Chapter 12 Equilibrium and elasticity

Equilibrium We already introduced the concept of equilibrium in Chapter 8: dU(x)/dx = 0 More general definition of equilibrium: Static equilibrium: Stable equilibrium: the body returns to the state of static equilibrium after having been displaced from that state. Unstable equilibrium: the state of equilibrium is lost after a small force displaces the body

Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the com) and its direction relative to the possible point(s) of rotation

Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the com) and its direction relative to the possible point(s) of rotation

Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the com) and its direction relative to the possible point(s) of rotation

Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the com) and its direction relative to the possible point(s) of rotation

Center of gravity Gravitational force on a body effectively acts on a single point, called the center of gravity If g is the same for all elements of a body (which is not always so: see for example Chapter 13) then the center of gravity of the body coincides with its center of mass

The requirements of equilibrium For an object to be in equilibrium, we should have two requirements met Balance of forces: the vector sum of all the external forces that act on the body is zero Balance of torques: the vector sum of all the external torques that act on the body, measured about any possible point, is zero

Equilibrium: 2D case If an object can move only in 2D ( xy plane) then the equilibrium requirements are simplified: Balance of forces: only the x- and y-components are considered Balance of torques: only the z-component is considered (the only one perpendicular to the xy plane)

Examples of static equilibrium

Chapter 12 Problem 67

Chapter 12 Problem 27

Indeterminate structures Indeterminate systems cannot be solved by a simple application of the equilibrium conditions In reality, physical objects are not absolutely rigid bodies Concept of elasticity is employed

Elasticity All real “rigid” bodies can change their dimensions as a result of pulling, pushing, twisting, or compression This is due to the behavior of a microscopic structure of the materials they are made of Atomic lattices can be approximated as sphere/spring repetitive arrangements

Stress and strain All deformations result from a stress – deforming force per unit area Deformations are described by a strain – unit deformation Coefficient of proportionality between stress and strain is called a modulus of elasticity stress = modulus * strain

Tension and compression Strain is a dimensionless ratio – fractional change in length of the specimen ΔL/L The modulus for tensile and compressive strength is called the Young’s modulus Thomas Young (1773 – 1829)

Tension and compression Strain is a dimensionless ratio – fractional change in length of the specimen ΔL/L The modulus for tensile and compressive strength is called the Young’s modulus

Shearing For the stress, force vector lies in the plane of the area Strain is a dimensionless ratio Δx/L The modulus for this case is called the shear modulus

Hydraulic stress The stress is fluid pressure p (Ch.14) Strain is a dimensionless ratio ΔV/V The modulus is called the bulk modulus

Answers to the even-numbered problems Chapter 12: Problem 8 (a) 2.77 kN; (b) 3.89 kN

Answers to the even-numbered problems Chapter 12: Problem kN

Answers to the even-numbered problems Chapter 12: Problem 14 (a) 49 N; (b) 28 N; (c) 57 N; (d) 29º

Answers to the even-numbered problems Chapter 12: Problem mJ

Answers to the even-numbered problems Chapter 12: Problem 48 (a) 50º; (b) 0.77 mg