Engineering Mechanics - Center of Gravity and Centroid - Discussion

Discussion :: Center of Gravity and Centroid - General Questions (Q.No.5)

5. 

Determine the distance to the centroidal axis of the beam's cross-sectional area.

[A]. = 112.3 mm
[B]. = 125.0 mm
[C]. = 100.0 mm
[D]. = 91.7 mm

Answer: Option D

Explanation:

No answer description available for this question.

Ahamed said: (Oct 8, 2011)  
Since the given figure is symmetric about y-axis

So therefore

x' = 0 and y' = (A1y1 + A2y2 + A3y3) / ( A1 + A2 +A3)

A1 = 350 X 50 = 17500 mm
A2 = 200 X 50 = 10000 mm
A3 = 200 X 50 = 10000 mm

y1 = 25
y2 = 50+100 = 150
y3 = 50+100 = 150

y' = (17500x25 + 10000x150 + 10000x150) / (17500 + 10000 + 10000)

y' = 91.6666666 mm

By rounding off we get,

y' = 91.7 mm.

Danica said: (Oct 10, 2015)  
Why does y1 didn't consider the 350?

Sai Lakshmi said: (Dec 4, 2015)  
Why didn't you do division for y2?

Ali said: (Aug 25, 2016)  
Please solve the question clearly by parts.

Ali J. said: (May 25, 2017)  
Good, Thanks @Ahamed.

Upasana said: (Apr 2, 2018)  
Please, solve the question clearly.

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