7. FORCE ANALYSIS. Fundamentals F C

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1 ME 352 ORE NLYSIS 7. ORE NLYSIS his chapter discusses some of the methodologies used to perform force analysis on mechanisms. he chapter begins with a review of some fundamentals of force analysis using vectors. hen a review of graphical and analytical methods of force analysis on stationary mechanisms, known as static force analysis, is provided. inally, force analysis of mechanisms in motion, known as dynamic force analysis, will be discussed. undamentals orce Vector force that acts on a point of a link carries the index of the point. or example P. P P Moment bout Point In planar systems, the moment of a force about an arbitrary point is a moment vector along an axis perpendicular to the plane (zaxis. or example, the moment of the force about O is a moment in the positive z-direction (W with a magnitude M O = h where h is the distance from O to the axis of the force, also called the moment arm. In the second example, the moment of about O is a moment in the negative z-direction (W with a magnitude M O = h h O O h M o O M o O he direction of a moment can be determined using our right-hand the thumb would indicate the direction of the moment when the other four fingers are curled about the point in the direction of the force. he moment of a force about a point can also be determined using the vector-product operation. or example, the moment of the force about O is determined as M O = R O short review of the vector-product operation is provided at the O end of this introductory section. orce ouples and orques wo parallel forces, equal in magnitude and opposite in direction, acting on two different points of a link form a couple. he moment of a couple, called a torque, is a vector h in the z-direction and its magnitude is = h where h is the distance between the two axes and = = is the magnitude of either force. he positive or negative direction of the torque can be determined based on the right-hand method. R O P.E. Nikravesh 7-1

2 ME 352 ORE NLYSIS ommon orces and orques orces and torques (moments that act on a link can be the result of gravity, springs, dampers, actuators, friction, etc. hese forces and torques can also be the result of reaction forces or reaction torques from other links. hese forces and torques can be categorized as applied, reaction, and friction. pplied forces and torques hese are either known constants (gravity for example, or functions of positions (springs, or functions of positions and velocities (dampers. Reaction forces and torques hese are functions of applied forces/torques in static problems, and functions of the applied forces/torques and accelerations in dynamics problems. riction forces and torques hese may appear in machines as viscous (wet or oulomb (dry. Viscous friction depends on velocities; therefore it can be categorized as an applied force/torque. oulomb friction depends on reaction forces and possibly velocities; therefore it can be categorized as a reaction force/torque. pplied orces and orques Gravity he weight of a body is applied as a force in the direction of gravity at the mass center. G m g Point-to-point spring he formula to determine the force of a linear spring is = k(l L 0 where k is the stiffness, L is the deformed length, and L 0 is the undeformed length of the spring. he deformed length, L, must be computed based on the instantaneous positions (coordinates of the two attachment points. If the computed force is negative, the spring is in compression. If the force is positive, the spring is in tension the pair of forces must be applied to the two links accordingly as shown. Point-to-point damper he formula to compute the force of a linear damper is = c L where c is the damping coefficient, and L is the time rate of change in the damper s length. L must be computed based on the velocities of the attachment points. s shown in the diagram, the relative velocity V = V V is first determined and then projected along the axis of the damper to obtain V. he magnitude of this vector is the magnitude of L. If V and R are in the same direction, L is positive (the L < 0 > 0 V V R V V V V P.E. Nikravesh 7-2

3 ME 352 ORE NLYSIS damper is increasing its length, otherwise L must be given a negative sign (the damper is shortening. he computed damper force is applied as a pair of forces to the attachment points, in the opposite directions, depending on the sign of as shown. < 0 > 0 Rotational spring (or damper rotational (also called torsional spring is attached between two links about the axis of a pin joint. wo axes originating from the center of the pin joint, one on each link, are defined and the angle between them is measured. he torque for a rotational spring is computed as = k(θ θ 0 where k is the stiffness, θ is the deformed angle, and θ 0 is the undeformed angle of the spring. M (i (i θ (j M (j pair of torques, one on each link, is applied in opposite directions. he formula to compute the torque of a rotational damper is = cθ where c is the damping coefficient, and θ is the time rate of change in the damper s angle. θ can be computed based on the angular velocities of the two bodies as θ = ω j ω i. Whether θ is positive (increasing angle or negative (decreasing angle, the pair of torques that are applied to the two bodies must oppose the motion. Reaction orces orques wo links connected by a kinematic joint apply reaction forces (and/or torques on one another. Pin joint wo links connected by a pin joint apply reaction forces on each other. he reaction forces are equal in magnitude and opposite in direction. he magnitude and (j directions that are shown on the free-body-diagrams are (i arbitrary they must be determined through an analysis. he reaction force on each link can be represented in term of its x and y components, such as ji( x and ji( y. or notational simplicity, we will use a single index to show each component; for example, 1 and 2. If the assigned direction to a component is determined to be correct through an analysis, the solution for that component will come out with a positive sign. Otherwise the solution will end up with a negative sign indicating that the assumed direction for the component must be reversed. Sliding joint wo links connected by a sliding joint apply reaction forces and torques on each other. he reaction forces, ij and ji, are equal in magnitude, opposite in direction, and act perpendicular to the axis of the joint. lthough these are distributed forces that act along the surfaces of contact, (i ji ij (i 1 2 (i 2 (j 1 (j (j P.E. Nikravesh 7-3

4 ME 352 ORE NLYSIS in our analyses we place them as concentrated forces at the center of the block. he reaction torques, ij and ji, are (i equal in magnitude and opposite in direction. In our free-body diagrams, each reaction force or torque is denoted with a single index for convenience. or example, 1 and 2. he correct direction for these components will be determined through an analysis. Pin-sliding joint wo links connected by a pin-sliding joint apply reaction forces on each other. On the free-body diagrams of the links, the reaction forces are shown equal in magnitude, opposite in direction, acting perpendicular to the axis of the joint. 2 (i 1 (j 1 2 (j 1 1 (j (i Scalar and Vector Products In static and dynamic analysis, we often encounter vector operations such as scalar product or vector product. In this section a short review on how to evaluate such products is presented. Scalar (dot product he scalar product of two vectors, such as and V, can be determined either analytically or graphically. nalytical: he analytical scalar product can be computed in two ways depending on how the vectors are defined: (a he magnitudes of the vectors are known as and V, and the angle between the two vectors is known as θ. he scalar product is computed as θ V = V cosθ V he angle can be measured from to V or vice versa. y (b he vectors are known in component form: = x, V = V y x V V y y V y he scalar product is computed as x V = x V x + y V x V x y Graphical: In order to determine the scalar product of two vectors graphically, one of the vectors must be projected onto the axis of the other vector. or example, is projected onto the axis of V to obtain. he magnitudes of and V are measured and denoted as and V. hen the scalar product is computed as: V = V If and V are in the same direction, the product is V considered positive, otherwise the product must be considered negative. P.E. Nikravesh 7-4

5 ME 352 ORE NLYSIS Vector (cross product he vector product of two vectors can be determined either analytically or graphically. nalytical: he vector product can be computed in two ways depending on how the vectors are defined: (a he magnitudes vectors R and are R and, and he angle between the two vectors is θ. he magnitude of the vector product is computed as: R = R sinθ he angle must be measured from R to, W. (b he vectors are known in component form: R = R x, = x θ R y R y he vector product is computed as: R = R = R x y R y x Graphical: In order to determine the vector product of vectors R and graphically, vector is projected onto an axis perpendicular to R to obtain. We measure the magnitudes of R and ; i.e., R and : (a or 0 θ 180 : R = R (b or 180 θ 360 : R = R 270 o θ R R θ 90 o R R Whether the vector product operation is carried out analytically or graphically, the resultant vector is perpendicular to the plane of the original two vectors. In planar problems, the resultant vector will be along the z-axis. Its direction depends on the angle between the two original vectors, measured from the first vector W towards the second vector. Example wo vectors are given as and with the magnitudes = 2.0 and = 3.0, and angles θ = 30 o and θ = 135 o. In component form these vectors can be described as 2.0 cos 30 = 2.0sin 30 = , = 3.0 cos sin135 = Scalar product he angle between these two vectors is θ = 105 o. he scalar product can be computed as = (2.0(3.0cos105 = 1.55 (a Using the component form of the vectors we have = (1.73( (1.0(2.12 = 1.55 (b We can project onto the axis of and measure the magnitude of the projected vector to be.77. Since this projected vector is in the opposite direction of, we have = (0.77(2 = 1.55 (c Vector product We compute the magnitude of the cross product of the two vectors as = (2.0(3.0sin105 = 5.79 (d Since the answer is positive, the resultant vector is in the positive z direction (coming out of the plane. We can also use the components to get = (1.73( (1.0( 2.12 = 5.79 (e Projecting onto an axis perpendicular to and measuring the magnitude of the projected vector provides = 2.89, which yields = (2.89(2 = 5.79 P.E. Nikravesh 7-5

6 ME 352 ORE NLYSIS Static orce nalysis Static Equilibrium Equations body is said to be in static equilibrium if under a set of applied forces and torques its translational (linear and rotational accelerations are zeros (a body could be stationary or in motion with a constant linear velocity. Planar static equilibrium equations for a single body that is acted upon by forces and torques are expressed as i( x i (se.1 (orce equation i( y j + M i (se.2 (Moment equation Equation (se.1 represents the sum of all the forces acting on the link, and Eq. (se.2 represents the sum of all the torques, j, and the moments, M i, caused by all the forces acting on the body with respect to any reference point. Example ssume three forces and a torque act on a body. wo of the forces, and, are known as shown. Determine and the applied torques in order for the body to stay in static equilibrium. onsider = 1.5, θ = 325 o, = 1.9, and θ = 40 o. Sum of forces Equation (se.1 can be solved either graphically or analytically. Graphical Equation (se.1 states that the sum of forces that act on the body must form a closed loop; i.e., + +. We construct this vector sum graphically in order to determine. Direct measurement from the figure yields = 2.7 and θ = 187 o. nalytical Equation (se.1 is projected onto the x- and y-axes to obtain two algebraic equations: n i cosθ i cosθ + cosθ + cosθ i=1 n i sinθ i sinθ + sinθ + sinθ i=1 o solve for, we write the equations as cosθ = cosθ cosθ sinθ = sinθ sinθ We square both sides and add the equations to get ( 1/2 = ( cosθ cosθ 2 + ( sinθ sinθ 2 hen sine and cosine of θ can be computed as P.E. Nikravesh 7-6

7 ME 352 ORE NLYSIS cosθ = ( cosθ + cosθ /, sinθ = ( sinθ + sinθ / hese expressions provide the value of θ in its correct quadrant. Numerically, the two known vectors are expressed in component form as = 1.23, 0.86 = hen vector can be computed as = = = he magnitude of this vector is = ( ( = 2.7. he vector can also be expressed as 2.68 / 2.7 = 2.7 = / or the angle of this vector we have cosθ = 0.99 and sinθ = hese values yield θ = 187 o. Note: If you use Matlab to compute the angle, take advantage of the function atan2. In this problem the function can be used as theta = atan2(-0.13, he function provides the angle in its correct quadrant. Sum of moments or equation (se.2, the sum of the torque and the moments of all three forces about any arbitrary point must be equal to zero. We take the sum of moments about ; i.e., + R + R Note that the moment arm for is a zero vector. his equation is re-written as = R R = R R R R We can also write the moment equation with respect to point, or any other point in the same way. or example, if we pick an arbitrary point, such as O, the moment equation becomes: R + R O + R O + R O O R his equation yields the unknown torque as: O O R = R O R O R O O ny of the above moment equations can be solved either graphically or analytically. In this example we R O take the moment equation about O. θ Graphical: θ We first need to determine the angle between every R O R vector and its corresponding vector. or the purpose of clarity, each pair of R and vectors is θ shown separately with the corresponding angle. he R O angles are measured (from R to W and the moment equation is evaluated as + R O sinθ + R O sinθ + R O sinθ rom the figure we measure the following magnitudes: R O = 1.1, R O = 1.8 and P.E. Nikravesh 7-7

8 ME 352 ORE NLYSIS R O = 1.5. he angles between each force vector and its corresponding position vector are measured to be: θ = 128 o, θ = 58 o and θ = 143 o. With these values the applied torque is determined to be = ((1.1(1.5sin128 + (1.8(1.9sin 58 + (1.5(2.7sin143= nalytical: n x-y frame is positioned at O. he x and y components of all the vectors are determined and denoted as x x x y =, =, y = y y R Ox R Ox R Ox R R O =, R R O =, R Oy R O = O O Oy R R Oy O he moment equation for this body is written as R Oy x + R Ox y R Oy x + R Ox y R Oy x + R Ox y he angle of position vectors are measured to be θ RO = 208 o, θ RO = 344 o and θ RO = 43 o. herefore the position vectors can be expressed in component form as R O = 0.97, R 0.52 O = 1.73, R 0.50 O = he applied torque can now be computed as = ( 0.52(1.23 ( 0.97( ( 0.50(1.46 (1.73(1.22 R O + (1.02( 2.68 (1.10( 0.36 = 6.66 Special ases If only two or three forces act on a body (and nothing else, and the body is in static equilibrium, the forces are balanced in such a way that they exhibit certain characteristics. We should take advantage of such cases when solving for unknown forces. wo-force member If only two forces act on a body that is in static equilibrium, the two forces are along the axis of the link, equal in magnitude, and opposite in direction. x hree-force member If only three forces act on a body that is in static equilibrium, their axes intersect at a single point. his knowledge can help us simplify the solution process in some problems. or example, if the axes of two of the forces are known, the intersection of those two axes can assist us in determining the axis of the third force. special case of the three-force member is when three forces meet at a pin joint that is connected between three links. When the system is in static equilibrium, the sum of the three forces must be equal to zero P.E. Nikravesh 7-8

9 ME 352 ORE NLYSIS Static Equilibrium nalysis of Mechanisms one degree-of-freedom planar mechanism may contain N moving links. If the mechanism is in static equilibrium, then 3 equilibrium equations can be written for each link, which results into 3N equations. In these equations, there are normally 3N - 1 unknown reaction force/torque components and 1 unknown applied force or torque. In a static force analysis, the main objective is to solve the equilibrium equations for the unknown applied force/torque. Depending on the method of solution, the reaction forces/torques may also be found in the process. In these notes, two methods for solving the static equilibrium equations are presented: the free-body-diagram (D method, and the power-formula (P method. ree-ody Diagram Method In this method the static equilibrium equations are constructed based on the free-body diagrams (D of each link. In each D, all of the reaction and applied forces/torques that act on that link are considered. or a mechanism of N moving links, the 3N equilibrium equations form a set of linear algebraic equations in 3N unknowns. lthough the equations are linear, due to their large number, a solution by hand may not be practical or may not be the best choice. In such cases, a numerical solution through the use of a computer program is recommended. However, if certain simplifications are made and the number of equations is reduced, then solving the reduced set of equations by hand, either analytically or graphically, may be practical. Example 1 his slider-crank mechanism is in static equilibrium in the shown configuration. known force acts on the slider block in the direction shown. n unknown torque acts on the crank. Our objective is to determine the magnitude and the direction of this torque in order to keep the system in static equilibrium. We construct the free body diagrams for each link. he reaction forces at the pin joints are unknown. Each reaction force is described in term of its x and y components, where the direction of each component is assigned arbitrarily. or notational simplification, simple numbered indices are used for all the components. or each link we construct three equilibrium equations. Link 2: Link 3: Link 4: a b c d a 3 + b 4 + (Sum of moments about c 5 + d 6 (Sum of moments about (Sum of moments about he moment arms are measured directly from the figure. Note that 7 and 8 are the reaction force and torque due to the sliding joint. P.E. Nikravesh 7-9

10 ME 352 ORE NLYSIS hese 9 equations can be put into matrix form. he unknowns and their coefficients are kept on the left-hand side and the only known quantity, the known 0 0 a b applied force is moved to the righthand side his set of 9 equations in = 0 unknowns can be solved by any preferred c d numerical method. If the arbitrarily assigned direction to a force component or a torque is not correct, the obtained 8 0 solution will be a negative quantity Numerical values for the link lengths are L 2 = and L 3 = 4.0. rom the figures we extract the following 10 2 measurements: a = 1.8, b = 1.0, c = 2.0, d = ssume the applied force is given to be = 10 units in 3 10 the negative direction. hese values are substituted in the 4 equilibrium equation. he solution to the unknowns is 5.56 determined to be as shown in array form. he applied 5 = 10 torque on the crank is = units. s a byproduct of this process, all the reaction forces/torques are also found Simplified D method 7 he connecting rod of this mechanism is a two-force 8 0 member. he reaction forces at and must be equal but in opposite directions. hese reaction forces are named 3 and given arbitrary directions. he D of links 2 and 4 can now be constructed. 3 or link 4 we write the sum of forces along the x-axis as 3( x +. Since the applied force is known, 3( x can be computed. hen based on the angle of 3, the 3 magnitude of 3 can be determined. or link 2 we write the moment equation about as 3 e 3 +. his equation yields the unknown applied 4 M 5 torque. Using numerical values for this example, we have 3 3( x = = makes a 27 o angle with the x-axis, e therefore we have 3 = 3( x / cos27 = Direct 1 measurement from the figure provides a value for the moment arm as e = 2.0, which yields = e 3 = Example 2 his example is the same slider-crank mechanism from Example 1, where we are asked to include gravitational forces on the links. he mass center for each link is positioned at the geometric center. he free body diagrams for the three links are constructed. Link 2: Link 3: Link 4: P.E. Nikravesh 7-10

11 ME 352 ORE NLYSIS a 4 b G 2 1 W c d G 3 6 W G 4 W W 2 a 3 + b 4 b 2 W W 3 c 5 + d 6 + d 2 W W 4 8 (Sum of moments about (Sum of moments about (Sum of moments about hese 9 equations are expressed in matrix form and solved for the unknowns. We note that the lefthand side of this matrix equation is 0 0 a b W 2 3 bw identical to that in Example 1. he 2 /2 difference is in the right-hand sides = W c d dw 3 / W Simplified D method! Since three forces act on the connecting rod, this link is no X longer a two-force member. We cannot assume that the G reaction forces at and are equal, in opposite directions, and 3 along the axis of the link. herefore, the simplified method is W not applicable to this problem. his is the case for most 3 examples when we include gravitational forces. X oulomb riction oulomb friction can be included between two contacting surfaces in a static force analysis. Given the static coefficient of friction, μ (s, the friction force can be described as the product of the coefficient of friction and the reaction force normal to the contacting surfaces. he friction force must act in the opposite direction of the tendency of any motion. Since the assumption is that the system is stationary, the tendency of motion must be considered for two cases. he process provides a range of values for the applied load while the system remains in equilibrium. Example 3 We repeat Example 1 with the assumption that dry friction exists between the slider block and the ground. We solve this problem with the D method. We may write the complete set of equilibrium equations or take advantage of a 2-force member in the system. Here we construct the complete set of equations. Since the D for the crank and the connecting rod are the same as before, we only show the D for the slider block. irst we assume that the block is about to move to the left. herefore, the friction force must be directed to the right. hen we assume that the slider block is about to move to the right. P.E. Nikravesh 7-11

12 ME 352 ORE NLYSIS herefore the friction force must be directed to the left: μ (s (he block tends to move to the left 8 + eμ (s 7 e 5 6 μ (s μ (s eμ (s 7 (he block tends to move to the right e 5 6 μ (s We construct two complete sets of equations for these two cases. he two sets are presented together as shown. We solve each set of equations twice: once with the positive sign in front of μ (s and once with the negative sign. Each solution yields a value for the unknown torque, left and right. s long as the applied torque stays in the range of left to right, the system remains in static equilibrium a b = c d ±μ (s ±eμ (s ssuming μ (s.25, e.5, and using the numerical values from Example 1, we solve the system of 9 equations in 9 unknowns twice. or the plus sign in front of the terms containing μ (s, the solution yields = 20.68, where the minus sign in front of those two terms yield = herefore, as long as the applied torque is in the range , the mechanism remains in static equilibrium. Example 4 In this example, a known force acts at point P an P unknown torque acts on the crank. It is assumed that y dry friction exists at the sliding joint. o solve this problem, we construct D s for P the links. lthough it is not necessary, in order to x simplify the process of projecting the reaction forces unto the x-y axes, the x-y axes are rotated in such a way that the x-axis is along the axis of link (4. O 4 Link 2: Link 3: Link 4: 3 4 P( x 6 b 6 μ(s a P( y μ (s 5 d c 8 7 O 4 P.E. Nikravesh 7-12

13 ME 352 ORE NLYSIS a 3 b 4 3 ± μ (s μ (s 5 + P( x P( y 6 c 5 + d P hese equations can be expressed in matrix form and solved twice to find the range of values for the applied torque. 0 0 a b ±μ (s μ (s P( x P( y c d P Power ormula Method he power formula method constructs one equation in one unknown for a one degree-offreedom mechanism regardless of its number of links and joints. Only applied forces and torques, including the unknown applied force or toque, appear in the equation. his means that the reaction forces and torques have been eliminated from the power formula. In order to use the power formula, although the system is in static equilibrium, we first assume that the system is in motion. We assign an arbitrary velocity (angular or linear to one of the links; e.g., the rotation of the crank, and determine the velocities that are needed in the power formula. hese are the velocities of the points were the applied forces act on, and the angular velocity of the links that a torques are applied to. he equation for the power formula is: (6 P iv P + ω (spf.1 (5 In this equation P is a typical applied force acting on a link 5 at point P, V P is the velocity (absolute of point P, is a typical applied torque acting on a link, and ω is the angular velocity of that link. If the torque and the angular velocity are in the same direction, the product ω is positive, otherwise the product is negative. s an example consider the six-bar mechanism shown. 2 (3 hree applied forces and three applied torques act on the (2 system. We need to find the imaginary velocities of points,, and, and the imaginary angular velocities of links 2, 4, and 5. We can find these velocities by assigning an (4 arbitrary value to the velocity of one of the degrees-offreedom, say to the angular velocity of link 2. We may apply 4 any preferred method to solve for the other velocities. he value of this arbitrarily assigned velocity or its direction will not change the result obtained from the power formula. Note: his method is also referred to as the energy method or the virtual work method. Since we have not discussed the concepts of virtual work and virtual displacements, the method is presented using velocities although the system is in static equilibrium. Note: Since reaction forces and torques do not appear in the power formula, this method is not applicable when oulomb friction is present. P.E. Nikravesh 7-13

14 ME 352 ORE NLYSIS Example 5 We apply the power formula to the slider-crank mechanism of Example 1. One external torque and one external force act on this system. We need to determine the imaginary angular velocity of link 2 and the velocity of the slider block. We assume link 2 rotates with an angular velocity ω 2 = 1 rad/sec, W. We construct the velocity polygon and determine the velocity of the slider block (point. onsidering the length of the crank as L 2 = 2.0, the polygon yields V = 2.4. (2 (3 V V V or this problem the power formula becomes iv + ω 2. Since and V are along the same axis but in opposite directions, the formula simplifies to V + ω 2, or = V / ω 2 = (10(2.4 / 1.0 = Example 6 We apply the power formula to the problem of Example 2. We assume an imaginary ω 2 = 1 rad/sec, W, and construct the velocity (as in the previous example. he velocities of the points where the applied forces act, including the gravitational forces, are determined. he P is then expressed as: iv + W 2 iv G2 + W 3 iv G3 + W 4 iv G4 + ω 2 he dot product between a velocity vector and a force vector can be obtained either analytically or graphically (the angle between the two vectors can be measured directly from the figure. he P yields the value for the unknown applied torque. G G 3 2 W 2 W 3 V G3 V G2 W 4 V O V =V G4 (4 gravity G 4 V G2 V G3 V G4 V W 2 W 3 W 4 Dynamic orce nalysis Dynamic Equilibrium Equations he dynamic equilibrium equations for a single body in planar motion are the revised version of the static equilibrium equations, expressed as m G i( x = m G( x I i = m G (de.1 G α G i( y = m G( y j + M i(g = I G α (de.2 he force equilibrium of Eq. (de.1 states that the sum of forces that act on a body must be equal to the mass of the body, m, times the acceleration of the mass center, G. his equation is P.E. Nikravesh 7-14

15 ME 352 ORE NLYSIS derived from Newton s second Law of motion. he moment equilibrium of Eq. (de.2 states that the sum of all the applied toques and the moments of all forces that act on a body with respect to its mass center must be equal to the body s moment of inertia, I G, times the angular acceleration of the body, α. he moment of inertia is defined with respect to an axis passing through the mass center and perpendicular to the plane. Equations (de.1 and (de.2 are also known as the translational and rotational equations of motion respectively. D lembert s Principle he D lembert s principle considers a rearranged form of the dynamic equilibrium equations as i m G (de.3 m G j + M i(g I G α (de.4 I G α G lthough this rearrangement may appear trivial, it has profound implications. One interpretation of this principle is that if m G and I G α are viewed as a force and a torque respectively, then the dynamic equilibrium equations become identical to the static equilibrium equations. he terms m G and I G α are called inertial force and inertial torque. Revised Rotational Equation D lembert s principle provides us a simple set of steps to revise Eq. (de.2 if the sum of moments is taken about a point that is not the mass center. If we consider m G and I G α as additional force and torque that act on the body, and consider point O as the reference point instead of the mass center, the rotational static equilibrium of Eq. (se.2 yields j + M i(o I G α mr GO G or, I G α R j + O M i(o = I G α + mr GO G (de.5 R R GO O mg his equation clearly shows that if the origin is not the O R mass center, then we must consider the moment caused by O the inertia force in the moment equation. Dynamic Equilibrium nalysis of Mechanisms he process of dynamic force analysis of mechanisms is only slightly different from that of the static force analysis, especially if we look at the dynamic equilibrium equations from the perspective of D lembert s principle. he main difference is that in a dynamic force analysis we need to include the linear and angular accelerations of each link in the process of solving the dynamic equilibrium equations. or the dynamic force analysis, as in the static force analysis, we consider the free-body diagram (D method and the power formula (P method. ree-ody Diagram Method he dynamic force analysis requires incorporating the applied and reaction forces and torques, and the inertial force and torque, in the D of each link. Inclusion of the applied and reaction forces and torques in this process is identical to that of the static force analysis. In order to include the inertial forces and torques in the process, we need to know the linear and angular accelerations for each link. P.E. Nikravesh 7-15

16 ME 352 ORE NLYSIS Example 7 he mass and moment of inertia for the links of this slider-crank are given. known force acts on the slider block, and an unknown torque acts on G G 3 2 the crank. In the depicted configuration, the angular velocity and acceleration of the crank are given. he objective is to find the magnitude and the direction of the unknown torque. ased on the given angular velocity and V acceleration of the crank, polygons are constructed V G2 G3 and the angular velocity and acceleration of all the links are found (magnitudes and directions ω 2, α V G4 O V 2 and α 3 are W, and ω 3 is W. hen, in a process similar to that of the static equilibrium (refer G3 to Example 1, the D for each link is constructed. Link 2: Link 3: Link 4: d α 2 4 a b G 2 G2 b a G3 c c G 3 α 3 d 6 5 G4 G4 G2 7 G 4 O 5 G = m 2 G2 ( x = m 3 G3 ( x 5 + = m 4 G = m 2 G2 ( y = m 3 G3 ( y a 1 b 2 a 3 + b 4 + = I G2 α 2 c 3 + d 4 + c 5 + d 6 = I G3 α 3 8 Note that,,, G2 ( x G 2 ( y G 3 ( x m and G2 ( x G3 are negative, ( y G m 2 2 G2 ( y where α 2 and α 3 are positive. a b a b I hese equations are put into G2 α 2 matrix form. his set of m 3 G3 ( x equations in 9 unknowns can be = m 3 G3 ( y solved by any preferred numerical 0 0 c d c d method I G3 α m 4 G oulomb riction oulomb friction can be included between two contacting surfaces in a dynamic force analysis. Given the dynamic (kinetic coefficient of friction, μ (k, the friction force can be described as the product of the coefficient of friction and the reaction force normal to the contacting surfaces. he friction force must act in the opposite direction of the motion. he process is illustrated through a simple example. P.E. Nikravesh 7-16

17 ME 352 ORE NLYSIS Example 8 We consider the slider-crank mechanism of Example 7 with the assumption that dry friction exists between the slider block and the ground. We construct the complete set of equations for this problem. Since the D for the crank and the connecting rod are the same as in Example 7, we only show the D for the slider block. Since, according to the velocity polygon the block is moving to the left, the friction force must be directed to the right. he equilibrium equations for the block are: μ (k 7 = m 4 G eμ (k 7 G4 e 6 5 μ (k 7 he complete set of equations m can now be presented in G2 ( x matrix form. he solution to m this set of equations yields all 2 2 G2 ( y the unknowns. a b a b I G2 α m 3 G3 ( x = m 3 G3 ( y 0 0 c d c d μ (k 0 0 I G3 α m 4 G eμ (k Power ormula Method he power formula (P for dynamic force analysis is a revised form of the formula from the static force analysis. If we consider the inertial forces and torques as any other applied forces and torques, the revised formula can be expressed as P iv P + ω = m i Gi iv Gi + I Gi α i ω i (dpf.1 In this formula, the velocities and accelerations are actual and not imaginary. If the angular velocity and acceleration are in the same direction, the product αω is positive, otherwise the product is negative. Example 9 We apply the P method to the slider-crank of Example 8. he formula for this system can be expressed as iv + ω 2 = m 2 G2 iv G2 + m 3 G3 iv G3 + m 4 G4 iv G4 + I G2 α 2 ω 2 + I G3 α 3 ω 3 + I G4 α 4 ω 4 In addition to the angular velocities and accelerations, from the velocity and acceleration polygons, the velocity and acceleration of the mass centers are determined. ll the known quantities are substituted in the formula to find the unknown applied torque. 7 8 V V G2 V G3 G4 G2 G3 V G4 P.E. Nikravesh 7-17

18 ME 352 ORE NLYSIS Shaking orce and Shaking orque/moment Dynamic forces that act on a mechanism cause the foundation (the ground to shake. he shaking force is (3 defined as the sum of reaction forces the links of a mechanism apply on the ground link. or a four-bar mechanism, since there are only two links that are (2 (4 directly pinned to the ground at and O 4, the sum of O reaction forces acting on the ground at these two points 2 O 4 is the shaking force denoted as s = ( (2 (4 where 1, 2, 7 and 8 could be positive or negative. 1 7 When link 2 is the input link (the crank, a motor O about the axis of must rotate it. he motor applies a torque on link 2 and an opposite torque on the ground. 2 he torque that is applied by the motor to the ground is 8 called the shaking torque (moment: s = 1 O 4 7 he shaking force tends to move the ground (the frame up and down, and left and right. he shaking torque tends to rock the frame about the vertical axis to the plane. Matlab Program Matlab program (fourbar_force.m for static P 3 and dynamic force analysis of a four-bar mechanism is provided. he four-bar is considered in its most G3 general form as shown. he program allows one (3 known force and one known torque to be applied on each link. he gravitational force may also be P 2 L 3 considered to act in the negative y-direction on the G 2 P 4 G 4 L 4 system. he program requires only one unknown (2 (4 L torque, which could be considered on only one of the 2 L links. 1 or the purpose of generality, the mass center of a O 4 link, G i, is positioned from the reference point of the link by an angle γ i and a length L Gi. he force P i application point, P i, is positioned by an angle β i and L Pi G i LGi a length L Pi. Note that the reference point for link 2 γ i L is, for link 3 is, and for link 4 is O 4. β i i θ i his program retrieves the data from a file named fourbar_force_data.m. he user is required to provide the following data in this file: onstant values for the link lengths ( L 1, L 2, L 3, L 4 ngle of the crank (θ 2 Estimates for the angles of the coupler and the follower (θ 3, θ 4 ngular velocity and acceleration of the crank (ω 2, α 2 onstant values for the position of P points ( L P2, L, L P3 P 4 O 4, β 2, β 3, β 4 P.E. Nikravesh 7-18

19 ME 352 ORE NLYSIS onstant values for the position of G points ( L G2, L, L G3 G 4 O 4, γ 2,γ 3, γ 4 Known applied forces ( P2, P3, P4 Known applied torques ( 2, 3, 4 Masses and moments of inertia ( m 2, m 3, m 4, I 2, I 3, I 4 Gravitational constant (g (the value of g determines the system of units for the analysis he program prompts the user for the following information: Should the gravitational force be included? o nswer y for yes or n for no he unknown torque is applied to which link? o nswer 2, 3 or 4 Static or dynamic force analysis? o nswer s for static or d for dynamic 4 3 (3 6 he program reports the results for the reaction forces at and on link 2, at and O 4 on link 4, and the unknown applied torque. he reaction forces acting at and on link 3 could easily be determined by the user. he program also reports the shaking force and the shaking torque. 1 (2 2 (4 5 7 O 4 8 P.E. Nikravesh 7-19