Answer:
a. 51.84Kj
b. 2808.99 W/m^2
c. 11.75%
Explanation:
Amount of heat this resistor dissipates during a 24-hour period
= amount of power dissipated * time
= 0.6 * 24 = 14.4 Watt hour
(Note 3.6Watt hour = 1Kj )
=14.4*3.6 = 51.84Kj
Heat flux = amount of power dissipated/ surface area
surface area = area of the two circular end + area of the curve surface
= 2.136 *10^-4
Heat flux = = 2808.99
fraction of heat dissipated from the top and bottom surface
=11.75%
Answer:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Explanation:
Given:
- The diameter of the pipe d = 5-in
- The pipe is supported every L = 9 ft of pipe in length
- The weight if the pipe + contents W = 10 lb/ft
Find:
determine the magnitude and location of the maximum bending moment in members AC and BC.
Solution:
- The figure (missing) is given in the attachment.
- We will first determine the external forces acting on each member:
Section: 9-ft section of pipe.
Sum of forces perpendicular to member AC = 0
F_d - 0.8*W*L = 0
F_d = 0.8*10*9 = 72 lb
Sum of forces perpendicular to member BC = 0
F_e - 0.6*W*L = 0
F_e = 0.6*10*9 = 54 lb
F_d = 72 lb , F_e = 54 lb
- Then we will determine the support reactions for each member AC point A and BC point B.
Section: Entire Frame.
Sum of moments about point B = 0
-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0
-A_y*(1.5625) + 15 + 39.375 = 0
A_y = 34.8 lb
Sum of forces in vertical direction = 0
A_y + B_y - 0.8*F_d - 0.6*F_e = 0
B_y = 0.8*(72) + 0.6*(54) - 34.8
B_y = 55.2 lb
Sum of forces in horizontal direction = 0
A_x + B_x - 0.6*F_d + 0.8*F_e = 0
A_x + B_x = 0
Section: Member AC
Sum of moments about point C = 0
F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0
72*2.5 - 34.8*12 - 9*A_x = 0
A_x = -237.6 / 9 = - 26.4 lb
B_x = - A_x = 26.4 lb
A_x = -26.4 lb , B_x = 26.4 lb
- Now we will calculate bending moment for each member at different sections.
Member AC:
From point A till just before point D
-0.6*A_x*x - A_y*0.8*x + M = 0
15.84*x - 27.84*x + M = 0
M = 12*x ..... max value at D, x = 12.25 in
M_max = 12*12.25/12 = 12.25 lb-ft
Member BC:
From point B till just before point E
-0.8*B_x*x + B_y*0.6*x + M = 0
-21.12*x + 33.12*x + M = 0
M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in
M_max = -12*8.75/12 = -8.75 lb-ft
- The maximum bending moments and their locations are:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Answer:
0.2 m/s
Explanation:
The velocity of a point on the edge of a disk rotating disk can be calculated as:
Where is the angular velocity and r the radius of the disk. This leads to:
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
