The stress profile shown below is applied to six different biological materials: Log Time (s) The mechanical behavior of each of the materials can be modeled as a Voigt body. In response to ơ,= 20 Pa applied to each of the six materials, the following responses are obtained: anl ssang
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- 100 80 60 40 20 0.002 0.004 0.006 0.008 0.01 0.012 Strain, in/in. FIGURE P1.17 1.18 Use Problem 1.17 to graphically determine the following: a. Modulus of resilience b. Toughness Hint: The toughness (u) can be determined by calculating the area under the stress-strain curve u = de where & is the strain at fracture. The preceding integral can be approxi- mated numerically by using a trapezoidal integration technique: u, = Eu, = o, + o e, - 6) %3D c. If the specimen is loaded to 40 ksi only and the lateral strain was found to be -0.00057 in./in., what is Poisson's ratio of this metal? d. If the specimen is loaded to 70 ksi only and then unloaded, what is the permanent strain? Stress, ksiThermodynamics is extensively used in engineering field. The following polynomial can be used to relate the zero-pressure specific heat of dry air c, kl/(kg K) to temperature (K) c, = 0.99403 + 1.671 x 104T +9.7215 x 10-872-9.5838 x 10-"T³ + 1.9520 x 10-4T Which of the following(s) is/are the correct method to determine the temperature that corresponds to a specific heat of 1.1 kl/(kg K). (Choose all that applyl) O A) Gauss-Seidel Method O B) Lagrange Interpolation O C) Newton-Divided Difference O 0) Bisection Method Newton MethodQ3/ In real life engineering application, it is required to design a spring in train, which one of the following materials in the table below should be choose, justify your answer ? used correct units? Poisson's Yield Strength Modulus of Elasticity 106 psi Material MPа psi GPa Ratio 30 0.30 120000 207 Steel alloy Brass alloy Aluminum alloy Titanium alloy 830 0.34 0.33 380 55000 97 14 275 40000 69 10 690 100000 107 15.5 0.34
- Give True or False for the following: 1.In liquids and gases, heat transmission is caused by conduction and convection 2.The surface geometry is the important factor in convection heat transfer 3. The heat transfer by conduction from heated surface to the adjacent layer of fluid, 4. The heat transfer is increased in the fin when &> 1 5.The unit of the thermal diffusivity is m²/s 6. Temperature change between the materials interfaces is attributed to the thermal contact resistance 7. A material that has a low heat capacity will have a large thermal diffusivity. 8. Heat conduction flowing from one side to other depends directly on thickness 9.Fin efficiency is the ratio of the fin heat dissipation with that of no fin 10.The critical radius is represented the ratio of the convicted heat transfer to the thermal conductivityIn the Fig. 2 below, let Ki = K2 = K and ti = t=t. %3D T -T X Fig. 2 (a) Let T= 0 °C and T= 200 °C. Solve for T: and unknown rates of heat flow in term of k and t. MEC_AMO_TEM_035_02 Page 2 of 11 Finite Element Analysis (MECH 0016.1) – Spring - 2021 -Assignment 2-QP (b) Let T- 400 °C and let fs have the prescribed value f. What are the unknowns? Solve for them in term of K, t, and f.CALCULATE THE FOLLOWING: We performed the experiment to measure MATERIAL 1 - BRASS the thermal conductivity of 2 materials (Brass & Steel) in the laboratory and measured the following tabulated values: Calculation for Brass Quantities Values Material 1- BRASS Calculated Power (Q') W (Diameter = 25mm) Area of cross section (A) m2 Power Difference in Temperature Temperature (°C) between two points (AT) °C Difference in distance between two points (Ax) m Q' 1 2 3 (W) Thermal conductivity of brass (kp) W/m°C 4 5 6 8 9. 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 33.6 MATERIAL 2 - STEEL Calculation for Steel Material 2 - STEEL %3D Quantities Values Calculated (Diameter = 25mm) %3D Power (Q') W Power Area of cross section (A) m2 Temperature (°C) Difference in Temperature between two points (AT) °C Q' 1 |(W) Difference in distance between two points (Ax) m 2 3 7 8 14.2 88.6 87.5 85 33.9 33.6 32.4 Thermal conductivity of steel (kg) W/m°C
- i. Derive the linear and quatratic approximation of the below resistance temperature readings(temperature from 40 °c to 80 °c). (. ii. find out the resistance of RTD at a temperature of 200°c using linear approximation? ( Temperature 40 45 50 55 60[TO] 65 70 75 80 Resistance ohm 115.239 120.521 167.096 254.966 384.13 554.587 766.34 1019.38 1313.724Consider a process to prepare a metal for a certain application. There are five parameters that must be considered: temperature, quenching rate, cooling time, carbon content, CO₂ concentration. It is desired to determine which of these parameters has the most influence on the process. There are two levels for each parameter as shown below. Temperature (°C) Quenching Rate(°C/s) Cooling time (s) Carbon Content (wt% C) CO₂ Concentration (%) Eight experiments were defined as follows: Experiment Carbon Content (wt% C) 1 2 3 4 5 6 7 8 1) What size orthogonal array should be used for evaluation (assume noise is negligible)? 2) Generate the array with the level values (i.e. Level 1 and Level 2). 1 1 1 1 6 6 6 6 Quenching Rate (°C/s) 35 35 140 140 35 35 140 140 Four trials were run for the experiments defined above: Experiment 1 2 3 4 5 6 7 8 Level 1 760 35 1 1 5 T1 68.00 69.84 74.36 71.71 91.27 54.39 64.65 60.31 Cooling Time 1 1 300 300 300 300 1 1 T2 61.41 64.76 61.30 58.42 90.89 Level 2 900…Q8): To find how much heat is required to bring a kettle of water to its boiling point, you are asked to calculate the specific heat of water at 61°C. The specific heat of water is given as a function of time in Table below. Temperature, T Specific heat, C₂ (°C) J kg-°C 22 42 52 82 100 4181 4179 4186 4199 4217 Determine the value of the specific heat at 7=61°C using the direct T method of interpolation and a third order polynomial. Find the absolute relative approximate error for the third order polynomial approximation (Lagrange Method).
- The heat transfer conducted through material is calculated from the equation: Q = KX AXTD/L Where K: Conductivity of material A: Area of heat transfer TD: Temperature difference across material L: Thickness of material A student measures the area, thickness and temperature difference and assumes that the error in conductivity is negligible. The student also estimates the uncertainty range for each variable. In estimating the maximum possible value of Q, the student should use the following formula: A B Q max= K x A max x TD max / L max Q max= K x A max x TD max / L nom Q max= Q nominal + dQ/dLmin Q max= K x A max x TD max / L mindated metncpdf Thermodynamics An Engineering X E Module2-chap2propertiesofpure x O File C:/Users/DANIEL/Desktop/300L%202ND%20SEMESTER%20MATERIALS/Module2-chap2propertiesofpuresubstances-130703012604 phpap. ID Page view A Read aloud V Draw H Highlight O Erase 40 MEC 451 - THERMODYNAMICS Faculty of Mechanical Engineering, UITM Supplementary Problems The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kRa. If the volume of the tire is 0.025 m3, Cetermine the pressure rise in the tire when the air temperature in the tire rises to 50°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure is 100 kPa. [ 26 kPa, 0.007 kg] 1. A 6 4 2:57 PM Lucky CORER 144 Tum lock 314 % 8 1/2 24 6 1/4 23 T K F pause BT'sec In helical spring experiment a student plotted the graph of T2 versus the oscillating mass (M), and Used the slope to find the spring :constant, the value of k is 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 O. O a. 13.3 N/m O b. 6.8N/m O c. 5 N/m O d. 3.3 N/m O e. 24 N/m 5 10 15 20 25 30 35 40 45 50 55 60 65 M (9) 70 75 80