Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 5
... parallel forces , having the same line of action , P1 and P2 , act on a body . We say that the two forces balance each other ; the body does not move , and the system of forces is in equilibrium . These two forces are concurrent ; if ...
... parallel forces , having the same line of action , P1 and P2 , act on a body . We say that the two forces balance each other ; the body does not move , and the system of forces is in equilibrium . These two forces are concurrent ; if ...
Página 7
... parallel forces , having the same line of action , P1 and P2 , act on a body . We say that the two forces balance each other ; the body does not move , and the system of forces is in equilibrium . These two forces are concurrent ; if ...
... parallel forces , having the same line of action , P1 and P2 , act on a body . We say that the two forces balance each other ; the body does not move , and the system of forces is in equilibrium . These two forces are concurrent ; if ...
Página 20
... Parallel Forces . The resultant of parallel forces may be found by constructing the funicular polygon . EXAMPLE EXAMPLE . The five forces shown in Fig . 19 ( a ) constitute a system of parallel forces . Determine the resultant ...
... Parallel Forces . The resultant of parallel forces may be found by constructing the funicular polygon . EXAMPLE EXAMPLE . The five forces shown in Fig . 19 ( a ) constitute a system of parallel forces . Determine the resultant ...
Contenido
CHAPTER | 1 |
Elements of a Force | 9 |
Equilibrant | 15 |
Derechos de autor | |
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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