Engineering Mechanics of Materials |
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Page 564
... joint 4 is assumed to be frictionless , which means that a reaction does not act in the global X direction at joint 4. This is stated under JOINT RELEASES in the program . MEMBER PROPERTIES are stated in the local coordinate system for ...
... joint 4 is assumed to be frictionless , which means that a reaction does not act in the global X direction at joint 4. This is stated under JOINT RELEASES in the program . MEMBER PROPERTIES are stated in the local coordinate system for ...
Page 567
... joint 1 ( the origin ) toward joint 2 , which is consistent with MEMBER INCIDENCES 112 STRUCTURE INDETERMINATE PLANE FRAME LOADING NO.ONE MEMBER FORCES #### MEMBER 12233 JO INT AXIAL FORCE -30.903 30.903 SHEAR FORCE 83.904 MOMENT ...
... joint 1 ( the origin ) toward joint 2 , which is consistent with MEMBER INCIDENCES 112 STRUCTURE INDETERMINATE PLANE FRAME LOADING NO.ONE MEMBER FORCES #### MEMBER 12233 JO INT AXIAL FORCE -30.903 30.903 SHEAR FORCE 83.904 MOMENT ...
Page 571
... JOINT LOADS , FREE JOINTS should all equal zero , since external forces are not applied to the arch at these joints . All of these output values , while not zero , are very small when compared to the member forces on a percentage basis ...
... JOINT LOADS , FREE JOINTS should all equal zero , since external forces are not applied to the arch at these joints . All of these output values , while not zero , are very small when compared to the member forces on a percentage basis ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
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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 Compute coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter elastic curve equal equation equilibrium Example factor of safety flexural stress free-body diagram Homework Problems k-ft k-in kN-m length M₁ material maximum shear stress MN/m² modulus of elasticity Mohr's circle moment of inertia neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal stresses r₁ ratio Refer to Figure respect rotation section a-a shaft shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress at point stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁