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## Q. 13.5

Table E13.5(a) gives the history of nominal stress S for Fig. 13.20. Estimate the local notch stress–strain response for stable behavior. Note that the value of stress concentration factor is $k_{t}$ = 2.40 and the material is 2024-T351 aluminum.

## Verified Solution

Constants for the material’s cyclic stress–strain curve $ε_{a} = f (σ_{a}$) in the form of Eq. 12.54 are available from Table 12.1 as E = 73,100MPa, $H^{\prime }$= 662MPa, and $n^{\prime }$= 0.070. Assume that the materials behavior is similar to that of a spring and slider rheological model. Use Neuber’s rule to approximate the load–strain function $ε_{a}$ = g($S_{a}$), specifically, by the combination of Eqs. 12.54 and 13.61. To analyze point A, assume loading that follows the cyclic stress–strain curve as if it were a monotonic one, and employ Eq. 13.61 in the same manner, to obtain

$S_{A} = ψ\frac{1}{k_{t} } \sqrt{σ^{2}_{A} + ψσ_{A} E\left(\frac{ψσ_{A} }{H^{\prime }}\right) ^{1/n^{\prime }} } , ε_{A} = \frac{σ_{A} }{E} + ψ\left(\frac{ψσ_{A} }{H^{\prime }} \right)^{1/n^{\prime }}$

$ε_{a} =\frac{ σ_{a} }{E} +\left(\frac{ σ_{a}}{H^{\prime }}\right) ^{1/n^{\prime } }$                                    (12.54)

$S = \frac{1}{k_{t}}\sqrt{ σ^{2} + σ E \left(\frac{σ}{H}\right)^{1/n}}, k_{t} S =\sqrt{σ^{2} + σ E\left( \frac{σ}{H}\right) ^{1/n} }$                          (a, b)          (13.61)

Table 12.1           Constants for Cyclic Stress–Strain Curves for Four Engineering Metals

 Material Yield $σ_{o}$ Ultimate $σ_{u}$ Cyclic σ-ε curve E $H^{\prime }$ $n^{\prime }$ RQC-100 steel 683 (99) 758 (110) 200,000 (29,000) 903 (131) 0.0905 AISI 4340 steel 1103 (160) 1172 (170) 207,000 (30,000) 1655 (240) 0.131 2024-T351 Al 379 (55) 469 (68) 73,100 (10,600) 662 (96) 0.070 7075-T6 Al 469 (68) 578 (84) 71,000 (10,300) 977 (142) 0.106

Notes: Units are MPa (ksi), except for dimensionless $n^{\prime }$. See Table 14.1 for additional properties and sources.

Table 14.1    Cyclic Stress–Strain and Strain–Life Constants for Selected Engineering Metals.¹

 Material Source Tensile Properties Cyclic σ-ε Curve Strain–Life Curve $σ_{o}$ $σ_{u}$ $\overset{\sim }{\sigma }f_{B}$ % RA E $H^{\prime }$ $n^{\prime }$ $σ^{\prime }_{f}$ b $ε^{\prime }_{f}$ c (a) Steels SAE 1015 (normalized) (8) 228 (33.0) 415 (60.2) 726 (105) 68 207,000 (30,000) 1349 (196) 0.282 1020 (148) −0.138 0.439 −0.51 Man-Ten² (hot rolled) (7) 322 (46.7) 557 (80.8) 990 (144) 67 203,000 (29,500) 1096 (159) 0.187 1089 (158) −0.115 0.912 −0.606 RQC-100 (roller Q & T) (2) 683 (99.0) 758 (110) 1186 (172) 64 200,000 (29,000) 903 (131) 0.0905 938 (136) −0.0648 1.38 −0.704 SAE 1045 (HR & norm.) (6) 382 (55.4) 621 (90.1) 985 (143) 51 202,000 (29,400) 1258 (182) 0.208 948 (137) −0.092 0.260 −0.445 SAE 4142 (As Q, 670 HB) (1) 1619 (235) 2450 (355) 2580 (375) 6 200,000 (29,000) 2810 (407) 0.04 2550 (370) −0.0778 0.0032 −0.436 SAE 4142 (Q & T, 560 HB) (1) 1688 (245) 2240 (325) 2650 (385) 27 207,000 (30,000) 4140 (600) 0.126 3410 (494) −0.121 0.0732 −0.805 SAE 4142 (Q & T, 450 HB) (1) 1584 (230) 1757 (255) 1998 (290) 42 207,000 (30,000) 2080 (302) 0.093 1937 (281) −0.0762 0.706 −0.869 SAE 4142 (Q & T, 380 HB) (1) 1378 (200) 1413 (205) 1826 (265) 48 207,000 (30,000) 2210 (321) 0.133 2140 (311) −0.0944 0.637 −0.761 AISI 4340² (Aircraft Qual.) (3) 1103 (160) 1172 (170) 1634 (237) 56 207,000 (30,000) 1655 (240) 0.131 1758 (255) −0.0977 2.12 −0.774 AISI 4340 (409 HB) (1) 1371 (199) 1468 (213) 1557 (226) 38 200,000 (29,000) 1910 (277) 0.123 1879 (273) −0.0859 0.640 −0.636 Ausformed H-11 (660 HB) (1) 2030 (295) 2580 (375) 3170 (460) 33 207,000 (30,000) 3475 (504) 0.059 3810 (553) −0.0928 0.0743 −0.7144 (b) Other Metals 2024-T351 Al (Prestrained) (1) 379 (55.0) 469 (68.0) 558 (81.0) 25 73,100 (10,600) 662 (96.0) 0.07 927 (134) −0.113 0.409 −0.713 2024-T4 Al3 (soln. tr. & age) (4) 303 (44.0) 476 (69.0) 631 (91.5) 35 73,100 (10,600) 738 (107) 0.08 1294 (188) −0.142 0.327 −0.645 7075-T6 Al (5) 469 (68.0) 578 (84) 744 (108) 33 71,000 (10,300) 977 (142) 0.106 1466 (213) −0.143 0.262 −0.619 Ti-6Al-4V (soln. tr. & age) (1) 1185 (172) 1233 (179) 1717 (249) 41 117,000 (17,000) 1772 (257) 0.106 2030 (295) −0.104 0.841 −0.688 Inconel X (Ni base, annl.) (1) 703 (102) 1213 (176) 1309 (190) 20 214,000 (31,000) 1855 (269) 0.120 2255 (327) −0.117 1.16 -0.749

Notes: ¹The tabulated values either have units of MPa (ksi), or they are dimensionless.  ²Test specimens prestrained, except at short lives, also periodically overstrained at long lives. ³For nonprestrained tests, use same constants, except $σ^{\prime }_{f}$  = 900(131) and b = −0.102.
Sources: Data in (1) [Conle 84]; (2) author’s data on the ASTM Committee E9 material; (3) [Dowling 73]; (4) [Dowling 89] and [Topper 70]; (5) [Endo 69] and [Raske 72]; (6)  [Leese 85]; (7) [Wetzel 77] pp. 41 and 66; (8) [Keshavan 67] and [Smith 70].

Table E13.5

 (a) Load History (b) Calculated Values Point (Y) S MPa Origin (X) Origin S MPa Direction ψ ΔS to Point Δσ MPa Δε Stress σ, MPa Strain ε A 414 — — +1 — — — 503.3 0.02683 B −69 A 414 -1 483 900.3 0.02042 −397.0 0.00642 C 345 B −69 +1 414 857.3 0.01575 460.3 0.02217 D −310 A 414 -1 724 983.4 0.04200 −480.0 −0.01517 E 310 D −310 +1 620 954.6 0.03173 474.5 0.01656 F −172 E 310 -1 482 899.8 0.02034 −425.3 −0.00378 G 172 F −172 +1 344 784.9 0.01188 359.6 0.00810 H −241 E 310 -1 551 930.6 0.02571 −456.0 −0.00915

where ψ = +1 if the direction is positive (as is the case here), or ψ = −1 if the direction is negative. Substitute $S_{A}$ = 414MPa into the first equation, solve iteratively to obtain $σ_{A}$ = 503.3MPa, and then substitute this value into the second equation to obtain $ε_{A}$ = 0.02683.
Analysis beyond point A can now be done by applying Eqs. 12.54 and 13.61 to ranges that correspond to smooth hysteresis loop curves, employing these in the forms Δε/2 = g(ΔS/2) and Δε/2 = f (Δσ/2). First, identify points in the history where the memory effect acts and stress–strain hysteresis loops are closed, which are also points where rainflow cycles are completed, here at points $B^{\prime }$, $F^{\prime }$, $E^{\prime }$, and $A^{\prime }$. Then apply Eqs. 12.54 and 13.61 for each closed loop, B-C, F-G, E-H, and A-D, and also for ranges locating the starting points of loops, A-B, D-E, and E-F.
The calculations are organized as shown in Table E13.5(b). For each peak or valley (Y) in the history, the origin point (X) of the smooth hysteresis loop curve is tabulated, along with its S value. (For example, when loop B-C closes at point $B^{\prime }$, the origin for the continuing curve is point A, so that range A-D is analyzed to locate point D.) Next, ψ is listed as +1 if the strain is increasing during the range, or −1 if it is decreasing. Then each load range $ΔS_{XY}$ is calculated, from which each corresponding stress and strain range, $Δσ_{XY}$ and $Δε_{XY}$ , can be determined:

$\Delta S_{XY} =|S_{Y} − S_{X} |$

$\frac{ΔS_{XY}}{2} = \frac{1}{k_{t} } \sqrt{\left(\frac{σ_{XY} }{2} \right) ^{2} + \frac{σ_{XY} E}{2} \left(\frac{σ_{XY} }{2H^{\prime } }\right) ^{1/n^{\prime } } }$

$\frac{\Delta ε_{XY}}{2} = \frac{\Delta σ_{XY} }{2E} +\left(\frac{\Delta σ_{XY} }{2H^{\prime }}\right) ^{1/n^{\prime } }$

For example, the range from $S_{A}$ = 414 to $S_{B}$ = −69 is   Δ$S_{AB}$ = 483MPa. Entering this value into the second equation just given and solving iteratively yields $Δσ_{AB} = 900.3 \ MPa.$
Substituting the latter value into the third equation then gives $Δε_{AB}$ = 0.02042.
Finally, determine the stress and strain values for all peaks and valleys in the history by starting from the initial point A and calculating the stress and strain at each subsequent point.
Do so by adding or subtracting the appropriate ranges, using the equations

$σ_{Y} = σ_{X} +ψ Δσ_{XY} , ε_{Y} = ε_{X} +ψ Δε_{XY}$

where the use of ψ causes $Δσ_{XY}$ to be added to or subtracted from $σ_{X}$ to obtain $σ_{Y}$ , and similarly for the strains, depending on whether the S is increasing or decreasing. For example, for points
B and C,

$σ_{A} = 503.3, \Delta σ_{AB}$ = 900.3,                  ψ= −1,               so that $σ_{B} = −397.0MPa$

$ε_{A} = 0.02683, \Delta ε_{AB}$ = 0.02042,             ψ= −1,                so that  $ε_{B}$ = 0.00642

$σ_{B} = −397.0, \Delta σ_{BC}$= 857.3,                 ψ= +1,           so that $σ_{C } = 460.3MPa$
$ε_{B} = 0.00642, \Delta ε_{BC}$= 0.01575,                     ψ= +1,                   so that $ε_{C} = 0.02217$

Comments               The stress and strain values for the peaks and valleys in the load history, A, B, C, etc., can be plotted on stress–strain coordinates as in Fig. 13.20(e). Equation 12.54 in the form Δε/2 = f (Δσ/2) can be employed to calculate a number of points along each smooth curve connecting peak–valley points, while observing the memory effect. (The equations at the end of Ex. 12.5 apply.) Or the peak–valley points can be used as a guide to sketch the curves by hand, noting that only one curve shape is needed.