Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 3)
3.
A short masonry pillar is 60 cm x 60 cm in cross-section, the core of the pillar is a square whose side is
Discussion:
15 comments Page 1 of 2.
Deepankar said:
3 years ago
Its a sq.colum as its dimensions are equal,
For square column,the shape of kern is also square with diagonal = d/3 = 60/3 = 20cm.
Now the Q asks about the side of the core not the diagonal of the core.
But we know that Area of square=1/2(diagonal)^2.
Also area of sq.=a^2,
Solving these two we get a=diagonal/(underroot of 2).
So a=20/(under root 2)= 14.14cm.
For square column,the shape of kern is also square with diagonal = d/3 = 60/3 = 20cm.
Now the Q asks about the side of the core not the diagonal of the core.
But we know that Area of square=1/2(diagonal)^2.
Also area of sq.=a^2,
Solving these two we get a=diagonal/(underroot of 2).
So a=20/(under root 2)= 14.14cm.
(5)
Ramya said:
5 years ago
Thank you @Yogesh.
Subhankar said:
6 years ago
Here, we can use ((B^2 +D^2)^.5)/6 for calculation.
(1)
Sumit said:
6 years ago
Thanks all for explaining.
(1)
Anand said:
6 years ago
The shape of the core in the square section is square and side value is b/3 & d/3 so the answer is 20 cm.
(1)
Gopal said:
6 years ago
Kern of a square section is also a square so the sides are b/3 & b/3 hence the answer should be 60/3= 20.
If it is a rectangle of 60x60 then the above-given approach is correct.
If it is a rectangle of 60x60 then the above-given approach is correct.
(1)
Neel said:
7 years ago
The shape of the core for the square column is square, not a rhombus.
so, ans is b/6 i.e. 60/3=20.
so, ans is b/6 i.e. 60/3=20.
(2)
Umesh said:
8 years ago
Thanks @Yogesh.
(1)
Yogesh said:
8 years ago
For a rectangular column, kernel core is a rhombus with diagonals B/3 & D/3 resp.
In the above Question, dimensions of the core are asked in terms of SIDE.
So, to apply Pythagoras theorem, we must consider a quarter of the rhombus, i.e., a triangle which has the sides B/6 & D/6
So, the side of the rhombus = hypotenuse of the triangle = sq root (10^2 + 10^2) = sq root(200) = 14.14
In the above Question, dimensions of the core are asked in terms of SIDE.
So, to apply Pythagoras theorem, we must consider a quarter of the rhombus, i.e., a triangle which has the sides B/6 & D/6
So, the side of the rhombus = hypotenuse of the triangle = sq root (10^2 + 10^2) = sq root(200) = 14.14
(1)
Fgf said:
8 years ago
Can you please explain it step by step?
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