Strength of Materials |
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Results 1-3 of 68
Page 127
... slopes are directed down to the right , i.e. , + slope = slope = we observe from Eq . ( 4-5 ) that the shear diagram in Fig . 4–20b must slope continuously down to the right . The inclination varies directly with the corresponding ...
... slopes are directed down to the right , i.e. , + slope = slope = we observe from Eq . ( 4-5 ) that the shear diagram in Fig . 4–20b must slope continuously down to the right . The inclination varies directly with the corresponding ...
Page 132
... slope is again increasingly steeper and directed downward to the right as the corresponding shear ordinates change their sign to negative . - At C , the shear ordinate changes abruptly from -4 to 12 kN , at which it remains until D. The ...
... slope is again increasingly steeper and directed downward to the right as the corresponding shear ordinates change their sign to negative . - At C , the shear ordinate changes abruptly from -4 to 12 kN , at which it remains until D. The ...
Page 269
... slope at C in the original cantilever , the fictitious shear must be zero at C ; therefore [ V = ( EY ) L ] 0 = V - A From this we see that the fictitious shear of the conjugate beam at B must equal the area A of the M / EI diagram ...
... slope at C in the original cantilever , the fictitious shear must be zero at C ; therefore [ V = ( EY ) L ] 0 = V - A From this we see that the fictitious shear of the conjugate beam at B must equal the area A of the M / EI diagram ...
Common terms and phrases
allowable stresses aluminum angle applied load assumed axial load beam carrying beam in Fig beam loaded beam shown bending moment bolt bronze cantilever beam centroid column compressive stress concentrated load continuous beam cross section deformation Determine the maximum diameter elastic curve end moments equal equilibrium equivalent exploratory section factor of safety fibers free-body diagram GN/m² Hence Hooke's law horizontal ILLUSTRATIVE PROBLEMS kN.m kN·m kN/m length load diagram loaded as shown M₁ M₂ maximum shearing stress midspan deflection mm² MN/m² Mohr's circle moment of inertia N/m² neutral axis obtain P₁ P₂ plane plate positive principal stresses proportional limit R₁ R₂ R2 Figure radius reaction resisting restrained beam resultant rivet segment shaft shear diagram shearing force shown in Fig simply supported beam Solution span statically indeterminate steel strain tangent drawn tensile stress three-moment equation torque torsional uniformly distributed load vertical shear weld ΕΙ