Engineering Mechanics of Materials |
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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 |
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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