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Phoenix [80]
4 years ago
14

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5 × 10-4

mm (1.969 × 10-5 in.) and a crack length of 4.5 × 10-2 mm (1.772 × 10-3 in.) when a tensile stress of 190 MPa (27560 psi) is applied?
Engineering
1 answer:
White raven [17]4 years ago
7 0

Answer:

The question is a problem that requires the principles of fracture mechanics.

and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.

Explanation:

Max Stress, σ = 2σ₀√(α/ρ)

where,

σ₀  = Tensile stress = 190MPa = 1.9x10⁸Pa

α  = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m

ρ  = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m

Max Stress, σ = 2 x  1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵

Max Stress, σ = 2 x  1.9x10⁸ x 6.708 Pa

Max Stress, σ = 2549MPa

Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa

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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
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
Who can help me with electric systems for cars?
hoa [83]

Answer: i can see if i can what is the problem

Explanation:

7 0
3 years ago
Which of the following answer options are your employer's responsibility?
tino4ka555 [31]

Answer:

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Explanation:

To find - Which of the following answer options are your employer's responsibility?  Select all that apply.

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Solution -

The correct options are -

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

All are the Responsibilities of an employer

Reason -

The most important duty of the employer is to stay alert and implement a correctly and efficiently written communication program related to hazards of the substances in the workplace.

He also has to maintain the program so that employees do not get affected.

3 0
3 years ago
Select the correct statement(s) regarding IEEE 802.16 WiMAX BWA. a. WiMAX BWA describes both 4G Mobile WiMAX and fixed WiMax b.
Gala2k [10]

Answer:

d. all of the statements are correct.

Explanation:

WiMAX Broadband Wireless Access has the capacity to provide service up to 50 km for fixed stations. It has capacity of up to 15 km for mobile stations. WiMAX BWA describes both of 4G mobile WiMAX and fixed stations WiMAX. OFMD is used to increase spectral efficiency of WiMAX and to improve noise performance.

5 0
3 years ago
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