MSE 230 Assignment 5 Solutions Fall 2013 1 6 20 We are asked to ascertain whether or not it is possible to compute for brass the magnitude of the load necessary to produce an elongation of 7 6 mm 0 30 in It is first necessary to compute the strain at yielding from the yield strength and the elastic modulus and then the strain experienced by the test specimen Then if test yield deformation is elastic and the load may be computed using Equations 6 1 and 6 5 However if test yield computation of the load is not possible inasmuch as deformation is plastic and we have neither a stress strain plot nor a mathematical expression relating plastic stress and strain We compute these two strain values as test l 7 6 mm 0 03 l 250 mm o and yield y E 275 MPa 0 0027 3 103 x 10 MPa Therefore computation of the load is not possible as already explained 2 This problem asks that we assess the four alloys relative to the two criteria presented The first criterion is that the material not experience plastic deformation when the tensile load of 27 500 N is applied this means that the stress corresponding to this load not exceed the yield strength of the material Upon computing the stress F A o F do 2 2 27500 N 10 x 10 3 m 2 2 6 2 350 x 10 N m 350 MPa Of the alloys listed in the table the Ti and steel alloys have yield strengths greater than 350 MPa Relative to the second criterion it is necessary to calculate the change in diameter d for these two alloys From Equation 6 8 x z d d o E 1 Now solving for d from this expression d d o E For the steel alloy d 0 27 350 MPa 10 mm 3 4 57 x 10 mm 3 207 x 10 MPa Therefore the steel is a candidate For the Ti alloy d 0 36 350 MPa 10 mm 3 11 8 x 10 mm 3 107 x 10 MPa Therefore the Ti alloy is not acceptable 3 This problem calls for us to make a stress strain plot for a ductile cast iron given its tensile loadlength data and then to determine some of its mechanical characteristics a The data are plotted below on two plots the first corresponds to the entire stress strain curve while for the second the curve extends just beyond the elastic region of deformation 400 Stress MPa 300 200 100 0 0 0 0 1 0 2 Strain 2 300 Stress MPa 200 100 0 0 000 0 001 0 002 0 003 0 004 0 005 0 006 Strain b The elastic modulus is the slope in the linear elastic region as E 100 MPa 0 psi 3 6 200 x 10 MPa 200 GPa 29 x 10 psi 0 0005 0 c For the yield strength the 0 002 strain offset line is drawn dashed It intersects the stress strain curve at approximately 280 MPa 40 500 psi 4 a In order to compute the final length of the brass specimen when the load is released it first becomes necessary to compute the applied stress thus F A o F do 2 2 6000 N 7 5 x 10 3 m 2 2 136 MPa 19 000 psi Upon locating this point on the stress strain curve Figure 6 12 we note that it is in the linear elastic region therefore when the load is released the specimen will return to its original length of 90 mm 3 54 in b In this portion of the problem we are asked to compute the final length after load release when the load is increased to 16 500 N 3700 lb Again computing the stress f 16500 N 373 MPa 52 300 psi 7 5 x 10 3 m 2 2 3 The point on the stress strain curve corresponding to this stress is in the plastic region We are able to estimate the amount of permanent strain by drawing a straight line parallel to the linear elastic region this line intersects the strain axis at a strain of about 0 08 which is the amount of plastic strain The final specimen length l may be determined from Equation 6 2 as i l l 1 90 mm 1 0 08 97 20 mm 3 82 in i o 5 F 15 000N 191 MPa Ai 0 005m 2 Step 1 Calculate T T Step 2 Solve for T ln T lnK nln T ln T ln T ln K ln 191 ln 315 n 0 54 T 0 396 Step 3 Solve for l0 T ln li 1 5 1 5 0 396 ln l 0 0 396 1 01 m l0 l0 e 6 6 51 a We are asked to compute the Brinell hardness for the given indentation It is necessary to use the equation in Table 6 5 for HB where P 1000 kg d 2 50 mm and D 10 mm Thus the Brinell hardness is computed as HB 2P D D D2 d 2 2 1000kg 10mm 10mm 2 2 50mm 2 10mm 200 5 b This part of the problem calls for us to determine the indentation diameter d which will yield a P 500 kg Solving for d from this equation in Table 6 4 gives 300 HB when d 2 D D 2P 2 HB D 4 7 2 500kg 10mm 2 10mm 2 2 11 mm 300 10mm For each of four pairs of polymers we are to do the following 1 determine whether or not it is possible to decide which has the higher tensile modulus 2 if so note which has the higher tensile modulus and then state the reasons for this choice and 3 if it is not possible to decide then state why a Yes it is possible The linear and isotactic polyvinyl chloride will display a greater tensile modulus Linear polymers are more likely to crystallize that branched ones In addition polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures Increasing a polymer s crystallinity leads to an increase in its tensile modulus In addition tensile modulus is independent of molecular weight the atactic branched material has the higher molecular weight b No it is not possible Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don t crystallize readily The atactic polypropylene probably also has a relatively low degree of crystallinity atactic structures also don t tend to crystallize and polypropylene has a more complex mer structure than does polyethylene Tensile modulus increases with degree of crystallinity and it is not possible to determine which polymer is more crystalline Furthermore tensile modulus is independent of molecular weight c Yes it is possible The linear polyethylene will have a higher tensile modulus Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don t crystallize readily Linear polyethylene has a high degree of crystallinity d …
View Full Document
Unlocking...