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dybincka [34]
4 years ago
13

Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1

0 pts) Calculate the von Mises stress and the factor of safety. 10. (10 pts) Of the two factors of safety computed, which one is more realistic? What failure theory should you use if you want to be conservative? 11. (10 pts) Suppose all the principal stresses are equal in magnitude and sign, and larger than Sy. What are the predicted safety factors by the maximum shear stress and distortion energy failure theories? Calculate your results and explain them. What do you think would happen in reality?

Engineering
1 answer:
WITCHER [35]4 years ago
7 0

Answer:

Detailed solution is given below:

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Ronny wants to calculate the mechanical advantage. He needs to determine the length of the effort arm and the length of the load
kakasveta [241]

Answer:

I hope it's helpful.

Explanation:

Simple Machines

Experiments focus on addressing areas pertaining to the relationships between effort force, load force, work, and mechanical advantage, such as: how simple machines change the force needed to lift a load; mechanical advantages relation to effort and load forces; how the relationship between the fulcrum, effort and load affect the force needed to lift a load; how mechanical advantage relates to effort and load forces and the length of effort and load arms.

Through investigations and models created with pulleys and levers, students find that work in physical terms is a force applied over a distance. Students also discover that while a simple machine may make work seem easier, in reality the amount of work does not decrease. Instead, machines make work seem easier by changing the direction of a force or by providing mechanical advantage as a ratio of load force to effort force.

Students examine how pulleys can be used alone or in combination affect the amount of force needed to lift a load in a bucket. Students find that a single pulley does not improve mechanical advantage, yet makes the effort applied to the load seem less because the pulley allows the effort to be applied in the direction of the force of gravity rather than against it. Students also discover that using two pulleys provides a mechanical advantage of 2, but that the effort must be applied over twice the distance in order to gain this mechanical advantage Thus the amount of work done on the load force remains the same.

Students conduct a series of experiments comparing the effects of changing load and effort force distances for the three classes of levers. Students discover that when the fulcrum is between the load and the effort (first class lever), moving the fulcrum closer to the load increases the length of the effort arm and decreases the length of the load arm. This change in fulcrum position results in an increase in mechanical advantage by decreasing the amount of effort force needed to lift the load. Thus, students will discover that mechanical advantage in levers can be determined either as the ratio of load force to effort force, or as the ratio of effort arm length to load arm length. Students then predict and test the effect of moving the fulcrum closer to the effort force. Students find that as the length of the effort arm decreases the amount of effort force required to lift the load increases.

Students explore how the position of the fulcrum and the length of the effort and load arms in a second-class lever affect mechanical advantage. A second-class lever is one in which the load is located between the fulcrum and the effort. In a second-class lever, moving the load changes the length of the load arm but has no effect on the length of the effort arm. As the effort arm is always longer than the load arm in this type of lever, mechanical advantage decreases as the length of the load arm approaches the length of the effort arm, yet will always be greater than 1 because the load must be located between the fulcrum and the effort.

Students then discover that the reverse is true when they create a third-class lever by placing the effort between the load and the fulcrum. Students discover that in the case of a third-class lever the effort arm is always shorter than the load arm, and thus the mechanical advantage will always be less than 1. Students also create a model of a third-class lever that is part of their daily life by modeling a human arm.

The CELL culminates with a performance assessment that asks students to apply their knowledge of simple machine design and mechanical advantage to create two machines, each with a mechanical advantage greater than 1.3. In doing so, students will demonstrate their understanding of the relationships between effort force, load force, pulleys, levers, mechanical advantage and work. The performance assessment will also provide students with an opportunity to hone their problem-solving skills as they test their knowledge.

Through this series of investigations students will come to understand that simple machines make work seem easier by changing the direction of an applied force as well as altering the mechanical advantage by afforded by using the machine.

Investigation focus:

Discover that simple machines make work seem easier by changing the force needed to lift a load.

Learn how effort and load forces affect the mechanical advantage of pulleys and levers.

8 0
3 years ago
A clean machine is a _______________ machine.
solniwko [45]
A clean machine is a clean machine :-)
4 0
3 years ago
Which option identifies the type of device the engineer will develop in the following scenario?
Stells [14]
It would be actuator
4 0
2 years ago
Read 2 more answers
The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current
Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}

\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

If i = 2A and g = 10 cm

\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97

f_{m}=-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N

b) Using the coenergy of the system:

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2}  }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N

8 0
3 years ago
A mixture of air and methane is formed in the inlet manifold of a natural gas-fueled internal combustion engine. The mole fracti
german

Answer:

The mass flow rate of the mixture in the manifold is 6.654 kg/min

Explanation;

In this question, we are asked to calculate mass flow rate of the mixture in the manifold

Please check attachment for complete solution and step by step explanation.

4 0
3 years ago
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