Ans. σ f = = σ (MPa) 100 f = 0.23 (mm/mm) Fig Solution 0.2% f 0.23 mm/mm, it fractures at point C. - PDF

E X A M P L E 3.1 A tension test for a steel alloy results in the stress strain diagram shown in Fig Calculate the modulus of elasticity and the yield strength based on a 0.% offset. Identify on

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E X A M P L E 3.1 A tension test for a steel alloy results in the stress strain diagram shown in Fig Calculate the modulus of elasticity and the yield strength based on a 0.% offset. Identify on the graph the ultimate stress and the fracture stress. σ (MPa) σ u 800 = σ f = σ YS = B A A A E E C 100 f = 0.3 (mm/mm) O Fig % Modulus of Elasticity. We must calculate the slope of the initial straight-line portion of the graph. Using the magnified curve and scale shown in color, this line extends from point O to an estimated point A, which has coordinates of approximately ( mm mm, 345 MPa). Therefore, ksi MPa E = = GPa ksi in. in. mm/mm Note that the equation of the line OA is thus 15(10 3 ). Yield Strength. For a 0.% offset, we begin at a strain of 0.% or mm mm and graphically extend a (dashed) line parallel to OA until it intersects the s P curve at A. The yield strength is approximately s YS = ksi MPa Ultimate Stress. This is defined by the peak of the s P graph, point B in Fig s u = ksi MPa Fracture Stress. When the specimen is strained to its maximum of f 0.3 mm/mm, it fractures at point C. Thus, s f = ksi MPa E X A M P L E 3. The stress strain diagram for an aluminum alloy that is used for making aircraft parts is shown in Fig If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, compute the modulus of resilience both before and after the load application. Permanent Strain. When the specimen is subjected to the load, it strain-hardens until point B is reached on the s P diagram, Fig The strain at this point is approximately 0.03 mm mm. When the load is released, the material behaves by following the straight line BC, which is parallel to line OA. Since both lines have the same slope, the strain at point C can be determined analytically. The slope of line OA is the modulus of elasticity, i.e., 450 MPa E = = 75.0 GPa mm mm From triangle CBD, we require E = BD CD = Pa = Pa CD CD = mm mm This strain represents the amount of recovered elastic strain. The permanent strain, P OC, is thus P OC = 0.03 mm mm mm mm = mm mm Note: If gauge marks on the specimen were originally 50 mm apart, then after the load is released these marks will be 50 mm mm = mm apart σ Y = 450 σ (MPa) O C 0.01 Y = A parallel OC B D Fig F (mm/mm) Modulus of Resilience. Applying Eq. 3 8, we have* 1u r initial = 1 s plp pl = MPa mm mm = 1.35 MJ m 3 1u r final = 1 s plp pl = MPa mm mm =.40 MJ m 3 By comparison, the effect of strain-hardening the material has caused an increase in the modulus of resilience; however, note that the modulus of toughness for the material has decreased since the area under the original curve, OABF, is larger than the area under curve CBF. *Work in the SI system of units is measured in joules, where 1 J = 1 N # m. E X A M P L E 3.3 An aluminum rod shown in Fig. 3 0a has a circular cross section and is subjected to an axial load of 10 kn. If a portion of the stress strain diagram for the material is shown in Fig. 3 0b, determine the approximate elongation of the rod when the load is applied. If the load is removed, what is the permanent elongation of the rod? Take E al = 70 GPa. 0 mm 15 mm A B C 10 kn 10 kn 600 mm 400 mm (a) σ (MPa) Fig. 3 0A F 50 σ Y = 40 parallel G O OG rec = BC = (b) Fig. 3 0 (mm/mm) For the analysis we will neglect the localized deformations at the point of load application and where the rod s cross-sectional area suddenly changes. (These effects will be discussed in Sections 4.1 and 4.7.) Throughout the midsection of each segment the normal stress and deformation are uniform. In order to study the deformation of the rod, we must obtain the strain. This is done by first calculating the stress, then using the stress strain diagram to obtain the strain. The normal stress within each segment is s AB = P A = N = MPa p m s BC = P A = N = MPa p m From the stress strain diagram, the material in region AB is strained elastically since s Y = 40 MPa MPa. Using Hooke s law, P AB = s AB E al = Pa Pa = mm mm The material within region BC is strained plastically, since s Y = 40 MPa MPa. From the graph, for s BC = MPa, P BC L mm mm The approximate elongation of the rod is therefore d = PL = mm mm = 18.3 mm When the 10-kN load is removed, segment AB of the rod will be restored to its original length. Why? On the other hand, the material in segment BC will recover elastically along line FG,Fig.3 0b. Since the slope of FG is E al, the elastic strain recovery is P rec = s BC E al = Pa Pa = mm mm The remaining plastic strain in segment BC is then P OG = = mm mm Therefore, when the load is removed the rod remains elongated by an amount d =P OG L BC = mm = 17.7 mm E X A M P L E 3.4 A bar made of A-36 steel has the dimensions shown in Fig. 3. If an axial force of P = 80 kn is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically. P = 80 kn y 1.5 m 50 mm x P = 80 kn 100 mm z Fig. 3 The normal stress in the bar is s z = P A = N 10.1 m10.05 m = Pa From the table on the inside back cover for A-36 steel, and so the strain in the z direction is P z = s z E st = Pa Pa = mm mm The axial elongation of the bar is therefore d z =P z L z = [ ]11.5 m = 10 mm Using Eq. 3 9, where n st = 0.3 as found from the inside back cover, the contraction strains in both the x and y directions are P x =P y = -n st P z = -0.3[ ] = -5.6 mm m Thus the changes in the dimensions of the cross section are E st = 00 GPa, d x =P x L x = -[ ]10.1 m = -.56 mm d y =P y L y = -[ ]10.05 m = -1.8 mm E X A M P L E 3.5 (MPa) 600 u = pl = A O γ pl = B γ u = 0.54 (a) Fig. 3 5A A specimen of titanium alloy is tested in torsion and the shear stress strain diagram is shown in Fig. 3 5a. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this material, shown in Fig. 3 5b, could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement? Shear Modulus. This value represents the slope of the straight-line portion OA of the t g diagram. The coordinates of point A are (0.008 rad, 360 MPa). Thus, MPa ksi γ (rad) G = = (10ksi 3 ) MPa rad The equation of line OA is therefore 45(10 3 ), which is Hooke s law for shear. Proportional Limit. By inspection, the graph ceases to be linear at point A. Thus, t pl = ksi MPa Ultimate Stress. 100 mm d 50 mm γ (b) Fig mm This value represents the maximum shear stress, point B. From the graph, V t u = ksi MPa Maximum Elastic Displacement and Shear Force. Since the maximum elastic shear strain is rad, a very small angle, the top of the block in Fig. 3 5b will be displaced horizontally: tan rad L rad = d = mmin. d d 50 in. mm The corresponding average shear stress in the block is pl 360 MPa. Thus, the shear force V needed to cause the displacement is t avg = V A ; V V 3605 MPa ksi = 13 (75 in.14 mm)(100 in. mm) V = kip kn E X A M P L E 3.6 An aluminum specimen shown in Fig. 3 6 has a diameter of d 0 = 5 mm and a gauge length of L 0 = 50 mm. If a force of 165 kn elongates the gauge length 1.0 mm, determine the modulus of elasticity. Also, determine by how much the force causes the diameter of the specimen to contract. Take G al = 6 GPa and s Y = 440 MPa. Modulus of Elasticity. The average normal stress in the specimen is s = P A = N = MPa 1p m and the average normal strain is d kn L 0 P= d L Since s 6 s Y = 440 MPa, modulus of elasticity is = 1.0 mm 50 mm the material behaves elastically. The E al = s P = Pa = 70.0 GPa Contraction of Diameter. First, we will determine Poisson s ratio for the material using Eq G = 6 GPa = = mm mm E 11 + n 70.0 GPa 11 + n n = Since P long = mm mm, then by Eq. 3 9, 165 kn Fig. 3 6 n = P lat P long = P lat mm mm P lat = mm mm The contraction of the diameter is therefore d = mm = mm
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