Engineering Mechanics of Materials |
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Page 141
... elastic range and it is measured by the modulus of elasticity . For example , steels with a modulus of elasticity of about 30 x 106 psi are stiffer than aluminums , with a modulus of elasticity of about 10 x 10 psi . YIELD POINT . The ...
... elastic range and it is measured by the modulus of elasticity . For example , steels with a modulus of elasticity of about 30 x 106 psi are stiffer than aluminums , with a modulus of elasticity of about 10 x 10 psi . YIELD POINT . The ...
Page 147
... modulus of elasticity . ་ ་ ( b ) Poisson's ratio . ( c ) The modulus of resilience . 3.17 A compressive force of 200,000 N is gradually applied to a prismatic bar whose cross section is 0.03 m × 0.04 m and whose length is 0.1 m . The ...
... modulus of elasticity . ་ ་ ( b ) Poisson's ratio . ( c ) The modulus of resilience . 3.17 A compressive force of 200,000 N is gradually applied to a prismatic bar whose cross section is 0.03 m × 0.04 m and whose length is 0.1 m . The ...
Page 427
... Elasticity In 1744 , Euler first derived his column buckling equation but not until over 200 years later was it fully understood in a general sense ... MODULUS OF ELASTICITY 427 Generalized Euler Equation Using the Tangent Modulus Elasticity.
... Elasticity In 1744 , Euler first derived his column buckling equation but not until over 200 years later was it fully understood in a general sense ... MODULUS OF ELASTICITY 427 Generalized Euler Equation Using the Tangent Modulus Elasticity.
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 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