Elements of Graphic StaticsThe author, 1907 - 127 pages |
Expressions et termes fréquents
applied center of gravity center of moments centroid closing string compression compressive stress concurrent forces cut the members dead load stresses determined diagonals draw the line End of Truss equilibrium find the stresses force polygon forces acting forces represented funicular polygon given force given system inertia joint U₁ kind of stress known left reaction line of action load stress diagram lower chord magnitude and direction maximum snow load maximum stresses members cut members X-1 method moment of inertia non-concurrent forces parallel forces point of application pounds Problem.-It is required purlins R₁ radius of gyration replacing the stresses represent in magnitude required to find resultant roof truss section p-p seen shown in Fig stresses in X-2 strings system of forces take the section tensile stress tension tion triangle truss on rollers unknown external force unknown forces unknown stresses wind load stress
Fréquemment cités
Page 42 - Let ab and be (Fig. 36) be the lines of action of the two parallel forces AB and BC, and let ac be the line of action of their resultant AC. Draw any line MN perpendicular to the lines of action of the given forces, intersecting them at the points M, O, and N. Since AC is the resultant of AB and BC, the moment of AC about any point in their plane is equal to the sum of the moments of AB and BC about the same point. If the origin of moments is taken on the line of action ac, then the moment of AC...
Page 34 - ... 1. Three forces of 30 pounds, 50 pounds, 90 pounds act at a point, the angle between the directions of any two forces being 120°. Find the magnitude of their resultant, and the angle which its direction makes with the greatest force. 2. Prove that the algebraic sum of the moments of two concurrent forces about any point in their plane is equal to the moment of their resultant about the same point. Forces 2P, P, P act along BC, CA, AB, the sides of an equilateral triangle ABC. Show that the resultant...
Page 34 - It will be shown that the moment of a couple is equal to the product of one of the forces into the perpendicular distance between the lines of action of p the forces. Let O (Fig. 26) be the origin of moments, and pMOMENT OF THE KESULTANT OF FORCES. 41 Then the moment of the couple is equal to — P...
Page 120 - SM — 0, the moment of the stresses in the members cut by the section must balance the moment of the external forces to the left of the section mm.
Page 34 - The sum of the moments of a number of parallel coplanar forces about any point in their plane is equal to the moment of the resultant about that point.
Page 54 - Steiner. is true for bath a plane laminar body and a thin three-dimensional body, and states that the moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through its...
Page 3 - Resolution of forces is the process of finding, for a given system of forces, an equivalent system having a greater number of forces than the given system. The process of finding two or more forces which are equivalent to a given force is the most important case of resolution. 20. Couple. A couple consists of two equal, parallel forces, opposite in direction, having different lines of action.
Page 81 - ... from the principles of statics we know that: 1. The vertical shear V equals the sum of the vertical forces on the left of the cut. 2. The total compression C equals the total tension T. 3. The sum of the moments of the tensile and compressive stresses on the portion shown equals the algebraic sum of the moment of the external forces to the left of the section. From experimental evidence it is known that the unit deformations vary directly as the distance of the fiber from the neutral surface;...
Page 51 - ART. 2. MOMENT OF INERTIA OF AREAS. 78. Moment of Inertia of an Area. The moment of inertia of an area about any axis is equal to the summation of the products of the differential areas composing the area into the squares of the distances of these differential areas from the axis. The moment of inertia of an area may be found by dividing the given area into elementary areas, treating these elements as parallel forces, and finding the moment of inertia of the forces.
Page 40 - The center of gravity of a triangle lies on a line drawn from a vertex to the middle point of the opposite side, and at a distance from that side equal to one-third of the length of the line; or it is at the intersection of lines drawn from the vertexes to the middle of the opposite sides.