230 F14 HW6 SOLS.pdf

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<ul><li><p>MSE 230 Assignment 6 Solutions Fall 2014 </p><p> 1*. a) = cos cos = the applied tensile stress = the resolved shear stress Slip initiates when the applied tensile stress is equal to the yield stress y. The resolved shear stress when =y is called the critical resolved shear stress or CRSS. CRSS = y cos cos, a = [001], b = [111], c = [0 1 1] cos = [001] [111]1 3 =</p><p>13 </p><p> cos = [00 1 ] [01 1]1 2 =</p><p>12 </p><p> CRSS = (7.8 MPa)</p><p>13</p><p>" # </p><p>$ % 12</p><p>" # </p><p>$ % </p><p>= 3.2 MPa b) The applied stress refers to the tensile stress applied during a mechanical test. The yield stress is the applied tensile stress needed to initiate plastic deformation (or slip). The shear stress is the component of the tensile stress in the slip direction and is calculated using =coscos. The critical resolved shear stress is the shear stress that is sufficient to initiate slip. Note that the critical resolved shear stress is never more than 50% of the yield stress. </p></li><li><p>2. In this problem, a=[101], and b and c vary depending on the given slip system. a Slip System cos cos Schmid Factor [101] (111) [10 1 ] 22 3 </p><p>02 2 0 </p><p> [011 ] " 12 2 -0.408 [1 10] " 12 2 -0.408 [101] 111( ) 101!" #$ 22 3 </p><p>02 2 0 </p><p> [110] " 12 2 0.408 [011] " </p><p>12 2 0.408 </p><p> 101[ ] 111!" #$= 0 no resolvable shear stress, so slip cannot occur along the 111( ) plane. </p><p> 101[ ] 111!" #$= 0no resolvable shear stress, so slip cannot occur along the 111( ) plane. </p><p> b) Slip is possible on any of the slip systems above that have a nonzero Schmid factor. The fact that some are negative does not matter, only the magnitude of the Schmid factor. This is because the plane above the slip plane is moving in one direction, and the plane below the slip plane is moving in the opposite (negative) direction. c) CRSS = y (cos cos) CRSS = 10 MPa, cos cos = 0.408 y = 24.5 MPa </p></li><li><p>3*. a) This is an example of solid solution strengthening. Lattice strains associated with the substitutional solid solution inhibit dislocation motion. See pgs. 231-232 of Callister. </p><p> b) This is an example of strain hardening or cold working. An increase in dislocation </p><p>density during deformation makes dislocation motion more difficult. See pgs. 232-235 of Callister. </p><p> c) Polycrystalline metals have grain boundaries. Grain boundaries inhibit the motion of </p><p>dislocations thereby strengthening the metal. See pgs. 229-231 of Callister. 4.* Please refer to Figure 7.19 in Callister. The goal is to have a metal with a minimum yield strength of 300 MPa and a minimum </p><p>ductility of 25%. If the metal does not meet the specification as is, we are allowed to use cold work to achieve the specs. </p><p> 1040 Steel: At 0% cold work 1040 steel has a y = 440 MPa and ductility = 25%. </p><p>Therefore it is ready to go right off of the shelf. Brass: At 0% cold work brass has a y = 190 MPa and ductility = 67%. We need to </p><p>increase y via cold work. Cold working brass to approximately 12% will increase y to 300 MPa, and decrease ductility to 35%. We can meet the specs with 12% cold work. </p><p> Copper: At 0% cold work copper has a y = 160 MPa and ductility = 43%. Again, we </p><p>must increase y via cold work. Cold working copper to approximately 37% will increase y to 300 MPa. However, this decreases ductility to under 10%. Therefore, we can't use copper. </p><p> 5. a) Deforming a brass rod in tension from a diameter of 1.5 cm to 1.128 cm </p><p> % cold work </p><p>=Ao AdAo</p><p>" </p><p># $ % </p><p>&amp; ' x 100</p><p>=(0.75cm)2 (0.564cm)2</p><p>(0.75cm)2" </p><p># $ </p><p>% </p><p>&amp; ' x 100</p><p>=1.767cm2 1cm2</p><p>1.767cm2" </p><p># $ </p><p>% </p><p>&amp; ' x 100 = 43.4% CW</p><p> According to Fig. 7.19, brass has a yield stress of approximately 400 MPa after 43% CW. b) Force = Ad. he minimum stress to allow yielding is y. Therefore, assume = y. </p></li><li><p> 400 x 106 Nm2! " </p><p># $ * m</p><p>2104cm2</p><p>! </p><p>" % </p><p># </p><p>$ &amp; *1 cm2 </p><p> F = 40,000 N c) No, you can't deform your material in one step because you cannot apply sufficient </p><p>force to complete the deformation process. If brass did not exhibit strain hardening we wouldn't have a problem. </p><p> If we wanted to determine the amount of cold work we could perform in one step, we </p><p>would need to convert the y vs. %CW curve in Fig. 7.19 to a Force vs. %CW curve using our sample cross-sectional area (which also changes with %CW). When the Force curve = 35,000 N we have performed the maximum %CW possible for one step. </p><p> d) Two choices: 1) Deform the rod as much as possible, then heat treat the rod at 475C (see Table </p><p>7.2) to make the brass recrystallize. This will reduce y and allow more cold work (see Fig. 7.22). </p><p> 2) Perform the deformation process at 475C or above so that recrystallization occurs simultaneously with deformation. This is called hot working. </p><p> 6*. </p><p> There are two major things two notice here. First, the dislocation density decreases with time. This is associated with recrystallization, a process that produces a microstructure consisting of new strain-free grains, and accompanies a decrease in yield strength and increase in ductility of the material. Second, the dislocation density does not go to zero, rather it reaches an equilibrium value. Virtually all crystalline materials have dislocations even in the absence of plastic deformation, and dislocations are the root cause of ductility of ductile metals. </p><p>DislocationDensity</p><p>time</p></li><li><p>7. First we need to calculate the critical flaw size at which fracture will occur rather than yielding. To do this, we set the fracture stress (c) equal to the yield stress. </p><p>K = a c =KICac</p><p>155 MPa = 4 MPa mac</p><p> ac = 2.12x104m or 0.212 mm</p><p> Since we can detect flaws as small as 0.10 mm, we can detect flaws small enough to cause fracture. To cause fracture, the flaw must be larger than 0.212 mm. Alternatively, we could have calculated the stress needed to cause fracture for a flaw size at our detection limit. </p><p>c =4MPa m(1x104m)</p><p> c = 225.7 MPa </p><p> Since c &gt; y, the vessel will yield before it ruptures. 8. </p><p> The gauge section of a tensile specimen sees a uniform tensile stress while a bend bar </p><p>sees a stress gradient ranging from compression to tension. For specimens of the same size, a greater volume of the tensile specimen is under tension </p><p>compared to the bend bar. Therefore, the probability that a critical flaw coincides with </p><p>Tensile Specimen: entire gauge lengthis in tension.</p><p>COMPRESSION</p><p>TENSIONNeutral AxisBend Specimen: Bottom half of the sample isin tension while the top half is in compression.</p></li><li><p>the maximum stress is greater in a tensile test. Hence the tensile specimens should have a lower strength. </p><p> 9.* a) No, according to the specification the material should have failed in a ductile manner. For a yield stress of 205 MPa and a fracture toughness of 50 MPam, the critical flaw size at the boundary between brittle and ductile behavior is: </p><p>ac =150MPa m205MPa</p><p># </p><p>$ % </p><p>&amp; </p><p>' ( </p><p>2=18.9 mm </p><p>Therefore, a crack 5.3 mm in size should not cause a brittle failure. You could also calculate the yield stress based on the 5.3 mm flaw size and state that the yield stress above which brittle failure occurs is much higher than the manufacturers specification. b) This can be done by using 5.3 mm as the critical flaw size. Also, note that we are assuming that the fracture toughness is still 50 MPam. </p><p>y =K1ca =</p><p>50MPa m 5.3103m( )</p><p>= 387.5MPa </p><p>If the yield stress is 387.5 MPa or greater, a flaw of size 5.3 mm can cause brittle failure. c) No. From part b) it appears that the yield stress is larger than the manufacturers specification. Therefore, it is likely that the stainless steel had undergone some cold working which would result in a larger yield stress than specified by the manufacturer d) Microstructural analysis. A cold worked sample would have nonequiaxed grains and have a texture associated with any cold working. Also, by etching one could measure the dislocation density. Perform a hardness test, then heat the material to the recrystallization temperature, cool to room temperature and repeat the hardness test. A significant drop in hardness after heating suggests that the material had been cold worked. 10.* It has to do with the role of probability in the fracture of brittle materials. The full scale sculpture has a greater probability of finding a larger flaw than the 1:8 scale model. Since the strength of brittle materials depends on the size of the largest flaw, this means that the full scale model will have a lower fracture stress than the 1:8 scale model. (see section 12.8 in Callister. Too bad that school didnt have anyone studying engineering.) 11.* From Figure 8.20, the fatigue limit stress amplitude for this alloy is 310 MPa. Engineering stress is defined as </p><p> = FA0</p><p>. For a cylindrical bar </p></li><li><p>A0 = d02</p><p># </p><p>$ % </p><p>&amp; </p><p>' ( 2 </p><p> Substitution for A0 into the equation leads to </p><p> = FA0</p><p> = F</p><p>d02</p><p>$ </p><p>% &amp; </p><p>' </p><p>( ) </p><p>2 = 4Fd0</p><p>2 </p><p> We now solve for d0, taking stress as the fatigue limit divided by the factor of safety. Thus </p><p>d0 =4F</p><p>N</p><p>$ </p><p>% &amp; </p><p>' </p><p>( ) </p><p>= (4)(22,000 N)</p><p>() 310 106 N /m2</p><p>2$ </p><p>% &amp; </p><p>' </p><p>( ) </p><p>= 13.4 103 m = 13.4 mm (0.53 in.) </p><p> 12.* In order to solve this problem, it is necessary to compute both the mean stress and stress amplitude for each specimen. Since from Equation 8.14, mean stresses are the specimens are determined as follows: </p><p>m =max + min</p><p>2 </p><p>m (A) =450 MPa + (350 MPa)</p><p>2 = 50 MPa </p><p>m (B) =400 MPa + (300 MPa)</p><p>2 = 50 MPa </p><p>m (C ) =340 MPa + (340 MPa)</p><p>2 = 0 MPa Furthermore, using Equation 8.16, stress amplitudes are computed as </p><p>a =max min</p><p>2 </p><p>a (A) =450 MPa (350 MPa)</p><p>2 = 400 MPa </p></li><li><p>a (B) =400 MPa (300 MPa)</p><p>2 = 350 MPa </p><p>a (C ) =340 MPa (340 MPa)</p><p>2 = 340 MPa On the basis of these results, the fatigue lifetime for specimen C will be greater than specimen B, which in turn will be greater than specimen A. This conclusion is based upon the following S-N plot on which curves are plotted for two m values. </p></li></ul>