Engineering Mechanics of Materials |
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Page 290
... beam - couple C applied at left end V -L 01 = -CL 3Elu CL v = 026Elu Case 4. Cantilever beam - uniformly distributed loading V 02 L W X 02 = -wL3 6Elu 3 v = Case 5. Cantilever beam - concentrated load at any point P 02 == -Pa2 2E1 2 V ...
... beam - couple C applied at left end V -L 01 = -CL 3Elu CL v = 026Elu Case 4. Cantilever beam - uniformly distributed loading V 02 L W X 02 = -wL3 6Elu 3 v = Case 5. Cantilever beam - concentrated load at any point P 02 == -Pa2 2E1 2 V ...
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 . 2 k 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 . 2 k FIGURE H6.35 6.36 A simply supported beam ... CANTILEVER PARTS 297 ...
Page 336
... beam deflections . P = 1000 lb 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 × 106 psi and the following ...
... beam deflections . P = 1000 lb 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 × 106 psi and the following ...
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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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