## 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 64

By convention , a normal stress is positive if it points in the

outward normal to the plane . Thus a positive normal stress produces tension and

a negative normal stress produces compression . A component of shear stress is

...

By convention , a normal stress is positive if it points in the

**direction**of theoutward normal to the plane . Thus a positive normal stress produces tension and

a negative normal stress produces compression . A component of shear stress is

...

Page 93

the assumption is made that strains in the z

strain condition exists ) , the element undergoes normal strains in the x and y

would ...

the assumption is made that strains in the z

**direction**are zero ( i . e . , a planestrain condition exists ) , the element undergoes normal strains in the x and y

**directions**as well as shear strains in the xy plane . A sketch of how the elementwould ...

Page 104

Similarly , if the uniaxial tensile stress is in the y

a ) , the resulting extensional deformation 8 , = £ , L , would be such that the

stress o , is related linearly to the strain & y , as shown in Fig . 2 . 24 ( b ) ,

provided ...

Similarly , if the uniaxial tensile stress is in the y

**direction**as shown in Fig . 2 . 24 (a ) , the resulting extensional deformation 8 , = £ , L , would be such that the

stress o , is related linearly to the strain & y , as shown in Fig . 2 . 24 ( b ) ,

provided ...

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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