In-tutorial Exercise # 6 MECH 321 Winter 2018 Question 1:

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1 In-tutorial Exercise # 6 MECH 321 Winter 2018 Question 1: (a) Why is the abrasive wear resistance of a material a function of its hardness? (b) Why is it difficult to use friction sawing on nonferrous metals? Explain. (a) Higher hardness indicates greater resistance to penetration, hence less penetration of the abrasive particles or hard protrusions into surfaces, and the grooves produced are not as deep. Thus, abrasive wear is a function of hardness (b) Nonferrous metals have a tendency to adhere to the blade, caused by adhesion at the high temperatures and attributable to the softness of these materials. Note also that these materials typically have high thermal conductivity, so if the metal has melted, it will quickly solidify and make the operation more difficult. Question 2: How can fatigue wear be reduced? (a) reducing the load and sliding distance and increasing the hardness (b) improving the quality of the contacting materials, such as eliminating inclusions, impurities, and voids; (c) improving the surface finish and integrity during the manufacturing process; (d) surface working, such as shot peening or other treatments; (e) reducing contact stresses; and (f) reducing the number of total cycles.

2 Question 3: For adhesive wear, the volume loss of the surface being worn away is given by Archard s equation as following: Where is a nondimensional wear coefficient, is the normal load, is the sliding distance, and is the hardness of the material being worn. Consider a flat face of a brass pin (rod) having a diameter of 20 mm is placed on a flat carbon steel disc under a normal load of 10 N and rotates in a track of a radius of 50 mm at 100 rpm for 100 h. As a result of wear during the test, the mass loss of the brass is 20 mg. Calculate (i) wear coefficient and (ii) wear depth for the brass. (hardness of brass = 0.8 GPa, and density of brass = 7.5 g/cm 3.) The wear coefficient and wear depth to be calculated. The volume loss of the rod can be calculated as follows: The sliding distance can be calculated as: Then, using Archad s equation: The wear depth is simply:

Question 4: A simple house table, shown in the figure, is to be designed. The top of table is made of an already chosen glass, however, a material for the legs is to be selected.

3 Question 4: A simple house table, shown in the figure, is to be designed. The top of table is made of an already chosen glass, however, a material for the legs is to be selected. The legs of the table must be as light as possible. They must also support the load,, on the table top without buckling. A leg is a slender column that has a material density, modulus, and a fixed length. The radius of the leg can be varied. (a) Drive the performance index equation. (b) Determine the slope of the line to be used in the performance chart. Note that the critical load to cause elastic buckling on a column of length and radius is given by: Where 4 is the second moment of area, and is the elastic modulus. (a) This question asks that the material index (performance index) equation to be derived for a leg. The function of the leg is to carry the load on the table top. The leg is a slender column of material density and modulus. Its length is, the maximum load must be carried without buckling. The objective is to minimize the weight. The free variable is the radius of the table. The constraints are: (i) fixed length and (ii) that the load should carried without buckling. To minimize the weight, the objective function can be written as: (1) While the constraint equation is (2)

4 The free variable is the radius. By substituting the value of from equation (2) into equation (1), the objective function becomes: Since the length and load are fixed values (constraints), the performance index equation is: (4) Where, the equation is written in such way that the performance index is to be maximized (instead of minimized). (3) (b) The slope of the line can be determined by taking the log of the performance index equation. Therefore, the line has slope of 2. 2 loglog

5 Question 5: A material for a floor joists (beam) is to be selected. The beam must be strong, stiff, and as cheap as possible. The length of the beam is fixed. The beam has a rectangular cross section with variable depth and width, but their proportion is fixed (i.e. fixed /). The cost of the beam is represented in terms of its mass,, times the cost per kg,. (a) Drive the performance index equation for the beam to be strong and cheap. (b) Drive the performance index equation for the beam to be stiff and cheap. (c) Determine the slopes in (a) and (b). Note that the failure load and critical bending stiffness of a beam are given by: Where and are constants, is the failure strength of the material of the beam, is the elastic modulus, is the distance from the neutral axis, and I is the moment of inertia (from rectangular cross section 12). This question asks that the performance index equations to be derived for a beam. The function of the beam is to carry bending load without much deflection (stiff) and without failure (strong). The beam has a rectangular cross section, fixed length L, material density, failure strength, and an elastic modulus. The objective is to minimize the cost. The free variable is the area of the beam. The constraints are: (i) fixed length, (ii) the beam should be stiff (no large deflections), and (iii) the beam should be strong (carry load without failure). The objective function of the cost can be written as: (1) (a) the performance index equation for the beam to be strong and cheap. The critical load to failure is given by:

6 (2) Since 12 12, the critical load can be written also as: (3) By substituting the value of from equation (3) into equation (1), the objective function of the cost becomes: / / / (4) Since the length, load, and / are fixed values (constraints), the performance index equation is then: / (5) (b) the performance index equation for the beam to be stiff and cheap. The same analysis in (a) can be followed here to arrive at the following performance index: / (6) (c) for strong and cheap index, the slope is 3/2, whereas for stiff and cheap index the slope is 2.

Materials Selection: Case Studies

MME445: Lecture 21 & 22 Materials Selection: Case Studies A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Learning Objectives Knowledge & Understanding Knowledge of how to use materials property