Engineering Mechanics of Materials |
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Page 84
... shear strains . For the sake of consistency with the equations developed earlier for plane stress , the sign convention adopted in this text is such that a positive shear strain represents an increase in the 90 ° angle , while a negative ...
... shear strains . For the sake of consistency with the equations developed earlier for plane stress , the sign convention adopted in this text is such that a positive shear strain represents an increase in the 90 ° angle , while a negative ...
Page 92
... strain ( i.e. , one of the three principal strains is zero ) , it follows that the labeling of the three principal ... shear strain occurs between two orthogonal axes bisecting the 90 ° angles between ɛ1 and ɛ3 . It is important to note ...
... strain ( i.e. , one of the three principal strains is zero ) , it follows that the labeling of the three principal ... shear strain occurs between two orthogonal axes bisecting the 90 ° angles between ɛ1 and ɛ3 . It is important to note ...
Page 98
... shear strain at the point , and define the orthogonal axes corresponding to this strain . 2.37 The equiangular ( 60 ° ) rosette is one in which the three strain axes make 60 ° with respect to each other . If one of the three axes is ...
... shear strain at the point , and define the orthogonal axes corresponding to this strain . 2.37 The equiangular ( 60 ° ) rosette is one in which the three strain axes make 60 ° with respect to each other . If one of the three axes is ...
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 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 strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁