Engineering Mechanics of Materials |
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Page 188
... obtained from these circles . Use G = 75 GPa . 4.41 A membrane of equilateral triangular shape with each side a = 0.05 m was subjected to a uniform pressure and the following results obtained : = 3.125 × 10-6 m3 volume under the ...
... obtained from these circles . Use G = 75 GPa . 4.41 A membrane of equilateral triangular shape with each side a = 0.05 m was subjected to a uniform pressure and the following results obtained : = 3.125 × 10-6 m3 volume under the ...
Page 489
... obtained from the few tests that are performed are a true representation of the behavior of that material . Furthermore , the properties obtained from a small sample are not necessarily the same as would be obtained from a full - size ...
... obtained from the few tests that are performed are a true representation of the behavior of that material . Furthermore , the properties obtained from a small sample are not necessarily the same as would be obtained from a full - size ...
Page 682
... obtained from Eq . 14.22 . Thus ( oak ) max = n ( at ) max = ( 6250 ) = 1250 psi The same answers may be obtained by transforming the aluminum into an equivalent amount of oak . Thus t · - - ( ) ( ) - = - n = 2.5 in . The equivalent ...
... obtained from Eq . 14.22 . Thus ( oak ) max = n ( at ) max = ( 6250 ) = 1250 psi The same answers may be obtained by transforming the aluminum into an equivalent amount of oak . Thus t · - - ( ) ( ) - = - n = 2.5 in . The equivalent ...
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