Mechanics And Biomechanics-Midterm-BMTS 2130 PDF

Mechanics and Biomechanics BMTS 2130 Dr. Sofiene MANSOURI [email protected] (Whatsapp:0533791123)  Plan 1 Fluid mechanics 2 Solid mechanics 3 Biomechanics Muscles and bones act as levers 2 Fluid mechanics...

Mechanics and Biomechanics BMTS 2130 Dr. Sofiene MANSOURI [email protected] (Whatsapp:0533791123)  Plan 1 Fluid mechanics 2 Solid mechanics 3 Biomechanics Muscles and bones act as levers 2 Fluid mechanics 3 Fluid mechanics ❖Pascal’s law ❖ The effect of a force F on a motionless liquid generates a pressure P within the liquid, which at any point acts equally in all directions. The pressure always acts perpendicular to the boundary surface S of the liquid. P=F/S [N/m² = Pa] Dimension? With 1N = 1 Kg m/s2 ❖[P] = [F]/[S] = [ M1L1T-2/L2] = [ M1L-1T-2] S ❖ All force and pressure processes in liquids are based on this law. 4 Exercise 1 In a hydraulic system, a piston with a cross-sectional area of 21 square centimeters pushes on an incompressible liquid with a force of 38 N. The far end of the hydraulic pipe connects to a second piston with a cross-sectional surface area of 100 square centimeters. What is the force on the second piston? 5 Solution Pascal’s principle gives In this case, you know that F1 = 38 N, A1 = 21 cm2 and A2 = 100 cm2. Solve the equation for F2 and insert the given values to find the force on the second piston: 6 Fundamental Principles of the Hydrostatic Fundamental theorem F1 , F2 : forces applied by the fluid on The basic surfaces S of the cylinder P is the cylinder weight and h is its height Equilibrium condition: - F1 + F2 – P = 0 Or F1 = P1.S and F2 = P2.S - P1.S + P2.S = P = mg g : gravitation acceleration 7  Fundamental Principles of the Hydrostatic Fundamental theorem -P1.S + P2.S = P = mg g : gravitation acceleration Or m = ρ.V = ρ.S.h ρ : density (P2 – P1). S = ρ.S.h.g => P2 – P1 = ρ.g.h Or h = Z1 – Z2 P2 – P1 = ρ.g.(Z1 – Z2) P1 + ρ.gZ1 = P2 + ρ.gZ2 P + ρ.gZ = Constant (1) (1) is the fundamental relatioship of hydrostatics or fundamental theorem P: (static pressure) ; & ρgZ : (gravity pressure). 8 Fundamental Principles of the Hydrostatic Proposal The pressure is the same in any point of an horizontal plan of a fluid in equilibrium Demonstration: Apply the fundamental principle of hydrostatic to point M and N PM + ρ.gZM = PN + ρ.gZN Or ZM = ZN Then : PM = PN 9  Hemodynamic of Ideal fluid: Bernoulli theorem Hemodynamic of Ideal fluid: Bernoulli theorem is expressed by the following expression in each point of the fluid defined by (P, Z, v) ; (P: static pressure, Z: level, v: velocity) : PM + ½ ρVM² + ρ.gZM = PN + ½ ρVN² + ρ.gZN P + ½ ρV² + ρgZ = C (C is a constant) http://scienceworld.wolfram.com/physics/bimg74.gif * N P: (static pressure) ; 1/2 ρv²: (dynamic pressure) ; * M ρgZ : (gravity pressure). 10 Bernoulli theorem ❖ From point A to point B, a particle of fluid sees: P + 1/2 ρv² + ρgZ = Constante its altitude decrease, the passage section of the flow increase. ❖ The decrease in altitude leads to a decrease in the potential energy of the particles: Epot ❖ The increase of the passage section leads to a decrease in the flow rate, and therefore to a decrease in the kinetic energy of the particles: Ec ❖ The result is that the pressure along the flow between A and B increases, since according to the theorem the total load "E" does not vary. Epr = P 11 Bernoulli theorem P + 1/2 ρv² + ρgZ = Constant If v = 0 (static fluid), according Berouilli theorem we have P + ρgZ = Constant Conclusion : Bernouilli therorem is applicable also at hydrostatic fluid 12 Equation of continuity Volume Flow rate: D or Q The flow rate is the product of the velocity by the surface: D=Sv D = Sv = constant Then if we have a flow in a tube with variable section S1 and S2 And respectively v1 and v2 velocity We can write : S1.v1 = S2.v2 This is the equation of continuity 13 Equation of continuity D = S v = constant  if S increases then v decreases  if S decreases then v increases 14 Exercise 2 ❖Calculate VB, With SA = 0.02 m2 and VA = 4 m/s VB = SAVA/SB =SAVA/2SA=VA/2=2m/s 15 Exercise 3 Consider the flowing tubes as following: VA and VB are the velocities respectively in SA and SB 1. Calculate VB in the case of a; b; c; and d. 16 Solution ❖a) VB/VA ; Equation of continuity: SAVA =SBVB or SB = 2SA then: SAVA =2 SA VB VB= ½ VA ❖b) VB = VA ❖c) VB = 2 VA ❖d) VB = 2/3 * VA 17 Exercise 4 The flow velocity in water in a horizontal tube of RA = 5 cm radius is VA = 10 m/s 1. Calculate the radius RB of the tube at the output if we want a water velocity VB = 1000 m/s Solution: RB = 0.5 cm = 0.005 m 18 Exercise 5 Suppose two vessels A and C horizontal with section S connected by a vessel of section S/10 as shown in the diagram: Consider the blood as a liquid without viscosity, calculate: 1. The hydrostatic (P1B) and dynamic pressure (P2B) in vessel B, Take: P1A = 100 mm Hg and P2A = 1 mmHg 19 Solution ❖P1A+P2A = P1B+P2B = P1C + P2C = 101 mmHg P1A= 100 mmHg P2A= 1 mmHg = ½ pVA2 then VA2 = 2/p mmHg P2B = ½ pVB2 Or SAVA =SBVB then VB = 10 VA P2B = ½ p 100. 2/p = 100 mmHg Then P1B=1 mmHg 20  Exercise 6 Suppose two vessels A and B as shown in the diagram: SA SB A With: SB =SA/3 B Take: P1A = 80 mm Hg, P2A = 5 mmHg and SA = 15 cm2 Calculate: 1. The radius RA and the radius RB 2. The hydrostatic and dynamic pressure in vessel B, respectively P1B and P2B 21 Solution RA= SAπ = 2.19 cm RB= SBπ = 1.26 cm Because SB = 5 cm2 The hydrostatic and dynamic pressure in vessel B, respectively P1B and P2B P2B = ½ p VB2 = 9* (1/2 p VA2) = 45 mmHg With SAVA =SBVB => VB = 3 VA P1B = 85 mmHg – 45 mmHg = 40 mmHg 22