ME108-PracticeExamSolutions

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ME108 Exam 1 October 6, 2021 Professor Grace O’Connell No calculators 50 points total ● The exam is scheduled for 50 minutes. ● Questions will not be answered during the exam. ● Please note any assumptions that you make and show all of your work . Problem Total points Awarded points 1 4 2 5 3 7 4 16 5 6 6 10 Total 48 Honor code: Read and sign the honor code prior to taking the exam. Students that do not follow the honor code will be removed from the course. "As a member of the UC Berkeley community, I act with honesty, integrity, and respect for others." Signature: ____________________________________________________ Print Name: ___Solutions_________________________________________________

1. (4 points) Type of Materials: a. (1 point) The Columbia Space Shuttle used carbon fibers on the external surface. What is the material type? Ceramic (composite accepted) b. (2 points) Describe two advantages to using this type of material in the design for the Space Shuttle. 1) Carbon fiber has high resistance to heat, which helps the design support re- entry into the atmosphere 2) High strength/high modulus or lightweight or high strength to weight ratio c. (1 point) Explain one disadvantage to NASA’s design of the Columbia Space Shuttle which used carbon fibers on the external surface. The material was very brittle, which caused catastrophic failure (fracture no yield) 2. (5 points) Bonds a. (3 points) List and describe the three primary bonds. i. Ionic- an electron is donated form one atom to another ii. Covalent- an electron is shared between two atoms iii. Metallic- all atoms share a “sea” of electrons b. (2 points) Explain how the type of bond can alter material properties without altering material composition. Using carbon as an example, both diamond and graphite have the same material compositions. Graphite: Each planar array is held together by covalent bonds, but the stacking of these planar arrays is held together by weaker secondary bonds (van der waals forces), making it easy for layers to break off and a very soft material. Meanwhile, diamond carbon atoms are arranged in an organized infinite network where the bonds are all covalent which accounts for diamonds extraordinary hardness. c. (2 points) For a low alloy steel with signs of large plastic deformation, is the structure more likely to be pearlite or martensite. Why? A low-alloy steel with large plastic deformation means the microstructure is more likely to be pearlite, as martensite is hard and brittle and would fracture before large deformations.

3. (7 points) Crystal Structures a. (2 points) Draw a BCC and an HCP unit cell as a reduced-sphere unit cell. BCC HCP b. (2 points) How many full cells are in a BCC and HCP unit cell? BCC _____ 2 ___________ HCP _______ 6 __________ BCC: 2 atoms , the one in the center and eight eighths from the corners HCP : 6 atoms per unit cell 12 corner atoms contributing 1/6 atom 12*(1/6)=2 2 base centered atoms which contribute ½ atom (1/2)*2=1 3 atoms within the volume of the unit cell c. (3 points) Imagine that you work for an alloying firm that would like to convert an element that naturally exists in BCC to an HCP crystal structure. What predicted impact would this have on density and material properties? Density: We are going from 2 atoms/unit cell to 6 atoms/unit cell, our density will increase since we have more material per unit of the material. Recall that theoretical density = ( # atoms per unit cell * Atomic weight)/(volume per unit cell * Avogadro’s #) The ductility of our material would decrease since HCP structure is more brittle than BCC.

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4. (16 points+) A Honda Odyssey was driving too fast while going over a speed hump resulting in a dent in the steel frame, which was designed with A36 steel treated to create a tempered martensite microstructure. To repair the frame, the auto body shop heated the steel to 825 o C with a blowtorch for 30 minutes to ensure the entire thickness of the frame was heated to be able to deform the material and repair the dent. The dent was repaired, and the customer was happy. a. (1 point) What type of heat treatment is described here? Annealing Recall that annealing is the slow cooling after heating a metal. This steel frame was most likely just air cooled (not dunked in oil or water) b. (3 points) What expected changes, if any, might occur to the microstructure of the frame? Explain why. The microstructure would change to austenite at 825 ° C, then as it slowly cools carbon starts diffusing out, creating alternating layers of Ferrite (mostly iron) and Cementite (iron + a lot of carbon), which is what we call pearlite. c. (4 points) On the following graph draw the expected stress strain curves for the A36 steel before and after the repair. Remember to label your axis’. Note: Drawing the percent elongation as follows would both be acceptable answers, - Percent elongation drawn beginning at the strain from yield point and up to fracture (accounts for plastic strain only) - Percent elongation drawn from from 0% strain to fracture strain (total percent elongation) d. (8 points) Label the following properties on the graph above and explain how the repair would impact the material property. i. Young’s modulus: Stays the same- heat treatment does not alter the modulus of elasticity ii. Yield Strength: Yield strength decreases with heat treatment iii. Percent elongation: Pearlite is more ductile than martensite, so the % elongation would increase with heat treatment

iv. Fracture strength: Fracture strength decreases with annealing e. ( Bonus point ) Describe a potential risk of such a repair and why the customer may not be happy in the long-run. You would have a material that would yield more easily, so next time it might dent with a smaller applied force 5. (6 points) After a long day of Avenging in New York City (NYC), the Hulk decides to knock over the Alamo Statue in the East Village in NYC to sit and rest. The statue is a 3m cube with a Young’s modulus of 100 GPa and a Poisson’s ratio of 0.3. If the cube experienced 0.1% strain in both transverse directions, how much does The Hulk weigh? Recall ࠵? = ࠵?࠵? !"#$%&’(%#)! = * + , where the weight is P. We can get the longitudinal/axial strain (in the direction it is being compressed) from Poisson’s ratio. ࠵? = − ࠵? &,)#-. ࠵? )/%)! ==> ࠵? )/%)! = − ࠵? &,)#-. ࠵? = − 0.001 0.3 A good tip for what goes in the numerator versus what goes in the denominator for Poisson’s ratio is that the numerator will be the strain in the direction perpendicular to the applied stress, while the denominator will be the strain in the direction parallel to the applied stress. Then we can insert the strain in the first equation, solved for the force. ࠵? = ࠵?࠵? = ࠵?࠵?࠵? )/%)! = ࠵?࠵? /− ࠵? &,)#-. ࠵? 0 = (3࠵?࠵?3࠵?)(100࠵?10 0 ࠵?࠵?) ∗ (− 0.001 0.3 ) P = -3000 MPa ࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? ࠵? = ࠵?࠵?࠵? )/%)! because we will assume that the hulk resting on the block will not cause plastic deformation.

6. (2 points each) True/False – Circle one. If False, explain why it is false a. True / False – True stress and strain are defined using the initial length and area of the piece. Explain: Engineering stresses are defined using initial length and area b. True / False – Forgetting to zero the displacement and load data at the beginning of a mechanical test, can result in compliance in the data. Explain: Compliance is linked to the deformation of the grips, which is not automatically “fixed” by zeroing load and displacement c. True / False – van der Waals bonds are the strongest of the three primary bond types between atoms. Explain: Van der Waals bonds are secondary bonds. d. True/ False - All metals, including cast irons, strengthen after yielding. Explain: Cast irons are brittle and fracture before yielding e. We always start heat treatments with a BCC phase because the interstitial spaces are larger. Explain: False, we start heat treatments at an FCC phase because the interstitial spaces are larger

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Constants in class: Avogadro’s Number, N A = 6.023X10 23 Possible ME108 Equations (Exam 1): Atomic and Crystal Structure, Heat Treatment: ࠵? !"" = # √% ࠵? ࠵? &"" = 4࠵?/√2 ’ ( = 1.633 Mechanical Testing: ࠵? = − ࠵? &,)#-.1,-1 ࠵? !"#$%&’(%#)! ࠵? 2 = 2࠵? 34! ࠵?࠵?࠵?࠵? d hkl = a √( h 2 + k 2 + l 2 ) L initial /d initial = 4

Phase Diagram for Steel