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 127
**m. Fig. 47 SOLUTION Internal Force. The freebody diagram of the tube and
rod, Fig. 476, shows that the rod is subjected to a tension of 80 kN and the tube
is subjected to a compression of 80 kN. Displacement. We will first determine the
...
**m. Fig. 47 SOLUTION Internal Force. The freebody diagram of the tube and
rod, Fig. 476, shows that the rod is subjected to a tension of 80 kN and the tube
is subjected to a compression of 80 kN. Displacement. We will first determine the
...
Page 294
SOLUTION Pari (a). The flexure formula is omax = Mc/I. From Fig. 627a, c = 6 in.
and crmax = 2 ksi. The neutral axis is defined as line NA, because the stress is
zero along this line. Since the cross section has a rectangular shape, the moment
...
SOLUTION Pari (a). The flexure formula is omax = Mc/I. From Fig. 627a, c = 6 in.
and crmax = 2 ksi. The neutral axis is defined as line NA, because the stress is
zero along this line. Since the cross section has a rectangular shape, the moment
...
Page 312
SOLUTION Internal Moment Components. By inspection it is seen that the y and z
axes represent the principal axes of inertia since they are axes of symmetry for
the cross section. As required we have established the z axis as the principal ...
SOLUTION Internal Moment Components. By inspection it is seen that the y and z
axes represent the principal axes of inertia since they are axes of symmetry for
the cross section. As required we have established the z axis as the principal ...
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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