Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 36)
36.
According to Euler's column theory, the crippling load for a column of length (l) with one end fixed and the other end free is __________ the crippling load for a similar column hinged at both the ends.
Discussion:
13 comments Page 1 of 2.
Ankit raj said:
6 years ago
For both the ends hinged it will be 'L' and for one end fixed and one hinged it will be 'L/√(2), therefore it is less than both ends fixed.
Yesha said:
6 years ago
The L is in the denominator so B is the correct option.
Yesha said:
6 years ago
(π * ^2 EI/L^2) and bothe end hinged is L and One end fixed and other end free is 2L.
So, B Is the right option.
So, B Is the right option.
Partha Sarkar said:
6 years ago
Option "B" is right, the equivalent length is inversely proportional to the crippling load.
Partha Sarkar said:
6 years ago
Ans "B" is right. because Crippling load is inversely proportional to the equivalent length of the column. So, if equivalent length increases then obviously crippling load decreases.
P= (π)^2*EI/(le)^2.
Where, le= effective length of the column, given by,
le=2l for fixed-free.
le=l for hinged-hinged.
P= (π)^2*EI/(le)^2.
Where, le= effective length of the column, given by,
le=2l for fixed-free.
le=l for hinged-hinged.
Phoring said:
6 years ago
@Raj.
How the last condition came 0.25?? Can you please explain?
How the last condition came 0.25?? Can you please explain?
Karthik raja said:
6 years ago
Option C is the correct answer.
Parry said:
6 years ago
Option C is right.
Because equivalent length for a column with one end is fixed and other is free : L= 2l.
And for both end is hinged then : L= l.
Therefore 2l> l hence c is right.
Because equivalent length for a column with one end is fixed and other is free : L= 2l.
And for both end is hinged then : L= l.
Therefore 2l> l hence c is right.
Bhunesh malav said:
8 years ago
But Euler formula is apply long column and long column fail in bukling.
Raj said:
8 years ago
Column pivoted in both ends : n = 1.
Both ends fixed : n = 4.
One end fixed, the other end rounded : n = 2.
One end fixed, one end free : n = 0.25.
Both ends fixed : n = 4.
One end fixed, the other end rounded : n = 2.
One end fixed, one end free : n = 0.25.
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