a. b. C. d. 4. The relationship between Young's modulus (E), Bulk modulus (K) and Poisson's ratio (u) is given by E-2K(1-2µ) E-3K(1-2μ) E-2K(1-2μ) E-2K(1-3μ)
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![a.
b.
C.
d.
4.The relationship between Young's modulus (E), Bulk modulus (K) and
Poisson's ratio (u) is given by
E-2K(1-2µ)
E-3K(1-2μ)
E-2K(1-2μ)
E-2K(1-3μ)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16c38ec6-8fe5-42fd-ac7a-d4a44479af0e%2F83697af6-4698-4a3a-8091-cc710cce533d%2Fof3sv_processed.jpeg&w=3840&q=75)
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- The graph shown below shows the Modulus of Elasticity (E) of different materials with respect to temperature (°C). 240 210 170 140 100 -Carbon Steel, C < 0.3% -Nickel Steels, Ni 2% - 9% 70 Cr-Mo Steels, Cr 2%-3% Copper -Leaded Ni-Bronze -Nickel Alloys - Monel 400 35 -Titanium rengineeringtoolbox.com -Aluminium -300 -200 -100 100 300 400 500 600 700 800 Temperature (degc) a) Which material is the most ductile with increasing temperature? b) Which material is the least ductile with increasing temperature? c) Discuss the change in ductility (brittle / ductile behavior) occurring for different materials. Also, comment on why Young's modulus changes with temperature. E-modulus of elasticity (GPa)ki= 15 N/m; k2 = 10 N/m; m, k3 = 12 N/m; %3D ci = 1 N.s/m; m2 k,Z c C2 = 2 N.s/m; C3 = 2 N.s/m; m3 mı = 25 kg; 7 kg; m2 = m3 = 15kg; solve the mathematical model.A:YV 36 ll 46.ll Test 3.pdf F1 F2 a F3 A d. C a F4 E Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3 = 8 kN; F4 = 9 kN; F5 = 10 kN; a = 30° 0 = 45° B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 111.6 kN.m b) 112.6 kN.m c) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m f) None of them 2] What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 kN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m f) None of them 3] What is the resultant moment of the five forces acting on the rod about point C? a) 32.6 kN.m b) - 21.1 kN.m c) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m f) None of them 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m f) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e)…
- Consider the loaded shaft illustrated in Figure Q4 (i). The shaft is simply supported at A and D. The shoulder fillet radius is 0.3 cm at B and C. If the shaft is subjected to a force of F=800 N, determine the following:a=2, b=1 A bullet is to be tested in the laboratory to determine the drag force on it. Dependent parameter the drag force D (Newton) depends on the velocity of the bullet V(m/s), the length of the bullet L(m), sound velocity c(m/s), density of fluid ρ (kg/m3) and dynamic viscosity µ(kg/ms). Solve the problem by making the necessary assumptions and drawing the schematic figure I-Determine the nondimensional p parameters using repeating variables ii-a bullet with a speed of 9a,b m/s in air may be modelled in a water tunnel with a test section velocity of 2ab cm/s. Determine the length of the model, if the length of the bullet is 5a,b mm. The air and water temperature is 20 oC degree at 1 atm. iii- if the drag force on the model is measured to be 2,ab N, then determine the expected drag force on the bullet. Comment on dynamic similarity equivalence?1. Aspring mass system serving as a shock absorber under a car s suspension, supports the M = 1000 kg mass or the car. For this shock absorber, k = 1x 10°/m and c = 2x 10' N s/m. Tne car drives over a corrugated road with rorce F = 2x 10' sin(@t) N. Use your notes to model the second order difrerential equation suited to this application. Simplity the equation with the coerricient of X" as one. Soive X (tne general solution) in terms or @ using the complimentary and particular solution method. In determining the coernicients of your particular solution, it will be required that you assume o -1xo or 1- w x -m. Do not use Matiab as its solution will not be identiriable in the solution entry. Do not determine the value of W. You must Indicate In your solution! 1. Tne simpliried dirrerential equation in terms or the displacement X you will be solving 2. Ine m equation and complimentary solution Xe 3. Tne choice ror the particular solution and the actual particular solution Xp 4. Express…
- 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/MTwo dimensional stress tensor at a point is given by a matrix Ox To 100 30 5x 0 30 20 The maximum shear stress is MPa MPa.Surface tension is due to uneven molecular attraction between immiscible fluid media and solid surface. Its importance is governed by dimensionless parameter Which of the Inertia force called Weber number given by We = Surface tension force™ following statements are correct? (i) Surface tension becomes important when the length scale L is small such as in ink jet printing, coating of thin film, oxygenation of water by falling rain drops. (ii) 111 Super-hydrophobicity can be achieved by micro/nano texturing of surface such as in lotus leave, butterfly and dragonfly wing. During oil spill in ocean, oil floats and spreads out due to gravity. By conservation of mass, the oil film becomes thinner as it spreads. Finally, the surface tension force becomes important causing the formation of tiny oil droplets. (A) (B) (C) (D) (i) and (ii) (i) and (iii) (ii) and (iii) All of the above
- Q7// State the Cowper-Symonds empirical equation which describes the effect of strain rate on the flow stress of metallic materials. Using the Cowper-Symonds equation, together with the material constants D and q in the equation as given in the Table below, describe the comparative high strain rate sensitivity of the different materials (assume e° = 3 4 s'). Material D (s') Copper 4.8 52 Stainless Steel 9.2 99 Titanium Alloy 9.2 122 Aluminium Alloy 4.2 6710 Mild Steel 5.1 45The 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 VMeasurements at a certain point of a pipe have been done where the following parameters were recorded: Fluid of density = 887 kg/m3, Fluid velocity = 4 m/s, Pressure= 11.3 KN/m2 If the total energy per unit weight at this point = 32 m, then the potential energy is: O Zero m 10 m 20 m 30 m &
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