NUC323M4A1 Version C

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Excelsior University *

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Mechanical Engineering

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Apr 3, 2024

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NUC 323 Material Science M4.4: Physical Properties and Material Testing (Version C) 1. Describe how the Knoop Hardness test is performed. (10 points) The Knoop Hardness test consists of indenting a material with a diamond indenter. With a force ranging from grams to kilograms. The load is applied for a specific duration to ensure consistency. After the specified duration, the load is removed, and the indention length is measured using a microscope. 2. A piece of steel has a grain diameter of 5.0 x 10 -3 mm and is heated to 350 ⁰C for 12 minutes. After 12 minutes, the grain size increases to 9.2 x 10 -3 mm. Calculate the grain diameter (in mm) when a specimen of the original material is heated to 350 ⁰C for 45 minutes. Assume the grain diameter exponent “n” is 2.0. (20 points) #2: wrong math (-5pt) First, we have to solve for K: d 2 d 0 2 = kt d 0 = 5 x 10 3 mm ¿ d = 9.2 x 10 3 mm d 2 d 0 2 = ( 9.2 x 10 3 ) 2 ( 5 x 10 3 ) 2 = K 12 k = 8.464 x 10 5 2.5 x 10 5 12 = 4.97 x 10 6 mm We can then use the above answer to solve for 45 minutes: 4.97 x 10 6 ( 45 ) + 5 x 10 3 d 2 = 5.224 x 10 3 d = 7.23 x 10 2 3. What is the angle between the [2,1,2] and [1,0.5,0] lattice directions in a cubic system? (20 points) cosδ = uu 1 + v v 1 + w w 1 ( u 2 + v 2 + w 2 ) ¿¿ Plug in and solve: ( 2 ) ( 1 ) + ( 1 ) ( 0.5 ) + ( 2 ) ( 0 ) ( ( 2 2 ) + ( 1 2 ) + ( 2 2 ) ) ( ( 1 ) 2 + ( 0.5 ) 2 + ( 0 ) 2 ) = ¿ cos = 2.5 9 1.25 => cos = 0.745 cos -1 (0.745) = 0.730 radians or 41.84 o
NUC 323 Material Science M4.4: Physical Properties and Material Testing (Version C) 4. A Knoop hardness (HK) test is performed per ASTM E384 on a gold foil. A load of 0.1 kg produces a pyramidal indentation whose outer legs measure l = 0.143 mm. Calculate the HK of the gold foil in Kg/mm 2 . (P in kg, D and l in mm) (20 points) Using the equation in module 4 we can plug in the knowns and solve: HK = 14.2 P/l 2 Given: P = 0.1 kg l = 0.143 mm HK = 14.2(0.1)/(0.143) 2 HK = 69.44Kg/mm 2 5. Calculate the resilience (ability of a material to absorb energy when it is deformed elastically and release that energy upon unloading in Pa) of tungsten that has a yield stress of 750 MPa (@Strain 0.200 %). (10 points) #5: read module notes "Resilience" and Table 3- 1 in the notes, the modulus of elasticity is denominator. (-5pt) U = 1 2 * y * y Given: = 750 MPa = 0.002 Plug in numbers U = ½*750*10 6 *0.002 U = 0.75*10 6 Pa 6. A bar of copper has 25 percent elongation and final cross-sectional area of 9 mm 2 following cold working. Using Figure 4-1 in the Module Notes in your solution and convert percent elongation (ductility) to percent cold work, calculate the initial (original or A o ) area (in mm 2 ) of the cross-section. (20 points) %CW = (A o -A d /A o )*100 From Figure 4-1, we know that a 25% elongation corresponds to approximately 12% cold work for copper. We can substitute these values into the formula and solve for A o 12 = (A o -9/A o )*100 A o = 9/(1-12/100) A o = 10.227 mm 2
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