At first glance you could say what would I need all of them for. Its all to do with bonding the pattern of bricks together in way which eliminate the possibility of having one joint directly above another; the professionals refer to this as having a straight joint. The effects of straight joints are to serve only to weaken a wall. Think about when you first drew the front of a house. I bet you drew straight lines across the page to show different rows of brick and then you naturally alternated the joints between each brick. So in essence the reason bricks are cut into different lengths and shapes are to help bond the bricks together with creating a straight joint; if this subject grabs you I suggest you take a more technical manual for further advice. It is possible to purchase a range of pre shaped bricks from specialist manufactures. Wonderful names such plinth, bullnose, squint and cant are among those on offer. These prefabricated shapes are expensive and are to be used on occasions when its difficult to cut the shape required and you need to see the same finish to the brick face form all angels. Again take a look at the text books on bricklaying to see examples of their use.

the bottom part (Fig. 18b). The top force is the resultant of compressive stresses acting over the upper portion of the beam, and the bottom force is the resultant of tensile stresses acting over the bottom part. The surface at which the stresses change from compression to tensionwhere the stress is zerois called the neutral surface. FIGURE 5.19 .Shear diagram for the beam with loads shown in Fig. 5.17. 5.5.4 Shear Diagrams The unbalanced external vertical force at a section is called the shear. It is equal to the algebraic sum of the forces that lie on either side of the section. Upward acting forces on the left of the section are considered positive, downward forces negative; signs are reversed for forces on the right. A diagram in which the shear at every point along the length of a beam is plotted as an ordinate is called a shear diagram. The shear diagram for the beam in Fig. 5.17 is shown in Fig. 5.19b. The diagram was plotted starting from the left end. The 2000-lb load was plotted downward to a convenient scale. Then, the shear at the next concentrated loadthe left supportwas determined. This equals 2000  200
12, or 4400 lb. In passing from must to the left of the support to a point just to the right, however, the shear changes by the magnitude of the reaction. Hence, on the right-hand side of the left support the shear is 4400  14,000, or 9600 lb. At the next concentrated load, the shear is 9600  200
6, or 8400 lb. In passing the 4000-lb load, however, the shear changes to 8400  4000, or 4400 lb. Proceeding in this manner to the right end of the beam, we terminate with a shear of 3000 lb, equal to the load on the free end there. It should be noted that the shear diagram for a uniform load is a straight line sloping downward to the right (see Fig. 5.21). Therefore, the shear diagram was completed by connecting the plotted points with straight lines. FIGURE 5.20 Shear and moment diagrams for a simply supported beam with concentrated

Suppose a rigid body is acted upon by a system of forces with a resultant R. Given a virtual displacement ds at an angle  with R, the body will have virtual work done on it equal to R cos  ds. (No work is done by internal forces. They act in pairs of equal magnitude but opposite direction, and the virtual work done by one force of a pair is equal but opposite in sign to the work done by the other force.) If the body is in equilibrium under the action of the forces, then R  0 and the virtual work also is zero. Thus, the principle of virtual work may be stated: If a rigid body in equilibrium is given a virtual displacement, the sum of the virtual work of the forces acting on it must be zero. FIGURE 5.49 Principle of virtual work applied to determination of a simple-beam reaction (a) and (b) and to the reaction of a beam with a suspended span (c) and (d ). As an example of how the principle may be used to find a reaction of a statically determinate beam, consider the simple beam in Fig. 5.49a, for which the reaction R is to be determined. First, replace the support by an unknown force R. Next, move that end of the beam upward a small amount dy as in Fig. 5.49b. The displacement under the load P will be x dy/L, upward. Then, by the principle of virtual work, R dy  Px dy/L  0, from which R  Px/L. The principle may also be used to find the reaction R of the more complex beam in Fig. 5.49c. The first step again is to replace the support by an unknown force R. Next, apply a virtual downward displacement dy at hinge A (Fig. 5.49d ). Displacement under load P is x dy/c, and at the reaction R, a dy/ (a  b). According to the principle of virtual work, Ra dy/ (a  b)  Px dy/c  0, from which reaction R  Px(a  b) /ac. In this type of problem, the method has the advantage that only one reaction need be considered at a time and internal forces are not involved. 5.10.2 Strain Energy When an elastic body is deformed, the virtual work done by the internal forces is equal to the corresponding increment of the strain energy dU, in accordance with the principle of virtual work. Assume a constrained elastic body acted upon by forces P1, P2, . . . , for which the corresponding deformations are e1, e2 . . . . Then, Pn den  dU. The increment of the strain energy due to the increments of the deformations is given by dU  de1  de2     e e 1 2 In solving a specific problem, a virtual displacement that is not convenient in simplifying the solution should be chosen. Suppose, for example, a virtual displacement is selected that affects only the deformation en corresponding to the load Pn, other deformations being unchanged. Then, the principle of virtual work requires that P de  de n n n en This is equivalent to