I am stuck on the calculation of forces on a certain system involving parallel springs. The schematic of the system is as follows:

enter image description here

The lower plane is fixed and the upper plane movable. There is a symmetric 2x2 matrix of springs between the upper and lower plane. All springs (same size and spring constant) are rigidly attached to the respective planes. The upper plane has a rigid extension (not shown in the image) where forces can be applied such that the upper plane moves/rotates. In a parallel setup of springs with a symmetric force acting in -z direction I know that I can just add up the spring constants. However what if the force is not parallel to the springs? In case of a single spring I would model the bending of the spring as beam, but what about a spring array?

Is there a way to compute the forces/ torques acting on the upper plane when I know the relative position (rotation and translation) of the upper plane to the lower plane?

Thank you in advance!

  • $\begingroup$ Check this out. The problem seems similar. study.com/academy/answer/… $\endgroup$ – Manu G May 7 at 1:32
  • $\begingroup$ Anyone with the solution, feel free to post. $\endgroup$ – Manu G May 7 at 1:33
  • $\begingroup$ @Manu G, thanks for the link! $\endgroup$ – rfn123 May 7 at 7:25

Found this answer here. Feel free to check it out for more details.

enter image description here

Given Data:

  • The stiffness of each spring is k
  • The total force on the plate is P
  • The length of the plate is 2a
  • The coordinate axis are x and y
  • The reaction force on spring A is ${R_A}$
  • The reaction force on spring B is ${R_B}$
  • The reaction force on spring C is ${R_C}$
  • The reaction force on spring D is ${R_D}$

The expression for force equilibrium,

${R_A} + {R_B} + {R_C} + {R_D} = P $

The expression for the moment is

$\begin{align*} {R_A} \times 2a + {R_B} \times 2a - P \times \left( {a - x} \right) &= 0........\left( 1 \right)\\ {R_A} + {R_B} &= \dfrac{{P\left( {a - x} \right)}}{{2a}}........\left( 2 \right)\\ {R_B} \times 2a + {R_C} \times 2a - P \times \left( {y + a} \right) &= 0.......\left( 3 \right)\\ {R_B} + {R_C} &= \dfrac{{P\left( {y + a} \right)}}{{2a}}........\left( 4 \right) \end{align*}$

The expression due to the uniformity,

$\begin{align*} {R_A} + {R_C} &= \dfrac{P}{2}.......\left( 5 \right)\\ {R_B} + {R_D} &= \dfrac{P}{2}.......\left( 6 \right) \end{align*}$

Substituting the value of ${R_C}$ in equation $\left( 3 \right)$ from equation $\left( 5 \right)$

$\begin{align*} {R_B} \times 2a + \left( {\dfrac{P}{2} - {R_A}} \right) \times 2a - P \times \left( {y + a} \right) &= 0\\ {R_B} - {R_A} &= \dfrac{{Py}}{{2a}}.........\left( 7 \right) \end{align*}$

By solving equation $\left( 2 \right)$ and equation $\left( 7 \right) $,

Expression for the reaction force in spring A is

${R_A} = - \dfrac{{Py}}{{4a}} + \dfrac{{P\left( {a - x} \right)}}{{4a}}........\left( 8 \right)$

Expression for the reaction force in spring B is

${R_B} = \dfrac{{P\left( {a - x} \right)}}{{4a}} + \dfrac{{Py}}{{4a}}........\left( 9 \right)$

Expression for the reaction force in spring C is

${R_C} = \dfrac{P}{2} - \dfrac{{Py}}{{4a}} + \dfrac{{P\left( {a - x} \right)}}{{4a}}........\left( 10 \right) $

Expression for the reaction force in spring D is

${R_D} = \dfrac{P}{2} - \dfrac{{P\left( {a - x} \right)}}{{4a}} + \dfrac{{Py}}{{4a}}........\left( 11 \right)$

Thus equation 8,9,10,11 are the expression for the forces in A,B,C,D respectively.

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