A plane wall of thickness 2L = 60 mm and thermal conductivity k= 5W/m.K experiences uniform volumetric heat generation at a rate
1 answer:
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Answer:
(a)
Plastic section modulus Z =92.72 in^3
Plastic moment Mp = 4636 kip-in
(b)
Elastic section modulus S = 80.88 in^3
Yield moment My = 4044 kip-in
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached files.i hope my explanation will help you in understanding this particular question.
Answer:
Explanation:
Floor Load:
Lo= 50psf
At= 25x25 = 625 square feet
= 13.1psf
%reduction= 13.1/50 = 26%
Fr= 3[(13.1psf)(25ft)(25ft)+(20psf)(25ft)(25ft)]= 62k
Answer:
See attachment
1512 ft
Explanation:
Since the acceleration is either constant or zero, the a−t curve is made of horizontal straight-line segments. The values of t2 and a4 are determined as follows:
Acceleration - Time
0 < t < 6: Change in v = area under a–t curve
V_6 - 0 = (6 s)(4 ft/s2 ) = 24 ft/s
6 < t < t2: Since the velocity increases from 24 to 48 ft/s,
Change in v = area under a–t curve
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: Since the velocity is constant, the acceleration is zero.
34 < t < 40: Change in v = area under a–t curve
0 - 42 = 6*a4
a4 = - 8 ft / s^2
The acceleration being negative, the corresponding area is below the t axis; this area represents a decrease in velocity.
Velocity - Time
Since the acceleration is either constant or zero, the v−t curve is made of straight-line segments connecting the points determined above.
Change in x = area under v−t curve
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Adding the changes in x, we obtain the distance from A to B:
d = x40 - 0 = 1512 ft
Answer:
Explanation:
For we to calculate the shear stress on the upper plate and give its direction. Sketch the variation of shear stress across the channel, I used hand in solving it, check attached file below
