# Calculate forces/ torques on a parallel spring array

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

• Check this out. The problem seems similar. study.com/academy/answer/… May 7, 2020 at 1:32
• Anyone with the solution, feel free to post. May 7, 2020 at 1:33
• @Manu G, thanks for the link! May 7, 2020 at 7:25

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

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.