Dynamics of Machines - Part II - IFS.pdf

Integrating twice in X, it gives:
EI d2 y(x)
dx 2 = M(x) = C1x + C2 (3)
Integrating again in X, one gets the rotation angle of the beam:
EI dy(x)
dx
x
= EIΘ(x) = C1
2
2 + C2x + C3
And finally, integrating the last time in X, one achieves the equation responsible for describing
the deflexion of the beam:
x
EI y(x) = C1
3
6
x
+ C2
2
2 + C3x + C4
The boundary conditions for the clamped-free beam are:
•y(x = 0) = 0
•Θ(x = 0) = 0
•M(x = L) = 0
•F(x = L) = −F (reaction)
⎫
⎪⎬
⎪⎭
After calculating the constants Ci using the boundary condition for the clamped-free beam case,
one gets
y(x) = − F
E I
dy(x)
dx
x 3
6
= Θ(x) = − F
E I
L x2
−
2
x 2
2
− L x
Using the relationship between applied force F and the induced deflexion at a given point along
the beam length, x = L for instance, one gets the equivalent stiffness as:
K = F
y(L)
= 3 EI
L 3
Suggestion (I): Change the beam boundary conditions, for example bi-supported at both ends,
and calculate the equivalent stiffness in the new case.
Suggestion (II): Change the beam boundary conditions, for example clamped-clamped at both
ends, and calculate the equivalent stiffness in the new case.
6
(4)
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