Engineering Mechanics of Materials4. 2 Solid Circular Shafts-Angle of Twist and Shearing Stresses 159 4. 3 Hollow Circular Shafts-Angle of Twist and Shearing Stresses 166 4. 4 Principal Stresses and Strains Associated with Torsion 173 4. 5 Analytical and Experimental Solutions for Torsion of Members of Noncircular Cross Sections 179 4. 6 Shearing Stress-Strain Properties 188 *4. 7 Computer Applications 195 5 Stresses in Beams 198 5. 1 Introduction 198 5. 2 Review of Properties of Areas 198 5. 3 Flexural Stresses due to Symmetric Bending of Beams 211 5. 4 Shear Stresses in Symmetrically Loaded Beams 230 *5. 5 Flexural Stresses due to Unsymmetric Bending of Beams 248 *5. 6 Computer Applications 258 Deflections of Beams 265 I 6. 1 Introduction 265 6. 2 Moment-Curvature Relationship 266 6. 3 Beam Deflections-Two Successive Integrations 268 6. 4 Derivatives of the Elastic Curve Equation and Their Physical Significance 280 6. 5 Beam Deflections-The Method of Superposition 290 6. 6 Construction of Moment Diagrams by Cantilever Parts 299 6. 7 Beam Deflections-The Area-Moment Method 302 *6. 8 Beam Deflections-Singularity Functions 319 *6. 9 Beam Deflections-Castigliano's Second Theorem 324 *6. 10 Computer Applications 332 7 Combined Stresses and Theories of Failure 336 7. 1 Introduction 336 7. 2 Axial and Torsional Stresses 336 Axial and Flexural Stresses 342 7. 3 Torsional and Flexural Stresses 352 7. 4 7. 5 Torsional, Flexural, and Axial Stresses 358 *7. 6 Theories of Failure 365 Computer Applications 378 *7. |
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Page 428
... solve Problem 8.16 ( p . 405 ) . 8.67 ( Sec . 8.3 ) Solve Problem 8.17 ( p . 405 ) for a steel alloy for which E 29 × 10 ° psi and Ор 45 ksi . = = 8.68 ( Sec . 8.3 ) Refer to Problem 8.20 ( p . 406 ) and let b = 0.375 in . and solve as ...
... solve Problem 8.16 ( p . 405 ) . 8.67 ( Sec . 8.3 ) Solve Problem 8.17 ( p . 405 ) for a steel alloy for which E 29 × 10 ° psi and Ор 45 ksi . = = 8.68 ( Sec . 8.3 ) Refer to Problem 8.20 ( p . 406 ) and let b = 0.375 in . and solve as ...
Page 465
... Solve Problem 9.38 ( p . 458 ) by the method of superposition . 9.57 Solve Problem 9.39 ( p . 458 ) by the method of superposition . 9.58 Solve Problem 9.40 ( p . 458 ) by the method of superposition . 9.59 Solve Problem 9.42 ( p . 459 ) ...
... Solve Problem 9.38 ( p . 458 ) by the method of superposition . 9.57 Solve Problem 9.39 ( p . 458 ) by the method of superposition . 9.58 Solve Problem 9.40 ( p . 458 ) by the method of superposition . 9.59 Solve Problem 9.42 ( p . 459 ) ...
Page 587
... solve it for h = 3 in . rather than h 6 in . = 12.62 ( Sec . 12.3 ) Refer to Problem 12.2 ( p . 567 ) and solve it for h = 0.20 m rather than for h 0.15 m . - 12.63 ( Sec . 12.3 ) Use an energy input of 20,000 lb - in . rather than ...
... solve it for h = 3 in . rather than h 6 in . = 12.62 ( Sec . 12.3 ) Refer to Problem 12.2 ( p . 567 ) and solve it for h = 0.20 m rather than for h 0.15 m . - 12.63 ( Sec . 12.3 ) Use an energy input of 20,000 lb - in . rather than ...
Contents
Stresses in Beams | 198 |
Deflections of Beams | 265 |
Combined Stresses and Theories of Failure | 336 |
Copyright | |
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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 constant coordinate cross section cross-sectional area cylinder deflection deformation depicted in Fig diameter elastic curve equal equation equilibrium Euler EXAMPLE factor of safety FIGURE flexural stress FORTRAN free-body diagram k-ft k-in kN-m lb/ft length longitudinal M₁ material maximum shear stress modulus of elasticity Mohr's circle moment of inertia neutral axis normal stress obtained perpendicular plane stress plane stress condition plot principal centroidal axis principal stresses r₁ radius ratio rectangular Refer to Fig rotation shaft shear force shear strain shown in Fig slope SOLUTION statically indeterminate steel stress element T₁ t₂ tensile Tmax torque torsional uniform load V₁ yield stress zero