Answer:
Detailed solution is given below:
6. You are evaluating flow through an airway. The current flow rate is 10 liters per minute with a fixed driving pressure (P1) o
1 answer:
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Answer:
See attached picture.
Explanation:
Answer:
1.505
Explanation:
cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.
stress is force per unit area
stress=P/A
A = πd^2/4.
uncertainty of axial force P= +/-.11
s=+/-.20, strength
d=+/-.04 diameter
fail load/max allowed
minimum design=fail load/max allowed
minimum design =s/(P/A)
sA/P
A=(.96d^2)/4, so Amin=
(because the diameter at minimum is (1-0.04=0.96)
minimum design=Pmax/(sminxAmin)
1.11/(.80*.96^2)=
1.505
Answer:
The change in length of the circular strut DC = 0.0028 in.
The vertical displacement of the rigid bar at point B = 0.00378 in.
Explanation:
We have the following parameters or information in the question given above:
=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.
The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC is given in the attached picture below.
According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.
Where FCD = p × [4 + 5] ÷ [sin Ф × 4].
kindly note that from the diagram sin Ф = 3/5, cos Ф = 4/5 and tan Ф = 3/4. Also p =5.
Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.
The next thing here is to determine the area and length of CD, say the area of CD is G, thus, G = π/4 × [ 15² - 14.4²] = 13.854 in².
The lenght of CD is = √[4² + 3²] = √[16 + 9] = 5ft. Thus, 5 × 12 = 60in.
Hence, the change in length of the circular strut DC = [18.75 × 60] ÷ 13.854 × 29000 = 0.0028 in.
The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.
We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.
(CV = vertical deflection of CD).
The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.
