Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 20
... intersect . This point of intersection is a point on the line of action of the resultant . Therefore , we draw the resultant , whose magnitude and direction are found in the force polygon , through the point of intersection of strings ...
... intersect . This point of intersection is a point on the line of action of the resultant . Therefore , we draw the resultant , whose magnitude and direction are found in the force polygon , through the point of intersection of strings ...
Página 29
... intersection is the point in common of the three forces ; therefore , a line drawn from this point of intersection to the point of support of the left reaction determines the direction of the left reaction . Return now to Fig . 26 ( b ) ...
... intersection is the point in common of the three forces ; therefore , a line drawn from this point of intersection to the point of support of the left reaction determines the direction of the left reaction . Return now to Fig . 26 ( b ) ...
Página 53
... intersect at a common point . Therefore , the lines of action of the 1,000 # load and right support are extended until they meet at point o . From this point of intersection a line is drawn to the left support , thus determining the ...
... intersect at a common point . Therefore , the lines of action of the 1,000 # load and right support are extended until they meet at point o . From this point of intersection a line is drawn to the left support , thus determining the ...
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