Dynamic Analyses of Whitworth Quick Return Mechanism

Abstract

Herein, we present a dynamic analysis on a Whitworth quick return mechanism, which was proposed as a class project to engage Mechanical Engineering students in the Machine Dynamics course. The objectives of the project are to teach (1) kinematic analysis on planar linkage mechanism using complex numbers and (2) kinetic analyses using distributed and lumped mass models. The analytical models were implemented using an engineering math software – Mathcad, and the results were validated by a multibody dynamics simulation software – Inventor. Loop closure equations were first built and solved to find link positions, and then velocities and accelerations were confirmed with numerical differentiation in Mathcad. Based on the solved kinematic quantities, a kinetic analysis was carried out using distributed and lumped mass models as well as in Inventor. The reactions and torques at the input link of the distributed mass model were found identical to those derived from Inventor. It was found the maximum reactions at pin joints were 11% to 100% different between the distributed mass model and the simplified lumped mass model. However, the shaking forces found in both models were nearly identical, and the peak values of the shaking moments were comparable. It is confirmed the lumped mass model can serve as a quick method to find the magnitude of the shaking forces and moments without solving simultaneous equation systems.

Keywords

complex number method kinematic pairs shaping machine mobility time ratio

Introduction

Whitworth quick return mechanisms are widely used in numerous industrial applications such as shapers, power-driven saws, and mechanical actuators. In those applications, a load-intensive working stroke is required with a low-load return stroke. The time spent in the working stroke divided by the return stroke defines a time ratio greater than unity. Several basic 4-link mechanisms feature quick-return actions, including crank-slider mechanisms and four-bar mechanisms. Rao1 designed a four-bar crank lever mechanism to achieve a given time ratio and angle of oscillation. Hsieh and Tsai² tested the feasibility of combining a generalized Oldham coupling with a slider-crank mechanism.

Whitworth quick return mechanism has six links consisting of a driving crank with a constant angular velocity and a ram with reciprocating actions featuring a quick return stroke. Research has been focused on designing the mechanism for various applications. Dwivedi³ modified the mechanism for use in high-velocity impact presses. Lin et al.⁴ introduced it as the actuating unit of pots for automatic cooking - the time ratio of cooking and pot retrieving is greater than 1.4 using their design. Ying and Cheng⁵ employed the mechanism to develop a bag-breaking device, whereas Endeshaw et al.⁶ explored the feasibility to mount a piezoelectric energy harvester on the Whitworth quick return mechanism. Recently, Shyam et al.⁷ used Whitworth quick return mechanisms to improve the efficiency of stationary concrete pumps. Although traditional analytical and graphical methods are widely used in mechanism design, nowadays, computer-aid programmes have streamlined the design process. Echempati⁸ implemented the Whitworth quick return mechanism project in Microsoft Excel, and the mechanism was modelled in a Catia CAE software. Zhou and Chen ⁹ modelled and simulated a sharper quick-return mechanism in Automated Dynamic Analysis of Mechanical System (ADAMS). Cheetancheri and Cheng¹⁰ developed an Excel-based mechanism analysis toolkit.

In this study, we utilized the analytical method along with the computer programmes (Mathcad 15, Inventor 2021) to solve the kinematic and kinetic quantities of the Whitworth quick return mechanism. The technical goal is to formulate kinematic and kinetic relationships as well as to model the mechanism in modern software. The analytical results were first derived and then validated against the numerical values. In addition, the difference between distributed mass model and lumped mass model was explored. We attempted to provide insight into the strength and limitations of the lumped mass model - a question that was not answered in a standard textbook.¹¹ The complete solution to the mechanism, including the Mathcad scripts and CAD files, are provided for researchers to modify and tailor for specific needs. The models can be adopted in the courses covering planar linkage mechanisms. At Penn State Behrend, the projects described in this work were assigned to mechanical engineering students in the ME-380 Machine Dynamics and completed within the first eight weeks. Our previous study¹² showed that introducing practical applications to students is beneficial to their comprehension and engagement. After completing these projects, students should be able to: (1) analyze the kinematics of planar linkage mechanisms, (2) use Mathcad to solve complex machine dynamics problems or systems, and (3) model/simulate planar linkage mechanisms in Inventor.

Methods

The Whitworth quick return mechanism is shown in Figure 1. Six links were numbered, and seven kinematic pairs were identified, including five turning pairs between links 1–2, 1–3, 2–4, 3–5, and 5–6, and two sliding pairs between links 3–4 and 1–6. The mobility was determined using Gruebler's equation $m = 3 (n - 1) - 2 j_{1}$ , where n is the total number of links and $j_{1}$ the number of single degree-of-freedom pairs. The mobility equals one, indicating the configuration of the mechanism can be uniquely determined based on the position of crank link 2, which rotates at a constant angular speed of $ω_{2}$ . The time ratio can be determined using the geometric relationship, $T_{R} = \frac{π - \arccos (\frac{r_{2}}{r_{1}})}{\arccos (\frac{r_{2}}{r_{1}})}$ . In the presence of an external force $F_{D}$ , the reactions of each link and the amount of torque needed to keep the mechanism at the static equilibrium position were determined. Based on the derived kinematic quantities, the dynamic loads on each link were calculated using distributed and lumped mass models. For simplicity, no external load was considered in the dynamic analysis. The reactions on the joints, along with the shaking forces and moments imparted on the ground link, were compared.

Figure 1.

(a) skeleton diagram and (b) vector loops for the Whitworth quick return mechanism. The x and y are the real and imaginary parts, respectively.

Kinematic analysis

Two loops were built, as shown in Figure 1, to traverse the Whitworth quick return mechanism along the links and through kinematic pairs. The primary loop covers the driving crank, the base links, and the slotted slider ( $R_{1}$ , $R_{2}$ and $R_{4}$ ), and the secondary loop tracks the position of the slider and the rest of the links ( $R_{3}$ , $R_{5}$ , $R_{6}$ , and $R_{7}$ ). Accordingly, loop closure equations can be written as

\begin{matrix} R_{2} = R_{1} + R_{4} \end{matrix}

(1)

\begin{matrix} R_{3} + R_{5} = R_{6} + R_{7} \end{matrix}

(2)

where the complex numbers

R_{n}

represent vectors shown in Figure 1, and Equations (1)-(2) can be resolved into real and imaginary axes, yielding

\begin{matrix} r_{2} \cos (θ_{2}) = r_{4} \cos (θ_{4}) \end{matrix}

(3)

\begin{matrix} r_{2} \sin (θ_{2}) = - r_{1} + r_{4} \sin (θ_{4}) \end{matrix}

(4)

\begin{matrix} r_{3} \cos (θ_{3}) + r_{5} \cos (θ_{5}) = - r_{6} \end{matrix}

(5)

\begin{matrix} r_{3} \sin (θ_{3}) + r_{5} \sin (θ_{5}) = r_{7} \end{matrix}

(6)

where the vector angle

θ_{n}

are defined from the real positive axis to the direction of the vector counterclockwise. The positions of all the links can be analytically obtained by solving Equations (3)-(6). The input angle

θ_{2}

serves as the only independent variable in the primary loop. Equation (3)-(4) can be first used to solve for two dependent unknowns

θ_{4}

and

r_{4}

\begin{matrix} θ_{4} = \arctan (\frac{r_{2} \sin (θ_{2}) + r_{1}}{r_{2} \cos (θ_{2})}) \end{matrix}

(7)

\begin{matrix} r_{4} = \frac{r_{2} \cos (θ_{2})}{\cos (θ_{4})} \end{matrix}

(8)

Using the condition

θ_{4} = θ_{3}

, the unknowns

θ_{5}

and

r_{6}

can be found by solving Equations (5)-(6),

\begin{matrix} θ_{5} = \arcsin (\frac{r_{7} - r_{3} \sin (θ_{3})}{r_{5}}) \end{matrix}

(9)

\begin{matrix} r_{6} = - r_{3} \cos (θ_{3}) - r_{5} \cos (θ_{5}) \end{matrix}

(10)

where

r_{6}

has a positive value (magnitude of the vector) when pointing towards the negative x-direction. Hence,

- r_{6}

denotes the horizontal position of the slider Link 6. A direct method to find the velocities is to take time derivatives from position expressions. However, multiple expressions

θ_{4}

and

θ_{5}

involve inverse trigonometric functions, complicating the derivatives.13 Instead, differentiation was performed on Equations (3)-(6), generating another set of equations,

\begin{matrix} - r_{2} \sin (θ_{2}) {\dot{θ}}_{2} = - r_{4} \sin (θ_{4}) {\dot{θ}}_{4} + {\dot{r}}_{4} \cos (θ_{4}) \end{matrix}

(11)

\begin{matrix} r_{2} \cos (θ_{2}) {\dot{θ}}_{2} = r_{4} \cos \cos (θ_{4}) {\dot{θ}}_{4} + {\dot{r}}_{4} \sin (θ_{4}) \end{matrix}

(12)

\begin{matrix} - r_{3} \sin (θ_{3}) {\dot{θ}}_{3} - r_{5} \sin (θ_{5}) {\dot{θ}}_{5} = - {\dot{r}}_{6} \end{matrix}

(13)

\begin{matrix} r_{3} \cos (θ_{3}) {\dot{θ}}_{3} + r_{5} \cos (θ_{5}) {\dot{θ}}_{5} = 0 \end{matrix}

(14)

where the four unknowns are

{\dot{r}}_{4}

{\dot{θ}}_{4} ({\dot{θ}}_{3})

{\dot{θ}}_{5}

, and

{\dot{r}}_{6}

. When solving these unknowns, elegant solution forms can be derived using an elimination method rather than a substitution method. To find

{\dot{r}}_{4}

and

{\dot{θ}}_{4}

, let

(11) \times \cos (θ_{4}) + (12) \times \sin (θ_{4})

and

(11) \times \sin (θ_{4}) - (12) \times \cos (θ_{4})

, giving

\begin{matrix} {\dot{r}}_{4} = r_{2} {\dot{θ}}_{2} \sin (θ_{4} - θ_{2}) \end{matrix}

(15)

\begin{matrix} {\dot{θ}}_{4} = \frac{r_{2} {\dot{θ}}_{2} \cos (θ_{2} - θ_{4})}{r_{4}} \end{matrix}

(16)

It can be seen the

{\dot{r}}_{4}

and

{\dot{θ}}_{4}

depend on the geometry

θ_{2}

θ_{4}

r_{2}

r_{4}

, and the crank speed

{\dot{θ}}_{2}

, while they are not influenced by the secondary loop. Such expressions provide an intuitive understanding of the mechanism working principle. Note that

{\dot{r}}_{4}

is a relative velocity of point

B_{2}

on link 2 with respect to

B_{3}

on link 3. Using the same elimination method,

{\dot{θ}}_{5}

and

{\dot{r}}_{6}

can be found from Eqs. (13) and (14).

\begin{matrix} {\dot{θ}}_{5} = - \frac{r_{3} \cos (θ_{3}) {\dot{θ}}_{3}}{r_{5} \cos (θ_{5})} \end{matrix}

(17)

\begin{matrix} {\dot{r}}_{6} = r_{3} \sin (θ_{3}) {\dot{θ}}_{3} + r_{5} \sin (θ_{5}) {\dot{θ}}_{5} \end{matrix}

(18)

where velocities expressions above depend on the solved positions. In other words, high order kinematic variables rely on lower order quantities. Acceleration terms can be solved using a similar procedure through differentiating Eqs. (11)-(14), generating

\begin{aligned} - r_{2} \sin (θ_{2}) {\ddot{θ}}_{2} - r_{2} \cos (θ_{2}) {({\dot{θ}}_{2})}^{2} = & - {\dot{r}}_{4} \sin (θ_{4}) {\dot{θ}}_{4} - r_{4} \cos (θ_{4}) {({\dot{θ}}_{4})}^{2} \\ - r_{4} \sin (θ_{4}) {\ddot{θ}}_{4} + {\dot{r}}_{4} \cos (θ_{4}) - {\dot{r}}_{4} \sin (θ_{4}) {\dot{θ}}_{4} \end{aligned}

(19)

r_{2} \cos (θ_{2}) {\ddot{θ}}_{2} - r_{2} \sin (θ_{2}) {({\dot{θ}}_{2})}^{2} = {\dot{r}}_{4} \cos (θ_{4}) {\dot{θ}}_{4} - r_{4} \sin (θ_{4}) {({\dot{θ}}_{4})}^{2} + r_{4} \cos (θ_{4}) {\ddot{θ}}_{4} + {\ddot{r}}_{4} \sin (θ_{4}) + {\dot{r}}_{4} \cos (θ_{4}) {\dot{θ}}_{4}

(20)

\begin{matrix} r_{3} \sin (θ_{3}) {\ddot{θ}}_{3} + r_{3} \cos (θ_{3}) {({\dot{θ}}_{3})}^{2} + r_{5} \sin (θ_{5}) {\ddot{θ}}_{5} + r_{5} \cos (θ_{5}) {({\dot{θ}}_{5})}^{2} = {\ddot{r}}_{6} \end{matrix}

(21)

\begin{matrix} r_{3} \cos (θ_{3}) {\ddot{θ}}_{3} - r_{3} \sin (θ_{3}) {({\dot{θ}}_{3})}^{2} + r_{5} \cos (θ_{5}) {\ddot{θ}}_{5} - r_{5} \sin (θ_{5}) {({\dot{θ}}_{5})}^{2} = 0 \end{matrix}

(22)

To solve

{\ddot{θ}}_{3}

(

{\ddot{θ}}_{4}

), let

(19) \times \sin (θ_{4}) - (20) \times \cos (θ_{4})

, we obtain

\begin{matrix} {\ddot{θ}}_{4} = \frac{r_{2} {\ddot{θ}}_{2} \cos (θ_{2} - θ_{4}) - r_{2} {({\dot{θ}}_{2})}^{2} \sin (θ_{2} - θ_{4}) - 2 {\dot{r}}_{4} {\dot{θ}}_{4}}{r_{4}} \end{matrix}

(23)

To sovle

{\ddot{r}}_{4}

, let

(19) \times \cos (θ_{4}) + (20) \times \sin (θ_{4})

, we find

\begin{matrix} {\ddot{r}}_{4} = r_{4} {({\dot{θ}}_{4})}^{2} - r_{2} {\ddot{θ}}_{2} \sin (θ_{2} - θ_{4}) - r_{2} {({\dot{θ}}_{2})}^{2} \cos (θ_{2} - θ_{4}) \end{matrix}

(24)

where

{\ddot{r}}_{4}

is the sliding acceleration of

B_{2}

with respect to

B_{3}

. It should not be confused with the absolute acceleration of slider 4, which includes the sliding, tangential, normal, and Coriolis.

To solve ${\ddot{r}}_{6}$ , let $(21) \times \cos (θ_{5}) + (22) \times \sin (θ_{5})$ , yielding

\begin{matrix} {\ddot{r}}_{6} = \frac{r_{3} {\ddot{θ}}_{3} \sin (θ_{3} - θ_{5}) + r_{3} {({\dot{θ}}_{3})}^{2} \cos (θ_{3} - θ_{5}) + r_{5} {({\dot{θ}}_{5})}^{2}}{\cos (θ_{5})} \end{matrix}

(25)

Equation (22) gives

\begin{matrix} {\ddot{θ}}_{5} = \frac{r_{3} \sin (θ_{3}) {({\dot{θ}}_{3})}^{2} + r_{5} \sin (θ_{5}) {({\dot{θ}}_{5})}^{2} - r_{3} \cos (θ_{3}) {\ddot{θ}}_{3}}{r_{5} \cos (θ_{5})} \end{matrix}

(26)

An alternative way to find the accelerations is to take 2nd order differentiation of the position functions or 1st order differentiation of the velocities, which were implemented in Mathcad to confirm the analytical solutions.

Static analysis

The driving torque from the crank is transmitted to the working slider through multiple kinematic pairs and links. In this static force analysis, it is assumed that the kinematic pairs were frictionless, and the links were rigid bodies. When a constant external force is applied on the slider (link 6), a torque $T_{12}$ is needed to balance the external load, keeping the mechanism in static equilibrium. Free-body diagrams (FBD) are depicted in Figure 2, where vector angles follow the same convention as kinematic analysis. Links 4 and 5 are two-force members subjected to a pair of forces equal in magnitude, co-linear and opposite in sense. Applying the force equilibrium condition to link 6, the summation of $\vec{F_{D}}$ , $\vec{F_{16}}$ , and $\vec{F_{56}}$ is zero.

\begin{matrix} \vec{F_{D}} = F_{D} {\begin{matrix} 1 \\ 0 \end{matrix}}, \vec{F_{16}} = F_{16} {\begin{matrix} 0 \\ - 1 \end{matrix}}, \vec{F_{56}} = F_{56} {\begin{matrix} \cos (θ_{5}) \\ \sin (θ_{5}) \end{matrix}} \end{matrix}

(27)

Figure 2.

(a) mechanism subject to an external force and (b) free-body diagrams of individual links.

Two unknown forces $F_{56}$ and $F_{16}$ can be found

\begin{matrix} F_{56} = - \frac{F_{D}}{\cos (θ_{5})} \end{matrix}

(28)

\begin{matrix} F_{16} = F_{56} \sin (θ_{5}) \end{matrix}

(29)

In general, 2D equilibrium conditions generate three governing equations, i.e., forces equilibrium in horizontal and vertical directions and zero net moments. Observations in Figure 2 revealed no link subject to more than three forces, implying a sequential solving procedure can be applied in a link-by-link manner to find all the unknown forces. The calculations were implemented in the supplemented Mathcad files.

\begin{matrix} F_{35} = F_{56} \end{matrix}

(30)

\begin{matrix} F_{43} = - \frac{r_{3}}{r_{4}} F_{35} \sin (θ_{5} - θ_{4}) \end{matrix}

(31)

\begin{matrix} F_{13, x} = F_{35} \cos (θ_{5}) + F_{43} \sin (θ_{4}) \end{matrix}

(32)

\begin{matrix} F_{13, y} = F_{35} \sin (θ_{5}) - F_{43} \cos (θ_{4}) \end{matrix}

(33)

\begin{matrix} \vec{F_{12}} = F_{43} {\begin{matrix} - \sin (θ_{4}) \\ \cos (θ_{4}) \end{matrix}} \end{matrix}

(34)

\begin{matrix} \vec{F_{42}} = F_{43} {\begin{matrix} \sin (θ_{4}) \\ - \cos (θ_{4}) \end{matrix}} \end{matrix}

(35)

\begin{matrix} T_{12} = (r_{2} F_{43}) \cos (θ_{4} - θ_{2}) \end{matrix}

(36)

Dynamic analysis – distributed mass model

Kinetic analyses of the Whitworth quick return mechanism are based on the governing equations of motion, requiring solved kinematic expressions. As shown in Figure 3, there are 13 unknowns from the FBD, namely, $F_{16}, F_{56, x}, F_{56, y}, F_{35, x}, F_{35, y}, F_{43}, F_{13, x}, F_{13, y}, F_{24, n}, F_{24, t}, F_{12, x}, F_{12, y}, T_{12}$ .

Figure 3.

(a) free-body diagrams and (b) kinetic diagrams of the distributed mass model.

In order to solve these unknowns, 13 equations must be derived from the kinetic diagrams (KD) to form a simultaneous linear equation system. The unbalanced inertia of the moving links causes the shaking forces defined as:

\begin{matrix} F_{s, x} = F_{12, x} - F_{13, x} \end{matrix}

(37)

and

\begin{matrix} F_{s, y} = F_{16} - F_{13, y} - F_{12, y} \end{matrix}

(38)

Since the shaking forces result from the inertia forces, alternatively,

\begin{matrix} F_{s, x} = - (m_{2} a_{2, x} + m_{3} a_{3, x} + m_{5} a_{5, x} + m_{6} a_{6} - m_{4} a_{4} \cos (θ_{2})) \end{matrix}

(39)

\begin{matrix} F_{s, y} = - (m_{2} a_{2, y} + m_{3} a_{3, y} + m_{5} a_{5, y} - m_{4} a_{4} \sin (θ_{2})) \end{matrix}

(40)

The shaking moment about the pivot

O_{2}

are resulted from the reaction of the driving torque onto the base link, the horizontal pin force at the base pivot

O_{3}

, and the normal force exerted from slider 6, expressed as

\begin{matrix} M_{s} = T_{12} + r_{1} F_{13, x} + r_{6} F_{16} \end{matrix}

(41)

Assume the links were uniform and made of the same homogeneous materials with a density

ρ

. Equation of motions were applied to link 6:

\begin{matrix} F_{56, y} - F_{16} = 0 \end{matrix}

(42)

\begin{matrix} F_{56, x} = m_{6} a_{6} \end{matrix}

(43)

link 5:

\begin{matrix} F_{35, x} - F_{56, x} = m_{5} a_{5, x} \end{matrix}

(44)

\begin{matrix} F_{35, y} - F_{56, y} = m_{5} a_{5, y} \end{matrix}

(45)

\begin{matrix} \frac{1}{2} r_{5} \sin (θ_{5}) F_{35, x} - \frac{1}{2} r_{5} \cos (θ_{5}) F_{35, y} + \frac{1}{2} r_{5} \sin (θ_{5}) F_{56, x} - \frac{1}{2} r_{5} \cos (θ_{5}) F_{56, y} = I_{G_{5}} {\ddot{θ}}_{5} \end{matrix}

(46)

where

a_{5, x} = a_{C, x} - \frac{1}{2} r_{5} {\ddot{θ}}_{5} \sin (θ_{5}) - \frac{1}{2} r_{5} ({\dot{θ}}_{5})^{2} \cos (θ_{5}) a_{5, y} = a_{C, y} + \frac{1}{2} r_{5} {\ddot{θ}}_{5} \cos (θ_{5}) - \frac{1}{2} r_{5} ({\dot{θ}}_{5})^{2} \sin (θ_{5}) I_{G_{5}} = \frac{1}{12} m_{5} r_{5}^{2} = \frac{1}{12} ρ r_{5}^{3}

(47)

link 3:

\begin{matrix} F_{13, x} - F_{35, x} - F_{43} \sin (θ_{4}) = m_{3} a_{3, x} \end{matrix}

(48)

\begin{matrix} F_{13, y} - F_{35, y} + F_{43} c o s (θ_{4}) = m_{3} a_{3, y} \end{matrix}

(49)

where

\begin{aligned} a_{3, x} = - \frac{1}{2} r_{3} ({\dot{θ}}_{4})^{2} \cos (θ_{4}) - \frac{1}{2} r_{3} {\ddot{θ}}_{4} \sin (θ_{4}) \\ a_{3, y} = - \frac{1}{2} r_{3} ({\dot{θ}}_{4})^{2} \sin (θ_{4}) + \frac{1}{2} r_{3} {\ddot{θ}}_{4} c o s (θ_{4}) m_{3} = r_{3} ρ \end{aligned}

\frac{1}{2} r_{3} \sin (θ_{4}) F_{35, x} - \frac{1}{2} r_{3} \cos (θ_{4}) F_{35, y} + \frac{1}{2} r_{3} \sin (θ_{4}) F_{13, x} - \frac{1}{2} r_{3} \cos (θ_{4}) F_{13, y} + (r_{4} - \frac{1}{2} r_{3}) F_{43} = I_{G_{3}} {\ddot{θ}}_{3}

where

I_{G_{3}} = \frac{1}{12} m_{3} r_{3}^{2} = \frac{1}{12} ρ r_{3}^{3}

link 2:

\begin{matrix} F_{24, t} - F_{43} \cos (θ_{2} - θ_{4}) = 0 \end{matrix}

(50)

\begin{matrix} F_{24, n} + F_{43} \sin (θ_{2} - θ_{4}) = m_{4} a_{4} \end{matrix}

(51)

where

a_{4} = r_{2} ({\dot{θ}}_{2})^{2}

\begin{matrix} F_{24, n} \cos (θ_{2}) + F_{24, t} \sin (θ_{2}) - F_{12, x} = m_{2} a_{2, x} \end{matrix}

(52)

\begin{matrix} F_{24, n} \sin (θ_{2}) - F_{24, t} \cos (θ_{2}) + F_{12, y} = m_{2} a_{2, y} \end{matrix}

(53)

where

a_{2, x} = - \frac{1}{2} r_{2} ({\dot{θ}}_{2})^{2} \cos (θ_{2})

a_{2, y} = - \frac{1}{2} r_{2} ({\dot{θ}}_{2})^{2} \sin (θ_{2})

, and

m_{2} = r_{2} ρ

\begin{matrix} - \frac{1}{2} r_{2} F_{24, t} - \frac{1}{2} r_{2} \sin (θ_{2}) F_{12, x} - \frac{1}{2} r_{2} \cos (θ_{2}) F_{12, y} + T_{12} = 0 \end{matrix}

(54)

A simplified way to find the shaking forces is to use a lumped mass model, which requires the mass of links to be allocated only at the joints. Link 5 becomes a weightless two-force member, as shown in Figure 4. This treatment guarantees the same total mass and the centre of mass; however, it changes the polar mass moment of inertia. Therefore, it is expected the reactions at links differ from the distributed mass model. Assume the all the links were uniform and masses were accolated at two end joints, giving

\begin{matrix} m_{D} = m_{6} + \frac{1}{2} m_{5}, m_{C} = \frac{1}{2} m_{5} + \frac{1}{2} m_{3}, m_{O_{3}} = \frac{1}{2} m_{3}, m_{B} = m_{4} + \frac{1}{2} m_{2}, m_{O_{2}} = \frac{1}{2} m_{2} \end{matrix}

(55)

Figure 4.

(a) free-body diagrams, and (b) kinetic diagrams of the lumped mass model.

Apply Equation of Motion to solve for link 6,

\begin{matrix} {\vec{F}}_{56} + {\vec{F}}_{16} = m_{D} {\vec{a}}_{6} \end{matrix}

(56)

\begin{matrix} - F_{56} {\begin{matrix} \cos (θ_{5}) \\ \sin (θ_{5}) \end{matrix}} + {\begin{matrix} 0 \\ - F_{16} \end{matrix}} = {\begin{matrix} m_{D} a_{6} \\ 0 \end{matrix}} \end{matrix}

(57)

leading to

\begin{matrix} F_{56} = - \frac{m_{D} a_{6}}{\cos (θ_{5})} \end{matrix}

(58)

\begin{matrix} F_{16} = - F_{56} \sin (θ_{5}) \end{matrix}

(59)

link 5

\begin{matrix} F_{65} = F_{35} = F_{56} \end{matrix}

(60)

link 3

\begin{matrix} F_{43} = \frac{I_{G_{C}} {\ddot{θ}}_{3}}{r_{4}} - F_{56} \frac{r_{3}}{r_{4}} \sin (θ_{5} - θ_{4}) \end{matrix}

(61)

\begin{matrix} F_{13, x} = F_{43} \sin (θ_{4}) - F_{53} \cos (θ_{5}) + m_{c} a_{c x} \end{matrix}

(62)

\begin{matrix} F_{13, y} = - F_{43} \cos (θ_{4}) - F_{53} \sin (θ_{5}) + m_{c} a_{c y} \end{matrix}

(63)

link 4

\begin{matrix} F_{24, t} = F_{43} \cos (θ_{2} - θ_{4}) \end{matrix}

(64)

\begin{matrix} F_{24, n} = m_{B} a_{4} - F_{43} \sin (θ_{2} - θ_{4}) \end{matrix}

(65)

link 2

\begin{matrix} F_{12, x} = F_{24, n} \cos (θ_{2}) + F_{24, t} \sin (θ_{2}) \end{matrix}

(66)

\begin{matrix} F_{12, y} = - F_{24, n} \sin (θ_{2}) + F_{24, t} \cos (θ_{2}) \end{matrix}

(67)

\begin{matrix} T_{12} = r_{2} F_{24, t} \end{matrix}

(68)

The unknowns listed above can be solved in sequential order, similar to the static analysis. The shaking forces are the summation of all forces acting on the base link 1:

\begin{matrix} F_{s, x} = F_{12, x} - F_{13, x} \end{matrix}

(69)

and

\begin{matrix} F_{s, y} = F_{16} - F_{13, y} - F_{12, y} \end{matrix}

(70)

The corresponding forms in terms of the inertia forces are,

\begin{matrix} F_{s, x} = - (m_{D} a_{6} - m_{B} a_{4} \cos (θ_{2}) - m_{c} r_{3} {\ddot{θ}}_{3} \sin (θ_{3}) - m_{c} r_{3} {({\dot{θ}}_{3})}^{2} \cos (θ_{3})) \end{matrix}

(71)

\begin{matrix} F_{s, y} = - (- m_{B} a_{4} \sin (θ_{2}) + m_{c} r_{3} {\ddot{θ}}_{3} \cos (θ_{3}) - m_{c} r_{3} {({\dot{θ}}_{3})}^{2} \sin (θ_{3})) \end{matrix}

(72)

The parameters of the mechanism were scaled from a Cincinnati Metal Shaper, which cuts the workpiece in a linear toolpath. The scaled lengths of the links are listed in Table 1. All links were assumed the same density, and the sliders 4 and 6 possessed the same amount of mass as link 2. The vertical distance of the pivot

O_{2}

to the path of the slider 6 is 12 in.

Table 1.

Scaled parameters used in simulating the Whitworth quick return mechanism.

$r_{O_{2} O_{3}}$ in	$r_{O_{2} B_{2}}$ in	$r_{O_{3} C}$ in	$r_{C D}$ in	${\dot{θ}}_{2}$ rad/s	$ρ$ lbm/in
12	8	24	14	20	10

Results and discussion

Model validation

The positions, velocities, and accelerations of slider 6 are presented in Figure 5a-c. It can be seen that the analytical values ( $r_{D x}$ , $v_{D x}$ and $a_{D x}$ ) are consistent with the numerical results, as well the Inventor simulations ( $r_{D i n t}$ , $v_{D i n t}$ and $a_{D i n t}$ ) with deviation less than 1%. As slider 6 moves in the negative x-direction, the $r_{D x}$ has a negative value in the working stroke. The slope (Figure 5(a)) becomes much steep after 0.2 s, responsible for a quick return action. The maximum value of $r_{D x}$ is 3.4044 in at $0.2777$ s, whereas the minimum is $- 28.5956$ in at $0.1936$ s. Figure 5(b) shows the velocity of slider 6 moves towards the negative x-direction before returns at 0.1936 s; subsequently, the slider retracted to have a positive velocity. The total amount of time in retraction is 0.0841 s, and the extension time is 0.3983 s. A time ratio of 2.7353 was found using the time spent in the extension stroke, divided by the retraction stroke. This value agrees with the formula $T_{R}$ of 2.735. Figure 6(d) shows a constant acceleration up to 2 s, corresponding to a constant speed in the working stroke. A sudden increase in the acceleration was experienced with the peak value above $10^{4} \frac{f t}{s e c^{2}}$ , at which point the slider changes the direction around the left limit position.

Figure 5.

Positions (a), velocities (b), and accelerations (c) of the slider 6; reactions in x-direction (d), y-direction (e) and input toque on (f) the crank link 2. The subscript ‘int’ denotes the results from Inventor, whereas ‘sol’ corresponds to the distributed mass model.

Figure 6.

Magnitude (a) and direction (b) of the vector $R_{4}$ in one period; the path of link 3 endpoint of in rectangular (c) and polar (c-d) coordinate systems.

Figure 5(d)-f compares distributed mass model with Inventor results through the reactions in the x-direction (d), y-direction (e), and input toque on the crank link 2. It can be seen from Figure 5(d), the reaction of link 2 in the x-direction has the same trend as the slider acceleration in Figure 5(c), indicating the reaction ( $F_{12, x}$ ) at link 2 is the major driving force for the acceleration of slider 6. As the slider motion is restricted in the horizontal direction, $F_{12, y}$ in Figure 5(e) is one magnitude smaller ( $10^{4}$ lbf) as compared to $F_{12, x}$ ( $10^{5}$ lbf). $F_{12, y}$ seems symmetric about 0.235 s due to the fact that, unlike the x-direction, there is no asymmetry of motion in the y directions; therefore, a similar magnitude of peak forces occurs in both working and return stroke. Figure 5(f) shows positive torques in the working stroke, leading to the acceleration of slider 6 and negative values in the return stroke. The maximum values are very close in both strokes, around $1 \times 10^{5}$ lbf⋅ft.

Kinematics

Some of the link positions were plotted in Figure 6. It can be seen from Figure 6(a), the maximum length of link 4 is 20 in, which occurs when slider 4 is right above the pivot $O_{2}$ , $r_{4, \max} = r_{1} + r_{2}$ . When the slider is below $O_{2}$ , a minimum value of 4 in is attained, $r_{4, \min} = r_{1} - r_{2}$ . The angular limit position of link 4 can be estimated graphically (not shown); the minimum angle is reached when link 2 is perpendicular to link 4 in the fourth quadrant giving a value $θ_{4, \min} = \frac{π}{2} - \arcsin (\frac{r_{2}}{r_{1}}) = 0.8411$ , close to 0.8410 as shown in Figure 6(b). The maximum angle of $θ_{4}$ occurs when link 2 is perpendicular to link 4 in the first quadrant, giving $θ_{4, \max} = \frac{π}{2} + \arcsin (\frac{r_{2}}{r_{1}}) = 2.3005$ as compared to 2.3000 from Mathcad. Rocker link 3 is oscillating around the pivot $O_{3}$ , with the tracepoint C moving in a circular path, as shown in Figure 6(c). The path in a diameter of 24 in can be visualized in a polar coordinate system in Figure 6(d).

Statics

When an external load of 5 lbf is applied on slider 6, the x-component of $F_{56}$ shown in Figure 7(a), balances the load, keeping the slider at static equilibrium. The polar diagram in Figure 7(b) indicates the reaction $F_{56}$ is in the second quadrant with an angle around 150–180 degrees. The maximum torque is 20 lbf·ft clockwise as shown in Figure 7(c), and Figure 7(d) indicates the maximum reaction forces $F_{12}$ has a value of 30 lbf. The statics analysis serves a function for validating the lumped mass model. When the same external load was applied to the lumped mass model, the same results were confirmed by setting zero accelerations. The calculations were provided in the Mathcad scripts.

Figure 7.

Reaction at joint D in the x-direction (a) and in a polar coordinate (b); the torque needed to balance the mechanism (c) and the reaction in at $O_{2}$ (d).

Kinetics – lumped versus distributed mass model

Pin forces and the shaking force were derived from the distributed and lumped mass models. It can be seen from Figure 8(a), the pin force from the lumped mass model tends to underestimate the reaction at point C except for the low angles around 0 degrees. Although both models predict maximum pin forces on the same magnitude of order, the maximum reaction derived from the lumped mass model P35 is about $1.9 \times 10^{4}$ lbf, 30% lower than that from distributed mass model F35 around $2.7 \times 10^{4}$ lbf. It appears in Figure 8(b) that the maximum reaction at joint D occurs at the same angle, where the lumped model overestimates the maximum reaction at $1.9 \times 10^{4}$ lbf, almost double of the distributed model, $1 \times 10^{4}$ lbf. Pin reactions at pin $O_{2}$ , $O_{3}$ , and $B_{2}$ show similar patterns in Figure 8(c)-e, where the lumped mass model tends to overestimate the reactions. The maximum values of reactions at pin $O_{2}$ , $O_{3}$ , and $B_{2}$ in the lumped mass model are $1.91 \times 10^{5}$ , $1.57 \times 10^{5}$ and $1.91 \times 10^{5}$ lbf, as compared to $1.72 \times 10^{5}$ , $1.38 \times 10^{5}$ and $1.72 \times 10^{5}$ lbf in the distributed mass model, giving relative errors of 11, 14 and 11%. The deviations are due to the fact that half of the masses of the individual links 2 and 3 were allocated to the joints, leading to a change of the mass moment of inertia. The results show that the lumped mass model deviated from the distributed mass model in predicting the maximum pin reactions by a factor varying from 0.7 to 2. Hence, a safety factor of 2 may be considered to achieve a safe design on the links. In contrast, the overall shaking forces in both models matched very well, as shown in Figure 8(f), indicating the lumped mass model excels in estimating the shaking forces.

Figure 8.

Reactions at joints C (a), D (b), $O_{2}$ (c), $O_{3}$ (d), and $B_{2}$ (e). The notation P corresponds to the force derived from the lumped mass model, and F the distributed mass model. The shaking forces from the lumped mass model were compared against the distributed mass model (f).

Despite the identical shaking forces, Figure 9a shows the shaking moment in the lumped mass model is slightly higher than the distributed model. The deviation is expected as the lumped model relaxed the constraint on the polar mass moment of inertia. Intuitively, half mass at a double distance will increase the polar mass moment of inertia as $(\frac{1}{2} m) (2 r)^{2} > m {(\frac{r}{2})}^{2}$ . At the same angles, the lumped mass model predicts a higher peak shaking moment of $3.09 \times 10^{4}$ lbf⋅ft, 33% higher than the distributed mass model ( $2.31 \times 10^{4}$ lbf⋅ft). The maximum torque at $O_{2}$ is very close as well, shown in Figure 9(b), demonstrating the lumped mass model is not only acceptable in predicting the shaking moments but also effective in solving reactions and torques on input links.

Figure 9.

Shaking moments (a) and torques (b) at pivot $O_{2}$ . The letter ‘d’ denotes the distributed model, and ‘l’ the lumped.

Conclusion and future works

A Whitworth quick return mechanism was modelled using a complex number method to analyze the kinematics. The analytical expressions derived from loop closure equations were exactly identical to the numerical results. Based on kinematic quantities, static and dynamic analyses were performed. The distributed mass model was directly validated by Inventor, whereas the lumped mass model was indirectly validated by reducing it to a static model. The static analysis assumed links 4 and 5 as two force members found a maximum torque of 20 lbf·ft is needed to withstand the 5 lbf external load. Pin reactions, shaking moments, and forces were compared in lumped and distributed mass models. The maximum deviations of the pin reactions occur at joint D, where the lumped model overestimates the maximum reaction at $1.9 \times 10^{4}$ lbf, almost double of the distributed model, $1 \times 10^{4}$ lbf, yielding a 100% relative error. The pin reactions have relative errors range from 11% to 33%. The shaking forces were found identical, and a 30% deviation in shaking moments was confirmed. The lumped mass model has advantages in that it circumvented the need to solve a linear equation system yet excelled in predicting the shaking forces. Although the model is less accurate in finding pin forces and torques of individual links, a safety factor of 2 can be recommended to achieve a conservative pin force estimation.

At Penn State, Mathcad and Inventor modelling are taught in a first-year engineering design class (EDSGN 100: Cornerstone Engineering Design), and they are proven with the ease of usability for engineering students to analyze any linkage mechanisms. Our future works will take aim at evaluating the course outcomes, i.e., after completing these projects, students should be able to: (1) analyze the kinematics of planar linkage mechanisms, (2) use Mathcad to solve complex machine dynamics problems or systems, and (3) model/simulate planar linkage mechanisms in Inventor. We also plan to assess student perceptions of the project-based learning (PBL), specifically on team-based PBL, to determine whether the proposed projects would promote active learning and help students develop engineering soft skills.

Footnotes

Acknowledgements

This work is supported by Miami University New Faculty Research Fund.

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) received no financial support for the research, authorship, and/or publication of this article.

ORCID iD

Jiawei Gong

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