Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 93
Page 84
... strain represents an increase in the 90 ° angle , while a negative shear strain represents a decrease . In Figure 2.13 ( c ) , for example , y ' is negative and y'in is positive . Note that as in the case of the normal strain , & , the ...
... strain represents an increase in the 90 ° angle , while a negative shear strain represents a decrease . In Figure 2.13 ( c ) , for example , y ' is negative and y'in is positive . Note that as in the case of the normal strain , & , the ...
Page 85
... strain in the direction of the applied stress ( normal longitudinal strain ) , but also a smaller normal strain in a direction perpendicular to the applied stress ( normal transverse strain ) . If the applied normal stress is tensile ...
... strain in the direction of the applied stress ( normal longitudinal strain ) , but also a smaller normal strain in a direction perpendicular to the applied stress ( normal transverse strain ) . If the applied normal stress is tensile ...
Page 92
... strains in a plane are positive . Since the problem under consideration is one of plane strain ( i.e. , one of the three principal strains is zero ) , it follows that the labeling of the three principal strains at the point must be such ...
... strains in a plane are positive . Since the problem under consideration is one of plane strain ( i.e. , one of the three principal strains is zero ) , it follows that the labeling of the three principal strains at the point must be such ...
Contents
T Shear and Moment at Specified Sections | 24 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
14 other sections not shown
Other editions - View all
Common terms and phrases
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