Mechanics of MaterialsFor undergraduate Mechanics of Materials courses in Mechanical, Civil, and Aerospace Engineering departments. Containing Hibbeler's hallmark student-oriented features, this text is in four-color with a photorealistic art program designed to help students visualize difficult concepts. A clear, concise writing style and more examples than any other text further contribute to students' ability to master the material. Click here for the Video Solutions that accompany this book. Developed by Professor Edward Berger, University of Virginia, these are complete, step-by-step solution walkthroughs of representative homework problems from each section of the text. |
From inside the book
Results 1-3 of 85
Page 10
... body is sectioned , it will be necessary to determine the reactions acting on the chosen segment of the body . Draw the free- body diagram for the entire body and then apply the necessary equations of equilibrium to obtain these reactions .
... body is sectioned , it will be necessary to determine the reactions acting on the chosen segment of the body . Draw the free- body diagram for the entire body and then apply the necessary equations of equilibrium to obtain these reactions .
Page 12
... free - body diagram . Free - Body Diagram . Passing an imaginary section perpendicular to the axis of the shaft through C yields the free - body diagram of segment AC shown in Fig . 1-5c . Equations of Equilibrium . + ΣΕ , = 0 ; + 1ΣΕ ...
... free - body diagram . Free - Body Diagram . Passing an imaginary section perpendicular to the axis of the shaft through C yields the free - body diagram of segment AC shown in Fig . 1-5c . Equations of Equilibrium . + ΣΕ , = 0 ; + 1ΣΕ ...
Page 290
... free - body diagram , Fig . 6-18b . Shear Diagram . At x = 0 , VA = +4.8 kN , and at x = 10 , VD = −11.2 kN , Fig . 6-18c . At intermediate points between each force the slope of the shear diagram will be zero . Why ? Hence the shear ...
... free - body diagram , Fig . 6-18b . Shear Diagram . At x = 0 , VA = +4.8 kN , and at x = 10 , VD = −11.2 kN , Fig . 6-18c . At intermediate points between each force the slope of the shear diagram will be zero . Why ? Hence the shear ...
Other editions - View all
Common terms and phrases
absolute maximum shear allowable bending stress allowable shear stress aluminum angle of twist Applying Eq average normal stress axial load beam's buckling caused centroid column compressive constant cross section cross-sectional area deformation Determine the maximum diameter displacement distributed load Draw the shear elastic curve element EXAMPLE factor of safety free-body diagram ft Prob Hooke's law in² kip/ft kN·m kN/m length linear-elastic loading shown located material maximum bending stress maximum in-plane shear maximum shear stress modulus of elasticity Mohr's circle moment of inertia neutral axis normal strain plane stress plastic principal strains principal stresses radius redundant sectional area segment shaft shear force shear strain shown in Fig slope SOLUTION Solve Prob statically indeterminate steel strain energy stress acting stress distribution stress-strain diagram tensile tensile stress torque torsional yield zero ΕΙ σχ