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SIMPLE STATIC FORCES
Statics is the branch of mechanics that is concerned with the analysis of loads (force and torque, or "moment") acting on physical systems that do not experience an acceleration ( a =0), but rather, are in static equilibrium with their environment. When in static equilibrium, the acceleration of the system is zero and the system is either at rest, or its center of mass moves at constant velocity. Elementary statics Consider the force F acting on an angle θ to the horizontal, as shown by Figure 1.3(a). Now as the force F is a vector, (i.e. it has magnitude and direction), it can be represented as being equivalent to its horizontal and vertical components, namely FH and FV respectively, as shown by Figure 1. 3 (b). These horizontal and vertical components are also vectors, as they have magnitude and direction.
Problem 1.1 Determine the forces in the plane pin-jointed framework shown below Solution Assume all unknown forces in each member are in tension, i.e. the internal force in each member is pulling away from its nearest joint, as shown below. Resolving forces vertically
arrowheads to the triangle so that the arrowheads are either all in a clockwise direction or, alternatively, all in a counter-clockwise direction. Applying the sine rule to the triangle of forces above, These forces can now be transferred to the joint A of the pin-jointed truss below, where it can be seen that the member with the load F, is in tension, and that the member with the load F2 is in compression. This is known as a free body diagram. COUPLES A couple can be described as the moment produced by two equal and opposite forces acting together, as shown in Figure 1.4 where, the moment at the couple = M = F x l (N.m) F = force (N) I = lever length (m)
It should be noted from Figure 1.4 that the lever can be described as the perpendicular distance between the line of action of the two forces causing the couple. Furthermore, in Figure 1.5, although the above definition still applies, the same value of couple can be calculated, if the lever is chosen as the perpendicular distance between the components of the force that are
Taking moments about B clockwise moments = counter-clockwise moments RA x (4+2) = 5 x 2 or RA = 10/ 16 Problem 1.3 Determine the values of the reactions of R, and R, for the simply-supported beam shown below in figure 1.7.
Problem 1.4 A uniform slab of mass 25000 kg is being raised as shown in Fig below. Determine all the forces on the slab. Figure 1. Figure 1. 9 Solution Consider the Free Body Diagram (FBD) of the slab, as shown in Fig above. In order to use only one equation to determine T , we locate D about which the other two unknowns P and N do not produce any moment.
Figure 1. Solution From the FBD of the walk-way shown in figure above. Figure 1.
TUTORIAL QUESTIONS
Problem 1 Determine the reactions RA and R, for the simply-supported beams. (a) RA = 3.333 KN RB =6.667 KN (b) RA = 9.6 KN RB= 6.4 KN (c) RA =4.625 KN RB =3.375 KN
Problem 3 A load of 20 kg is supported as shown in Fig. 1.12. Assuming the pulley to be ideal, determine the length of CD required to realize the equilibrium configuration shown. Also, determine the tension in the cable CD. [ FCD = 150.2 KN, CD = 1568 mm]
