Engineering Mechanics of Materials |
From inside the book
Results 1-3 of 42
Page 288
B. B. Muvdi, J. W. McNabb. 6.17 A linearly varying loading is applied to the simply supported beam of Figure H6.17 . For an origin at point A , substitute the loading function w ( x ) into Eq . 6.12 and derive equations for shear ...
B. B. Muvdi, J. W. McNabb. 6.17 A linearly varying loading is applied to the simply supported beam of Figure H6.17 . For an origin at point A , substitute the loading function w ( x ) into Eq . 6.12 and derive equations for shear ...
Page 290
... Simply supported beam - uniformly distributed loading v -w [ 3 01 = W x 24Elu 02-01 WX V = ( L3 − 2Lx2 + x3 ) 24Elu + 01 02 L Case 2. Simply supported beam - concentrated load at any point ע V = -5wL4 384Elu Maximum at center 01 ...
... Simply supported beam - uniformly distributed loading v -w [ 3 01 = W x 24Elu 02-01 WX V = ( L3 − 2Lx2 + x3 ) 24Elu + 01 02 L Case 2. Simply supported beam - concentrated load at any point ע V = -5wL4 384Elu Maximum at center 01 ...
Page 595
... simply supported beam loaded by a uniform load as shown in Figure 12.14 ( a ) will resist increasing loads until a plastic hinge forms at the center under the action of the fully plastic load , w ,. A larger force cannot be applied to the ...
... simply supported beam loaded by a uniform load as shown in Figure 12.14 ( a ) will resist increasing loads until a plastic hinge forms at the center under the action of the fully plastic load , w ,. A larger force cannot be applied to the ...
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