Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
1.5 Calculating Stiffness Matrices – Beam Theory<br />
Two Different Lengths for Applying Forces – To facilitate the understanding <strong>of</strong> steps<br />
which will be presented, one can introduce the follow nomenclature (see figure 3(b)):<br />
• L ∗ = L1 or L ∗ = L2 – length where the force F is applied.<br />
• x = L1 or x = L2 – length where the displacement is measured.<br />
Taking into account two different points for applying the forces and measuring the displacements<br />
<strong>of</strong> the beam, one works with the following set <strong>of</strong> equations<br />
x [0, L ∗ ]<br />
and<br />
x [L ∗ , L]<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
y(x) = − F<br />
E I<br />
dy(x)<br />
dx<br />
= − F<br />
E I<br />
<br />
x3 6 − L∗ x2 <br />
2<br />
<br />
x2 2 − L∗ <br />
x<br />
y(x) = y(L ∗ ) + dy(x)<br />
dx<br />
dy(x)<br />
dx<br />
= dy(x)<br />
dx<br />
<br />
<br />
x=L ∗<br />
<br />
<br />
<br />
x=L∗ · (x − L∗ )<br />
which are responsible for describing the deflection <strong>of</strong> the beam, considering the loading on<br />
different coordinates.<br />
Let us introduce an example <strong>of</strong> a system with two points <strong>of</strong> force application. Assuming in case<br />
(I) the force F is applied to the first coordinate L ∗ = L1. One can measure and/or calculate<br />
the beam deflection at the coordinates x = L1 and x = L2 through equations (10) and (11):<br />
y1 = y(L1) = F · L3 1<br />
3 · EI<br />
and<br />
y2 = y(L2) = F<br />
6EI · (2L3 1 + 3L 2 1(L2 − L1)) (13)<br />
Assuming in case (<strong>II</strong>) that the force F is applied to the second coordinate L ∗ = L2, one can<br />
measure and/or calculate the follow beam displacements at the coordinates x = L1 and x = L2<br />
through the equations (10) and (11):<br />
y1 = y(L1) = F<br />
6EI · (2L3 1 + 3L 2 1(L2 − L1)) (14)<br />
and<br />
y2 = y(L2) = F · L3 2<br />
3 · EI<br />
7<br />
(10)<br />
(11)<br />
(12)<br />
(15)