Engineering Mechanics of Materials |
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Page 274
... equation for v and division by EI , yields the equation of the elastic curve as a function of x for the given beam and loading . Thus v = u 1 ( wL EI , 12 x3 - W 24 ተ - WL3 X 24 ( 0 ≤ x ≤ L ) The equation for the slope of the elastic ...
... equation for v and division by EI , yields the equation of the elastic curve as a function of x for the given beam and loading . Thus v = u 1 ( wL EI , 12 x3 - W 24 ተ - WL3 X 24 ( 0 ≤ x ≤ L ) The equation for the slope of the elastic ...
Page 322
... Equation 6.29 may also be interpreted to apply to moments and slopes as well as to forces and deflections . If P ; is replaced by M , and v ; is replaced by 0 ;, then the equation may be used to obtain rotations . This use of the equation ...
... Equation 6.29 may also be interpreted to apply to moments and slopes as well as to forces and deflections . If P ; is replaced by M , and v ; is replaced by 0 ;, then the equation may be used to obtain rotations . This use of the equation ...
Page 458
... Equations ( b ) and ( c ) represent two simultaneous algebraic equations with the three unknown A , B ,, and M ̧ . One more equation is needed to complete the solution . This equation is obtained from knowledge of the deformation charac ...
... Equations ( b ) and ( c ) represent two simultaneous algebraic equations with the three unknown A , B ,, and M ̧ . One more equation is needed to complete the solution . This equation is obtained from knowledge of the deformation charac ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
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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