## 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 174

Once again , the center of the circle lies at the origin of the coordinate system and

the radius of the circle

principal normal strains , & z and £3 , each have magnitudes

Once again , the center of the circle lies at the origin of the coordinate system and

the radius of the circle

**equals**Tr / 2JG . In the special case of pure torsion , theprincipal normal strains , & z and £3 , each have magnitudes

**equal**to Tr / 2JG ...Page 377

Assume the material to be steel for which the yield strengths , 00 , in tension and

compression are numerically identical and

diameter of the shaft so that it does not fail by yielding , using the maximum shear

...

Assume the material to be steel for which the yield strengths , 00 , in tension and

compression are numerically identical and

**equal**to 30 , 000 psi . Determine thediameter of the shaft so that it does not fail by yielding , using the maximum shear

...

Page 649

EO . less than or

or

. Since two - valued logic is used , a logical expression is either true or false .

EO . less than or

**equal**to less than**equal**to not**equal**to greater than greater thanor

**equal**to . NE . . GT . . GE . The logical operators are . AND . , . OR . , and . NOT. Since two - valued logic is used , a logical expression is either true or false .

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### Common terms and phrases

acting angle applied Assume axes axis beam bending cantilever circular column compressive Consider constant Construct coordinate cross section cross-sectional area curve deflection deformation depicted in Fig Determine developed diagram diameter differential direction distance elastic element equal equation equilibrium EXAMPLE expressed factor FIGURE fixed flexural force free-body diagram function given inertia length limit load magnitude material maximum shear stress method modulus Mohr's circle moment moments neutral axis normal stress Note obtained plane plane stress plot positive principal stresses Problem properties ratio reactions Refer to Fig relation represents respect rotation shaft shear strain shear stress Show shown in Fig slope SOLUTION Solve steel strain stress condition subjected Substitution supported surface tensile torque twist unit vertical Write yield zero