Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 47
... moments , the sum of the moments of the two forces about this point is ( P1 X x ) + ( P2 × 0 ) , which , of course , equals P1 Xx . The moment of P2 about point a is ( P2 × 0 ) because the lever arm is zero . The moment of a couple is ...
... moments , the sum of the moments of the two forces about this point is ( P1 X x ) + ( P2 × 0 ) , which , of course , equals P1 Xx . The moment of P2 about point a is ( P2 × 0 ) because the lever arm is zero . The moment of a couple is ...
Página 54
... moments and write an equation of moments of all the forces acting on the beam , positive moments on one side of the equation and negative moments on the other . The force tending to produce clockwise rotation about R2 is the force R1 ...
... moments and write an equation of moments of all the forces acting on the beam , positive moments on one side of the equation and negative moments on the other . The force tending to produce clockwise rotation about R2 is the force R1 ...
Página 130
... moment at any section in the length of a beam equals the moments of the reactions minus the moments of the loads to the left of the section . Particular attention is called to the similarity between this rule and the rule for ...
... moment at any section in the length of a beam equals the moments of the reactions minus the moments of the loads to the left of the section . Particular attention is called to the similarity between this rule and the rule for ...
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