Answer:
Please see the attached picture for the complete answer.
Explanation:
Answer:
Max shear = 8.15 x 10^7 N/m2
Explanation:
In order to find the maximum stress for a solid shaft having radius r, we will be applying the Torsion formula which can be written as;
Allowable Shear Stress = Torque x Radius / pi/2 x radius^4
Putting the values we have;
T = 2000 N/m
Radius = Diameter/2 = 0.05 / 2 = 0.025 m
Putting values in formula;
Max shear = 2000 x 0.025 / 3.14/2 x (0.025)^4
Max shear = 8.15 x 10^7 N/m2
Answer:
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Answer:
According to the standard Safe Practices for Motor Vehicle Operations, ANSI/ASSE Z15.1, defensive driving skills involves "driving to save lives, time, and money, in spite of the conditions around you and the actions of others."
Explanation:
Defensive driving is a type of training provided to the learners eligible to drive motor vehicles. It involves the basics of driving and rules involved in driving. The learners learn about the methods of driving safely with the minimum expense of time and money. It also helps in reducing the risks of collision in adverse situations. The methods to rescue oneself from dangerous situations are also trained. Preventing oneself from the mistakes of others is also termed under defensive driving.
Answer:
36.0 kpsi2.
Explanation:
From the question
Sut=110 kpsi
Se’=0.5(110)=55 kpsi
For surface factor ka,
a=2.70, b= -0.265, ka=a(Sut)b=2.70(110)-0.265=0.777
Assuming the worst case for size factor kb, and since 0.11 less than or equal to d less than or equal to 2in,
Therefore:
kb=0.879d-0.107 = 0.879(1.5)-0.107=0.842
Hence, the endurance strength is Se= ka kb Se’ = 36.0 kpsi2.