Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 5
... common , and the resultant of the two forces will pass through this common point . The resultant of two non - parallel forces may be found graphically by constructing a parallelogram of forces . The components of a force are the two or ...
... common , and the resultant of the two forces will pass through this common point . The resultant of two non - parallel forces may be found graphically by constructing a parallelogram of forces . The components of a force are the two or ...
Página 7
... common , and the resultant of the two forces will pass through this common point . The resultant of two non - parallel forces may be found graphically by constructing a parallelogram of forces . The components of a force are the two or ...
... common , and the resultant of the two forces will pass through this common point . The resultant of two non - parallel forces may be found graphically by constructing a parallelogram of forces . The components of a force are the two or ...
Página 76
... Common structural . J. & P.-B. & S . 2,150 145 455 J. & P.-B. & S . 1,900 100 400 1,400 1,450 95 380 1,750 1,600,000 1,500,000 1,250 1,500,000 Select structural . P. & T . 455 1,750 1,600,000 Structural .. P. & T . 400 1,400 1,500,000 ...
... Common structural . J. & P.-B. & S . 2,150 145 455 J. & P.-B. & S . 1,900 100 400 1,400 1,450 95 380 1,750 1,600,000 1,500,000 1,250 1,500,000 Select structural . P. & T . 455 1,750 1,600,000 Structural .. P. & T . 400 1,400 1,500,000 ...
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