# 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.

(4)

Ramya said:
5 years ago

Thank you @Yogesh.

Subhankar said:
5 years ago

Here, we can use ((B^2 +D^2)^.5)/6 for calculation.

Sumit said:
5 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:
6 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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