Simplified Mechanics and Strength of MaterialsWiley, 1951 - 275 páginas |
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Página 27
... draw the force polygon of the external forces . SOLUTION . The force polygon is begun by drawing the two forces ab and be as shown in Fig . 24 ( b ) . The pole o is selected , and the rays oa , ob , and oc are drawn . Since there is a ...
... draw the force polygon of the external forces . SOLUTION . The force polygon is begun by drawing the two forces ab and be as shown in Fig . 24 ( b ) . The pole o is selected , and the rays oa , ob , and oc are drawn . Since there is a ...
Página 31
... draw a line parallel to BJ . The point j will be somewhere on this line . The next force is JI ; through point i draw a line parallel to JI . Since the point j is on this line , as well as on the line through b parallel to BJ , it must ...
... draw a line parallel to BJ . The point j will be somewhere on this line . The next force is JI ; through point i draw a line parallel to JI . Since the point j is on this line , as well as on the line through b parallel to BJ , it must ...
Página 53
... drawn to the left support , thus determining the direction of the left support . To draw the force polygon first draw to scale ab , the 1,000 # load , Fig . 44 ( g ) . From point b draw a line parallel to BC , and from point a draw a ...
... drawn to the left support , thus determining the direction of the left support . To draw the force polygon first draw to scale ab , the 1,000 # load , Fig . 44 ( g ) . From point b draw a line parallel to BC , and from point a draw a ...
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