11.6 Potential-Energy Criterion for Equilibrium System having One Degree of Freedom When the displacement of a frictionless connected system is infinitesimal, from q to q + dq dU = V (q) – V (q + dq) OrdU = -dV If the system undergoes a virtual displacement δq, rather than an actual displacement dq, δU = -δV For equilibrium, principle of work requires δU = 0, provided that the potential function for the system is known, δV = 0

11.6 Potential-Energy Criterion for Equilibrium System having One Degree of Freedom dV/dq = 0 When a frictionless connected system of rigid bodies is in equilibrium, the first variation or change in V is zero Change is determined by taking first derivative of the potential function and setting it to zero Example To determine equilibrium position of spring and block

11.6 Potential-Energy Criterion for Equilibrium System having One Degree of Freedom dV/dy = -W + ky = 0 Hence equilibrium posyion y = y eq y eq = W/k Same results obtained by applying ∑F y = 0 to the forces acting on the FBD of the block

11.6 Potential-Energy Criterion for Equilibrium System having n Degree of Freedom When the system of n connected bodies has n degrees of freedom, total potential energy stored in the system is a function of n independent coordinates q n, V = V (q 1, q 2, …, q n ) In order to apply the equilibrium criterion, δV = 0, determine change in potential energy δV by using chain rule of differential calculus δV = (∂V/∂q 1 )δq 1 + (∂V/∂q 2 )δq 2 + … + (∂V/∂q n )δq n Virtual displacements δq 1, δq 2, …, δq n are independent of one another

11.6 Potential-Energy Criterion for Equilibrium System having n Degree of Freedom Equation is satisfied ∂V/∂q 1 = 0, ∂V/∂q 2 = 0, ∂V/∂q n = 0 It is possible to write n independent equations for a system having n degrees of freedom

11.7 Stability of Equilibrium Once the equilibrium configuration for a body or system of connected bodies are defined, it is sometimes important to investigate the type of equilibrium or the stability of the configuration Example Consider a ball resting on each of the three paths Each situation represent an equilibrium state for the ball

11.7 Stability of Equilibrium When the ball is at A, it is at stable equilibrium Given a small displacement up the hill, it will always return to its original, lowest, position At A, total potential energy is a minimum When the ball is at B, it is in neutral equilibrium A small displacement to either the left or right of B will not alter this condition The balls remains in equilibrium in the displaced position and therefore, potential energy is constant

11.7 Stability of Equilibrium When the ball is at C, it is in unstable equilibrium A small displacement will cause the ball’s potential energy to be decreased, and so it will roll farther away from its original, highest position At C, potential energy of the ball is maximum

11.7 Stability of Equilibrium Types of Equilibrium 1. Stable equilibrium occurs when a small displacement of the system causes the system to return to its original position. Original potential energy is a minimum 2. Neutral equilibrium occurs when a small displacement of the system causes the system to remain in its displaced state. Potential energy remains constant 3. Unstable equilibrium occurs when a small displacement of the system causes the system to move farther from its original position. Original potential energy is a maximum

11.7 Stability of Equilibrium System having One Degree of Freedom For equilibrium, dV/dq = 0 For potential function V = V(q), first derivative (equilibrium position) is represented as the slope dV/dq which is zero when the function is maximum, minimum, or an inflexion point Determine second derivative and evaluate at q = q eq for stability of the system

11.7 Stability of Equilibrium System having One Degree of Freedom If V = V(q) is a minimum, dV/dq = 0 d 2 V/dq 2 > 0 stable equilibrium If V = V(q) is a maximum dV/dq = 0 d 2 V/dq 2 < 0 unstable equilibrium

11.7 Stability of Equilibrium System having One Degree of Freedom If d 2 V/dq 2 = 0, necessary to investigate higher- order derivatives to determine stability Stable equilibrium occur if the order of the lowest remaining non-zero derivative is even and the is positive when evaluated at q = q eq Otherwise, it is unstable For system in neutral equilibrium, dV/dq = d 2 V/dq 2 = d 3 V/dq 3 = 0

11.7 Stability of Equilibrium System having Two Degree of Freedom A criterion for investigating the stability becomes increasingly complex as the number for degrees of freedom for the system increases For a system having 2 degrees of freedom, equilibrium and stability occur at a point (q 1eq, q 2eq ) when δV/δq 1 = δV/δq 2 = 0 [(δ 2 V/δq 1 δq 2 ) 2 – (δ 2 V/δq 1 2 )(δ 2 V/δq 2 2 )] 0 or δ 2 V/δq 2 2 >0

11.7 Stability of Equilibrium System having Two Degree of Freedom Both equilibrium and stability occur when δV/δq 1 = δV/δq 2 = 0 [(δ 2 V/δq 1 δq 2 ) 2 – (δ 2 V/δq 1 2 )(δ 2 V/δq 2 2 )] < 0 δ 2 V/δq 1 2 > 0 or δ 2 V/δq 2 2 >0

11.7 Stability of Equilibrium Procedure for Analysis Potential Function Sketch the system so that it is located at some arbitrary position specified by the independent coordinate q Establish a horizontal datum through a fixed point and express the gravitational potential energy V g in terms of the weight W of each member and its vertical distance y from the datum, V g = Wy Express the elastic energy V e of the system in terms of the sketch or compression, s, of any connecting spring and the spring’s stiffness, V e = ½ ks 2

11.7 Stability of Equilibrium Procedure for Analysis Potential Function Formulate the potential function V = V g + V e and express the position coordinates y and s in terms of the independent coordinate q Equilibrium Position The equilibrium position is determined by taking first derivative of V and setting it to zero, δV = 0

11.7 Stability of Equilibrium Procedure for Analysis Stability Stability at the equilibrium position is determined by evaluating the second or higher-order derivatives of V If the second derivative is greater than zero, the body is stable If the second derivative is less than zero, the body is unstable If the second derivative is equal to zero, the body is neutral

11.7 Stability of Equilibrium Example 11.5 The uniform link has a mass of 10kg. The spring is un-stretched when θ = 0°. Determine the angle θ for equilibrium and investigate the stability at the equilibrium position.

11.7 Stability of Equilibrium Solution Potential Function Datum established at the top of the link when the spring is un-stretched When the link is located at arbitrary position θ, the spring increases its potential energy by stretching and the weight decreases its potential energy

11.7 Stability of Equilibrium Solution

11.7 Stability of Equilibrium Solution Equilibrium Position For first derivative of V, or Equation is satisfied provided

11.7 Stability of Equilibrium Solution Stability For second derivative of V, Substituting values for constants

11.7 Stability of Equilibrium Example 11.6 Determine the mass m of the block required for equilibrium of the uniform 10kg rod when θ = 20°. Investigate the stability at the equilibrium position.

11.7 Stability of Equilibrium Solution Datum established through point A When θ = 0°, assume block to be suspended (y w ) 1 below the datum Hence in position θ, V = V e + V g = 9.81(1.5sinθ/2) – m(9.81)(Δy)

11.7 Stability of Equilibrium Solution Distance Δy = (y w ) 2 - (y w ) 1 may be related to the independent coordinate θ by measuring the difference in cord lengths B’C and BC

11.7 Stability of Equilibrium Solution Equilibrium Position For first derivative of V, For mass,

11.7 Stability of Equilibrium Solution Stability For second derivative of V, For equilibrium position,

11.7 Stability of Equilibrium Example 11.7 The homogenous block having a mass m rest on the top surface of the cylinder. Show that this is a condition of unstable equilibrium if h > 2R.

11.7 Stability of Equilibrium Solution Datum established at the base of the cylinder If the block is displaced by an amount θ from the equilibrium position, for potential function, V = V e + V g = 0 + mgy y = (R + h/2)cosθ + Rθsinθ Thus, V = mg[(R + h/2)cosθ + Rθsinθ] View Free Body Diagram

11.7 Stability of Equilibrium Solution Equilibrium Position dV/dθ = mg[-(R + h/2)sinθ + Rsinθ +Rθcosθ]=0 =mg[-(h/2)sinθ + Rθsinθ] = 0 Obviously, θ = 0° is the equilibrium position that satisfies this equation

11.7 Stability of Equilibrium Solution Stability For second derivative of V, d 2 V/dθ 2 = mg[-(h/2)cosθ + Rcosθ - Rθsinθ] At θ = 0°, d 2 V/dθ 2 = - mg[(h/2) – R] Since all the constants are positive, the block is in unstable equilibrium If h > 2R, then d 2 V/dθ 2 < 0

Chapter Summary Principle of Virtual Work The forces on a body will do virtual work when the body undergoes an imaginary differential displacement or rotation For equilibrium, the sum of virtual work done by all the forces acting on the body must be equal to zero for any virtual displacement This is referred to as the principle of virtual work, and it is used to find the equilibrium configuration for a mechanism or a reactive force acting on a series of connected members

Chapter Summary Principle of Virtual Work If this system has one degree of freedom, its position can be specified by one independent coordinate q To apply principle of virtual work, use position coordinates to locate all the forces and mechanism that will do work when the mechanism undergoes a virtual movement δq The coordinates are related to the independent coordinate q and these expressions differentiated to relate the virtual coordinate displacements to δq

Chapter Summary Principle of Virtual Work Equation of virtual work is written for the mechanism in terms of the common displacement δq, then it is set to zero By factoring δq out of the equation, it is possible to determine either the unknown force or couple moment, or the equilibrium position q

Chapter Summary Potential Energy Criterion for Equilibrium When a system is subjected only to conservative forces, such as the weight or spring forces, then the equilibrium configuration can be determined using the potential energy function V for the system This function is established by expressing the weight and spring potential energy for the system in terms of the independent coordinate q Once it is formulated, first derivative, dV/dq = 0

Chapter Summary Potential Energy Criterion for Equilibrium The solution yields equilibrium position q eq for the system Stability of the system can be investigated by taking the second derivative of V If this is evaluated at q eq and d 2 V/dq 2 > 0, stable equilibrium occurs If this is evaluated at q eq and d 2 V/dq 2 < 0, unstable equilibrium occurs If all higher derivatives equals zero, the system is in neutral equilibrium

Chapter Review