Which type of engineer often works with radar, communication, and navigation systems? electrical engineer chemical engineer indu
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Answer:
stress_ac = 5.333 MPa
shear stress_c = 1.763 MPa
Explanation:
Given:
- The missing figure is in the attachment.
- The dimensions of member AC = ( 6 x 25 ) mm x 2
- The diameter of the pin d = 19 mm
- Load at point A is P = 2 kN
Find:
- Find the axial stress in AE and the shear stress in pin C.
Solution:
- The stress in member AE can be calculated using component of force P along the member AE as follows:
stress_ac = P*cos(Q) / A_ae
Where, Angle Q: A_E_B and A_ac: cross sectional area of member AE.
cos(Q) = 4 / 5 ..... From figure ( trigonometry )
A_ae = 0.006*0.025*2 = 3*10^-4 m^2
Hence,
stress_ae = 2*(4/5) / 3*10^-4
stress_ae = 5.333 MPa
- The force at pin C can be evaluated by taking moments about C equal zero:
(M)_c = P*6 - F_eb*3
0 = P*6 - F_eb*3
F_eb = 0.5*P
- Sum of horizontal forces for member AC is zero:
P - F_eb - F_c = 0
F_c = 0.5*P
- The shear stress of double shear bolt is given by an expression:
shear stress = shear force / 2*A_pin
Where, The area of the pin C is:
A_pin = pi*d^2 / 4
A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2
Hence,
shear stress = 0.5*P / 2*A_pin
shear stress = 0.5*2 / 2*2.8353*10^-4
shear stress = 1.763 MPa
Question
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.
If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?
Answer:
a. 4.52W/m
b. 13mm
Explanation:
Given
Diameter of electrical wire = 2mm
Wire Thickness = 2-mm
Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)
Thermal contact resistance = 3E-4m².K/W
Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,
Temperature of the ambient air = 20°C.
Maximum Allowable Sheet Temperature = 50°C.
From the thermal circuit (See attachment), we my write
E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)
= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))
Where r in,i = D/2
= 2mm/2
= 1 mm
= 0.001m
r in,o = r in,i + t = 0.003m
T in, i = Tmax = 50°C
Hence
q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]
= 30/[(Ln3/0.26π) + 1/0.06π)]
= 30/[(1.34) + 5.30)]
= 30/6.64
= 4.52W/m
The critical radius is unaffected by the constant resistance.
Hence
Critical Radius = k/h
= 0.13/10
= 0.013m
= 13mm
