Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 18)
18.
The Centre of gravity of a 10 x 15 x 5 cm T section from its bottom, is
Discussion:
8 comments Page 1 of 1.
Lakshmi said:
4 years ago
Iy = (A1*y1+A2*y2)/(A1+A2).
= {(10*5*(10/2)) + (10*5*(10+5/2)}/(10*5+10*5).
= 8.75 cm.
= {(10*5*(10/2)) + (10*5*(10+5/2)}/(10*5+10*5).
= 8.75 cm.
(1)
Baloch said:
9 years ago
How?
Y= 2.5 (75) + 7.5 (50)
-------------------------
75 + 50
= 4.5 cm.
Y= 2.5 (75) + 7.5 (50)
-------------------------
75 + 50
= 4.5 cm.
Vaibhav said:
8 years ago
Answer C is correct.
Kanhaiya said:
8 years ago
Please explain again.
RR27 said:
8 years ago
Horizontal section (flange) = length, 100mm, height 50mm.
Same applies to its vertical section Because the total height of both horizontal and vertical section is 150mm.
Therefore, Horizontal section (A1) = Area, 100 x 50 = 5000mm2, Centroid to bottom = 125mm (i.e.half of 50mm + 100mm),
Vertical section (A2), Area = 100 x 50=5000mm2, centroid to bottom = 50mm (i.e half of 100mm).
Centre of gravity, y = (5000 x 125) + (5000 x 50), i.e. sum of areas x centroid, divided by (5000+5000, i.e.sum of areas), ans = 87.5mm.
Same applies to its vertical section Because the total height of both horizontal and vertical section is 150mm.
Therefore, Horizontal section (A1) = Area, 100 x 50 = 5000mm2, Centroid to bottom = 125mm (i.e.half of 50mm + 100mm),
Vertical section (A2), Area = 100 x 50=5000mm2, centroid to bottom = 50mm (i.e half of 100mm).
Centre of gravity, y = (5000 x 125) + (5000 x 50), i.e. sum of areas x centroid, divided by (5000+5000, i.e.sum of areas), ans = 87.5mm.
Rishi said:
6 years ago
Thanks for explaining.
Ihsan khan said:
5 years ago
Thanks all for the explanation.
Fenil said:
4 years ago
No, @Rr27.
The dimection is 10*15*5 cm.
The dimection is 10*15*5 cm.
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