EML3005_HW01_Solution.pdf

# EML3005_HW01_Solution.pdf - EML 3005 Homework 1 Solution...

• Homework Help
• 5

This preview shows page 1 out of 5 pages.

Unformatted text preview: EML 3005: Homework # 1 Solution 1-8 CA = CB, 10 + 0.8 P = 60 + 0.8 P - 0.005 P2 P2 = 50/0.005 P = 100 parts Ans. 1-9 Max. load = 1.10 P Min. area = (0.95)2A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be nd 1.10 1.43 0.85(0.95) 2 S 16(1000) 25(103 ) d 0.799in. nd d3 2.5 Ans. 1-12. Table A-17: 7 d in. 8 Ans. n Factor of safety: Ans. S 25(103 ) 3.29 16(1000) 7 Ans. 3 8 1-14 From Eq. (1-6), x 1 N k fx i 1 i i 1 (39125.973) 198.609 198.6kpsi 9 Ans. From Eq. (1-7), k sx fx i 1 2 i i Nx N 1 2 18433.628 9.69kpsi 196 Ans. EML 3005: Homework # 1 Solution 2-­‐6 (a) A0 = π (0.503)2/4 = 0.1987 in2, σ = Pi / A0 For data in elastic range, ı = Δ l / l0 = Δ l / 2 A Δl l − l0 l For data in plastic range, Ú = = = − 1 = 0 − 1 l0 l0 l0 A On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1-­‐7. Figure (b) shows data points 1-­‐12. Figure (c) shows the complete range. Note: The exact value of A0 is used without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be σ = 30.5(106)ı − 61 000 (1) The equation for the line between data points 8 and 9 is σ = 7.60(105)ı + 42 900 (2) Solving Eqs. (1) and (2) simultaneously yields σ = 45.6 kpsi which is the 0.2 percent offset yield strength. Thus, Sy = 45.6 kpsi Ans. The ultimate strength from Figure (c) is Su = 85.6 kpsi Ans. The reduction in area is given by Eq. (2-­‐12) is A0 − Af 0.1987 − 0.1077 R= (100 ) = (100 ) = 45.8 % Ans. A0 0.1987 Data Point Pi Δl, Ai ı σ 1 2 0 0 0 1000 2000 3000 4000 7000 8400 8800 9200 8800 9200 9100 13200 15200 17000 16400 14800 0.0004 0.0006 0.001 0.0013 0.0023 0.0028 0.0036 0.0089 0.1984 0.1978 0.1963 0.1924 0.1875 0.1563 0.1307 0.1077 0.00020 0.00030 0.00050 0.00065 0.00115 0.00140 0.00180 0.00445 0.00158 0.00461 0.01229 0.03281 0.05980 0.27136 0.52037 0.84506 0 5032 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 10065 15097 20130 35227 42272 44285 46298 44285 46298 45795 66428 76492 85551 82531 74479 Stress (psi) 50000 40000 y = 3.05E+07x -­‐ 1.06E+01 30000 20000 Series1 10000 Linear (Series1) 0 0.000 0.001 0.001 0.002 Strain Stress (psi) (a) Linear range 50000 Y 45000 40000 35000 30000 25000 20000 15000 10000 5000 0 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (b) Offset yield 90000 U 80000 Stress (psi) 70000 60000 50000 40000 30000 20000 10000 0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Strain (c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, Sut = 82 kpsi, and R = 40 %. Ans. 2-­‐7 To plot σ true vs.ε, the following equations are applied to the data. P σ true = A Eq. (2-­‐4) l ε = ln for 0 ≤ Δl ≤ 0.0028 in (0 ≤ P ≤ 8 400 lbf ) l0 A0 ε = ln for Δl > 0.0028 in (P > 8 400 lbf ) A π (0.503)2 = 0.1987 in 2 where A0 = The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log σ vs log ε 4 The curve fit gives m = 0.2306 log σ0 = 5.1852 ⇒ σ 0 = 153.2 kpsi Ans. S y = 153.2(0.002) 0.2306 = 36.54 ksi (Note: Sy from offset = 45.6 ksi) For 20% cold work, Eq. (2-­‐14) and Eq. (2-­‐17) give, A = A0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2 A 0.1987 ε = ln 0 = ln = 0.2231 A 0.1590 Eq. (2-18): S yʹ′ = σ 0ε m = 153.2(0.2231)0.2306 = 108.4 kpsi Ans. Eq. (2-19), with Su = 85.6 from Prob. 2-6, Suʹ′ = Su 85.6 = = 107 kpsi 1 − W 1 − 0.2 P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800 Δl 0 0.000 4 0.000 6 0.001 0 0.001 3 0.002 3 0.002 8 A Ans. ε σtrue 0.198 7 0 0 0.198 7 0.000 2 5 032.71 0.198 7 0.000 3 10 065.4 0.198 7 0.000 5 15 098.1 0.198 7 0.000 65 20 130.9 0.198 7 0.001 15 35 229 0.198 7 0.001 4 42 274.8 0.198 4 0.001 51 44 354.8 0.197 8 0.004 54 46 511.6 0.196 3 0.012 15 46 357.6 0.192 4 0.032 22 68 607.1 0.187 5 0.058 02 81 066.7 0.156 3 0.240 02 108 765 0.130 7 0.418 89 125 478 0.107 7 0.612 45 137 419 log ε -­‐3.699 -­‐3.523 -­‐3.301 -­‐3.187 -­‐2.940 -­‐2.854 -­‐2.821 -­‐2.343 -­‐1.915 -­‐1.492 -­‐1.236 -­‐0.620 -­‐0.378 -­‐0.213 log σtrue 3.702 4.003 4.179 4.304 4.547 4.626 4.647 4.668 4.666 4.836 4.909 5.036 5.099 5.138 ...
View Full Document

• Spring '09
• NagarajArakere
• Trigraph, Eqs

{[ snackBarMessage ]}

###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern