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Unformatted text preview: Homework # 8 ENMA 300/ENME 382 Fall 2011 Due Monday, November 7 1. Dislocations A metal crystal has a Burger’s vector of length 0.212 nm. If the lattice parameter, a , is 0.300 nm, is the metal crystal structure FCC or BCC? HINT: Determine the length from one atom to the next in the slip direction for FCC and BCC. The Burgers vector is ( a /2)*[110] in FCC structures and ( a /2)*[111] in BCC structures, where [xyz] donates a vector of length sqrt(x 2 +y 2 +z 2 ) along the [xyz] crystal direction. The Burgers vector lengths are sqrt(2) a /2 and sqrt(3) a /2, in FCC and BCC structures, respectively. 0.212 = sqrt(2)*0.300/2. It is an FCC. 2. Critical Resolved Shear Stress a) The critical resolved shear stress for nickel is 60 MPa (8.75 ksi). Determine the minimum possible yield strength for a single crystal of nickel pulled in tension. In order to determine the minimum possible yield strength for a single crystal of nickel pulled in tension, we simply employ Equation 8.5 as σ y = 2 τ crss = (2)(60 MPa) = 120 MPa (17.5 ksi) b) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] direction. If the magnitude of this stress is 30 MPa, compute the resolved shear stress in the [111] direction on each of the (110), (011), and (101) planes. This part of the problem asks, for a BCC metal, that we compute the resolved shear stress in the [111] direction on each of the (110), (011), and (101) planes. In order to solve this problem it is necessary to employ Equation 8.2, which means that we first need to solve for the for angles λ and φ for the three slip systems. For each of these three slip systems, the λ will be the same— i.e., the angle between the direction of the applied stress, [100] and the slip direction, [111]. This angle λ may be determined using Equation 8.6. cos λ ൌ ݄ ଵ ݄ ଶ ݇ ଵ ݇ ଶ ݈ ଵ ݈ ଶ ඥ݄ ଵ ଶ ݇ ଵ ଶ ݈ ଵ ଶ ඥ݄ ଶ ଶ ݇ ଶ ଶ ݈ ଶ ଶ where (for [100]) h 1 = 1, k 1 = 0, l 1 = 0, and (for [111]) h 2 = 1, k 2 = 1, l 2 = 1. Therefore, cos λ is determined as cos ߣ ൌ ሺ1ሻሺ1ሻ ሺ0ሻሺ1ሻ ሺ0ሻሺ1ሻ √1 √3 ൌ √3 3 Let us now determine cos φ for the angle between the direction of the applied tensile stress (i.e., the [100] direction) and the normal to the (110) slip plane (i.e., the [110] direction). Again, using Equation 8.6 where h 1 = 1, k 1 = 0, l 1 = 0 (for [100]), and h 2 = 1, k 2 = 1, l 2 = 0 (for [110]), cos φ is equal to cos Ԅ ൌ ݄ ଵ ݄ ଶ ݇ ଵ ݇ ଶ ݈...
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 Spring '12
 Christou
 Strength of materials, Tensile strength, Work hardening, MPa, critical resolved shear

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