Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 79
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 1 wL W w13 v = x3 - EI 12 Χ 24 24 ( 0 ≤ x ≤ L ) The equation for the slope of the elastic curve may ...
... 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 1 wL W w13 v = x3 - EI 12 Χ 24 24 ( 0 ≤ x ≤ L ) The equation for the slope of the elastic curve may ...
Page 289
... Equation 6.6 , EI , ( d2v / dx2 ) = M „ , with M , a function of x , is an ordinary , second- order differential equation . An ordinary differential equation is said to be linear if the dependent variable and its derivatives appear only ...
... Equation 6.6 , EI , ( d2v / dx2 ) = M „ , with M , a function of x , is an ordinary , second- order differential equation . An ordinary differential equation is said to be linear if the dependent variable and its derivatives appear only ...
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 ...
Contents
T Shear and Moment at Specified Sections | 24 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
14 other sections not shown
Other editions - View all
Common terms and phrases
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