Chapter6 hw

Chapter6 hw - Chapter 6, Problem 13 Calculate the percent...

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Unformatted text preview: Chapter 6, Problem 13 Calculate the percent cold reduction when an aluminum wire is cold-drawn from a diameter of 5.25 111111 to a diameter of 2.30 111111. Chapter 6, Solution 13 initial area —fi11al area % cold reduction = _ _ , inttial area :| X10091) _[(a:,«’4)(5.25 111111): —(a:_/4)(2.30 111111): a 2 x1009?) = 80.8% (JI,-*4)(S.25 111111) Chapter 6, Problem 14 A 0.15-in.-dia1neter 99.5% copper wire is to be cold-drawn with a 30 percent cold reduction. What 11111stl1e the final diameter of the wire? Chapter 6, Solution 14 . . filo Af . 91:: cold reduction = 4 x l 000 11 I 0 E(0.1511192 —£d2 4 4 f 30.0: K7 X10091) — 0.15111. 3 4( ) (if 20.1255111 Chapter 6, Problem 18 Calculate the engineering stress in SI units on a 2.00-0111—dia111eter rod that is subjected to a load of 1300 kg. Chapter 6, Solution 18 First, the load must be converted to a force. F = ma = (1300 kg)(9.81 111/52) = 12. 753 N . The engineering stress is then. ’1' = = 406x106 Pa = 40.6 MPa 1 AG $0.02 m)2 a: Chapter 6, Problem 23 A tensile specimen of cartridge brass sheet has a cross section of 0.320 in. X 0.120 in. and a gage length of 2.00 in. Calculate the engineering strain that occurred during a test if the distance between gage markings is 2.35 in. after the test. Chapter 6, Solution 23 3—2,, _ 2.35 le.-2.00 in. _ to 2.00 in. engineering strain 8 — 0.175 Chapter 6, Problem 24 A 0.505-in.-dia1nete1‘ rod of an alumian alloy is pulled to failure in a tension test. If the final diameter of the rod at the fiactured surface is 0.440 in, what is the percent reduction in area of the sample due to the test? Chapter 6, Solution 24 Chapter 6, Problem 25 The following engineering stress-strain data were obtained for t 0.3% C‘ plain-carbon steel. ((2) Plot the engineering stress-strain curve. ([3) Determine the ultimate tensile strength of the alloy. (C) Determine the percent elongation at fracture. Engineering Engineering Engineering Engineering Stress stress strain strain (in..-'in.) (ksi) [ks-i) (in. fin.) 0 0 7 6 0.08 3 0 0. 00 I T 5 0. l 0 5 5 0.002 7 3 0. 12 60 0. 005. 69 0. l4 6 8 0.0 l 6 5 0. I 6 T2 0 . 02 5 6 0. I 8 T4 0.04 51 (Fracture) 0.19 T 5 0.06 Chapter 6, Solution 25 Engineering Engineering Engineering Engineering Prob. 6.25: Stress v5. strain Stress Strain Stress Strain (ksi) (infinJ (ksi) (in..-"in.) n 0 0 76 0.08 g 30 0.001 75 0.10 E 5.5. 0.003 T3 0.13 0:) 60 0.005 69 0.14 68 0.010 65 0.16 T2 0.020 56 0.13 T4 0.040 51 0.19 0.00 0.05 0.10 0.15 020 T5 0.060 (Fracture) Strain tinfin.) (a) See stress-strain plot above. (b) The ultimate tensile strength. based on the stress-strain curve. is 76 ksi. (c) % elongation = engineering strain X 100% = 0.19 X10094: = 19% . Chapter 6, Problem 29 A 0.505-in.-dia1neter aluminum alloy test bar is subjected to a load of 25 .000 lb. If the diameter of the bar is 0.490 in. at this load. determine (a) the engineering stress and strain and (b) the true stress and strain. Chapter 6, Solution 29 3012 . - . '1 Area at stafthO = 40 2%(0505111)2 2 0.200111‘ 2rd}; 2: Area under loadil. = (0.490 in.)2 = 0.1886 in2 4:] Assuming no volume change during extenslon. 14030 = 147.31. 01‘ If. (0 = :10 .9" _ I F 25,000 lb f A Engineering stress 2 — = +2 2 125, 000 psi AD 0200 111 - 2 B Engineering strain — a — E’ 30 — AG 1 — 0200 1.11 2 2’0 A,- 0.1886 111 F 25,000 lb); True stress 2 0'?- =— 2 —I2 = A?- 0.1886111 . r 2 ' 3 True strain — SI — 1n 2’ — in A0 — 1n 0' 00 If] 2 —h1(1.0604) = 0.0587 30 A. 0.1886111 I l — 1.0604 —1 = 0.060 132, 600 psi Chapter 6, Problem 46 A stress of 75 MPa is applied in the [O 0 1] direction on an FCC single crystal. Calculate (a) the resolved shear stress acting 011 the (111) O 1] slip system and. (b) the resolved shear stress acting on the (111) 1 0 ] slip system. Chapter 6, Solution 46 The resolved shear stress is calculated as: If, I 0' cos 21008 = 75 cos 27., cos . we must therefore determine the values of the angles for A and (a) The slip plane and direction of the (1 1 1) [ 1 01 ] slip system are shown on the next page in Figures (a) through (c). Specifically. Fig. (a) illustrates the (1 1 1 ) slip plane. L. the angle between the 01 ] slip direction and the applied axial stress direction. [0 0 l] . Referring to Fig. (13). the (001) plane is bisected by line AE and thus. l=R EAD=45°. Since we are concerned with the cubic crystal system. the direction normal to the slip plane is simply the Miller indices of that plane. For the present case. the direction normal to the (1 1 1 ) is thus [ l l l Referring to Fig. (c). rectangle ABij shows between the I: l l l ] normal and the applied axial stress direction. [0 0 l] . 3y geometry. ” =i=0.5774. 8:54.7" JEJE I", : (75 MPa)(cos 45°)(cos 54.70) : 30.6 NIPa c0395 = [1 D 1] Slip Direction [1 1 1] Normal to Slip Plane (11 1) Slip Plane (b) 3. on (001) Plane [111] (c) a 011(110) Plane (b) Referring to Fig. ((31). the slip plane and direction of the (111) l O] slip system are shown. As depicted by this sketch. the direction indices of the direction normal to the (l l 1) plane are [1 1 1], and the angle 5‘. is 90". The angle 11?! is shown below in Fig. (e) to lie between the axial [0' '0 l ] direction and the normal to the slip plane. {andillllu From rectangle ABC). this angle is calculated as: COS¢ = a {IS/Ea] =1f-J.— = 0.5774, [ 1— 1 0 ] Thus. 1}. : (75 MPa)( cos 90°)(cos 54.70) : 0. [0 0 1] Axial Stress Direction _ 0 [111] (111) (p SlipPlan [001] Slip Direction (e) (b 011(110) Plane ((1) Slip System Chapter 6, Problem 47 A stress of 55 MPa is applied in the [001] direction of a BC‘C single crystal. Calculate (a) the resolved shear stress acting on the (101) [ system and (b) the resolved shear stress acting on the (110) [ l l system. Chapter 6, Solution 47 (a) First. the direction normal to the (101) plane is shown below in Fig. (a) to be the [101] direction. From Fig. (b). the value of angle It. between the applied stress direction [0 0 1] and the 1 l] slip direction. is dictated by the geometry ofrectangle ABC‘D: COS A = COS (I r". [£0] = 0.5774, A = 54.70. Finally. the angle lies between the [0 0 1] stress direction and the [101] normal and is thus equal to 45°. Substituting into Eq. (5.15). 1', = (55 MPa)(cos 54.?O)( cos 450) = 22.5 NIPa. (a) (b) [0 01] _ _,_ =80 b = “an, =Ce.-\i.hz1.| K.R'| _,__ =6 11 =CEa-ar. =C.ét‘:l-.‘1H: ml [101] Normal .1' (c) (101) Slip Plane (b) The direction normal to The (l 10) slip plane. [I l U] . and the 1 1] slip direction are shown below in Fig. ((1). From Fig. (e). the value of angle 1. is calculated :19: C08 /1 : COS {.1 [J50] : 0.5774. 21- : 54.70. Finally. the angle lies between the [0 01] stress direction and the [1 1 0] normal. and is thus equal to 90°. Substituting into Eq. (5.15). If : (55 MPa)(cos 54.7O)(cos 900) : 0. (d) : T [l] 1] Slip Direction (6) [I11] ([10) Slip Plane [110]Nor‘11m1 Chapter 6, Problem 48 Determine the tensile stress that must be applied to the [l l 0] axis of a high-purity copper single clystal to cause slip on the (l T T )[0 T 1] system. The resolved shear stress for the ctystal is 0.85 MPa. Chapter 6, Solution 48 Referring to Fig. (a) below. the angle l. between the [l T 0] stress application and the [0T 1] slip direction is 3.- : COS_1[(COS 450)2] = 600. The angle . as depicted in Fig. (‘0). lies between the [l 0] stress direction and the [l T normal and is calculated as: COSQ = V60 fir: = 0.8165. = 35.30. Thus. the tensile stress is: a Z 1:}. Z 0.83 MPa 2 2.08 MPa cos it cos (3’) (cos 6OO)(cos 35 .30) (3,) [0T 1] Slip (:3) Direction Nonnal Chapter 6, Problem 63 An oxygen-free copper rod must have a tensile strength of 50.0 ksi and a final diameter of 0.250 in. (a) What amount of cold work must the rod undergo (see Fig. 6.45)? (b) What must the initial diameter of the rod be? 60 ‘Ji a-l-L 'H-r Yield strength Tensile strength and yield strength (Rail [10 E so I" i 20 =10 .E '5 an a .3: I'll El] "- 0 0' l 1'} 20 31'} 4'0 50 fill Percenl nil-Id wurk Figure 6.45 Chapter 6, Solution 63 (a) From Fig. 6.45, to attain a tensile strength of 50.0 ksi, the amount of cold work must be 40 percent. (b) The initial diameter, based on 40 percent cold work is: 5d? —£(0.250 in.)2 0.40 :% _d2 4 1 d12—-0.40d12 = 0.0625 1112, (21 = 0.323 in. Chapter 6, Problem 65 A F096 Cu—3 0% Zn brass wire is cold-drawn 20 percent to a diameter of 2.80 111111. The wire is then further cold- drawn to a diameter of 2.45 nun. (0) Calculate the total percent cold work that the wire undergoes. (5) Estimate the Wire‘s tensile and yield strengths and elongation from Fig. 6.46. 120 70% Ctr—30% Zn Tensile strength and yield strength (ksi) Elongation percent 0 10 20 30 4O 50 60 Percent cold work Chapter 6, Solution 65 (a) To calculate the total pcrc out cold work. the initial wire diameter must be dctermincd: Edi —£(2.80 mm)2 0.20 = 4 4% _d12 4 0’12— 0.20032 = ?.84 111mg, «’1 = 3.13 nun The total cold work is thus. 3.13 2 — 2.45 2 , 9.79?—6.003 total % cold work — ( mm) ( _ 2 mm) x100% = —X'100% (3.13 111.111) 9.?97 = 38.7 % b From Fig. 6.46. for ere ent cold work of a roxiinatel ’ 39%. read: . P PP fr Ultimate Tensile Strength :5 T5 ksi: Yield Strength 2: 62 ksi: and Elongation 2“: 5 9/0. ...
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This homework help was uploaded on 03/20/2008 for the course MSE 200 taught by Professor Wholedepartment during the Spring '08 term at N.C. State.

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Chapter6 hw - Chapter 6, Problem 13 Calculate the percent...

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