Engineering Mechanics of Materials |
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Page 601
... load of 2P = 250 kN and the right span will be subjected to P = 125 kN , which is well below the load of 150 kN ... uniform load associated with first yield of the simply supported beam ; ( b ) uniform load associated with collapse of ...
... load of 2P = 250 kN and the right span will be subjected to P = 125 kN , which is well below the load of 150 kN ... uniform load associated with first yield of the simply supported beam ; ( b ) uniform load associated with collapse of ...
Page 604
... load P , for Mp = 800 k - in . -20 ft B Pp +10 ft- FIGURE H12.23 12.24 A two - span continuous beam is loaded with ... uniform load associated with the first yield of the simply supported beam . ( b ) The uniform load associated with ...
... load P , for Mp = 800 k - in . -20 ft B Pp +10 ft- FIGURE H12.23 12.24 A two - span continuous beam is loaded with ... uniform load associated with the first yield of the simply supported beam . ( b ) The uniform load associated with ...
Page 691
... uniform load of intensity w and the allowable stresses are 2000 psi and 25,000 psi for concrete and steel , respectively , find w . This value of w would include the weight of the beam . Deduct this weight per foot to determine w ...
... uniform load of intensity w and the allowable stresses are 2000 psi and 25,000 psi for concrete and steel , respectively , find w . This value of w would include the weight of the beam . Deduct this weight per foot to determine w ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
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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