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
31 A tension test was performed on a steel specimen having an original
diameter of 0.503 in. and a gauge length of 2.00 in. The data is listed in the table.
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 but a strain scale of 1 in. = 0.001 in. *34 A tension test was performed on a
steel specimen ...
31 A tension test was performed on a steel specimen having an original
diameter of 0.503 in. and a gauge length of 2.00 in. The data is listed in the table.
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 but a strain scale of 1 in. = 0.001 in. *34 A tension test was performed on a
steel specimen ...
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 material tends to recover
elastically, following along line CD. Since elastic behavior occurs, we can
superimpose on the stress distribution in Fig. 545a a linear stress distribution
caused by applying the plastic torque TP in the opposite direction, Fig. 5456.
Here the maximum shear stress t>, computed for this stress distribution, is called
the modulus of rupture ...
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 material tends to recover
elastically, following along line CD. Since elastic behavior occurs, we can
superimpose on the stress distribution in Fig. 545a a linear stress distribution
caused by applying the plastic torque TP in the opposite direction, Fig. 5456.
Here the maximum shear stress t>, computed for this stress distribution, is called
the modulus of rupture ...
Page 851
... 26 Kinematic condition, 137 ksi (kilopounds per square inch), 26 Lap joints, 36
Lateral contraction, 107 Least moment of inertia, 659 Linear distributed load, 3
Linearly elastic material, 88 Linear variation: in normal strain, 290 in normal
stress, 290 in sheer strain, 182 in sheer stress, 182 Load: axial, 11978 critical,
65456, 659 Euler, 658, 662 external, 2, 3 linear distributed, 3 loaddisplacement
relationship, 137 plastic, 164 Loaddisplacement relationship, 137 Localized
distortions, ...
... 26 Kinematic condition, 137 ksi (kilopounds per square inch), 26 Lap joints, 36
Lateral contraction, 107 Least moment of inertia, 659 Linear distributed load, 3
Linearly elastic material, 88 Linear variation: in normal strain, 290 in normal
stress, 290 in sheer strain, 182 in sheer stress, 182 Load: axial, 11978 critical,
65456, 659 Euler, 658, 662 external, 2, 3 linear distributed, 3 loaddisplacement
relationship, 137 plastic, 164 Loaddisplacement relationship, 137 Localized
distortions, ...
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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