1、材料科学基础课后习题答案9Solutions for Chapter 91. FIND: What considerations are required for use of materials in the named applications?SOLUTION: Heart valve - biocompatible, long life (fatigue endurance), ease of installation: Turbine blade - able to withstand high temperature and stress without creep, long l
2、ife (fatigue endurance): Leaf spring - low cost, stress without creep, long life (fatigue endurance): Coffee mug - not contaminate contents, able to withstand 0 - 100 cycles and gradients: Golf club shaft: able to withstand, amplify, and transmit forces with little loss, long life (fatigue endurance
3、), aesthetically and tactily attractive: Suture - biocompatible, long life (fatigue endurance)COMMENTS: There are many other requirements, and we could begin to specify materials properties that are required to meet the applications needs.2. Find: Determine Youngs Modulus.Data: Upper yield point str
4、ess of 207 MPa, Yield point strain of 0.001.Solution: Recall that Youngs Modulus is the slope of the stress/strain curve in the elastic portion of the stress strain curve. That is: E=Stress/StrainSubstituting values into the above equation yields: E = 207/0.001 = 207,000 MPa = 207 GPa.3. FIND: Calcu
5、late the strain-to-fail of silicate glasses.GIVEN: The modulus is 107 psi and the strengths of three samples are 5, 50, and 500 ksi.ASSUMPTIONS: The material behaves in a linear elastic manner up to failure.SKETCH: The beginnings of the three curves lie on top of one another. The higher the strain-t
6、o-fail the longer the curve extends to the upper right. The xs indicate the points of failure. SOLUTION: Hookes law states: = E.Rearranging, = /E.Substitution for each of the samples: = 5000 psi / 107 psi = 0.05% = 50,000 psi / 107 psi = 0.5% = 500,000 psi / 107 psi = 5%. COMMENTS: The smallest valu
7、e represents that of ordinary window glass. The largest value is characteristic of an optical fiber.4. Find: Determine if the stress is above or below the yield stress. If the stress is below the yield stress compute Youngs Modulus.Data: The yield stress for the mild steel is 207 MPa. A specimen has
8、 a diameter of 0.01m and a length of 0.1m. It is loaded in tension to 1000N and deflects 6.077 x 10-6m.Solution: To solve this problem, we must first determine the applied tensile stress. Recall that the tensile stress is given by the formula: Stress=Force (P)/Area Normal to force (A). The cross sec
9、tional area is: A= D2/4 = 3.1416x(0.01)2/4 = 7.85x10-5The stress is: Stress = 1000/7.85x10-5 = 12.7MPaThe applied stress is much less than the yield stress. To obtain the Modulus recall the definition: E=/. We must thus compute the strain in order to determine the Modulus:=l/lo = 6.077x10-6/0.1 = 6.
10、077x10-5Thus E=12.7/6.077x10-5 = 2.09x105 MPa.5. Find: Compute deflection of specimens.Data: EAl=70,460MPa, ECu=122,500MPa, EW=388,080MPa, A=0.01m x 0.01m=10-4m2, Length(lo)=1m, Load(P)=5000N.Solution: Start with the relationship between stress and strain for linear elastic behavior: =E (1)Note that
11、: =P/A (2) =l/lo (3)Substitute (2) & (3) into (1): Thus P/A=El/lo (4)Rearranging to solve for l we have:l=Plo/AE=5000 x 1/(10-4E)=5x107/E (5)Note that for l to be in meters we require E to be expressed in Pa. Using (5) and E in Pa we get the following deflection values:Al=7.09x10-4m, Cu=4.08x10-4m,
12、W=1.29x10-4m.6. Find: Deflection at 5000N.Data: ENylon6/6=2.08GPa, Load(P)=5000N, lo=1m, A=10-4m2.Solution: Using the formula developed in the preceding problem we have: lo=Plo/(AE).Thus: lo=(5000 x 1)/(10-4x2.08x109)=2.4x10-2m.Comment: Note that the deflection is two orders of magnitude greater tha
13、n for steels.7. FIND: Calculate the strength of a round textile fiber.GIVEN: The fiber diameter is 10 micrometers and the load-at-failure is 25 g.ASSUMPTIONS: The fiber need not be in the elastic region at failure. SOLUTION: = F/A = = 3.1 GPa.8. Find: Determine the shear strain at yield.Data: For a
14、particular steel: =0.295, E = 205,000MPa, ys=300MPa and ys = 1/2ys.Solution: Recall Hookes Law in shear: ys= Gys. Also, it is stated that ys = 1/2ys. Substitution yields: 1/2ys = Gys or ys= ys/2G. The problem now is to determine G. Recall that G=E/(2(1+)= 205,000/(2(1+0.295)=79,151 MPa. Substituting
15、 we have: ys= 300/(2x79,151)=1.895x10-3.9. Find: Compute the 0.2% offset yield strength and the strain at yield.Data: EAl=69x103 MPa and =2950.1.Solution: We require the intersection point of the two curves shown below to solve the problems since the 0.2% offset yield is defined as the point where a
16、 line whose slope is equal to the modulus, and which is displaced 0.002 on the strain axis, intersects the stress/strain curve. To find the intersection point we must first get the equation for 0.2% offset yield. The general form of the equation is: =m + b, where m=slope and b=intercept. We know tha
17、t the point (0.002,0) is on the line thus: 0=69x103(0.002) + b, therefore b= -1.38x102. The equation of the 0.2% offset line is = 69x103 - 1.38x102. The intersection is obtained by rearranging and substituting the former equation with the first equation: 2950.1 = 69x103-1.38x102. This equation is be
18、st solved by trial and error or by writing a short computer program to check for the equality of the right and left hand sides of the equation for various values of . At a strain of 0.00448 the difference between the right and left-hand sides of the equation is negligible. Substituting this strain v
19、alue into the equation for stress yields an estimate for the yield stress of 172 MPa.10. Find: (a) Poissons ratio, (b) % volume change at ys/2 and (c) % volume change at ys.Data: Aluminum specimen with the following properties: E=69x103 MPa loaded such that l = 1.25x10-3 and v = 4.17x10-4 .Solution:
20、 (a) Recall that Poissons ratio is defined as = -v/l. Thus, = -(-4.17x10-4 /1.25x10-3)=0.333. (b) The volume change is given by v/vo=x+y+z. But y = l, x,z = v and v =-l. Thus, v/vo=x + l +l= l(1-2). In the preceding problem it was shown that the yield strain was 4.48x10-3. At half the yield the stra
21、in is 2.24x10-3. Then v/vo = (2.24x10-3)(1-0.666) = (7.48x10-4)x100=7.48x10-2.(c) At yield stress: v/vo x 100=4.48x10-3 x 0.334 x 100 = 1.50 x 10-1.11. Find: Predict the behavior of amorphous polymers.Solution: Steels have a definite yield point because dislocations are pinned by carbon atoms that r
22、eside in the interstitial positions. When stress rises to the point necessary for dislocations to break free from the carbon atoms, plastic deformation occurs due to dislocation movement. Yield stress is the critical stress necessary for freeing the dislocations from the carbon atoms. In Al and Cu,
23、that have an FCC structure, dislocation mobility increases gradually as the stress is increased. Therefore, there is no single stress level at which the dislocations begin to move suddenly. In these materials, yielding occurs gradually, dislocations motion may be impeded, but the dislocations are no
24、t pinned. In amorphous polymers, plastic deformation occurs by sliding between adjacent molecular chains. The sliding will commence at a definite stress level. Therefore, we expect a definite yield point to occur in amorphous polymers.12. Find: Determine the ultimate tensile strength.Data: The stres
25、s-strain behavior is given by = Kn.Solution: Let engineering stress be designated by S and engineering strain by e. The following equations relate engineering stress and strain to true stress and true strain: = S(1+e) (1)and = ln(1+e) (2)We are given = Kn (3)Substituting (1) and (2) into (3), we get
26、: S(1+e) = Kln(1+e)n. At the ultimate tensile stress point, dS/de = 0. Thus, dS/de = (d/de) (K/(1+e) (lu(1+e)n = 0, or - (K/(1+e)2 (ln(1+e)n+Kn/(1+e)2(ln(1+e)n-1 = 0or ln(1+e) = n (3a)If we designate e by eu at the ultimate tensile point, u = Kln(1+eu)n (4)Substituting equation (4) into (3) we get u
27、 = Kln(1+eu)n, or Su(1+eu) = Knn. Solving for Su yields: Su = K/(1+eu) nn = Knn-1 (4a)Su is the ultimate tensile strength.Comment: Equation (3a) is frequently used to estimate n if a complete stress-strain curve is not available. Further, equation (4a) can be used to obtain the strength coefficient,
28、 K.13. Find: 1) The physical basis for the observation and 2) show =0.5 during plastic deformation.Data: Volume of a crystalline material remains constant during plastics deformation.Solution: 1) During purely plastic deformation interatomic distance is not changing and the atoms eventually slide ov
29、er one another. for this reason the volume is constant during plastic deformation.2) Recall that /o = x + y + z = (1-2) for uniaxial deformation. If = 0 , then =1/2.14. Find: Compute the relative load bearing capacities of an Al alloy (UTS = 400 MPa, = 2.7 gm/cm3 ) and polypropylene (UTS = 40 MPa, =
30、0.9 gm/cm3 ).Data: Relative load capacity for constant weight is proportional to the strength divided by the density.Solution: To have consistent units, convert density to kg/m3.Al= 2.7x103 kg/m3 PP= 0.9x103 kg/m3 Al= 400x106/2.7x103(Pa/kg/m3)=148x103(Pa m3/kg)Recall Pa=N/m2, therefore Al= 148x103 (
31、N m/kg) and pp=40x106/0.9x103=44.4x103 (N-m/kg).16. Find: Plot the true strain to fracture, being sure to place Cu and steel from the preceeding problem in the graph.Data: A range of %RA from 0 to 70%.Solution: Recall that the true strain may be computed from the %RA using the formula: f = ln(100/(1
32、00-%RA). See graph for correlation of fracture strain and %RA.17. Find: Determine the engineering strain at which the difference between the true strain and the engineering strain is = ln (1+e). Data: Equation relating engineering strain to true strain.Solution: We can write a series expansion for the right hand side: = ln (1+e) = e-(e2/2!)+(e3/3!
