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 79
Page 10
... ft - long vertical cable supports a weight of 5000 lb at its lower end . The cable weighs 4 lb / ft and is supported ... k Fourth 100 k Third 120 k | Second Cross section Reinforced concrete column FIGURE P1.1 First 4 m 20 kN 3 m 3 m R ...
... ft - long vertical cable supports a weight of 5000 lb at its lower end . The cable weighs 4 lb / ft and is supported ... k Fourth 100 k Third 120 k | Second Cross section Reinforced concrete column FIGURE P1.1 First 4 m 20 kN 3 m 3 m R ...
Page 11
... k 50 k 10 k B A 4 m ft | 5n | 61 | 811— | —8 11 8 11 — Side view I Cross section 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 ...
... k 50 k 10 k B A 4 m ft | 5n | 61 | 811— | —8 11 8 11 — Side view I Cross section 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 ...
Page 12
... ft , wg = 8 lb / ft , and we = 4 lb / ft . Draw appropriate WB free - body diagrams to determine the internal axial ... k 8 k A B 5 k D 20 K E -10 ft 8 ft 10 ft 10 ft- FIGURE P1.12 B FIGURE P1.15 = dT q = = 10 CT 80/3 dT dT. 12 Ch . 1 ...
... ft , wg = 8 lb / ft , and we = 4 lb / ft . Draw appropriate WB free - body diagrams to determine the internal axial ... k 8 k A B 5 k D 20 K E -10 ft 8 ft 10 ft 10 ft- FIGURE P1.12 B FIGURE P1.15 = dT q = = 10 CT 80/3 dT dT. 12 Ch . 1 ...
Page 19
... ft k - ft / ft T * * * ) ) ) ) ) ) ) ) ) ) ) ( 4 , 10 ) -10 ft ( a ) B 10 , 10 TB ( 4 , 26.7 ) ( c ) ( b ) FIGURE 1.11 ( 10 , 86.7 ) x ( ft ) x ( ft ) At x = 4 , T = 30. This set. Sec . 1.5 / Variable Torsional Loading - Torque ...
... ft k - ft / ft T * * * ) ) ) ) ) ) ) ) ) ) ) ( 4 , 10 ) -10 ft ( a ) B 10 , 10 TB ( 4 , 26.7 ) ( c ) ( b ) FIGURE 1.11 ( 10 , 86.7 ) x ( ft ) x ( ft ) At x = 4 , T = 30. This set. Sec . 1.5 / Variable Torsional Loading - Torque ...
Page 20
... k - ft . Shear and Bending Moment in Beams = Consider a member. = dT q = = 10 CT 80/3 dT dT = = dx 10dx 10 x ୮ ་ dx T – 80 = 10 ( x − 4 ) T = 10x - 40 PROBLEMS 1.17 A beam along the ... ft 8 k - ft 8 k. 20 Ch . 1 Internal Forces in Members.
... k - ft . Shear and Bending Moment in Beams = Consider a member. = dT q = = 10 CT 80/3 dT dT = = dx 10dx 10 x ୮ ་ dx T – 80 = 10 ( x − 4 ) T = 10x - 40 PROBLEMS 1.17 A beam along the ... ft 8 k - ft 8 k. 20 Ch . 1 Internal Forces in Members.
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