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 80
Page 5
... slope = dF dx ( d ) -F + dF + x ( b ) FIGURE 1.4 F2 Fi equals ordinate change F2 - Fi ( L , P ) Length units " equations sign convention " will be established such that. Sec . 1.3 / Variable Axial Loading - Internal Force Relationships 5 ...
... slope = dF dx ( d ) -F + dF + x ( b ) FIGURE 1.4 F2 Fi equals ordinate change F2 - Fi ( L , P ) Length units " equations sign convention " will be established such that. Sec . 1.3 / Variable Axial Loading - Internal Force Relationships 5 ...
Page 6
... slope of the F - x curve at the same point . tion : Multiply Eq . 1.1 by dx and write dF = ƒ dx . Then , integrate both sides of the equa- Px2 F2 [ " dF = [ " ) f dx JF1 EXAMPLE 1.2 Jx1 The left side of this equation may be integrated ...
... slope of the F - x curve at the same point . tion : Multiply Eq . 1.1 by dx and write dF = ƒ dx . Then , integrate both sides of the equa- Px2 F2 [ " dF = [ " ) f dx JF1 EXAMPLE 1.2 Jx1 The left side of this equation may be integrated ...
Page 7
... slope of the F - y curve in Fig . 1.5 ( d ) given by ΔΕ 400 Ay 10 = 4 lb / ft — = To better understand Eq . 1.2 , consider rod A. The difference in ordinates to the F - y curve between the ends of this rod is given by 130 40 90 lb ...
... slope of the F - y curve in Fig . 1.5 ( d ) given by ΔΕ 400 Ay 10 = 4 lb / ft — = To better understand Eq . 1.2 , consider rod A. The difference in ordinates to the F - y curve between the ends of this rod is given by 130 40 90 lb ...
Page 8
... P = 80 N = FR resisting force ( b ) F ( N ) ( 0 , 80 ) Slope dF dx 0.1 ( 0.4 , 0 ) ( 0.4 , -f1 ) x ( m ) ( c ) x ( m ) ( d ) FIGURE 1.6 -100 fi 2 0.8 Then Integrating yields dFR = f dx FR 0.4 F2. 8 Ch . 1 Internal Forces in Members.
... P = 80 N = FR resisting force ( b ) F ( N ) ( 0 , 80 ) Slope dF dx 0.1 ( 0.4 , 0 ) ( 0.4 , -f1 ) x ( m ) ( c ) x ( m ) ( d ) FIGURE 1.6 -100 fi 2 0.8 Then Integrating yields dFR = f dx FR 0.4 F2. 8 Ch . 1 Internal Forces in Members.
Page 9
... slope and its equation is given by x = - 400 0.4 X = - 1000x f = 0.4 Equation 1.1 will be written for x = 0.1 m , dF dF f = dx fo.1 = – 1000 ( 0.1 ) = = 100 N / m 500x2 = 1000x F80 dF dx dx 0.1 = — 1000 ( 0.1 ) = = - 100 N / m At x ...
... slope and its equation is given by x = - 400 0.4 X = - 1000x f = 0.4 Equation 1.1 will be written for x = 0.1 m , dF dF f = dx fo.1 = – 1000 ( 0.1 ) = = 100 N / m 500x2 = 1000x F80 dF dx dx 0.1 = — 1000 ( 0.1 ) = = - 100 N / m At x ...
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