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 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 |
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acting allowable angle of twist applied Assume axes axis beam bending cantilever centroidal circle column components compressive Compute Consider constant construct coordinate cross section curve deflection deformation depicted in Figure Determine developed diameter direction discussed elastic element energy equal equation equilibrium Example expressed factor failure flexural force free-body diagram function given inertia joint length limit load material maximum shear stress method modulus moment moments neutral axis normal stress Note obtained plane plot positive principal stresses Problem properties quantity ratio reactions Refer to Figure relation represents resist respect rotation segment shaft shown in Figure slope Solution Solve static steel strain strength structural subjected Substitution supported surface tensile tension theory tion torque unit yield zero