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 1-5 of 79
Page 3
... vertical axis positive upward as shown in Fig . 1.2 ( a ) and write Σ F , = 0 in order to find P , the reaction at the base of the member . 10 - 2 ( 6 ) + 8 - · 2 ( 2 ) — P = 0 P = 2 kN This value of P will be used for a check on other ...
... vertical axis positive upward as shown in Fig . 1.2 ( a ) and write Σ F , = 0 in order to find P , the reaction at the base of the member . 10 - 2 ( 6 ) + 8 - · 2 ( 2 ) — P = 0 P = 2 kN This value of P will be used for a check on other ...
Page 10
... vertical cable supports a weight of 5000 lb at its lower end . The cable weighs 4 lb / ft and is supported at its upper end . Determine the vertical sup- porting force required at its upper end and construct the axial force diagram for ...
... vertical cable supports a weight of 5000 lb at its lower end . The cable weighs 4 lb / ft and is supported at its upper end . Determine the vertical sup- porting force required at its upper end and construct the axial force diagram for ...
Page 12
... vertically as preparations for drilling are made . Choose an origin at the top or bottom of the rod and express the ... vertical member AB of a metallic truss is shown in Fig . P1.15 . The metal weighs 163 lb / ft3 x 30 ft A = 20 in . 2 ...
... vertically as preparations for drilling are made . Choose an origin at the top or bottom of the rod and express the ... vertical member AB of a metallic truss is shown in Fig . P1.15 . The metal weighs 163 lb / ft3 x 30 ft A = 20 in . 2 ...
Page 22
... x2 of X 15 k - in . 60 k - in . 180 k - in . x ( ft ) 40 in . B 60 in . 60 in.60 in . C D 50 in . E F FIGURE P1.27 FIGURE P1.30 1.32 Refer to Fig . P1.32 and determine the vertical. 22 Ch . 1 Internal Forces in Members.
... x2 of X 15 k - in . 60 k - in . 180 k - in . x ( ft ) 40 in . B 60 in . 60 in.60 in . C D 50 in . E F FIGURE P1.27 FIGURE P1.30 1.32 Refer to Fig . P1.32 and determine the vertical. 22 Ch . 1 Internal Forces in Members.
Page 27
... ft- 1200 lb B R2 8 ft ( b ) 533 lb + 4 ft B R2 A ( 8 ) = 5.33 ft 2.67 ft ( c ) FIGURE 1.16 200 lb / ft -x x = ( 200 ) 133.3 lb / f MA Sum the vertical forces acting on the beam . 1. Sec . 1.7 Shear and Moment at Specified Sections 27.
... ft- 1200 lb B R2 8 ft ( b ) 533 lb + 4 ft B R2 A ( 8 ) = 5.33 ft 2.67 ft ( c ) FIGURE 1.16 200 lb / ft -x x = ( 200 ) 133.3 lb / f MA Sum the vertical forces acting on the beam . 1. Sec . 1.7 Shear and Moment at Specified Sections 27.
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