C)(122°C) = 1.28x10-3 m^3
Donde:                                                          
γ = 39 x 10-6
Vo = 0.27m3
Vf = Área final
To = 0
Tf = 122°C

[pic 15]
Ti=165.2 °F
Espesor= 0.065ft
Transferencia 1400 BTU/ft2
Coeficiente del hule: 0.00025 BTU*inc/ft^2
T2 = [{(0.065ft) (1400 BTU/ft^2)}/(- 0.00025 BTU*inc/ft^2)] + 165.2 °F
T2 = -363834.8 °F

[pic 16]

R= 2 ohm

Ƿ = 9.5x10-6 ohm/cm

L= 65cm

Ƿ = RA/l

A= (ǷL)/R

A= (65 cm* 9.5x10-6 ohm/cm)/(2 ohm) = 3.0875*10-4 cm^2

[pic 17] = l/RA      

[pic 18] = (65 cm)/{(3.0875*10-4 cm^2)(2 ohm)} = 105,263Ω-1·cm-1

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