Rotating Masses Balancing, Dynamic Balancing, and Linkage Shaking Forces

Fundamentals and Classification of Balancing

• Balancing is categorized into two major domains in mechanical engineering:

  • Rotating Masses Balancing

  • Reciprocating Masses Balancing

• Rotating Masses Balancing is divided based on geometric and dynamic requirements:

  • Static Balancing: Requires balancing in a single plane ($1$ plane).

  • Dynamic Balancing: Requires balancing in two distinct planes ($2$ planes).

• Imbalance forces and couples (moments) are evaluated through two primary methodologies:

  • Analytical Method: Involves resolving force and moment vectors into orthogonal Cartesian components ($x$ and $y$).

  • Graphical Method: Employs closed vector polygons (force polygons and moment polygons) constructed to scale.

Mathematical Derivation of Centrifugal Imbalance Forces

• Governing equations of motion originate from Newton's Second Law of Motion:

  • $F = m a$

  • Under circular rotation, acceleration is purely normal (centripetal):

  • $F = m a_n$

• Relational derivation from tangential velocity ($v$) and angular speed ($\omega$):

  • $a_n = \frac{v^2}{r}$

  • $v = \omega r$

  • $F = m \frac{(\omega r)^2}{r}$

  • $F = m r \omega^2$

• Formulating dynamic centrifugal force in terms of weight ($W$):

  • $W = m g \implies m = \frac{W}{g}$

  • Centrifugal imbalance force for a single rotating mass:

  • $F = \frac{W}{g} r \omega^2 = \frac{\omega^2}{g} W r$

• Multi-mass centrifugal imbalance formulation:

  • For a system of multiple rotating masses in a single plane:

  • $F = \frac{\omega^2}{g} \sum W r$

• Core engineering objective for complete balance:

  • The goal is to make the net centrifugal force vanish:

  • $F \approx 0$

  • For any designated operating speed $\omega$, the factor $\frac{\omega^2}{g}$ is constant and non-zero:

  • $\frac{\omega^2}{g} = \text{Constant} \neq 0$

• Consequently, dynamic force equilibrium requires the summation of mass-radius products (or weight-radius products) to equal zero:

• $\sum W r \approx 0 \implies F \approx 0$

Static Balancing of Several Masses in a Single Plane (Example 1)

• System parameters and specifications:

  • Mass 1: $m_1 = 1.2\,\text{kg}$, radius $r_1 = 0.2\,\text{m}$, angle $\theta_1 = 134^\circ$

  • Mass 2: $m_2 = 2.3\,\text{kg}$, radius $r_2 = 0.3\,\text{m}$, angle $\theta_2 = 58^\circ$

  • Mass 3: $m_3 = 1.8\,\text{kg}$, radius $r_3 = 0.25\,\text{m}$, angle $\theta_3 = 15^\circ$

  • Standard gravitational acceleration: $g = 9.81\,\text{m/s}^2$

• Objectives:

  • Determine the net imbalance force vector.

  • Determine the required resultant balance counterweight and radial location.

• Individual weight-radius products ($W r$):

  • $(W r)_1 = m_1 g r_1 = 1.2 \times 9.81 \times 0.2 = 2.35\,\text{Nm} \approx 2.4\,\text{Nm}$

  • $(W r)_2 = m_2 g r_2 = 2.3 \times 9.81 \times 0.3 = 6.77\,\text{Nm} \approx 6.8\,\text{Nm}$

  • $(W r)_3 = m_3 g r_3 = 1.8 \times 9.81 \times 0.25 = 4.41\,\text{Nm} \approx 4.4\,\text{Nm}$

• Analytical solution procedure:

  • Tabulated orthogonal components:

  • Mass 1: $m = 1.2\,\text{kg}$, $r = 0.20\,\text{m}$, $W = 11.8\,\text{N}$, $W r = 2.4\,\text{Nm}$, $\theta = 134^\circ$

    • $(W r)_x = W r \cos(\theta) = 2.4 \cos(134^\circ) = -1.6\,\text{Nm}$

    • $(W r)_y = W r \sin(\theta) = 2.4 \sin(134^\circ) = 1.7\,\text{Nm}$

  • Mass 2: $m = 2.3\,\text{kg}$, $r = 0.30\,\text{m}$, $W = 22.6\,\text{N}$, $W r = 6.8\,\text{Nm}$, $\theta = 58^\circ$

    • $(W r)_x = W r \cos(\theta) = 6.8 \cos(58^\circ) = 3.6\,\text{Nm}$

    • $(W r)_y = W r \sin(\theta) = 6.8 \sin(58^\circ) = 5.8\,\text{Nm}$

  • Mass 3: $m = 1.8\,\text{kg}$, $r = 0.25\,\text{m}$, $W = 17.7\,\text{N}$, $W r = 4.4\,\text{Nm}$, $\theta = 15^\circ$

    • $(W r)_x = W r \cos(\theta) = 4.4 \cos(15^\circ) = 4.3\,\text{Nm}$

    • $(W r)_y = W r \sin(\theta) = 4.4 \sin(15^\circ) = 1.14\,\text{Nm}$

  • Summation of orthogonal components:

  • $\sum(W r)_x = -1.6 + 3.6 + 4.3 = 6.3\,\text{Nm}$

  • $\sum(W r)_y = 1.7 + 5.8 + 1.14 = 8.64\,\text{Nm}$

  • Application of the Parallelogram Law of Forces:

  • $(W r)_R^2 = \left(\sum(W r)_x\right)^2 + \left(\sum(W r)_y\right)^2 + 2 \sum(W r)_x \sum(W r)_y \cos(90^\circ)$

  • $(W r)_R = \sqrt{\left(\sum(W r)_x\right)^2 + \left(\sum(W r)_y\right)^2}$

  • $(W r)_R = \sqrt{6.3^2 + 8.64^2} = \sqrt{39.69 + 74.6496} \approx 10.3\,\text{Nm}$

  • Resultant angular direction:

  • $\tan(\theta_R) = \frac{\sum(W r)_y}{\sum(W r)_x} = \frac{8.64}{6.3}$

  • $\theta_R = \arctan\left(\frac{8.64}{6.3}\right) \approx 54^\circ$

  • Balancing condition:

  • Magnitude of balancing vector: $(W r)_R = 10.3\,\text{Nm}$

  • Balancing angular position (diametrically opposite): $\theta_{\text{balance}} = \theta_R + 180^\circ = 54^\circ + 180^\circ = 234^\circ$

• Graphical solution procedure:

  • Scale setting: $0.20\,\text{Nm} = 1\,\text{mm}$.

  • Vectors are plotted tip-to-tail sequentially:

  • Vector 1: Length representing $2.4\,\text{Nm}$ parallel to $F_1$

  • Vector 2: Length representing $6.8\,\text{Nm}$ parallel to $F_2$

  • Vector 3: Length representing $4.4\,\text{Nm}$ parallel to $F_3$

  • The closing vector drawn from the origin of the first vector to the head of the last vector defines the resultant imbalance force; drawing in reverse defines the balancing force vector:

  • $(W r)_R = 10.3\,\text{Nm}$

  • $m_R g r_R = 10.3\,\text{Nm}$

  • If balancing mass $m_R$ is pre-selected: $r_R = \frac{10.3}{m_R g}$

  • If balancing radius $r_R$ is pre-selected: $m_R = \frac{10.3}{g r_R}$

Single Plane Axial Balancing Along a Shaft (Example 2)

• System configuration with masses distributed longitudinally along the axis:

  • Mass 1: $m_1 = 2.3\,\text{kg}$, $W_1 = 2.3 \times 9.81 = 22.6\,\text{N}$, radius $r_1 = 0.18\,\text{m}$

  • Mass 2: $m_2 = 4.5\,\text{kg}$, $W_2 = 4.5 \times 9.81 = 44.2\,\text{N}$, radius $r_2 = 0.13\,\text{m}$

  • Mass 3: $m_3 = 4.5\,\text{kg}$, $W_3 = 4.5 \times 9.81 = 44.2\,\text{N}$, radius $r_3 = 0.25\,\text{m}$

• Axial distances measured relative to a Reference Plane passing through Mass 1 ($m_1$):

  • $a_1 = 0\,\text{m}$

  • $a_2 = a = 0.13\,\text{m}$

  • $a_3 = 2a = 0.31\,\text{m}$

• Angular alignment:

  • Masses 1 and 3 are aligned in the positive upward radial direction ($0^\circ$).

  • Mass 2 is aligned in the negative downward radial direction ($180^\circ$).

• Analytical determination of resultant force and couple:

  • Mass 1:

  • Force product: $W r = 22.6 \times 0.18 = 4.0\,\text{Nm}$

  • Moment product: $W r a = 4.0 \times 0 = 0\,\text{N}\cdot\text{m}^2$

  • Mass 2:

  • Force product: $W r = -(44.2 \times 0.13) = -5.8\,\text{Nm}$

  • Moment product: $W r a = -5.8 \times 0.13 = -0.75\,\text{N}\cdot\text{m}^2$

  • Mass 3:

  • Force product: $W r = 44.2 \times 0.25 = 11.1\,\text{Nm}$

  • Moment product: $W r a = 11.1 \times 0.31 = 3.44\,\text{N}\cdot\text{m}^2$

• Net system summations:

  • Resultant radial force product:

  • $\sum(W r) = 4.0 - 5.8 + 11.1 = 9.3\,\text{Nm}$

  • Resultant couple moment product about the reference plane:

  • $\sum(W r a) = 0 - 0.75 + 3.44 = 2.7\,\text{N}\cdot\text{m}^2$

• Axial location of the balancing plane ($a_R$):

  • The moment of the balancing resultant about the reference plane must balance the system moments:

  • $\sum(W r a) = \left(\sum(W r)\right) \cdot a_R$

  • $a_R = \frac{\sum(W r a)}{\sum(W r)} = \frac{2.7}{9.3} = 0.29\,\text{m}$

  • Placement conclusion: The single balancing plane must be located at an axial distance of $0.29\,\text{m}$ to the right of Mass 1.

Dynamic Balancing Across Multiple Planes (Example 3)

• System configuration with four masses ($A, B, C, D$) rotating in four separate parallel planes:

  • Mass $A$: $m_A = 7.5\,\text{kg}$, $W_A = 7.5 \times 9.81 = 73.6\,\text{N}$, radius $r_A = 0.030\,\text{m}$, $W r_A = 2.21\,\text{Nm}$

  • Mass $B$: Unknown mass $m_B$, unknown weight $W_B$, radius $r_B = 0.036\,\text{m}$, $W r_B = 0.036 W_B\,\text{Nm}$

  • Mass $C$: $m_C = 5\,\text{kg}$, $W_C = 5 \times 9.81 = 49.1\,\text{N}$, radius $r_C = 0.039\,\text{m}$, $W r_C = 1.91\,\text{Nm}$

  • Mass $D$: $m_D = 4\,\text{kg}$, $W_D = 4 \times 9.81 = 39.3\,\text{N}$, radius $r_D = 0.033\,\text{m}$, $W r_D = 1.30\,\text{Nm}$

• Axial longitudinal spacing defined relative to Reference Plane passing through Mass $B$:

  • Plane $A$: $a_A = -0.4\,\text{m}$ (to the left of Reference Plane)

  • Plane $B$: $a_B = 0\,\text{m}$ (Reference Plane)

  • Plane $C$: $a_C = +0.5\,\text{m}$ (to the right of Reference Plane)

  • Plane $D$: $a_D = 0.5 + x\,\text{m}$

• Angular positions and orientations:

  • Mass $C$ acts horizontally along the $+x$-axis ($0^\circ$).

  • Mass $A$ acts vertically along the $+y$-axis ($90^\circ$).

  • Angle between Mass $A$ and Mass $D$ is designated as $\theta_1$.

  • Angle between Mass $D$ and Mass $B$ is designated as $\theta_2$.

• Tabulated parameters for dynamic equilibrium:

  • Plane $A$: $W = 73.6\,\text{N}$, $r = 0.030\,\text{m}$, $W r = 2.21\,\text{Nm}$, $a = -0.4\,\text{m}$, $W r a = -0.884\,\text{N}\cdot\text{m}^2$

  • Plane $B$: $W = W_B$, $r = 0.036\,\text{m}$, $W r = 0.036 W_B\,\text{Nm}$, $a = 0\,\text{m}$, $W r a = 0\,\text{N}\cdot\text{m}^2$

  • Plane $C$: $W = 49.1\,\text{N}$, $r = 0.039\,\text{m}$, $W r = 1.91\,\text{Nm}$, $a = +0.5\,\text{m}$, $W r a = +0.955\,\text{N}\cdot\text{m}^2$

  • Plane $D$: $W = 39.3\,\text{N}$, $r = 0.033\,\text{m}$, $W r = 1.30\,\text{Nm}$, $a = 0.5 + x\,\text{m}$, $W r a = 0.65 + 1.3 x\,\text{N}\cdot\text{m}^2$

• Step 1: Solving for $x$ and $\theta_1$ using the Moment Vector Polygon:

  • Scale: $0.01\,\text{N}\cdot\text{m}^2 = 1\,\text{mm}$.

  • Couple vector for $C$: Plotted in the positive $x$-direction (parallel to $F_C$) with length corresponding to $0.955\,\text{N}\cdot\text{m}^2$.

  • Couple vector for $A$: Since $W r a_A = -0.884\,\text{N}\cdot\text{m}^2$, it is directed opposite to $F_A$ (vertically downwards along $-y$-axis).

  • Closing vector represents the balancing/equilibrium couple produced by Mass $D$: $\sum(W r a)_D = 0.65 + 1.3 x$.

  • Graphical measurement from the closed moment polygon yields:

  • $0.65 + 1.3 x = 1.326$

  • $1.3 x = 1.326 - 0.65 = 0.676$

  • $x = \frac{0.676}{1.3} = 0.52\,\text{m}$

  • Measurement using a protractor from the vector polygon yields the angular orientation of Mass $D$:

  • $\theta_1 = 47^\circ$

• Step 2: Solving for $m_B$ and $\theta_2$ using the Force Vector Polygon:

  • Scale: $0.05\,\text{Nm} = 1\,\text{mm}$.

  • Known force vectors are drawn tip-to-tail:

  • $(W r)_C = 1.91\,\text{Nm}$ parallel to $F_C$ along $+x$.

  • $(W r)_A = 2.21\,\text{Nm}$ parallel to $F_A$ along $+y$.

  • $(W r)_D = 1.30\,\text{Nm}$ parallel to $F_D$ at angle $\theta_1 = 47^\circ$ relative to $F_A$.

  • The closing vector of the polygon represents the force product of Mass $B$ ($0.036 W_B$):

  • $0.036 W_B = 3.3\,\text{Nm}$

  • $W_B = \frac{3.3}{0.036} = 91.67\,\text{N}$

  • $m_B g = 91.67\,\text{N}$

  • $m_B = \frac{91.67}{9.81} \approx 9.4\,\text{kg}$

  • Direct measurement using a protractor determines the orientation angle $\theta_2$ of Mass $B$ relative to Mass $D$.

Shaking Forces in Linkage Mechanisms

• Theoretical definition:

  • Dynamic forces developed by accelerating links within a mechanism are transmitted into the supporting structure via ground connections.

  • The vector summation of all dynamic forces acting directly on the ground plane is defined as the Shaking Force ($F_s$).

  • Unbalanced shaking forces induce structural vibrations in the machine foundation.

• Governing mathematical relationship:

  • $F_s = F_{21} + F_{41} + F_{61}$

  • By Newton's Third Law (action-reaction pairs between mechanism links and ground link 1):

  • $F_{21} = -F_{12}$

  • $F_{41} = -F_{14}$

  • $F_{61} = -F_{16}$

• Planar components of link reactions on ground:

  • Link 2 reaction on ground ($F_{21}$):

  • $F_{21x} = 1990.37\,\text{N}$

  • $F_{21y} = 999.27\,\text{N}$

  • Link 4 reaction on ground ($F_{41}$):

  • $F_{41x} = -4.139\,\text{N}$

  • $F_{41y} = -566.15\,\text{N}$

  • Slider (Link 6) reaction on ground ($F_{61}$):

  • $F_{61x} = 76.16\,\text{N}$

  • $F_{61y} = 253.97\,\text{N}$

• Component-wise summations:

  • Net horizontal ground reaction:

  • $\sum F_{Rx} = F_{21x} + F_{41x} + F_{61x}$

  • $\sum F_{Rx} = 1990.37 - 4.139 + 76.16 = 2062.4\,\text{N}$

  • Net vertical ground reaction:

  • $\sum F_{Ry} = F_{21y} + F_{41y} + F_{61y}$

  • $\sum F_{Ry} = 999.27 - 566.15 + 253.97 = 687.09\,\text{N}$

• Resultant shaking force calculation:

  • $F_R^2 = \left(\sum(F_R)_x\right)^2 + \left(\sum(F_R)_y\right)^2 + 2 \sum(F_R)_x \sum(F_R)_y \cos(90^\circ)$

  • $F_R = \sqrt{\left(\sum(F_R)_x\right)^2 + \left(\sum(F_R)_y\right)^2}$

  • $F_R = \sqrt{2062.4^2 + 687.1^2} = \sqrt{4253493.76 + 472106.41} \approx 2173.84\,\text{N}$

• Directional orientation angle ($\phi$):

  • $\tan(\phi) = \frac{\sum(F_R)_y}{\sum(F_R)_x} = \frac{687.1}{2062.4} \approx 0.333$

  • $\phi = \arctan(0.333) = 18.4^\circ$

Shaking Torque and Shaking Moments in Linkage Mechanisms

• Shaking Torque ($T_s$):

  • While the shaking force tends to displace the ground structure in the transverse direction, the reaction against the driving link introduces a shaking torque.

  • Shaking torque is the net reaction torque felt by the ground that acts to rock the machine foundation about the driveline axis:

  • $T_s = T_{21} = -T_{12}$

  • Here, $-T_{12}$ denotes the negative of the driving input torque supplied by the motor through link 2.

  • Mechanism value:

  • $T_s = -143.491\,\text{Nm}$

• Shaking Moment ($M_s$):

  • The total shaking moment of a linkage mechanism represents the summation of the drive reaction torque $T_{21}$ and all shaking couples generated by ground forces acting at finite lever arms:

  • $M_s = T_{21} + (\mathbf{R}_1 \times \mathbf{F}_{41}) + (\mathbf{R}_2 \times \mathbf{F}_{61})$

  • Parameter definitions:

  • $T_{21}$: Reaction torque on ground ($143.491\,\text{Nm}$)

  • $\mathbf{R}_1$: Position vector from driver ground pivot ($O$) to link 4 ground pivot

  • $\mathbf{F}_{41}$: Force exerted by link 4 onto the ground

  • $\mathbf{R}_2$: Position vector from driver ground pivot ($O$) to slider ground guide

  • $\mathbf{F}_{61}$: Force exerted by slider onto the ground

• Geometric and kinematic coordinates:

  • Position components for link 4 ground pivot:

  • $R_{1x} = 0.5478\,\text{m}$

  • $R_{1y} = 0.1164\,\text{m}$

  • Angular position: $\theta = 12^\circ$

  • Position components for slider ground interface:

  • $R_{2x} = 0.951\,\text{m}$

  • $R_{2y} = 0.035\,\text{m}$

  • Transmitted ground forces:

  • $F_{41x} = 4.139\,\text{N}$, $F_{41y} = 566.135\,\text{N}$

  • $F_{61x} = 76.16\,\text{N}$, $F_{61y} = 253.87\,\text{N}$

• Expanded scalar cross-product formulation:

  • $M_s = T_{21} + (R_{1x} \times F_{41y}) - (R_{1y} \times F_{41x}) - (R_{2y} \times F_{61x}) - (R_{2x} \times F_{61y})$

  • Numerical substitution:

  • $M_s = 143.491 + (0.5478 \times 566.135) - (0.1164 \times 4.139) - (0.035 \times 76.16) - (0.951 \times 253.87)$

  • Step-by-step evaluation:

  • Torque term: $143.491\,\text{Nm}$

  • Link 4 vertical couple: $0.5478 \times 566.135 = 310.129\,\text{Nm}$

  • Link 4 horizontal couple: $0.1164 \times 4.139 = 0.482\,\text{Nm}$

  • Slider vertical couple: $0.035 \times 76.16 = 2.666\,\text{Nm}$

  • Slider horizontal couple: $0.951 \times 253.87 = 241.430\,\text{Nm}$

  • Net calculation:

  • $M_s = 143.491 + 310.129 - 0.482 - 2.666 - 241.430 = 209.04\,\text{Nm}$

• Physical effect on machinery:

  • The resultant shaking moment generates a cyclic rocking effect about the rotational axis of the motor drive shaft, which must be resisted by machine mountings and vibration isolators.