Engineering Mechanics of Materials |
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Page 56
... direction of the coordinate axis and if the outward normal to its plane is also in the positive direction of the corresponding axis . If , however , the outward normal is in the negative direction of the coordinate axis , a positive ...
... direction of the coordinate axis and if the outward normal to its plane is also in the positive direction of the corresponding axis . If , however , the outward normal is in the negative direction of the coordinate axis , a positive ...
Page 85
... directions . Thus , in order to achieve a condition of plane strain , say in the xy plane , the strain in the z direction must be prevented by physical restraint ( i.e. , by the application of a stress in the z direction ) . In such a ...
... directions . Thus , in order to achieve a condition of plane strain , say in the xy plane , the strain in the z direction must be prevented by physical restraint ( i.e. , by the application of a stress in the z direction ) . In such a ...
Page 101
... directions , and therefore no distinction is necessary between the modulus of elasticity in the x direction and that in the y direction . Thus E1 = E1 = E , and in subsequent work , the modulus of elasticity for a given material will be ...
... directions , and therefore no distinction is necessary between the modulus of elasticity in the x direction and that in the y direction . Thus E1 = E1 = E , and in subsequent work , the modulus of elasticity for a given material will be ...
Contents
Internal Forces in Members | 1 |
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 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 shaft shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress at point stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁