How are eras different from decades?
1 answer:
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Answer:
Rate of corrosion = 24.95 mpy
Rate of corrosion = 0.63 mm/yr
Explanation:
given data
steel sheet area = 150 in²
weight loss = 485 g
density of steel = 7.9 g/cm³
time taken = 1 year
to find out
rate of corrosion in (a) mpy and (b) mm/yr
solution
we get here the rate of corrosion that is express as
rate of corrosion = (k × W) ÷ (D × A × T) ..................1
here k is constant and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel
so put her value in equation 1 we get
Rate of corrosion =
Rate of corrosion = 24.95 mpy
and
Rate of corrosion =
Rate of corrosion = 0.63 mm/yr
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> = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j
Which gives;
The <em>equivalent force</em> ≈ (<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: