Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 43)
43.
The region of the cross-section of a column in which compressive load may be applied without producing any tensile stress, is known as the core of the cross-section. In circular columns the radius of the core, is
Discussion:
13 comments Page 1 of 2.
LAXMIKANTH said:
5 years ago
The answer is correct because in question asked for the radius of the core so eccentricity itself.
So, e=D/8 therefore the radius of core = r/4.
So, e=D/8 therefore the radius of core = r/4.
(3)
Keshav Kaushik said:
7 years ago
D/4 is the diameter of the core.
D= 2R
=> 2R/4
= R/2.
So, A is the answer.
D= 2R
=> 2R/4
= R/2.
So, A is the answer.
(1)
Harsha said:
7 years ago
Limit of eccentricity for circular column <= 8,
But the core of circular column is d/4 hence radius is r/2 only.
Option A is correct.
But the core of circular column is d/4 hence radius is r/2 only.
Option A is correct.
(1)
Asif wazir said:
6 years ago
Z = I/y= pai x d3 /32.
A = PAI x d2 /4.
So, e = z/A = d/ 8 = r/ 4.
A = PAI x d2 /4.
So, e = z/A = d/ 8 = r/ 4.
(1)
Irfan said:
5 years ago
Thanks @Asif Wazir.
(1)
Inayat Ullah Kakar said:
9 months ago
e <= d/8,
e <= 2r/8,
Then e <= r/4.
e <= 2r/8,
Then e <= r/4.
(1)
Shra1 said:
8 years ago
e<Z/A
Z= π/32 * D^3.
A= π/4 * D^2.
e<= D/8-radius.
Dia D/4.
Z= π/32 * D^3.
A= π/4 * D^2.
e<= D/8-radius.
Dia D/4.
Rahul said:
8 years ago
Should be 1/2 of radius.
Umair Sabeeh Swati said:
7 years ago
KERN core for the circular section is D/4.
As D=2R.
So in term of radius it is R/2.
A is the correct answer.
As D=2R.
So in term of radius it is R/2.
A is the correct answer.
Nrj said:
7 years ago
The middle-third rule states that no tension is developed in a wall or foundation if the resultant force lies within the middle third of the structure. (For circular foundations a different rule, known as the Middle Quarter Rule applies) so core dia will be D/4 abd radius will be D/8.
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