3. Using your answers to the two previous questions, evaluate the Mratio

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS MECHANICS AND MATERIALS II HOMEWORK NO. 4 Distributed: Friday, April 2, 2004 Due: Friday, April 9, 2004 Problem 1 (20 points) Note: for reference material, consult te laboratory write up on elastic plastic beam bending Consider te square cross section beam sown, of dimensions by, subject to diamondorientation bending in te plane sown (neutral axis: plane y = 0). Te beam can be considered to be composed of an elastic/perfectly plastic material aving Young s modulus E, and tensile yield strengt σ y. 1. Using te standard assumptions of engineering beam teory, evaluate te magnitude of applied moment, M y, just sufficient to bring te most igly stressed region to te verge of yielding. Express your answer in terms and of material properties, as appropriate. (Aside: are you surprised by te value you got I = for y 2 da in tis orientation?) 2. If te applied curvature is increased to very large values, te elastic/plastic boundaries (tension and compression sides) in tis geometry, like tose in te bending of rectangular cross sections studied earlier, will move inward, toward te neutral axis. At infinite curvature, te boundaries will reac opposite sides of y te = 0 surface, resulting in tensile yielding stress values of magnitude σ y in one triangle alf of te cross section, and compressive yielding stress values of magnitude σ y in te oter triangular alf of te cross section. At tis point, te bending moment carried by te cross section reaces a limiting value, M L. Evaluate M L for tis section.. Using your answers to te two previous questions, evaluate te Mratio L /M y for bending of tis section. How does tis value compare wit te ratio for bending of tis same cross section, but on rotated axes, so tat te cross section appears as a square? (Our usual orientation for bending.) 4. Compare M y for te diamond cross section wit te corresponding M y for te square orientation. Wat is te ratio of tese first yield bending moments? Explain wy tey differ in te way tat tey do. Evaluate te same ratio for te corresponding limit moments, and M L, and comment on reasons wy tey differ. Wic axes sould be used for applying bending moments to a square section, and wy? 1

2 5. Discuss te residual stress state wen te diamond orientation is unloaded M to = 0 immediately after being deformed to large curvature M at = M L. How does tis residual stress state compare or contrast wit te state for unloading of te square orientation from its limit value M? of Can any negative moment be applied to te diamond cross section after unloading from limit load, witout causing furter plasticity? Discuss y x Figure 1: Square cross section of beam, oriented for bending along diamond orientation. Problem 2 (0 points) A great deal of te mecanisms and penomenology of te strengtening of metallic crystals can be summarized in te following prase: Smaller is stronger... Discuss tree specific examples of strengtening mecanisms, and explain ow and wy te aporism smaller is stronger applies to eac strengtening mecanism. 2

3 Problem (0 points) Standard cylindrical compression specimens ave an initial eigt to diameter ratio of H 0 /D 0 = 2. It is desired to conduct a compression test in a demonstration lab, and to compress te specimen to a final eigt H of = H 0 /2. From prior testing, it is known tat te material as Young s modulus E = 200 GP a, Poisson ratio ν = 0., and its plasticity can be well caracterized by an initial value of tensile/compressive yield strengt as 0 = 500 MP a, along wit a constant ardening modulus, = 2 GP a, governing te evolution of uniaxial flow strengt, s, wit equivalent plastic strain, ɛ p, according to ds = = constant. d ɛ p In turn, tis expression can be integrated to express te current value of strengt, for any given value of ɛ p 0, as s( ɛ p ) = s 0 + ɛ p. Te load cell on te testing macine to be used for te compression test as a maximum load capacity of 100 kn. You are asked to provide an answer to te following question: Wat is te largest allowed value of initial diameter in a compression specimen of tis material D( 0(max) ) tat can be safely compressed to alf its initial eigt in te testing macine? In particular: (10 points) Explain wy te elastic strain is not an important feature in answering tis problem. Tat is, explain wy, for tis application, you may assume tat te material is rigid/plastic, so tat te total strains and strain rates are essentially equal to te plastic strains and strain rates, respectively. (20 Points) Wat is te largest diameter tat can safely be used for te compression specimen, under te imposed conditions? HINTS: Remember, for active yielding in uniaxial compression, te axial [true]stress, σ, is negative, so te yield criterion becomes s = σ = σ.. For monotonic loading in compression, te plastic portion of te [true] axial strain, ɛ = ɛ (p), is negative, and is tus related to te equivalent plastic strain ɛ (p). by = ɛ = ɛ p.

4 Problem # 4 (20 points) Long bars of an alloy steel are available in stock of rectangular cross section, wit [initial] tickness t 0 = 25 mm and widt w 0 = 100 mm. It is desired to use tese bars as tensileloaded truss members, and to be able to apply tensile loads Pup max to = 1.1M N witout causing plastic yielding in te bars. Te initial tensile yield strengt of te steel σ y = is 50 M P a. Can te as received bars support a load of magnitude P max = 1.1M N witout yielding? How muc tensile load can it support witout yielding? It is known tat te tensile flow strengt, s, of tis steel increases wit equivalent tensile plastic strain, ɛ p, according to ( ɛ ) p N s( ɛ p ) = σy 1 +, c were te strain ardening exponent N is = 0.14, and te constant c = Someone suggests tat it may be possible to cold roll te bar stock to a new cross sectional sape, of reduced tickness t, but essentially te same widt, w = w 0, and in te process generate enoug equivalent plastic strain and associated strain ardening so tat te rolled bar stock can be used as truss members tat can support tensile loads up to P max = 1.1M N witout [furter] plastic yielding, even toug te rolling reduces te tickness and cross sectional area of te bar. We will explore tis possibility. First note tat te equivalent plastic strain increment, dɛ p, can be expressed in terms of te cartesian components of te plastic strain increment tensor, dɛ (p) ij, by 2 d ɛ p = dɛ (p) dɛ (p) ij ij. i=1 j=1 Let te rolling direction (along te lengt of te bar) be cartesian direction number 1, let te troug tickness direction be 2, and let te breadt direction be. In te process of rolling, tere is an incremental reduction in tickness, dt < 0, so tat dɛ (p) = dt 22 < 0. t. As noted above, tere is negligible transverse plastic straining in rolling, dɛ (p) so = 0. Assume furter tat rolling introduces no cange in plastic sear strains dɛ (i.e., (p) = dɛ (p) = dɛ (p) = 0). Obtain an expression for d ɛ p in terms of t and dt, and sow ow tis expression can be integrated to give ( ) 2 t0 ɛ (p) = ln. t HINT: someting needs to be done about evaluating dɛ (p)

5 Wat is te maximum rolling reduced bar tickness, t = t max, wic gives a strain ardened strengt s and rolling reduced tickness t = t max combination suc tat te cold rolled bar stock does, indeed, support tensile load P mx = 1.1MN witout furter yielding? Note: tis part of te problem may best be solved by performing a set of numerical evaluations, for different values of tickness, and finding out wic t value answers te question. 5