Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 23)
23.
The slenderness ratio of a vertical column of a square cross-section of 2.5 cm sides and 300 cm length, is
Discussion:
33 comments Page 1 of 4.
Tanmoy Ghosh said:
5 years ago
Slenderness Ratio =Le/K,
K=√(I/A).
So,
Le=300,
K=√(2.5^4/12)/(2.5^2),
K=0.7216,
Le/K=300/0.7216,
Slenderness Ratio=415.69.
K=√(I/A).
So,
Le=300,
K=√(2.5^4/12)/(2.5^2),
K=0.7216,
Le/K=300/0.7216,
Slenderness Ratio=415.69.
(5)
Amit said:
5 years ago
We Know that Slenderness Ratio is Equal to Length upon Radius of Gyration = L/r.
we have a value of L that is given in question 300.
So, now we need to calculate the value of r that is equal to under root I/A means r= under root I/A.
I Means Moment of Inertia, here for Square Section that is equal to a^4/12 and for A= a^4.
Now, you will get the answer 416.
Thank You.
we have a value of L that is given in question 300.
So, now we need to calculate the value of r that is equal to under root I/A means r= under root I/A.
I Means Moment of Inertia, here for Square Section that is equal to a^4/12 and for A= a^4.
Now, you will get the answer 416.
Thank You.
(2)
Manikanta nelluri said:
5 years ago
@All.
2.5^4÷12 /2.5 * 2.5 = 3.25 ÷ 6.25 = 0.52 this is the correct value.
So, 0.721 this value is wrong.
2.5^4÷12 /2.5 * 2.5 = 3.25 ÷ 6.25 = 0.52 this is the correct value.
So, 0.721 this value is wrong.
(3)
Susanta sarkar said:
5 years ago
Slenderness ratio=L/k =300/.72=416.
K= √(I/A).
I=bd^3/12 b=d.
I=d^4/12=(2.5) ^4/12/(2.5x2.5) =.52.
K=√.52/6.25=.72.
K= √(I/A).
I=bd^3/12 b=d.
I=d^4/12=(2.5) ^4/12/(2.5x2.5) =.52.
K=√.52/6.25=.72.
(1)
DEVARAJ said:
6 years ago
Slenderness ratio = effective length / radius of gyration.
Radius of gyration = Sqrt(moment of inertia/area of Cross section).
Moment of inertia (I) = b^4/12 (column is a square Cross section).
Assume given length as effective length put All values in the above equations. You will get answer.
Radius of gyration = Sqrt(moment of inertia/area of Cross section).
Moment of inertia (I) = b^4/12 (column is a square Cross section).
Assume given length as effective length put All values in the above equations. You will get answer.
(1)
Sachin said:
6 years ago
Thanks all for explaining.
Kumud Chaudhary said:
6 years ago
I = A * r.
2.5^4/12=2.5 * 2.5 * r.
r = 0.72168.
Now slenderness ratio= l/r.
= 300/0.72168.
= 415.74.
So answer is correct.
2.5^4/12=2.5 * 2.5 * r.
r = 0.72168.
Now slenderness ratio= l/r.
= 300/0.72168.
= 415.74.
So answer is correct.
Pantlok said:
6 years ago
Effec length =300,
Radius of gyration=((a^4/12)/a^2),
S.R=effec. length/radius of gyration (ans: 415.69)
Radius of gyration=((a^4/12)/a^2),
S.R=effec. length/radius of gyration (ans: 415.69)
Someshwar said:
7 years ago
S.R =eff. length/radius of gyration.
=l/k.
K= √ Imin/A.
Imin=bd^3/12=2.5*2.5^3/12=3.25.
A=2.5*2.5=6.25,
=√3.25/6.25 =0.721.
S.R=300/0.721 =416.08.
=l/k.
K= √ Imin/A.
Imin=bd^3/12=2.5*2.5^3/12=3.25.
A=2.5*2.5=6.25,
=√3.25/6.25 =0.721.
S.R=300/0.721 =416.08.
Ritz said:
7 years ago
Slend. ratio = L(eff)/r(min).
r(min) = ( I(min) / A )^0.5 = 2.5 / 2(3)^0.5
Therefore, S.R = 300 * 2(3)^0.5 / 2.5 = 415.2 closest to 416.
r(min) = ( I(min) / A )^0.5 = 2.5 / 2(3)^0.5
Therefore, S.R = 300 * 2(3)^0.5 / 2.5 = 415.2 closest to 416.
(1)
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