Engineering Mechanics of Materials |
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Page 9
... lb / ft and wg = 4 lb / ft this becomes WA FA = 6y - 20 ( 10 ≤ y ≤25 ) FA is also a linear function of y which varies from 40 lb when y = 10 ft to 130 lb when y = 25 ft . A B These linear functions for F , and F , are plotted versus y ...
... lb / ft and wg = 4 lb / ft this becomes WA FA = 6y - 20 ( 10 ≤ y ≤25 ) FA is also a linear function of y which varies from 40 lb when y = 10 ft to 130 lb when y = 25 ft . A B These linear functions for F , and F , are plotted versus y ...
Page 51
... ft to the right of A and compute the axial force , shear , and moment acting at this section by considering both ... lb / ft and B weighs 20 lb / ft . Choose an origin at the bottom of the bar and direct a positive x axis upward . Plot ...
... ft to the right of A and compute the axial force , shear , and moment acting at this section by considering both ... lb / ft and B weighs 20 lb / ft . Choose an origin at the bottom of the bar and direct a positive x axis upward . Plot ...
Page 242
... ft 2000 lb -3 ft 200 lb / ft FIGURE H5.33 3 ft 1000 lb 100 lb / ft 5.34 The beam shown in Figure H5.34 ( a ) has the cross - sectional area shown in Figure H5.34 ( b ) such that the 5 - in . dimension of the trapezoidal section is ...
... ft 2000 lb -3 ft 200 lb / ft FIGURE H5.33 3 ft 1000 lb 100 lb / ft 5.34 The beam shown in Figure H5.34 ( a ) has the cross - sectional area shown in Figure H5.34 ( b ) such that the 5 - in . dimension of the trapezoidal section is ...
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