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max2010maxim [7]
2 years ago
9

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

ume that E = 10,000 ksi and that pinned connections are used at each end of the column. Determine the slenderness ratio and the Euler buckling load for the column.
Engineering
1 answer:
slega [8]2 years ago
3 0

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

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At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge
ale4655 [162]

Answer:

<em>a) 50 J/kg</em>

<em>b) 721 67 KW</em>

<em></em>

Explanation:

The velocity of the wind v = 10 m/s

diameter of the blades d = 70 m

efficiency of the turbine η = 30%

density of air ρ = 1.25 kg/m^3

The area of the blade A = \pi d^2/4

A = \frac{3.142 * 70^2}{4} = 3848.95 m^2

The mechanical energy air per unit mass is gives as

e = v^2/2 = \frac{10^2}{2} = <em>50 J/kg</em>

<em></em>

Theoretical Power of the turbine P = ρAve

where

ρ is the density of air

A is the area of the blade

v is the velocity of the wind

e is the energy per unit mass

substituting values, we have

P = 1.25 x 3848.95 x 10 x 50 = 2405593.75 W

Actual power = ηP

where η is the efficiency of the turbine

P is the theoretical power of the turbine

Actual power = 0.3 x 2405593.75 = 721678.1 W

==> <em>721 67 KW</em>

7 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
2 years ago
AC motor characteristics require the applied voltage to be proportionally adjusted by an AC drive whenever the frequency is chan
Margarita [4]
The answer is false
6 0
3 years ago
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Which statement about cathode ray tubes (CRTs) is true?
yuradex [85]

Answer:

CRTs consume less power than LCDs.

hope it helps (^^)

# Cary on learning

6 0
2 years ago
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
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