MSE 230 1 Assignment 6 Solutions Fall 2013 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 cos 001 111 1 1 3 3 cos 00 1 01 1 1 1 2 2 b 111 c 0 11 1 1 CRSS 7 8 MPa 3 2 MPa 3 2 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 cos cos 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 2 In this problem a 121 and b and c vary depending on the given slip system a Slip System 121 111 10 1 cos 121 Schmid Factor 4 6 3 0 6 2 0 011 1 6 2 0 272 1 10 1 6 2 0 272 2 6 3 2 6 2 0 272 110 3 6 2 0 408 011 1 6 2 0 136 2 6 3 1 6 2 0 136 101 2 6 2 0 272 011 3 6 2 0 408 1 11 101 121 cos 11 1 1 1 0 121 1 1 1 0 no resolvable shear stress so slip cannot occur along the 1 1 1 plane b Slip first occurs on the 1 11 in the 110 and on the 11 1 in the 011 both have the largest Schmid factor c CRSS y cos cos y 24 5 MPa CRSS 10 MPa cos cos 0 408 3 a This is an example of solid solution strengthening Lattice strains associated with the substitutional solid solution inhibit dislocation motion See pgs 213 215 of Callister b This is an example of strain hardening or cold working An increase in dislocation density during deformation makes dislocation motion more difficult See pgs 215218 of Callister c Polycrystalline metals have grain boundaries Grain boundaries inhibit the motion of dislocations thereby strengthening the metal See pgs 212 213 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 ductility of 25 If the metal does not meet the specification as is we are allowed to use cold work to achieve the specs 1040 Steel At 0 cold work 1040 steel has a y 440 MPa and ductility 25 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 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 Copper At 0 cold work copper has a y 160 MPa and ductility 43 Again we 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 5 a Deforming a brass rod in tension from a diameter of 1 5 cm to 1 128 cm A Ad x 100 o Ao 0 75cm 2 0 564cm 2 x 100 cold work 0 75cm 2 1 767cm 2 1cm2 x 100 43 4 CW 1 767cm2 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 N m2 400 x 106 2 4 2 1 cm2 m 10 cm F 40 000 N c No you can t deform your material in one step because you cannot apply sufficient force to complete the deformation process If brass did not exhibit strain hardening we wouldn t have a problem If we wanted to determine the amount of cold work we could perform in one step we 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 d Two choices 1 Deform the rod as much as possible then heat treat the rod at 475 C see Table 7 2 pg 222 to make the brass recrystallize This will reduce y and allow more cold work see Fig 7 22 2 Perform the deformation process at 475 C or above so that recrystallization occurs simultaneously with deformation This is called hot working 6 Dislocation Density time 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 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 K a 155 MPa c 4 MPa m a c K IC a c a c 2 12x10 4 m or 0 212 mm 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 c 4MPa m 1x10 4 m c 225 7 MPa Since c y the vessel will yield before it ruptures 8 Tensile Specimen entire gauge length is in tension COMPRESSION Neutral Axis TENSION Bend Specimen Bottom half of the sample is in tension while the top half is in compression The gauge section of a tensile specimen sees a uniform tensile stress while a bend bar 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 compared to the bend bar Therefore the probability that a critical flaw coincides with the maximum stress is greater in a tensile test Hence the tensile specimens should have a lower strength 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 MPa m the critical flaw size at the boundary between brittle and ductile behavior is 2 …
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