High masonry dams Author:Edward Sherman Gould Purchase of this book includes free trial access to www.million-books.com where you can read more than a million books for free. This is an OCR edition with typos. Excerpt from book: CHAPTER III. Unequally Distributed Unit Stress. Previous calculations relating to fig. 4 show that the resultant weight of the trapezoidal wall, when there... more » is no water pressure against it, acts vertically downward on a line passing through its centre of gravity and cutting the base at a point distant 11.30 ft. from the inner toe, A. When the wall sustains a full head of water, the horizontal thrust, combined with the weight of the wall, produces an oblique resultant which cuts the base 14 feet from the outer toe, G. These two pressures, the one acting vertically downward when there is no water behind the wall, and the other obliquely downward and outward when there is a full head behind it, are essentially different in character. The first produces compression only; the second,, compression and shearing stress. This second pressure, namely, the resultant of a vertical downward force and a horizontal thrust, may be resolved at the point P back to its components, producing a downward vertical pressure exactly equal to the weight of the dam, and a horizontal one exactly equal to the thrust of the water. The first produces the .crushing stress, and the second the shearing stress already mentioned. The horizontal, or shearing stress, is rightly or wrongly wholly ignored, and the vertical component, or crushing stress, .only considered. The water pressure is therefore supposed to have no other .effect than to move the point of the application, or point of mean pressure of the weight of the dam, from its position nearer the point A to a position nearer the point G. Now it has already been recognized (fig. 7) that, when the point of application of the weight of the mass does not cut the middle of the base, the uniformity of the distribution of the stress is destroyed, be...« less