Coursehero >> New York >> Cornell >> MSE 261

Course Hero has millions of student submitted documents similar to the one below including study guides, homework solutions, papers, exam answer keys and textbook solutions.

MSE 2610 Introduction to the Mechanical Properties of Materials Fall 2008 Homework #4 Solutions Due: 02 October 2008 1.) A true stress vs. true strain curve modeled as the function 0 < < 0.002 (80 GPa) = 0.3 0.002 < (3.58 GPa) The original unloaded sample had a diameter of 10 mm. a.) What is the ultimate tensile strength of this material? T @ UTS = m = 0.3 UTS = (3.58 GPa )(0.3) = 2.49 GPa b.) What is the reduced cross-sectional area of the sample at strains of 0.01, 0.05, and 0.1? A = A o exp(- T ) 0.3 A o = r 2 = (0.005 m) 2 = 7.85 10 -5 m 2 Ao (m2) T (no units) A (m2) 7.85 x 10-5 0.01 7.78 x 10-5 -5 7.85 x 10 0.05 7.47 x 10-5 7.85 x 10-5 0.1 7.11 x 10-5 2.) An extremely long hollow steel tube has an internal pressure of 1 MPa. The internal pressure is increased to some amount P2 such that the circumference of the tube increases by 0.06%. The modulus of the material is 200 GPa and the thickness of the tube is 0.5 cm. The thickness of the tube is 1 % of the diameter [Note: This allows for treatment of the problem as a thin-walled tube.] From an uniaxial tension test of the same material, it was found that the steel had a yield strength of 180 MPa. a.) Find P2. -A 0.06% increase in circumference corresponds to a strain in the material of 0.0006. Neglecting Poisson strain in the thickness of the pressure vessel wall, one can write: 0.5 cm d1 = = 50 cm 0.01 r1 = d1 / 2 = 25 cm r2 = r1 (1 + ) = (25 cm)(1 + 0.0006) = 0.25015 m Pr = E t Pr (1 MPa )(25 cm) 1 = = = 2.5 10 -6 Et (200 GPa )(0.5 cm) = P2 - P1 = E 2 t E 1 t - = Et 2 - 1 r r2 r1 2 r1 (2.5 10 - 6 + 0.0006) 2.5 10 - 6 = 2.40 MPa P2 - P1 = 200 GPa )(0.005 m) - 0.25015 m 0.25 m P2 = 3.4 MPa Note that assuming a constant radius does not change the result that much. b.) Find the stresses in the tube at 1 MPa and at P2. P r (1.0 MPa )(0.25 m) 1 = 1 = = 50 MPa t (0.005 m) P r (3.4 MPa )(0.25015 m) 2 = 2 = = 170 MPa t (0.005 m) c.) At what pressure does yield first occur in the tube? y t (180 MPa )(0.005 m) = = 3.6 MPa Py = r (0.25 m) d.) If the material obeys the Tresca yield criterion, how thick would the walls of the tube need to be to avoid failure if the internal pressure was raised to 9 MPa. -Tresca Criterion for shear: - 3 max = 1 1 - 3 y 2 The greatest stress (1) in a thin-walled cylindrical pressure vessel is always the hoop stress. The least stress is zero because it is a thin-walled pressure vessel. Therefore: Pr Pr (9 MPa )(0.25 m) y = t= = = 1.25 cm t y (180 MPa ) e.) How thick would the walls need to be if the internal pressure was raised to 9 MPa if the yield criterion follows von Mises. - 1 is equal to the hoop stress, 2 is equal to the longitudinal stress 3 is equal to zero 1 = Pr t 2 = Pr 2t 3 = 0 1 2 2 2 = (1 - 3 ) + (1 - 2 ) + ( 2 - 3 ) 2 1 Pr 2 Pr 2 Pr 2 = + + 2 t 2 t 2 t y < 1/ 2 [ ] 1/ 2 = 3 Pr 2 t 3 Pr 2 t 3 Pr 3 (9 MPa )(0.25 m) t= = = 1.08 cm 2 y 2 (180 MPa ) 3.) What class of materials was the Tresca failure criterion developed for? Why would the Tresca failure criterion not accurately predict when a ceramics would fail? - The Tresca failure criteria was primarily developed for ductile metals. Because of their atomic structure, ductile metals yield and plastically deform via shear significantly before failure. - The Tresca failure criteria is a shear criteria, meaning it can only predict when a material will fail due to shear. Ceramics, due to their brittleness, typically fracture before yielding. Fracture is due to normal stresses rather than shear stress. 4.) If the defects below were introduced into an otherwise perfect BCC iron lattice, what lattice stresses would predominate around these defects? Why? - Overall, there are tensile and compressive stress es present in all of these scenarios. A long way from the defect, the crystal must return to perfect periodicity. The best way to consider which predominates is to think about: 1.) Does adding or subtracting these atoms force more than the equilibrium number of atoms into the same volume (compression) or does it force fewer than the equilibrium number of atoms in the same volume (tension) 2.) Is the defect stress field spherically homogeneous? Does the defect affect different parts of the crystal in different ways? a.) Interstitial carbon atom - compressive stress ; the equilibrium size of the interstitial site is smaller than the radius of the carbon atom. b.) Substitutional tin atom - compressive stress ; the tin atom has a larger atomic radius than the iron atoms. c.) Iron atom vacancy - Tensile stress; the missing causes the bonds around it to length to try to close the gap in the material. d.) Edge dislocation - Tensile and compressive stress ; unlike the other defects, a dislocation is not spherically symmetric. In the area around the inserted half-plane, there is compressive stress . Below the half-plane, there is tensile stress. (See figure from the text below) Compressive Tensile Burgers vector Start End Note that the Burgers vectors for all edge dislocations are perpendicular to their sense vectors. The Burgers vectors for screw dislocations are either parallel (RHS) or antiparallel (LHS) to their sense vectors. Segment A-B B-C C-D D-E E-F F-G G-H H-I I-J J-A Dislocation Type LH Screw Edge RH Screw Edge Edge RH Screw Edge LH Screw Edge Edge Direction of Movement + x1 + x2 - x1 - x2 0 - x1 - x2 + x1 0 - x2 This means the loop expands. This makes sense because the slipped part of the lattice is a lower strain energy than the unslipped portion, so increase the area inside the loop will result in an overall decrease in the energy of the system.

Find millions of documents here - Study Guides, Homework Solutions, Papers, Exam Answer Keys and more. Course Hero has millions of course related materials that will enable you to learn better, faster and get an A in all your courses.
Below is a small sample set of documents: