Engineering Mechanics of Materials |
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Page 100
... limits . The upper limit for σ , for which the relationship between stress and strain is linear is known as the proportional limit . For values of stress above the proportional limit , the relation between stress and strain is a topic ...
... limits . The upper limit for σ , for which the relationship between stress and strain is linear is known as the proportional limit . For values of stress above the proportional limit , the relation between stress and strain is a topic ...
Page 141
... limit , σ ,, is represented by the ordinate to point A in Figures 3.12 and 3.13 . ELASTIC LIMIT . The elastic limit , σ , for a given material is the value of stress beyond which the material experiences a permanent deformation even ...
... limit , σ ,, is represented by the ordinate to point A in Figures 3.12 and 3.13 . ELASTIC LIMIT . The elastic limit , σ , for a given material is the value of stress beyond which the material experiences a permanent deformation even ...
Page 147
... limit . ( b ) The changes in diameter and in length at the pro- portional limit . 12,000 0.0040 19,500 0.1500 12,400 0.0050 20,000 0.2000 12,600 0.0070 20,000 0.2500 12,800 0.0100 18,600 0.3200 12,800 0.0130 17,000 0.4000 12,600 0.0150 ...
... limit . ( b ) The changes in diameter and in length at the pro- portional limit . 12,000 0.0040 19,500 0.1500 12,400 0.0050 20,000 0.2000 12,600 0.0070 20,000 0.2500 12,800 0.0100 18,600 0.3200 12,800 0.0130 17,000 0.4000 12,600 0.0150 ...
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 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 at point stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁