Strength of Materials |
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Page 12
... Elastic limit Proportional limit Strain € = FIG . 1-9 . Stress - strain diagram . 6/2 Proportional Limit . From the origin O to a point called the proportional limit , Fig . 1-9 shows the stress - strain diagram to be a straight line ...
... Elastic limit Proportional limit Strain € = FIG . 1-9 . Stress - strain diagram . 6/2 Proportional Limit . From the origin O to a point called the proportional limit , Fig . 1-9 shows the stress - strain diagram to be a straight line ...
Page 349
... elastic curve EI day dx2 His analysis is = M. As we know now , such an analysis is valid only up to the stress at the proportional limit . In Euler's time , neither the concept of stress nor the limiting stress at the proportional limit ...
... elastic curve EI day dx2 His analysis is = M. As we know now , such an analysis is valid only up to the stress at the proportional limit . In Euler's time , neither the concept of stress nor the limiting stress at the proportional limit ...
Page 355
... limiting slenderness ratio can be found easily for any material by substituting in Eq . ( 11-5 ) the known values of the proportional limit and the modulus of elasticity for the material . For example , for steel that has a proportional ...
... limiting slenderness ratio can be found easily for any material by substituting in Eq . ( 11-5 ) the known values of the proportional limit and the modulus of elasticity for the material . For example , for steel that has a proportional ...
Contents
SIMPLE STRESS | 1 |
RIVETED AND WELDED JOINTS | 39 |
TORSION | 65 |
Copyright | |
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acting AISC formula allowable stresses angle applied area-moment method assumed axes axial load beam shown bending bending moment C₁ C₂ Carry-over centroidal column formula compressive stress Compute the maximum concrete continuous beam critical load cross-section deflection deformation Determine the maximum diameter elastic curve element end moments equal equivalent Euler's Euler's formula factor of safety fixed end flange flexural stress ft long ft-lb Hence ILLUSTRATIVE PROBLEMS in.¹ lb-ft³ lb/ft length M₁ M₂ maximum shearing stress maximum stress midspan Mohr's circle moment of inertia moments of inertia neutral axis normal stress obtain P₁ plane plate principal stresses PROB product of inertia proportional limit R₂ radius resultant rivet shaft shear center shear diagram shown in Fig simply supported slenderness ratio slope Solution span steel strain tangent tensile stress three-moment equation torsional vertical shear whence zero ΕΙ