Solve Problem 15.7 if all specified conditions remain the same but radiant energy exchange from glass to the surroundings at the
1 answer:
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Answer:
Hello the table which is part of the question is missing and below are the table values
For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1
Answers:
belt size = 140 in with diameter of 20.1n
actual speed of belt = 288.49 in/s
actual center distance = 49.345 in
Explanation:
Given data :
Electric motor (driver sheave) speed (w1) = 950 rpm
Driven sheave speed (w2) = 250 rpm
pick D1 ( diameter of driver sheave) = 5.8 in ( from table )
To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first
VR = = 950 / 250
also since the speed of belt would be constant then ;
Vb = w1r1 = w2r2 ------- equation 1
r = d/2
substituting the value of r into equation 1
equation 2 becomes : = VR
Appropriate belt size ( d2) can be calculated as
d2 = =
= 22.04
From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value
next we have to determine the belt length /size
inputting all the values into the above equation including the value of C as calculated below
L ≈ 140 in
Calculating the center distance
we use this equation to get the ideal center distance
22.04 < c < 3 ( 5.8 + 20.1 )
22.04 < c < 77.7
the center distance is between 22.04 and 77.7 but taking an average value
ideal center distance would be ≈ 48 in
To calculate the actual center distance we use
-------- equation 3
B =
inputting all the values into (B)
B = 140(4) - 2( 20.01 + 5.8 )
B ≈ 399.15 in
inputting all the values gotten Back to equation 3 to get the actual center distance
C = 49.345 in ( actual center distance )
Calculating the actual belt speed
w1 = 950 rpm = 99.48 rad/s
belt speed ( Vb) = w1r1 = w1 *
= 99.48 * 5.8 / 2 = 288.49 in/s
Answer:
a) 0.978
b) 0.9191
c) 1.056
d) 0.849
Explanation:
Given data :
Stiffness of each bolt = 1.0 MN/mm
Stiffness of the members = 2.6 MN/mm per bolt
Bolts are preloaded to 75% of proof strength
The bolts are M6 × 1 class 5.8 with rolled threads
Pmax =60 kN, Pmin = 20kN
<u>a) Determine the yielding factor of safety</u>
------ ( 1 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
Input the given values into the equation above
equation 1 becomes ( np ) = = 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
<u>b) Determine the overload factor of safety</u>
------- ( 2 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
input values into equation 2 above
hence : = 0.9191
<u>C) Determine the factor of safety based on joint separation</u>
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence = 1.056
<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>
nf = 0.849
attached below is the detailed solution .
