Consider a cantilever beam of length l and uniform circular solid cross-section of radius r loaded by a concentrated axial force po at the center of its free end.
Answers
Answered by
0
ption (A) is correct.
Since EI dx
d y
2
2
= M
Integrating EIdx
dy = + mx C1 ...(i)
At x = 0, dx
dy = 0
So EI( )0 = + M C (0) 1 & C1 = 0
Hence Eq.(i) becomes
EIdx
dy = mx
Again integrating EIy mx C 2
2
= + 2 ...(ii)
At x = 0, y = 0, EI( )0 m( ) C 2
0 2
= + 2 & C2 = 0
Then Eq. (ii) becomes EIy Mx
2
2
=
y EI
Mx
2
2
= y EI
Since EI dx
d y
2
2
= M
Integrating EIdx
dy = + mx C1 ...(i)
At x = 0, dx
dy = 0
So EI( )0 = + M C (0) 1 & C1 = 0
Hence Eq.(i) becomes
EIdx
dy = mx
Again integrating EIy mx C 2
2
= + 2 ...(ii)
At x = 0, y = 0, EI( )0 m( ) C 2
0 2
= + 2 & C2 = 0
Then Eq. (ii) becomes EIy Mx
2
2
=
y EI
Mx
2
2
= y EI
Similar questions