Engineering Mechanics of Materials |
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Page 213
... substituted into Eq . 5.1i , we obtain In = x I , cos2 a +1 , sin2 a - y xy - 2P , sin a cos α , Substitution of appropriate trigonometric identities leads to ( 5.1j ) In Ix + Iy = + - 2 5 ( lx -1 , ) cos 2x - Px , sin 2x xy ( 5.1k ) ...
... substituted into Eq . 5.1i , we obtain In = x I , cos2 a +1 , sin2 a - y xy - 2P , sin a cos α , Substitution of appropriate trigonometric identities leads to ( 5.1j ) In Ix + Iy = + - 2 5 ( lx -1 , ) cos 2x - Px , sin 2x xy ( 5.1k ) ...
Page 273
... substituted into = which implies that C2 the equation for v yields EI , ( 0 ) = 0 - 0 + 0 + C2 0. The second condition , x = EI , ( 0 ) = WL4 12 - WL4 24 + C1L + 0 Solving , we obtain C1 = wĽ 24 2 Back substitution of these values for C ...
... substituted into = which implies that C2 the equation for v yields EI , ( 0 ) = 0 - 0 + 0 + C2 0. The second condition , x = EI , ( 0 ) = WL4 12 - WL4 24 + C1L + 0 Solving , we obtain C1 = wĽ 24 2 Back substitution of these values for C ...
Page 408
... Substitution of this effective length into Eq . 8.21 yields a critical load : Pe = 1 π ? ΕΙ 4 L ( 8.23 ) The column with one end free and the other end fixed will buckle elastically at a load that is one - fourth as large as the load ...
... Substitution of this effective length into Eq . 8.21 yields a critical load : Pe = 1 π ? ΕΙ 4 L ( 8.23 ) The column with one end free and the other end fixed will buckle elastically at a load that is one - fourth as large as the load ...
Contents
Internal Forces in Members | 1 |
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 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