Engineering Mechanics of Materials |
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Page 55
... normal stress acts in a direction perpendicular to the plane on which it acts and it can be either tensile or compressive . A tensile normal stress is one that tends to pull the material particles away from each other , while a compressive ...
... normal stress acts in a direction perpendicular to the plane on which it acts and it can be either tensile or compressive . A tensile normal stress is one that tends to pull the material particles away from each other , while a compressive ...
Page 85
... normal stress , it experiences not only a normal 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 ...
... normal stress , it experiences not only a normal 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 ...
Page 117
... normal stress σx x = ox = F n A = F B A = P A ( 3.1 ) Thus , in Figure 3.1 ( b ) , for example , the resultant force FB can be replaced by a stress system σ , as shown in Figure 3.1 ( d ) , which , when multiplied by the area A , yields ...
... normal stress σx x = ox = F n A = F B A = P A ( 3.1 ) Thus , in Figure 3.1 ( b ) , for example , the resultant force FB can be replaced by a stress system σ , as shown in Figure 3.1 ( d ) , which , when multiplied by the area A , yields ...
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