Interactive Lecture
Reciprocating Mechanisms and Vibration Reduction
by Dr Arnaldo Delli Carri
Reciprocating masses — components that move back and forth along a linear path — are found in piston engines, compressors, pumps, and other machinery. Unlike rotating masses that produce centrifugal forces, reciprocating masses generate reciprocating inertia forces that change direction periodically, causing vibrations and potentially damaging the machine and its foundation.
Reciprocating masses produce primary and secondary unbalanced forces that simple counterweights cannot fully eliminate. The goal is to minimise these forces to acceptable levels, reducing vibration, noise, and mechanical wear.
The slider-crank mechanism converts rotary motion into reciprocating motion (or vice versa). It consists of a crank rotating about a fixed pivot, connected via a connecting rod to a piston that slides in a cylinder. This is the fundamental mechanism in internal combustion engines.
Key Parameters:
The position of the piston relative to the crankshaft centre is given by an exact expression, but for practical analysis, we use an approximation that separates primary and secondary harmonics:
Piston Displacement:
\[x = r\left[(1-\cos\theta) + \frac{1}{2n}(1-\cos 2\theta)\right]\]
Where the first term is the primary motion and the second term is the secondary correction.
Piston Acceleration:
\[a = r\omega^2\left[\cos\theta + \frac{1}{n}\cos 2\theta\right]\]
This acceleration produces two components of inertia force.
The inertia force of the reciprocating mass has two harmonic components:
Frequency: Same as crankshaft speed (\(\omega\))
\[F_p = m_r r \omega^2 \cos\theta\]
Acts along the line of piston motion
Direction: Maximum towards crankshaft at TDC, away at BDC
Frequency: Twice the crankshaft speed (\(2\omega\))
\[F_s = m_r r \omega^2 \frac{1}{n}\cos 2\theta\]
Also acts along line of piston motion
Magnitude: Smaller than primary, typically \(\frac{1}{4}\) to \(\frac{1}{5}\) of \(F_p\)
Primary forces can be partially balanced using rotating counterweights attached to the crankshaft. However, a complete balance is impossible without creating an equal unbalanced force in the perpendicular direction.
Rotating Counterweight:
A counterweight of mass \(m_c\) at radius \(r_c\) rotating with the crank produces a centrifugal force:
\[F_c = m_c r_c \omega^2\]
To balance primary force horizontally: \(m_c r_c = c \cdot m_r r\)
where \(c\) is the balance fraction (typically 0.5 to 0.67 for single-cylinder engines)
Partial balancing (c = 0.5 to 0.67) trades horizontal vibration for vertical vibration. It lowers the vector sum of the horizontal and vertical unbalanced forces but eliminates neither, so choosing c is an optimisation between the two directions.
Secondary forces cannot be balanced with simple rotating counterweights because they oscillate at twice the crank frequency. Solutions include:
A critical limitation of rotating counterweights is that they can only balance forces that oscillate at the same frequency as the crankshaft rotation (ω). The counterweight creates a centrifugal force \(F_c = m_c r_c \omega^2\) that rotates with the crank, producing components:
Counterweight Force Components:
Horizontal: \(F_{c,h} = m_c r_c \omega^2 \cos\theta\)
Vertical: \(F_{c,v} = m_c r_c \omega^2 \sin\theta\)
Both oscillate at frequency ω (same as crank speed)
However, the reciprocating mass produces both primary force (at ω) and secondary force (at 2ω):
\[F_{\text{total}} = \underbrace{m_r r \omega^2 \cos\theta}_{\text{Primary at } \omega} + \underbrace{m_r r \omega^2 \frac{1}{n}\cos 2\theta}_{\text{Secondary at } 2\omega}\]
When the balance fraction c = 1, the counterweight completely eliminates the primary force component, but the secondary force remains unaffected. This is why even with "complete" balancing (c = 1), there is still residual vibration equal to the secondary force, typically 20-25% of the original unbalanced primary force for typical rod ratios (n = 4 to 5).
The Lanchester balancer, invented by Frederick Lanchester in 1904, is an elegant solution for eliminating secondary forces. It consists of two identical counter-rotating shafts positioned parallel to the crankshaft, each carrying eccentric masses.
The two shafts counter-rotate at exactly twice the crankshaft speed (2ω), so their vertical force components cancel while their horizontal components add. The combined horizontal force is set to match and oppose the secondary force from the reciprocating mass.
Lanchester Shaft Configuration:
Each shaft has mass \(m_b\) at radius \(r_b\), rotating at \(2\omega\)
Shaft 1 angle: \(\phi_1 = 2\theta\)
Shaft 2 angle: \(\phi_2 = -2\theta\) (counter-rotating)
Forces produced:
Shaft 1: \(\vec{F}_1 = m_b r_b (2\omega)^2 [\cos(2\theta)\hat{i} + \sin(2\theta)\hat{j}]\)
Shaft 2: \(\vec{F}_2 = m_b r_b (2\omega)^2 [\cos(-2\theta)\hat{i} + \sin(-2\theta)\hat{j}]\)
Resultant:
Horizontal: \(F_{b,h} = 2m_b r_b (2\omega)^2 \cos(2\theta) = 8m_b r_b \omega^2 \cos(2\theta)\)
Vertical: \(F_{b,v} = 0\) (components cancel)
By setting \(m_b r_b = \frac{m_r r}{8n}\), the Lanchester shafts produce a horizontal force that exactly cancels the secondary force. The beauty of this system is that the vertical components always cancel, so no additional vertical vibration is introduced.
Lanchester balancers are driven by gears or chains at a 2:1 ratio to the crankshaft and must be precisely phased. The combined horizontal force has to peak when the piston is at TDC or BDC (θ = 0° or 180°), where the secondary force is maximum. Modern applications include inline-4 engines (Mitsubishi Silent Shaft, Mercedes M264 engine) and some inline-3 engines.
Advantages:
Disadvantages:
In multi-cylinder engines, pistons are arranged with specific phase angles. The combined effect of multiple reciprocating masses can achieve better balance:
Primary Force Balance Condition:
\[\sum_{i=1}^{N} m_{ri} r_i \cos\theta_i = 0\]
Secondary Force Balance Condition:
\[\sum_{i=1}^{N} m_{ri} r_i \frac{1}{n_i}\cos 2\theta_i = 0\]
| Engine Configuration | Primary Balance | Secondary Balance | Typical Use |
|---|---|---|---|
| Single Cylinder | Partial (50-67%) | Unbalanced | Motorcycles, small engines |
| Inline-2 (180°) | Balanced | Additive (worst) | Rarely used |
| Inline-4 (Even firing) | Balanced | Unbalanced | Most common car engine |
| Inline-6 | Balanced | Balanced | Premium cars, trucks |
| V8 (90° cross-plane) | Balanced | Balanced | Performance vehicles |
| Flat-4 (Boxer) | Balanced | Unbalanced | Porsche, Subaru |
Even when forces are balanced, unbalanced couples (moments) can cause pitching vibration. The couple depends on the spacing between cylinders and requires consideration in engine design.
Primary Couple:
\[C_p = \sum_{i=1}^{N} m_{ri} r_i \omega^2 \cos\theta_i \cdot d_i\]
where \(d_i\) is the distance of cylinder \(i\) from a reference plane.
Observe the slider-crank mechanism in action. Adjust the parameters to see how crank radius, connecting rod length, and reciprocating mass affect the primary and secondary forces. The force vectors are shown in real-time, and you can add a counterweight to see partial balancing in action.
| Concept | Equation | Description |
|---|---|---|
| Rod Ratio | \(n = \frac{l}{r}\) | Connecting rod length to crank radius |
| Crank Angle | \(\theta = \omega t\) | Angular position from TDC |
| Piston Displacement | \(x = r\left[(1-\cos\theta) + \frac{1}{2n}(1-\cos 2\theta)\right]\) | Position relative to crankshaft |
| Piston Acceleration | \(a = r\omega^2\left[\cos\theta + \frac{1}{n}\cos 2\theta\right]\) | Linear acceleration of piston |
| Primary Inertia Force | \(F_p = m_r r \omega^2 \cos\theta\) | Force at crankshaft frequency |
| Secondary Inertia Force | \(F_s = m_r r \omega^2 \frac{1}{n}\cos 2\theta\) | Force at twice crankshaft frequency |
| Total Inertia Force | \(F = F_p + F_s\) | Combined reciprocating force |
| Counterweight Force | \(F_c = m_c r_c \omega^2\) | Centrifugal force of balance weight |
| Balance Condition | \(m_c r_c = c \cdot m_r r\) | Partial balance with fraction c |
| Horizontal Unbalance | \(F_h = m_r r \omega^2(1-c)\cos\theta\) | Remaining horizontal force |
| Vertical Unbalance | \(F_v = c \cdot m_r r \omega^2 \sin\theta\) | Introduced vertical force |
| Optimal Balance | \(c_{\text{opt}} = 0.5\) | Equal horizontal and vertical residuals; minimises the peak resultant force |
A single-cylinder engine has a reciprocating mass of 2 kg, crank radius of 60 mm, and operates at 3000 RPM. Calculate the maximum primary inertia force.
Show Solution
Solution:
Given:
Step 1: Convert RPM to rad/s
\[\omega = \frac{2\pi N}{60} = \frac{2\pi \times 3000}{60} = 314.16 \text{ rad/s}\]
Step 2: Calculate maximum primary force
The primary force is: \(F_p = m_r r \omega^2 \cos\theta\)
Maximum occurs when \(\cos\theta = 1\):
\[F_{p,\text{max}} = m_r r \omega^2\]
\[F_{p,\text{max}} = 2 \times 0.06 \times (314.16)^2\]
\[F_{p,\text{max}} = 2 \times 0.06 \times 98696.5\]
\[F_{p,\text{max}} = 11844 \text{ N} = 11.84 \text{ kN}\]
Answer: The maximum primary inertia force is 11.84 kN
For the same engine in Problem 1, if the connecting rod length is 240 mm, calculate the maximum secondary inertia force and compare it to the primary force.
Show Solution
Solution:
Given:
Step 1: Calculate rod ratio
\[n = \frac{l}{r} = \frac{0.24}{0.06} = 4\]
Step 2: Calculate maximum secondary force
The secondary force is: \(F_s = m_r r \omega^2 \frac{1}{n}\cos 2\theta\)
Maximum occurs when \(\cos 2\theta = 1\):
\[F_{s,\text{max}} = \frac{m_r r \omega^2}{n}\]
\[F_{s,\text{max}} = \frac{11844}{4} = 2961 \text{ N} = 2.96 \text{ kN}\]
Step 3: Compare to primary force
\[\frac{F_{s,\text{max}}}{F_{p,\text{max}}} = \frac{1}{n} = \frac{1}{4} = 0.25 = 25\%\]
Answer: The maximum secondary force is 2.96 kN, which is 25% of the primary force. With a typical rod ratio of 4, the secondary force is significant but smaller than the primary force.
Design a counterweight to achieve 60% balance of the primary force for the engine in Problem 1. If the counterweight is placed at a radius of 80 mm from the crankshaft axis, what mass is required?
Show Solution
Solution:
Given:
Balance Condition:
\[m_c r_c = c \cdot m_r r\]
Solve for counterweight mass:
\[m_c = \frac{c \cdot m_r r}{r_c}\]
\[m_c = \frac{0.60 \times 2 \times 0.06}{0.08}\]
\[m_c = \frac{0.072}{0.08} = 0.9 \text{ kg}\]
Verification:
Counterweight moment: \(m_c r_c = 0.9 \times 0.08 = 0.072\) kg·m
Required moment: \(c \cdot m_r r = 0.60 \times 2 \times 0.06 = 0.072\) kg·m ✓
Answer: A counterweight of 0.9 kg placed at 80 mm radius will provide 60% balance of the primary force.
A two-cylinder inline engine has pistons 180° out of phase (when one is at TDC, the other is at BDC). If each cylinder has identical reciprocating mass of 1.5 kg and crank radius of 50 mm, determine if the primary forces are balanced.
Show Solution
Solution:
Given:
Primary Force Balance Condition:
\[\sum F_p = m_{r1} r_1 \omega^2 \cos\theta_1 + m_{r2} r_2 \omega^2 \cos\theta_2\]
Substituting values:
\[\sum F_p = 1.5 \times 0.05 \times \omega^2 \cos\theta + 1.5 \times 0.05 \times \omega^2 \cos(\theta + 180°)\]
Using trigonometric identity: \(\cos(\theta + 180°) = -\cos\theta\)
\[\sum F_p = 1.5 \times 0.05 \times \omega^2 \cos\theta + 1.5 \times 0.05 \times \omega^2 \times (-\cos\theta)\]
\[\sum F_p = 1.5 \times 0.05 \times \omega^2 \cos\theta - 1.5 \times 0.05 \times \omega^2 \cos\theta = 0\]
Primary forces are BALANCED! ✓
Secondary Force Analysis:
\[\sum F_s = \frac{m_r r \omega^2}{n}\cos 2\theta + \frac{m_r r \omega^2}{n}\cos 2(\theta + 180°)\]
Since \(\cos 2(\theta + 180°) = \cos(2\theta + 360°) = \cos 2\theta\):
\[\sum F_s = \frac{m_r r \omega^2}{n}\cos 2\theta + \frac{m_r r \omega^2}{n}\cos 2\theta = \frac{2m_r r \omega^2}{n}\cos 2\theta\]
Secondary forces are ADDITIVE (worst case)! ✗
Answer: The 180° two-cylinder configuration achieves perfect primary force balance but has the worst secondary force balance, with both cylinders' secondary forces adding together. This configuration is rarely used in practice due to severe secondary vibration.