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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acting allowable angle of twist applied Assume axes axis beam bending cantilever centroidal circle column components compressive Compute Consider constant construct coordinate cross section curve deflection deformation depicted in Figure Determine developed diameter direction discussed elastic element energy equal equation equilibrium Example expressed factor failure flexural force free-body diagram function given inertia joint length limit load material maximum shear stress method modulus moment moments neutral axis normal stress Note obtained plane plot positive principal stresses Problem properties quantity ratio reactions Refer to Figure relation represents resist respect rotation segment shaft shown in Figure slope Solution Solve static steel strain strength structural subjected Substitution supported surface tensile tension theory tion torque torsional unit vertical yield zero