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3 years ago

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Line of Action of the Axial Forces for a Uniform Stress Distribution

1 answer:

nikklg [1K]3 years ago

4 0

Answer:

Line of action of axial force for a uniform stress distribution should pass through the centroid of the cross sectional area.

Explanation:

If the line of action of the force is along the centroidal axis of the cross sectional area there is no eccentricity in the line of application of force hence no moment is generated in the cross sectional area hence we get a uniform stress distribution as theorized by hookes law.

In case of eccentric force there is an additional moment in addition of the force. This induced moment induces bending in the section thus giving a non uniform stress distribution in the section.

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The normal stress at gage H calculated in Part 1 includes four components: an axial component due to load P, σaxial, P, a bendin

Degger [83]

Answer:

hello your question has some missing information attached to the answer is the missing component

Answer : αaxial,p = -6.034 ksi ( compressive )

αbend,p = 19.648 ksi ( tensile )

Explanation:

αaxial, p = $\frac{-p}{A}$ equation 1

αbend, p = $\frac{(P*A)*\frac{d}{2} }{I_{z} }$ equation 2

P = load = 35 kips

A = area of column = 5.8 $in^{2}$

d = column cross section depth = 9.5 in

$I_{Z}$ = 55.0 $in^{4}$

Hence equation 1 becomes

αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )

equation 2 becomes

αbend, p = $\frac{(35*6.5)(\frac{9.2}{2}) }{55}$ = + 19.648 ksi ( tensile )

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3 years ago

What should you use to turn the water off and open the door?

galina1969 [7]

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3 years ago

Read 2 more answers

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

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A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False

tiny-mole [99]

The answer is : True

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Suppose within your web browser you click on a link to obtain a web page. The IP address for the associated URL is already cache

aleksandrvk [35]

Answer:

All the detailed steps are mentioned in pictures.

Explanation:

See attached pictures.

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3 years ago