材料物理性能（Physical properties of materials）Physical properties of MaterialsExercises and SolutionsWu QishengYancheng Institute of Materials Engineering2007, 3Directory1 mechanical properties of materials 22 thermal properties of materials 123 Optical properties of materials 174 Material conductivity 205 Magnetic properties of materials 296 Functional conversion performance of materials 371 Mechanical properties of materials1-1 diameter of a round rod is 2.5 mm, 25cm and subjected to 4500N axial tensile force, if the diameter is drawn to 2.4mm, and the volume of the round rod after tensile deformation is unchanged, the true stress, true strain, nominal stress and nominal strain under this tensile force are obtained, and the results are discussed in comparison.Solution: According to the following tableThe related parameter table of the round bar before and after stretchingVolume v/mm3Diameter d/mmRound area s/mm2Before stretching1227.22.54.909After stretching1227.22.44.524The results are as follows: True stress is greater than nominal stress and true strain is less than nominal strain.1-2 A specimen long 40cm, wide 10cm, thick 1cm, under the stress of the 1000N pull, the young modulus of 3.5x109 min, how many centimeters can be stretched?Solution：1-3 the Youngs modulus at room temperature was 3.5x108 min, Poissons ratio is 0.35, the shear modulus and volume modulus of the material are calculated.解：根据 可知：The 1-4 test shows that the area under the stress-strain curve is proportional to the work done by the tensile specimen.Certificate：1-5 a ceramic with a volume percent of 95% $literal (e = $number gpa) and 5% of the glass phase (E = $number gpa), try to calculate its upper and lower modulus of elasticity. If the ceramic contains 5 pores, the upper and lower modulus of elasticity are estimated.Solution: Make E1=380gpa, E2=84gpa, v1=0 $number, V2=0 05. Then thereWhen the ceramic contains 5% pores, the =. 05-Generation empirical formula E=e0 (1-1.9p+0.9p2) can be obtained, the upper and lower limits of the elastic modulus respectively into $number. 3 GPA and $number 1 GPA.The relationship between stress relaxation and strain creep and time was drawn in 1-6, and the ordinate expressions of T = 0, t = t = are calculated respectively.Solution: The Maxwell model can simulate the stress relaxation process well:The Voigt model can simulate the strain creep process well:The above two models describe the simplest case, in fact, due to the complexity of the mechanical properties of the material, we will use multiple springs and a number of sticky pots through series and parallel combination of complex models. The creep process of a linear polymer is represented by a four component model.1-7 the influence of the temperature and the external force frequency on the mechanical loss angle tangent of the polymer is described and the corresponding temperature spectrum and frequency spectrum are drawn.Explanation: (see book).1-8 a specimen was subjected to tensile stress of 1.0x103 min, 10 seconds after the specimen length of 1.15 times times the original length, the length of the specimen after the removal of the original length of 1.1 times times, if the single Maxwell model to describe, to find its relaxation time tau value.Solution: According to the Maxwell model:Recoverable unrecoverableAccording to the following:So relaxation time =/e=1.0x105/2x104=5 (s).1-9 creep behavior of amorphous polymers can be described in series by a Maxwell model and a Voigt model, if the tensile stress of T is 1.0x104 min to 10 hours, the strain is 0.05, and the response strain after removing the stress can be described as T is the hour, please estimate the four parameter values of the mechanical model.Solution: According to the question, the E1, E2, 2 and ETA 34 parameters are obtained. As shown in the figurewhich 1 immediately reply, 2 gradually reply, 3 can not reply.The response equation for the Voigt is: here T is from the reply, and the question T is starting from the beginning of the stretching,So the answer equation for this question is:To eliminate the strain after immediate recovery, the response equation of the strain can be written1-10 when the TG is taken as the reference temperature, the c1=17 in log, c2=51.6, the constant C1 and C2 in the Wlf equation with Tg+50 as the reference temperature.Solution：1-11 A cylindrical $literal crystal is subjected to axial tensile F, and if its critical shear strength is f to $number MPA, the minimum tensile value needed to produce slippage in the sliding system along the direction shown in the graph is obtained, and the normal stress of the slip surface is obtained.Solution：1-12 drawing a sample to obtain the data of the following table, try to make a graph, an