Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 63
Page 213
... substituted into Eq . 5.1i , we obtain a I1 = I , cos2 a + 1 , sin2 x - 2Px , sin x cos x , n I , a Substitution of appropriate trigonometric identities leads to ( 5.1j ) In = I + 1 , 2 1 y + ( 1 - 1 , ) cos 2x - P ̧ , sin 2x Pxy ( 5.1k ) ...
... substituted into Eq . 5.1i , we obtain a I1 = I , cos2 a + 1 , sin2 x - 2Px , sin x cos x , n I , a Substitution of appropriate trigonometric identities leads to ( 5.1j ) In = I + 1 , 2 1 y + ( 1 - 1 , ) cos 2x - P ̧ , sin 2x Pxy ( 5.1k ) ...
Page 273
... substituted into the equation for v yields WL4 WL4 EI ( 0 ) = + C2L + O 12 24 Solving , we obtain C1 = wĽ 24 1 Back substitution of these values for C1 and C2 SEC . 6.3 / BEAM DEFLECTIONS - TWO SUCCESSIVE INTEGRATIONS 273.
... substituted into the equation for v yields WL4 WL4 EI ( 0 ) = + C2L + O 12 24 Solving , we obtain C1 = wĽ 24 1 Back substitution of these values for C1 and C2 SEC . 6.3 / BEAM DEFLECTIONS - TWO SUCCESSIVE INTEGRATIONS 273.
Page 274
... substitution of the value of C1 into the equation for dv / dx developed above after the first of the two successive integrations . This slope equation becomes dv 1 / wL = dx EI 4 W wL x2 - 6 ( 0 ≤ x ≤ L ) 24 Example 6.3 Refer to the ...
... substitution of the value of C1 into the equation for dv / dx developed above after the first of the two successive integrations . This slope equation becomes dv 1 / wL = dx EI 4 W wL x2 - 6 ( 0 ≤ x ≤ L ) 24 Example 6.3 Refer to the ...
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