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erastova [34]
2 years ago
15

The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e

quivalent force-couple system at the centroid of the beam cross-section at the cantilever O.

Engineering
1 answer:
snow_lady [41]2 years ago
4 0

The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A

The <em>equivalent force couple</em> system at O due to force <em>F</em> are;

Force, F =  (<u>8.65·i - 4.6·j</u>) KN

Couple, M₀ ≈ <u>40.9 </u>k kN·m

The reason the above values are correct is as follows:

The known values for the <em>cantilever</em> are;

The <em>height </em>of the beam = 0.65 m

The <em>magnitude of </em>the applied <em>force</em>, F = 9.8 kN

The <em>length </em>of the beam = 4.9 m

The <em>angle </em>away from the vertical the force is applied = 26°

The required parameter:

The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever

Solution:

The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The <em>equivalent force</em> \overset \longrightarrow F = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The <em>equivalent force</em> \overset \longrightarrow F ≈ (<u>8.65·i - 4.6·j</u>) KN

The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;

The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>

The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em>  ≈ 40.9 kN·m

The couple acting at O, due to F,  M₀ ≈ <u>40.9 kN·m</u>

The equivalent force couple system acting at point O due the force, F, is as follows

F =  (8.65·i - 4.6·j) KN

M₀ ≈ <u>40.9 </u>k kN·m

Learn more about equivalent force systems here:

brainly.com/question/12209585

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2 years ago
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate
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Answer:

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Explanation:

The given parameters are;

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The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

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The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

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The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

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A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
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Lets take heat transfer is Q ,then entropy change can be written as

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Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

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