Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 72
Page 9
... lb / ft and wB = 4 lb / ft this becomes FA = 6у - 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 as ...
... lb / ft and wB = 4 lb / ft this becomes FA = 6у - 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 as ...
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
T Shear and Moment at Specified Sections | 24 |
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 components 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 longitudinal 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 segment shaft shear force shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ vertical yield strength yield stress zero