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 672
Hint: Assume the top deflects 8. Show that cPv/dx2 + (PIEI)v = (P/EI)8. The
solution is of the form v = C\ sin(v7V£7.v) + C2 cos (VPIEIx) + 8. 1343 Consider
an ideal column as in Fig. 131 2c, having both ends fixed. Show that the critical
load on ...
Hint: Assume the top deflects 8. Show that cPv/dx2 + (PIEI)v = (P/EI)8. The
solution is of the form v = C\ sin(v7V£7.v) + C2 cos (VPIEIx) + 8. 1343 Consider
an ideal column as in Fig. 131 2c, having both ends fixed. Show that the critical
load on ...
Page 774
SOLUTION External Force P. A vertical force P is placed on the beam at B as
shown in Fig. 14426. Internal Moments M. A single x coordinate is needed for
the solution, since there are no discontinuities of loading between A and B. Using
the ...
SOLUTION External Force P. A vertical force P is placed on the beam at B as
shown in Fig. 14426. Internal Moments M. A single x coordinate is needed for
the solution, since there are no discontinuities of loading between A and B. Using
the ...
Page 781
SOLUTION I The y axis is placed along the axis of symmetry so that x = 0, Fig. A
4a. To obtain y we will establish the x axis (reference axis) through the base of
the area. The area is segmented into two rectangles as shown, and the centroidal
...
SOLUTION I The y axis is placed along the axis of symmetry so that x = 0, Fig. A
4a. To obtain y we will establish the x axis (reference axis) through the base of
the area. The area is segmented into two rectangles as shown, and the centroidal
...
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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