1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gre4nikov [31]
3 years ago
6

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv

ers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is 3/8 in.
A) Determine the maximum shear stress developed in the shaft.

TMAX = ?


B) Also, what is the "wind up," or angle of twist in the shaft at full power?

φ = ?
Engineering
1 answer:
Debora [2.8K]3 years ago
8 0

Answer:

Tmax = 2.934ksi

φ = 4.58°

Explanation:

Given.

Length = 100 ft

Maximum power = 2590 hp

Angular rotate = 1700 rpm .

Outer diameter of the shaft = 8 in. Wall thickness = 3/8 in.

Calculating angular velocity.

= 1700rev/min * 2πrad/rev * 1min/60s

= 56.67πrad/s

Convert power to ft.lb/s

2590hp = 2590 * 550

= 1424500ft.lb/s

At this point, we calculate torque.

Torque = Power/Angular Velocity

Torque = 1424500/56.67π

Torque = 8000.24lb.ft

Using the torsion formula, we'll calculate maximum stress due to the shear force acting on the body:

Tmax = ½(Tc/J)

Tmax =½ of (8001.24 * 12 * 4 )/ (π/4)(4⁴ - 3.625⁴)

Tmax = ½ * 5868.656298363693

Tmax = 2934.328149181846

Tmax = 2.934ksi

Calculating the angle twist

φ = TL/JG

φ = (8001.24 * 12 * 100 * 12 )/( (π/4)(4⁴ - 3.625⁴)(4⁴ - 3.625⁴)(11)(10^6))

φ = 0.08002 rad -- convert to degrees

φ = 4.58° --- approximated

You might be interested in
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De
Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0
4 years ago
What are the characteristic features of stress corrosion cracks?
Oduvanchick [21]

Answer and Explanation:

The crack formation growth that takes place in an environment corrosive.

Stress corrosion cracks can be defined as the spontaneous failures of the metal alloy as a result of the combined action of corrosion and high tensile stress.

Some of the characteristic features of stress corrosion cracks are:

  • These occur at high temperatures.
  • Occurrence of failures in metals mechanically.
  • Occurrence of sudden and unexpected failures under tensile stress.
  • The rate of work hardening of the metal alloy is high.
  • Time
  • An environment that is specific for stress corrosion cracking.
5 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
Which term describes a Cloud provider allowing more than one company to share or rent the same server?
svlad2 [7]

Answer:

multitenancy is the term.

6 0
3 years ago
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0
4 years ago
Other questions:
  • If 1 uF capacitor is fully charged with 120 V across it, how much energy is stored in it? (a) 7.2 kJ (b) 7.2 mJ (c) 0.12 mJ (d)
    6·1 answer
  • The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
    5·1 answer
  • What is pneumatic troubleshooting
    6·1 answer
  • In the designation of wrought Al alloys, eg. 3m, what does the first digit-3- refer to? A. The main alloying element B. Carbon p
    15·1 answer
  • When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is
    5·1 answer
  • What do you do if building doesn't have enough water pressure for sprinklers? a. Increase pipe size b. Adjust budget to accommod
    15·1 answer
  • Random question, does anyone here use Lego, do not answer unless that is a yes
    15·2 answers
  • Analyze the example of this band saw wheel and axle. The diameter of the wheel is 14 inches. The diameter of the axle that drive
    9·1 answer
  • Write a paragraph on computer 473
    5·1 answer
  • 1. What did observations between 1912 and 1917 show?_____
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!