Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 11
... equilibrium . The force required to maintain equilibrium is force E , shown by the dotted line . E , the equilibrant , is equal to the resultant in magnitude and opposite in direction and has P Ꭱ . P2 FIG . 11 . the same line of action ...
... equilibrium . The force required to maintain equilibrium is force E , shown by the dotted line . E , the equilibrant , is equal to the resultant in magnitude and opposite in direction and has P Ꭱ . P2 FIG . 11 . the same line of action ...
Página 17
... equilibrium . The system might be altered to provide equilibrium by moving the force BC to a parallel position in which its line of action passes through the intersection of oc and ob . The system , thus modified , would have both the ...
... equilibrium . The system might be altered to provide equilibrium by moving the force BC to a parallel position in which its line of action passes through the intersection of oc and ob . The system , thus modified , would have both the ...
Página 18
... equilibrium . The system might be altered to provide equilibrium by moving the force BC to a parallel position in which its line of action passes through the intersection of oc and ob . The system , thus modified , would have both the ...
... equilibrium . The system might be altered to provide equilibrium by moving the force BC to a parallel position in which its line of action passes through the intersection of oc and ob . The system , thus modified , would have both the ...
Contenido
CHAPTER | 1 |
Elements of a Force | 9 |
Equilibrant | 15 |
Derechos de autor | |
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Términos y frases comunes
allowable axial load allowable load allowable unit stress angle bars center of moments centroid column compressive stresses compressive unit stress Compute the maximum concentrated load cross section cross-sectional area deflection deformation determine diameter distance double bearing elastic limit EXAMPLE EXAMPLE factor of safety fillet weld flexure formula force polygon free body diagram funicular polygon hence indicated in Fig inertia intersection length line of action linear foot magnitude material maximum bending maximum shear modulus of elasticity moment of inertia neutral surface parallel parallelogram of forces pier plate pounds per linear pounds per square pressure PROBLEMS R₁ radius of gyration reactions reinforced concrete resisting respect resultant rivet rods section modulus shaft shear diagram shearing stress shearing unit stress shown in Fig simple beam single bearing slenderness ratio SOLUTION span square inch stirrups Table tensile stresses thickness three forces truss uniformly distributed load weight width zero