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- 2. What is the equivalent stiffness of the system of Fig. 1-19 using the displacement of the block as the generalized coordinate? E = 210 x 10 x 10ʻ N 1= 1.5 x 10 mª 2.5 m 2 x 10 N 3 x 10 N Fig. 1-19 Answer: 4,690,000 N/MPROBIEM 1 AssoHE A BONDARY LAYER FOR A Fluid OVER A F|AT PIATE CAN BE APPROKIMATED BY WHERE U= CONSTANT is THE FREE STREAM VelociTy AND S. S(x) is THE THICKNESS Of THE BON DARY LAYER. ) WHICH BOUNDARY CONDITIONS ARE SATISFIED AND WHICH ONES ARE NOT ? 2) calculATE S* AND g* 3) WRITE AN EXPRESSION FOR THEe wall STRESS, AS A PUNCTION OF U, S AND . WHERE Y is THE Visaosire OF THe Fluid. INDICATE THE DIRECTION OF THE STRESS. 4) OBTAIN AN EQJATION FOR S«) FROM THE MOMENTUM INTEGRAI EQUATION 5) OBTAIN S«)Imagine a 3D state of strain in a mechanical part, measured perhaps by digital volume correlation or other means, given in Cartesian coordinates by €rx = a(x² + y²), €yy = a(y² +2²), Exy=b(xyz), Exz = €yz = €zz = 0, where a and b are constants and x, y, z are coordinates in space. Determine whether this measured strain field is a possible state of strain for a continuum that remains intact and continuous after deformation.
- **Problem 1.15 Suppose you wanted to describe an unstable particle, that spon- taneously disintegrates with a "lifetime" t. In that case the total probability of finding the particle somewhere should not be constant, but should decrease at (say) an exponential rate: too P(t) = | V(x,1)1²dx = e=1/*. -0- A crude way of achieving this result is as follows. In Equation 1.24 we tacitly assumed that V (the potential energy) is real. That is certainly reasonable, but it leads to the "conservation of probability" enshrined in Equation 1.27. What if we assign to V an imaginary part: V = Vo – ir, where Vo is the true potential energy and r is a positive real constant? (a) Show that (in place of Equation 1.27) we now get dP 21 = --P. dt (b) Solve for P(1), and find the lifetime of the particle in terms of r.The pressure drops 4P = P₁ P2 through a long section of round pipe can be written in terms of the shear stress Tw along the wall. Shown in Figure 3 is the shear stress acting by the wall on the fluid. The shaded region is a control volume composed of the fluid in the pipe between axial locations 1 and 2. Using the method of repeating variables, generate a relationship for pressure drop as a function of all other parameters Final ans AP = P μ (₁ AP= {(₂17₂) P₁ CV UPL м = (²1 ²² an! VPL O P.H. L VG1/ in the Pipe (A) is equal (PA= 25 Lit15), the diameter is equal ( DA - 75mm, Do = Fig.shown, the Flow rate in 6o mm, Dg = Dc%=30 mm), and the velocit Pipe (D) is equal (U0=5m/s), the relation between flow rate in Pipe (B) and pipe (c) %3D Find the Value af . 2- Un, UB, Uc. water Os = 30 mm input A QA - 25 Lit/sec = 75 mm fo mm %3D Inpud = A outPut = B,C,D accumule tion =o.
- 1 A SPHERICAL PORCELAIN BALL, WITH RADIUS R, IS DROPPED WITHOUT INITIAL VELOCITY IN A TEST TUBE CONTAINING GLYCERIN. DURING FALL, THE BALL IS SUBJECT TO THE FOLLO... A spherical porcelain ball, with radius r, is dropped without initial velocity in a test tube containing glycerin. During fall, the ball is subject to the following forces: P its weight; A the thrust of Archimedes; fluid friction force f = k v with k = 6 πrn; n: viscosity (Pa s) of Glycerin, V, velocity Density of Glycerin pg= 1.3 g/mL; density of porcelain pp = 2.3 g/mL; r ball radius= 1.0 cm; n = 1.0 s.Pa; g = 10 N/kg; volume of a sphere V = 4/3 πr³. The speed limit (in m/s) is: 0.11 0.22 0.44 0.33 0.55A- Womersley number (a) of a human aorta is 20 and for the rabbit aorta is 17, the blood density is approximately the same across the species. The values of viscosity were 0.0035 Ns/m² for the human and 0.0040 Ns/m² for the rabbit. The diameter of the aorta is 2.0 cm for the man, and 0.7 cm for the rabbit, estimate the heart rate beats per minute (bpm) for both species3) ol Cag. - thing shownin Fig.2.2 the tetal hand loss ChL) from Pant 1 to pant Lis soft, hind the presune atpont 2 0.84)s Hlawnginapipeyender the Gand Port 1 D,z 6in こ 10.70 At Point 2 Dzz 9in Pz z 2 4.00Pt Reference detum
- 65 I 5:11 docs.google.com/forms 2 2.4 m/s Clear selection A simple pendulum is made of a 50 cm-string and a bob of mass m. At t = O, the pendulum is at its equilibrium position and is given an initial velocity v = 0.2 m/s. The maximum angular speed, O'max, is: 0.4 rad/s 0.1 rad/s 0.2 rad/s 0.8 rad/s 0.05 rad/s A mass-spring system oscillates on a frictionless horizontal surface in simple harmonic motion with an.B2) An automotive engine can be assumed as rectangular block with approximation size 0.4 m high, 0.6 m wide, and 0.7 m long as shown in Figure 6. The ambient air is at 1 atm and 15°C where air density (P) = 1.225 kg/m and kinematic viscosity (v) = 1.47x105 m2/s. Air 85km/h 15°C Engine Block Figure 6 a) Calculate the friction coefficient (C) and investigate the force acting on the bottom surface of the engine block as the car travels at the velocity of 85 km/h. Assume the flow to be turbulent (Cr = 0.074/Re5) over the entire surface because of the constant agitation of the engine block. b) An engine oil at 40°C with kinematic viscosity (v) = 2.485×104 m³/s is flowing over a long flat plate with velocity of 4 m/s. Determine the critical distance xer from the leading edge of the plate where the flow become turbulent (Recr = 5x105), and analyze the thickness of the boundary layers over a length of 2xer. Given boundary layer thickness for laminar flow is, öx = 4.91x/Re,2 and turbulent flow,…igure, PROBLEM 509 In the figure shown, determine the maxımum value of the applied force P to impend motion on the 2KN wedge if the angle of fnction for all contact surfaces is 15°. Figure, 2-KN WEDGE 1000-KN BLOCK 750