The expression for crippling load for a column with both ends hinged
● The load at which the column just buckles (or bends) is called crippling load. Consider a column AB of length l and uniform crosssectional area, hinged at both of its ends A andB.
● Let P be the crippling load at which the column has just buckled. Due to the crippling load, the column wiII deflect into a curved form ACB as shown in Figure below:
But moment =
EI Equating the two moments, we have
EI=−P.y or
EI+P.y=0 or
+.y=0 The solution of the above differential equation is
y=C1.cos(x√) +C2.sin(x√) ...(i)
Where
C1 and
C2 are the constants of integration and the valuesare obtained as follows:
At A, x = 0 and y = 0.
Substituting these values in equation (i), we get
0=C1.cos0°+C2sin0°=C1×1+C2×0 (∵ cos 0° = 1 and sin 0° = 0)
=
C1∴
C1=0. ...(ii)
(ii) At B, x = l and y = 0 (See Fig. above)
Substituting these values in equation (i), we get
0=C1.cos(l×√) +C2.sin(l×√)0=C2.sin(l×√) [∵
C1 = 0 from equation (ii)]
=C2sin(l√) ...(iii)
From equation (iii), it is clear that either
C2=0sin(l√)=0As if
C1 = 0, then if
C2 is also equals to zero, then from Eqn no. (i), we will find that y = 0. This means that the bending of the column will be zero or the column will not bend at all, which is not true.
∴
sin(l√)=0= sin 0 or sin π or sin 2π or sin 3π or...
or
l√=0, or π or 2π or 3π or....
Taking the least practival value.
l√=π, P=.