Method Of Sections PDF

Moment  De inition In addition to the tendency to move a body in the direction of its ap- plication, a force can also tend to rotate a body about an axis. The axis may be any line which neither intersects nor is parallel to the line of action of the force. This rotational...

Moment  De inition In addition to the tendency to move a body in the direction of its ap- plication, a force can also tend to rotate a body about an axis. The axis may be any line which neither intersects nor is parallel to the line of action of the force. This rotational tendency is known as the moment M of the force. Moment is also referred to as torque. M=r×F f Varignon’s Theorem Consider the force R acting in the plane of the body shown in the gure. The forces P and Q represent any two non-rectangular components of R. The One of the most useful moment of R about point O is principles of mechanics MO = P + Q is Varignon’s theorem, Because R = P + Q, we may write which states that the MO = r × (P + Q) = r × P + r × Q moment of a force about which says that the moment of R about O equals the any point is equal to the sum of the moments about O of its components P and sum of the moments of Q the components of the force about the same point. fi  Couple The moment produced by two equal, opposite, and noncollinear forces is called a couple. Consider the action of two equal and opposite forces F and −F a distance d apart, as shown in the gure. These two forces cannot be combined into a single force because their sum in every direction is zero. Their only effect is to produce a tendency of rotation. The combined moment of the two forces about an axis normal to their plane and passing through any point such as O in their plane is the couple M. The combined moment about point O of the forces forming the couple is M = rA × F + rB × (−F) = (rA − rB) × F fi Example problem 1 The rigid structural member is subjected to a couple consisting of the two 100-N forces. Replace this couple with an equivalent couple consisting of the two forces P and -P, each of which has a magnitude of 400 N. Determine the proper angle θ. Example problem 2 Replace the horizontal 80-lb force acting on the lever with an equivalent system consisting of a force at O and a couple. Example problem 3 Determine the moment associated with the couple applied to the rectangular plate. Exercise problem 1 Exercise problem 2 Exercise problem 3 Plane Trusses  De inition A framework composed of members joined at their ends to form a rigid structure is called a truss Bridges, roof supports, derricks, and other such structures are common examples of trusses When the members of the truss lie essentially in a single plane, the truss is called a plane truss f  Assumptions We assume all members to be two-force members. Each member of a truss is normally a straight link joining the two points of application of force. The two forces are applied at the ends of the member and are necessarily equal, opposite, and collinear for equilibrium. When we represent the equilibrium of a portion of a two-force member, the tension T or compression C acting on the cut section is the same for all sections. The weight of the member is small compared with the force it supports.  When welded or riveted connections are used to join structural members, we may usually assume that the connection is a pin joint if the centerlines of the members are concurrent at the joint. We also assume in the analysis of simple trusses that all external forces are applied at the pin connections.  Method of Joints This method for nding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint. fi  Method of Sections When analyzing plane trusses by the method of joints, we need only two of the three equilibrium equations. We can take advantage of the third or moment equation of equilibrium by selecting an entire section of the truss for the free body. Method of sections has the basic advantage that the force in almost any desired member may be found directly from an analysis of a section which has cut that member. Thus, it is not necessary to proceed with the calculation from joint to joint until the member in question has been reached. In choosing a section of the truss, not more than three members whose forces are unknown should be cut since there are only three available independent equilibrium relations.  Procedure Let us determine the force in the member BE, for example. This section has cut three members whose forces are initially unknown. In order for the portion of the truss on each side of the section to remain in equilibrium, it is nec- essary to apply to each cut member the force which was exerted on it by the member cut away. For simple trusses composed of straight two-force members, these forces, either tensile or compressive, will always be in the direction of the respective members. The left-hand section is in equilibrium under the action of the applied load L, the end reaction R , and the three forces exerted on the cut 1 members by the right-hand section, which has been removed.  The equation of moments about joint B eliminates three forces from the relation, and EF can be determined directly. The force BE is calculated from the equilibrium equation for the y-direction. Finally, we determine BC by balancing moments about point E. Calculate the force in member DJ of the roof illustrated. Neglect any horizontal components of force at the supports.