Mechanics of MaterialsThis text provides a clear, comprehensive presentation of both the theory and applications of mechanics of materials. The text examines the physical behaviour of materials under load, then proceeds to model this behaviour to development theory. The contents of each chapter are organized into welldefined units that allow instructors great flexibility in course emphasis. writing style, cohesive organization, and exercises, examples, and free body diagrams to help prepare tomorrow's engineers. The book contains over 1,700 homework problems depicting realistic situations students are likely to encounter as engineers. These illustrated problems are designed to stimulate student interest and enable them to reduce problems from a physical description to a model or symbolic representation to which the theoretical principles may be applied. The problems balance FPS and SI units and are arranged in an increasing order of difficulty so students can evaluate their understanding of the material. 
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Page 102
Plot the stressstrain diagram and determine approximately the modulus of
elasticity, the ultimate stress, and the rupture stress. Use a scale of 1 in. = 15 ksi
and 1 in. = 0.05 in./in. Redraw the linear elastic region, using the same stress
scale ...
Plot the stressstrain diagram and determine approximately the modulus of
elasticity, the ultimate stress, and the rupture stress. Use a scale of 1 in. = 15 ksi
and 1 in. = 0.05 in./in. Redraw the linear elastic region, using the same stress
scale ...
Page 247
Also, y\ is large enough so that the radius of the elastic core is assumed to
approach zero, that is, y\ >> yy. If TP is removed, the ... 545a a linear stress
distribution caused by applying the plastic torque TP in the opposite direction, Fig
. 5456.
Also, y\ is large enough so that the radius of the elastic core is assumed to
approach zero, that is, y\ >> yy. If TP is removed, the ... 545a a linear stress
distribution caused by applying the plastic torque TP in the opposite direction, Fig
. 5456.
Page 851
... 23943 elasticplastic torque, 24042 maximum elastic torque, 240 ultimate
torque, 24243 Inertia: least moment of, 659 ... 36 Lateral contraction, 107 Least
moment of inertia, 659 Linear distributed load, 3 Linearly elastic material, 88
Linear ...
... 23943 elasticplastic torque, 24042 maximum elastic torque, 240 ultimate
torque, 24243 Inertia: least moment of, 659 ... 36 Lateral contraction, 107 Least
moment of inertia, 659 Linear distributed load, 3 Linearly elastic material, 88
Linear ...
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allowable shear stress aluminum angle of twist Applying Eq assumed average normal stress average shear stress axes axial force axial load beam's buckling caused centroid column compressive stress computed constant cross section crosssectional area deflection deformation Determine the maximum distributed load Draw the shear elastic curve element EXAMPLE factor of safety freebody diagram Hooke's law inertia internal loadings kip/ft length linearelastic loading shown located material maximum bending stress maximum inplane shear maximum shear stress modulus of elasticity Mohr's circle neutral axis normal strain plastic positive principal stresses radius reactions sectional area segment shaft shear center shear force shear strain shown in Fig SOLUTION Solve Prob statically indeterminate steel strain energy stress acting stress components stress developed stress distribution stressstrain diagram Tanow tensile tensile stress thickness tion torque torsional tube vertical yield zero