Engineering Mechanics of Materials |
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Page 141
... material is the value of stress beyond which the material experiences a permanent deformation even after the stress is removed . Thus , if the material is loaded to any level of stress within the elastic limit and the load is then ...
... material is the value of stress beyond which the material experiences a permanent deformation even after the stress is removed . Thus , if the material is loaded to any level of stress within the elastic limit and the load is then ...
Page 680
... material B may be transformed into an equivalent amount of material A by changing the actual width b of material B into an equivalent width be of material A. It is evident from Eq . 14.23b that the relation between the actual and the ...
... material B may be transformed into an equivalent amount of material A by changing the actual width b of material B into an equivalent width be of material A. It is evident from Eq . 14.23b that the relation between the actual and the ...
Page 684
... material A and material B , where v1 = VB . For purposes of discussion , assume that EB < E , so that n is less than unity . It follows from Eq . ( b ) that at the junction ( i.e. , at v1 = VB ) , OB is less than σ . A stress ...
... material A and material B , where v1 = VB . For purposes of discussion , assume that EB < E , so that n is less than unity . It follows from Eq . ( b ) that at the junction ( i.e. , at v1 = VB ) , OB is less than σ . A stress ...
Contents
T Shear and Moment at Specified Sections | 24 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
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absolute maximum shear aluminum angle of twist applied Assume axes axial force axially loaded beam shown bending C₁ cantilever beam Castigliano's second theorem column components compressive Compute coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter elastic curve equal equation equilibrium Example factor of safety flexural stress free-body diagram Homework Problems k-ft k-in kN-m length longitudinal M₁ material maximum shear stress MN/m² modulus of elasticity Mohr's circle moment of inertia neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal stresses r₁ ratio Refer to Figure respect rotation section a-a segment shaft shear force shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ vertical yield strength yield stress zero