Set P = 45 kN, P 25 kN. (Figure 1) %3! Figure 1of
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- Two uniform spheres are positioned as shown. Determine the gravitational force F which the titanium sphere exerts on the copper sphere. The value of R is 30 mm. Assume a = 5.1, b = 2.6, 0=33°. y Copper T R aR Titanium Ө bR --xElement 1 Ук ** Answer in MPa. 2 cm Element 2 Assume the reaction forces at the origin were solved to be Rf = [325, 100, 0] N and = Rm [10, -2, 5] Nm. We will consider a hypothetical cut at the connection of the dolly to the axle. Z is positive out of the page. What is the value of normal stress in the x-direction Ox for element 2? Answer options: a) Ox < -300 b) -300 ≤ Ox < -100 c) -100 ≤ Ox < -200 d) 0 ≤ Ox < 100 e) 100 ≤ Ox < 300 f) 300 ≤ OxAn infismall element io plven on plane, where z = 0. 120uPa 120MPQ Modulus of elosticities are 60 G la ond * and y ohrectuons, respectively. 24 GPa in The radrd's 0.2 and o.4 ore parson in x ond y direc thens, respectively. Determine the stroin in x ond y directhens.
- Kindly solve this problem with complete solution so that I can Understand. Correct Final answer: 1400 or 1.4x10^3Q3: For sysem shown in Figure, Determine final temperature if R= 0297 KPam’ and cp = 1.04 Kikg:I just need the last problem at the bottom of the picture. When I submit the last problem by itself my question gets rejected because the value of the mass is required so I assume that you are supposed to refer to the previous problems to see that M=100Nm.
- Q3/ A patient who suffers from a problem in the knee joint and needs a joint replacement. As a Biomaterials engineer, you have three metal materials, any of these materials in the table 1. you prefer for the replacement process with mentioning the reason. If you know that the joint is exposed to compression resistance resulting from the weight of the person, and the load applied on the joint is 80 kg, with a cross-sectional area of 40 mm Table 1. The materials properties Materials Compressive strength (Mpa) Туре I 17.6 Туре II 30.2 Туре I 6.5The answer is already given. I just need the solution. Also, please make the solution neat and readable. Thank you. Answer: R= 193 lb, A= 74 lb3. Assume the plane stress state: ơ1 = 0, 02 = 20, T12 = 15 (all stresses in ksi). If oz alone is increased until the maximum shear stress t1, in the material reaches ±20 ksi, what is the value of o, at this point? HINT: this is relatively easy to figure out if you construct a Mohr's circle for the initial stress state and then consider what happens if o, is replaced with Ao2 and you solve for A when the maximum shear stress (from the circle geometry) reaches ±20 ksi.
- H.W 1. Consider a two degree of freedom bar elements as shown in figure. Using finite element method to formulate the equilibrium equation of it. If the cross sectional area is 12 mm and E=200 GN/m². 20KN +30KN 15KN 300 * 600- + 350 All Dimension in mm 2. Consider a two degree of freedom bar element as shown in figure. Using finite element method to formulate the equilibrium equation of it, and then estimate the stress distributions. If Esteel-200 GN/m, Ecopper 110GN/m and EAL= 120 GN/m?. d=3 Steel AL Соpper 20KN - 30KN →15KN 300 * 200 - 400 * 350 All Dimension in mmTheee blocks with masses Mz=12Kg by strings ouer pulleys. m, = 3.0 kg 18kg respectIvely ,are atteche %3D cnd my=l %3DFigure 1 shows a system of three bars each made of different materials, and connected together and the temperature is initially kept at 12°C. Thereafter, the temperature of the system was raised to 50°C. Steelbno abm Est = 200 GPa Epr = 100 GPa ag = 12(10-)/°C abr = Brass Сopper Ecu 120 GPa %3D 21(10-6)/°C acu = 17(10-6)/°C %3D %3D Acu = 515 mm? Ast = 200 mm? Abr = 450 mm² - 300 mm- -200 mm 100 mm Figure 1: An assembly of three bars 1.1. Determine the thermal stress raising in each bar if the assembly is placed between two rigid walls. 1.2. Determine the force exerted on the right rigid support if an allowance of 0.05 mm is provided between the left end of the steel bar and the left wall.