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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Results 1-3 of 76
Page 175
... psi and 0.30 , respectively , so that E G = 2 ( 1 + μ ) 29.0 × 106 2 ( 1 + ... Fig . 4.9 ( c ) . Points X and Y with coordinates ( 0 , 8120 ) and ( 0 ... psi in this case , and the principal normal stresses are σ1 = 8120 psi and σ3 - 8120 ...
... psi and 0.30 , respectively , so that E G = 2 ( 1 + μ ) 29.0 × 106 2 ( 1 + ... Fig . 4.9 ( c ) . Points X and Y with coordinates ( 0 , 8120 ) and ( 0 ... psi in this case , and the principal normal stresses are σ1 = 8120 psi and σ3 - 8120 ...
Page 354
... Fig . 7.8 ( a ) in relation to x - v coordinate system of Fig . 7.7 . The stresses ox and t , are computed as follows : By Eq . 5.10 , Ox = M I ν u = - Fd I u = ( 8000 ) ( 10 ) ( 2 ) ( π / 64 ) ( 4 ) 4 = 12,732 psi By Eq . 4.19 , Τρ Fec ...
... Fig . 7.8 ( a ) in relation to x - v coordinate system of Fig . 7.7 . The stresses ox and t , are computed as follows : By Eq . 5.10 , Ox = M I ν u = - Fd I u = ( 8000 ) ( 10 ) ( 2 ) ( π / 64 ) ( 4 ) 4 = 12,732 psi By Eq . 4.19 , Τρ Fec ...
Page 513
... psi ) cantilever beam having a circular cross section and a span of 18 ft is to carry a load that varies linearly ... Fig . P10.34 ( a ) . The beam is to be constructed by gluing two rectangular pieces to form a T section , as shown in Fig ...
... psi ) cantilever beam having a circular cross section and a span of 18 ft is to carry a load that varies linearly ... Fig . P10.34 ( a ) . The beam is to be constructed by gluing two rectangular pieces to form a T section , as shown in Fig ...
Contents
Stresses in Beams | 198 |
Deflections of Beams | 265 |
Combined Stresses and Theories of Failure | 336 |
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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