Oct 2006 w solutions

Oct 2006 w solutions - Applied Science 278 Midterm...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Applied Science 278 Midterm Examination - October 18, 2006 1 THE UNIVERSITY OF BRITISH COLUMBIA Department of Materials Engineering APSC 278 Engineering Materials MID-TERM EXAMINATION, October 18, 2006 Name: Student #: This is a closed book examination. Use of relevant material stored in programmable calculators prohibited. Answer all questions. Show all work and units. The complete exam is 8 pages in length
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Applied Science 278 Midterm Examination - October 18, 2006 2 (7) A standard tensile sample (meeting ASTM standards) of an aluminum alloy was tested with the resulting engineering stress-strain curve shown below. Determine the following parameters: i. the modulus of elasticity ii. 0.2% yield stress iii. the ultimate tensile stress iv. uniform elongation a) b) Figure 1 – Engineering stress-strain curve for aluminum alloy AA6111. Note: b) is an expanded view of a) in the low strain regime. The gauge length of the sample was 50 mm. Solution: i. According to the Fig. 1b : GPa E 67 003 . 0 200 = = ε σ , accepted in the range of 60- 75GPa ii. According to the Fig. 1b : MPa offset y 300 290 ) % 2 . 0 ( iii. According to the Fig. 1a: UTS 340MPa iv. According to the Fig. 1a: Uniform elongation 0.085-0.09 or 8.5-9% v. According to the Fig. 1a: Total elongation 0.1025 or 10.25% vi)due to lack of a material, subsequent tests had to conducted on tensile samples which had a gauge length of 10 mm. Based on the data shown in Figure 1, estimate the total elongation that you would expect for these samples. Solution: Assume: uniform elongation 0.085-0.09 Necking strain 0.1025-(0.085 or 0.09) 0.0175-0.0125 0 100 200 300 400 00 . 0 5 0 . 1 strain stress (MPa) 0 100 200 300 400 0 0.005 0.01 0.015 0.02 strain iii ii i v iv 0.002
Background image of page 2
Applied Science 278 Midterm Examination - October 18, 2006 3 For ASTM standard gage length (50mm): l/ l 0 = ε then l (uniform)=50×(0.085 to 0.09)= 4.25 to 4.5 mm and l (necking)=50×(0.0175 to 0.0125)=0.875 to 0.625 mm For 10mm gage length: l (uniform)=10×(0.085 or 0.09)/50= 0.85 to 0.9and
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

Oct 2006 w solutions - Applied Science 278 Midterm...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online