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t4145s08PracProbSolnRR - ENGR 145 — Chemistry of...

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ENGR 145 — Chemistry of Materials Page 1 of 7 Practice Problems for Exam 4, Spring 2008 Question 1: Optical Properties in Transmission a) (5 points) OGC problem 20.19 — The name of the organic dye “indanthrene brilliant orange” indicates the color of light transmitted through it. In what range of wavelengths is the absorption band of this dye? b) (5 points) A solution of indanthrene brilliant orange is placed in a test tube with a diameter of 1.0 cm. Assume that absorption due to the test tube, and reflection from both the tube and the solution, are negligible. A measurement finds that 20% of the incident light at a particular wavelength is transmitted through the specimen. Compute the optical absorption coefficient of the solution at this wavelength. Solution: a) Transmitting orange light when white light is incident arises from absorption of blue (the complementary color to orange). So the absorption band for this compound would fall in the range of 450-500 nm. b) Use the Lambert-Beer law (C&R eq. 19.18): I t I i = exp ! " x [ ] = 20% x = 1.0 cm # " = ! ln(0.20) 1.0 cm = 1.6 cm ! 1
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ENGR 145 — Chemistry of Materials Page 2 of 7 Practice Problems for Exam 4, Spring 2008 Question 2: Elastic Behavior of Materials Show your work in the blank space below and transfer your final answers to the table. A rod, 25.0 cm long and with circular cross section, is to support a load in uniaxial tension along its length. For each of the candidate materials listed below: a) (6 points) Compute the minimum diameter of the cylinder to ensure that a load of 44.4 kN will not exceed the listed yield strength. b) (6 points) At a stress equal to the yield strength of each material, compute the elongation of the cylinder. c) (8 points) Compute the modulus of resilience for each material. Solution: material elastic modulus, GPa yield strength, MPa diameter, cm elongation, mm modulus of resilience, MPa steel 207 355 1.26 0.429 0.304 aluminum oxide 303 337* 1.30 0.278 0.187 nylon 2.69 55 3.21 5.11 0.562 *) fracture strength a) σ = F/A < σ y . A F/ σ y (1 pt) A = π d 2 /4 d (4F/ π σ y ) 1/2 (1 pt). (1 pt) per answer in table above. b) Although yielding has begun (for the metal and the polymer) if the stress reaches the yield strength, the yield strength is often taken to approximate the stress at which yielding just begins, and Hooke’s law ( σ =E ε ) is estimated to hold up to the yield strength. Under these assumptions, elongation = l o ×ε (1 pt) = l o ×σ /E where stress here is taken to equal σ y (1 pt). Then (1 pt) per answer in the table above.
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