made in hollow blocks 14 inches thick, the weight varies from 65 to 85 pounds per cubic foot, the smaller pieces weighing the most. For pieces 12" x 18" or larger on the face, 70 pounds per cubic foot will probably be a fair average. For the exterior facing of fire-proof buildings, terra-cotta is now considered as the most suitable material available. CHAPTER VII. STABILITY OF PIERS AND BUTTRESSES. A PIER or buttress may be considered stable when the forces acting upon it do not cause it to rotate or "tip over," or any course of stones or brick to slide on its bed. When a pier has to sustain only a vertical load, it is evident that the pier must be stable, although it may not have sufficient strength. It is only when the pier receives a thrust such as that from a rafter or an arch, that its stability must be considered. In order to resist rotation, we must have the condition that the moment of the thrust of the pier about any point in the outside of the pier shall not exceed the moment of the weight of the pier about the same point. To illustrate, let us take the pier shown in Fig. 1. Let us suppose that this pier receives the foot of a rafter, which exerts a thrust 7' in the direction AB. The tendency of this thrust will be to cause the pier to rotate about the outer edge b; and the moment of the thrust about this point will be T× ab1, a,b, being the arm. Now, that the pier shall be just in equilibrium, the moment of the weight of the pier about the same edge must just equal Tab. The weight of the pier will, of course, act through the centre of gravity of the pier (which in this case is at the centre), and in a vertical direction; and its arm will be be, or one-half the thickness of the pier. Hence, to have equilibrium, we must have the equation, But under this condition the least additional thrust, or the crushing off of the outer edge, would cause the pier to rotate: hence, to have the pier in safe equilibrium, we must use some factor of safety. This is generally done by making the moment of the weight equal to that of the thrust when referred to a point in the bottom of the pier, a certain distance in from the outer edge. This distance for piers or buttresses should not be less than onefourth of the thickness of the pier. Representing this point in the figure by b, we have the necessary equation for the safe stability of the pier, Tx ab = W × it, t denoting the width of the pier. We cannot from this equation determine the dimensions of a pier to resist a given thrust; because we have the distance ab. . and W, all unknown quantities. Hence, we must first guess at the size of the pier, then find the length of the line ab, and see if the moment of the pier is equal to that of the thrust. If it is not, Graphic Method of determining the Stability of a Pier or Buttress. - When it is desired to determine if a given pier or buttress is capable of resisting a given thrust, the problem can easily be solved graphically in the following manner. Let ABCD (Fig. 2) represent a pier which sustains a given thrust Tat B. To determine whether the pier will safely sustain this thrust, we proceed as follows. Draw the indefinite line BX in the direction of the thrust. Through the centre of gravity of the pier (which in this case is at the centre of the pier) draw a vertical line until it intersects the line of the thrust at e. As a force may be considered to act anywhere in its line of direction, we may consider the thrust and th: weight to act at the point e; and the resultant of these two forces can be obtained by laying off the thrust T from e on eX, and the weight of the pier W, from on the line eY, both to the same scale (pounds to the inch), completing the parallelogram, and drawing the diagonal. If this diagonal prolonged cuts the base of the pier at less than one-fourth of the width of the base from the outer edge, the pier will be unstable, and its dimensions must be changed. The stability of a pier may be increased by adding to its weight (by placing some heavy material on top), or by increasing its width at the base, by means of "set-offs," as in Fig. 3. Figs. 3 (A and B) show the method of determining the stability of a buttress with offsets. The first step is to find the vertical line passing through the centre of gravity of the whole pier. This is best done by dividing the buttress up into quadrilaterals, as ABCD, DEFG, and GHIK (Fig. 3A), finding the centre of gravity of each quadrilateral by the method of diagonals, and then measuring the perpendicular distances X1, X2, X3, from the different centres of gravity to the line KI. Multiply the area of each quadrilateral by the distance of its centre of gravity from the line KI, and add together the areas and the products. Divide the sum of the latter by the sum of the former, and the result will be the distance of the centre of gravity of the whole buttress from KI. This distance we denote by Xo. EXAMPLE I.-Let the buttress shown in Fig. 3A have the dimensions given between the cross-marks. Then the area of the quadrilaterals and the distances from their centres of gravity to KI would be as follows: 1st area 35 sq. ft. X, = 0.95 1st area X = 33.25 Total area, 69 sq. ft. Total moments, 155.55 The sum of the moments is 155.55; and, dividing this by the total area, we have 2.25 as the distance X. Measuring this to the scale of the drawing from KI, we have a point through which the vertical line passing through the centre of gravity must pass. After this line is found, the method of determining the stability of the pier is the same as that given for the pier in Fig. 2. Fig. 3B also illustrates the method. If the buttress is more than one foot thick (at right angles to the plane of the paper), the cubic contents of the buttress must be obtained to find the weight. It is easier. however, to divide the real thrust by the thickness of the buttress, which gives the thrust per foot of buttress. Line of Resistance. — Definition. The line of resistance or of pressures, of a pier or buttress, is a line drawn through the centre of pressure of each joint. The centre of pressure of any joint is the point where the resultant of the forces acting on that portion of the pier above the joint cuts it. The line of pressures, or of resistance, when drawn in a pier, shows how near the greatest stress on any joint comes to the edges of that joint. It can be drawn by the following method. Let ABCD (Fig. 4) be a pier whose line of resistance we wish B to draw. First divide the pier in height, into portions two or three feet high, by drawing horizontal lines. It is more convenient to make the portions all of the same size. Prolong the line of the thrust, and draw a vertical line through the centre of gravity of the pier, intersecting the line of thrust at the point a. From a lay off to a scale the thrust T and the weights of the different portions of the pier, commencing with the weight of the upper portion. Thus, w, represents the weight of the portion above the first joint; we represents the weight of the second portion; and so on. The sum of the w's will equal the whole weight of the pier. 201 102 203 204 A Fig. 4. Having proceeded thus far, complete a parallelogram, with T and w for its two sides. Draw the diagonal, and prolong it. Where it cuts the first joint will be a point in the line of resistance. Draw another parallelogram, with 7 and ww2 for its two sides. Draw the diagonal intersecting the second joint at 2. Proceed in |