Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 63
Page 290
... beam - couple C applied at left end v 02 -L- 01 = -CL 3Elu CL X 02 6Elu Case 4. Cantilever beam - uniformly distributed loading 02 L W Χ −Pb ( L2 — b2 ) 3/2 - υ 9√3 EluL - Pb [ 4 ( x − a ) 3 + ( L2 — b2 ) x - x3 ] Maximum – b2 - 6ET ...
... beam - couple C applied at left end v 02 -L- 01 = -CL 3Elu CL X 02 6Elu Case 4. Cantilever beam - uniformly distributed loading 02 L W Χ −Pb ( L2 — b2 ) 3/2 - υ 9√3 EluL - Pb [ 4 ( x − a ) 3 + ( L2 — b2 ) x - x3 ] Maximum – b2 - 6ET ...
Page 297
... beam is loaded as shown in Figure H6.34 . Determine the slope and deflection at the free end using the method of superposition and appro- priate formulas from Table 6.1 . FIGURE H6.35 6.36 A simply supported beam ... CANTILEVER PARTS 297 ...
... beam is loaded as shown in Figure H6.34 . Determine the slope and deflection at the free end using the method of superposition and appro- priate formulas from Table 6.1 . FIGURE H6.35 6.36 A simply supported beam ... CANTILEVER PARTS 297 ...
Page 336
... beam deflections . Example 6.19 A cantilever beam of length L = 10 ft has the cross section shown in Figure 6.27 , which is an angle 9 x 4 x 1 in . The beam is fabricated of steel for which E = 30 x 106 psi and the following moments of ...
... beam deflections . Example 6.19 A cantilever beam of length L = 10 ft has the cross section shown in Figure 6.27 , which is an angle 9 x 4 x 1 in . The beam is fabricated of steel for which E = 30 x 106 psi and the following moments of ...
Contents
Internal Forces in Members | 1 |
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 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