A rotator has a weight of 100lb with a centroidal radius of gyration of 9 in. Determine the moment of inertia about the center o
1 answer:
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Answer:
hello your question is incomplete below is the missing part of the question and attached is the missing diagram
In the simply-supported beam shown in the figure below, d1=14 ft, d2=7 ft, and F=15 kips. so find Az, Ay, By.(in kips)
answer :
Reaction force at B = 10 kips
Reaction force in the y axis = 5 kips
Reaction force in the Z direction = 0 kips
Explanation:
Taking moment about point A
∑ Ma = 0
By + ( d1 + d2 ) - F*d1 = 0
By ( reaction force at B ) = ( 15 * 14 ) / ( 21 ) = 10 kips
Applying equilibrium forces in the Y-axis
∑ Fy = 0
Ay - F + By = 0
where : F = 15, By = 10
hence ; Ay = 5 kips
applying equilibrium forces in the Z-direction
∑ Fz = 0
Az = 0 kips
Answer:
The heat transfer q = 18.32W
Explanation:
In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K
Please check attachment for complete solution and step by step explanation
Answer:
The diameter of the shaft is 80.5 mm.
Explanation:
Torsion equation is applied for the diameter of the solid shaft.
Step1
Given:
Power of the shaft is 100 kw.
Revolution per minute is 160 RPM.
Allowable shear stress is 70 Mpa.
Maximum torque is 20% more than the mean torque.
Step2
Mean torque is calculated as follows:
T=5968.31 N-m
Step3
Maximum torque is calculated as follows:
T_{max}=7161.97 N-m
Step4
Apply torsional equation for diameter of shaft as follows:
d=0.0805 m
or,
d=80.5 mm
Thus, the diameter of the shaft is 80.5 mm.