Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 67
Page 265
... compressive flexural stresses in the beam . 5.65 The beam shown in Figure H5.32 has the cross- sectional area illustrated in Figure H5.20 . Assume the loads to act vertically downward and their plane to be normal to the 0.12 - m side of ...
... compressive flexural stresses in the beam . 5.65 The beam shown in Figure H5.32 has the cross- sectional area illustrated in Figure H5.20 . Assume the loads to act vertically downward and their plane to be normal to the 0.12 - m side of ...
Page 380
... compressive load greater than 1771.8 kN would lead to a compressive stress σ3 [ left - hand side of Eq . ( b ) ] , which is greater than the compressive property of 250 MPa for this brittle material . Thus the compressive property is ...
... compressive load greater than 1771.8 kN would lead to a compressive stress σ3 [ left - hand side of Eq . ( b ) ] , which is greater than the compressive property of 250 MPa for this brittle material . Thus the compressive property is ...
Page 444
... compressive restraining force is most conve- niently accomplished by assuming that one of the two supports , say ... compressive force F as shown in Figure 9.3 ( c ) . The compressive force would need to be of sufficient magnitude to ...
... compressive restraining force is most conve- niently accomplished by assuming that one of the two supports , say ... compressive force F as shown in Figure 9.3 ( c ) . The compressive force would need to be of sufficient magnitude to ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
14 other sections not shown
Other editions - View all
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 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