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. |
From inside the book
Results 1-5 of 85
Page 2
... FIGURE 1.1 acting inside a member such as the one shown in Fig . 1.1 ( a ) . In order to investigate these internal ... Figure 1.1 ( b ) : Figure 1.1 ( c ) : Figure 1.1 ( d ) : FA - P1 = 0 1 FA P1 = FB + P2- P1 = 0 2 1 FB = P1 - P2 Fc + ...
... FIGURE 1.1 acting inside a member such as the one shown in Fig . 1.1 ( a ) . In order to investigate these internal ... Figure 1.1 ( b ) : Figure 1.1 ( c ) : Figure 1.1 ( d ) : FA - P1 = 0 1 FA P1 = FB + P2- P1 = 0 2 1 FB = P1 - P2 Fc + ...
Page 4
... Figure 1.2 ( b ) : ↑ 10 F = 0 ― A FA = 10 kN ( T ) Figure 1.2 ( c ) : ↑ 10 - 2 ( 6 ) — FB = 0 FB - 2 kN ( C ) Figure 1.2 ( d ) : Note the choice of the bottom free body in this case . Of course , the top free body would yield the same ...
... Figure 1.2 ( b ) : ↑ 10 F = 0 ― A FA = 10 kN ( T ) Figure 1.2 ( c ) : ↑ 10 - 2 ( 6 ) — FB = 0 FB - 2 kN ( C ) Figure 1.2 ( d ) : Note the choice of the bottom free body in this case . Of course , the top free body would yield the same ...
Page 11
... FIGURE P1.5 ft . 0 40 kN FIGURE P1.7 1.8 The bar shown in Fig . P1.8 is attached to a column at its right end and loaded as shown . Find the reaction R required to maintain equilibrium of the bar . Draw the free - body diagrams required ...
... FIGURE P1.5 ft . 0 40 kN FIGURE P1.7 1.8 The bar shown in Fig . P1.8 is attached to a column at its right end and loaded as shown . Find the reaction R required to maintain equilibrium of the bar . Draw the free - body diagrams required ...
Page 12
... FIGURE P1.10 Construct an axial force diagram for the member shown in Fig . P1.11 . Choose an origin at point A and measure a length coordinate along the axis of the member . Sketch all free - body diagrams required to evaluate the ...
... FIGURE P1.10 Construct an axial force diagram for the member shown in Fig . P1.11 . Choose an origin at point A and measure a length coordinate along the axis of the member . Sketch all free - body diagrams required to evaluate the ...
Page 14
... Figure 1.7 : Tx = 0 4 − T1 + T2 − T3 + T4 = 0 Regarding the torque T as unknown , we may determine it from the preceding equation : T1 = T1 T2 + T3 Figure 1.7 ( b ) : Σ T = 0 Figure 1.7 ( c ) : T = 0 Figure 1.7 ( d ) : T = 0 - T1 = T1 ...
... Figure 1.7 : Tx = 0 4 − T1 + T2 − T3 + T4 = 0 Regarding the torque T as unknown , we may determine it from the preceding equation : T1 = T1 T2 + T3 Figure 1.7 ( b ) : Σ T = 0 Figure 1.7 ( c ) : T = 0 Figure 1.7 ( d ) : T = 0 - T1 = T1 ...
Contents
Stresses in Beams | 198 |
Deflections of Beams | 265 |
Combined Stresses and Theories of Failure | 336 |
Column Theory and Analyses | 384 |
Statically Indeterminate Members | 432 |
Introduction to Component Design | 484 |
Analysis and Design for Inelastic Behavior | 523 |
Analysis and Design for Impact and Fatigue Loadings | 552 |
Selected Topics | 590 |
13 7 | 625 |
APPENDIX | 647 |
Index | 687 |
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Common terms and phrases
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 plane stress plane stress condition plot principal centroidal axis principal stresses r₁ radius ratio 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 σ₁