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By Parviz Ghavami

This ebook, framed within the procedures of engineering research and layout, provides suggestions in mechanics of fabrics for college kids in two-year or four-year courses in engineering know-how, structure, and development building; in addition to for college students in vocational colleges and technical institutes. utilizing the foundations and legislation of mechanics, physics, and the basics of engineering, Mechanics of fabrics: An advent for Engineering expertise may also help aspiring and practising engineers and engineering technicians from throughout disciplines—mechanical, civil, chemical, and electrical—apply innovations of engineering mechanics for research and layout of fabrics, buildings, and computer elements. The ebook is perfect for these looking a rigorous, algebra/trigonometry-based textual content at the mechanics of fabrics.

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As an example, the relations among three forces in equilibrium could be shown by forces F1 ¼ F2 ¼ 6,000 lb (Fig. 34). These forces are concurrent forces since their lines of action meet at point M. Then, what is the direction and magnitude of a horizontal force F3 exerted at the same point M such that F1, F2, and F3 will all be in equilibrium? F1=6 k F1 M F3 30º 30º R F2 F2=6 k Fig. 8 Mathematical Statement of Equilibrium 35 Force F3, called the equilibrant (or balancing force) of the two forces, is the closing line of the force triangle MNP shown in Fig.

1 Moment of a Force Moment of a force F about a given point O is the product of the force F and its perpendicular distance r from the line of action of the force to the center of rotation O. # Springer International Publishing Switzerland 2015 P. 1007/978-3-319-07572-3_3 51 52 3 Moment of a Force Point O is the O moment center ° r Mo m arm ent F 90 Line of action of force Fig. m. 2 Sign Convention of a Moment In the figure shown (Fig. 2), it is evident that the 100-lb force tends to rotate about point A in one direction, and about point B in the opposite direction.

52 46 2 Force Systems on Structures 7. F2=270 lb Y 45º F1=150 lb X 28º F3=320 lb Fig. 53 8. Using the law of cosines and the law of sines, find the resultant of the given force triangle (Fig. 54). F1=50 lb 65º F2=150 lb Fig. 54 9. Using the method of components, find the resultant of the force system shown (Fig. 55). F2=100 N Y F1=500 N 3 2 25º X F3=1200 N Fig. 55 Problems 47 10. Draw a force polygon diagram to show the resultant of the forces shown (Fig. 56). F1=100 lb F2=150 lb 6 Y 3 1 1 X F3=200 lb F4=350 lb Fig.

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