ME 4733: Deformation and Fracture of Engineering Materials Problem Set 1

ME 4733: Deformation and Fracture of Engineering Materials

Spring 2002

Problem Set 1

1) Hertzberg, 1.3 A 200-mm-long rod with a diameter of 2.5mm is loaded with a 2000-N weight. If the diameter decreases to 2.2 mm, compute the following: (a) The final length of the rod. (b) The true stress and true strain at this load. (c) The engineering stress and strain at this load. Solution: = = (a) Initial state: diameter do 25.mm, l0 200mm = Final state: diameter d f 22.mm, lf Assume this is a constant-volume process d 2 d 2 ππ0 ⋅=l f ⋅l 440 f  2 2 d 25.  ⇒ The final length of the rod is l =  0  ⋅=l   ⋅=200mm 258. 26 mm f   0    d f  22.

P 2000N (b) The true stress is σ = = = 526. 4MPa; true d 22(.22mm ) π ⋅ f 314. ⋅ 4 4 l ε ==f 258. 26 = The true strain is true ln ln 0. 2556. l0 200

P 2000N (c) The engineering stress is σ = = = 407. 4MPa; true d 22(.25mm ) π ⋅ 0 314. ⋅ 4 4 ll− − ε = f 0 = 258. 26 200 = The engineering strain is true 0. 2913. l0 200 Note: (1) In tensile test, the true stress is always higher than engineering stress. How about in compressive state? (2) In tensile test, the true strain is always lower than engineering strain. 2) Hertzberg, 1.8 A 5-cm-long circular rod of 1080 as-rolled steel (diameter=1.28cm) is loaded to failure in tension. What was the load necessary to break the sample? If 80% of the total elongation was uniform in character prior to the onset of localized deformation, computer the true stress at the point of incipient necking. ( refer to table 1.6a in Hertzburg textbook ) Solution: (.128cm )2 The initial cross-sectional area of the rod is A ==π 1. 287cm2. 0 4 σ The tensile strength, ts, is defined as the maximum load divided the initial cross- sectional area, which is at the top point of the engineering stress-strain curve. For 1080 σ = as-rolled steel ts 965MPa (from Hertzburg, table 1.6a). The load, P, necessary to break the sample should reach this top point, so =⋅=σ ×64 ⋅ ×− = ` PAts 0 965 10 1.. 287 10NN 124 2 k = The initial length of the rod is l0 5cm. From Hertzburg, table 1.6a, the total elongation is 12% for 1080 as-rolled steel. This total elongation consists mainly of two parts, uniform elongation (80%) and localized necking (20%), elastic deformation is small and can be neglected. At the point of incipient necking, the elongation is uniformly distributed. After that, deformation contributes only (or mostly) to localized necking. So, at the point of the incipient necking, the rod length is =+ × ⋅= × = ll(.1 0 12 0 .). 80 1 096 5cm 5 . 48 cm. At the point of the incipient necking the rod arrives at the state to bear the maximum load, which is Pk=124.N 2 . The plastic deformation can be approximated as constant- volume process, and the rod has uniform elongation for the process before necking, so the cross-sectional area at the point for incipient necking is l 5 AA==0 1. 287cm22 ×=1. 174 cm . 0 l 548. The true stress at the point of incipient necking is P 124. 2kN σ == =1058MPa. true A 1. 174× 10−42m

Note: The elongation at the point of the incipient necking is the uniform part of the elongation during necking. 3) Hertzberg, 1.11 A platform is suspended by two parallel rods, as shown in the sketch, with each rod being 1.28cm in diameter. Rod A is manufactured from 4340 steel [()]QT+°650 ( E = 210GPa, σ = = ys 855MPa); rod B is made from 7075-T6 aluminum alloy ( E 70GPa, σ = ys 505MPa). a) What uniform load can be applied to the platform before yielding will occur? b) Which rod will be the first to yield? (Hint: Both rods experience the same elastic strain)

Solution: b) For rod A, the strain at the yielding point is σ A 6 855× 10 Pa − εA ==ys =×4. 071 10 3 ys E A 210× 109 Pa For rod B, the strain at the yielding point is σ B 6 505× 10 Pa − εB ==ys =×7. 214 10 3 ys E B 70× 109 Pa The platform needs to be always kept horizontal, that means both rods experience εεA < B the same elastic strain. Since ys ys, so rod A will be the first to yield. b) The strain the system can experience without yielding is εεε==×A B −3 min(ys ,ys )4 . 071 10 The load applied to the two rods is PE=+()AB EAε So, −22 − (.1 280× 10 ) PNkN=+××××()210 70 1093 4 . 071 10 π =147 4 Note: εεε==×A B −3 Here we can see the elastic strain, for 4340 steel min(ys ,ys )4 . 071 10 is much smaller than plastic strain, with is in the order of 0.10. When strain is very small, there is no difference between engineering strain and true strain. 4) Hertzberg, 1.12 A 100-mm-long rod of 1340 steel [()]QT+°205 is subjected to a load of 50,000N. If the diameter of the rod is 10 mm, what strain would the rod experience? What strain would remain if the load were removed? Solution: The stress is σ ==P 50000N = 2 636. 6MPa A π 001. 4 The yield strength of 1340 steel [()]QT+°205 is 1590MPa. The current stress state is below the yielding point, so the elastic theory can apply. Since Young’s modulus does not change mush for different treatment, we can use the Young’s modulus for steel from the last problem as E = 210GPa. Using Hooke’s law, the strain in the rod is 6 σ 636. 6× 10 Pa − ε == =×303. 103 E 210× 109 Pa For elastic deformation the strain returns to zero when load removed.

Note: Here because the elastic strain is small, there is no difference between engineering and true strain. So in the problem, only “strain” is mentioned.