127. - Stresses In A Derrick

Fig 274 is intended to represent a boom derrick, with a load of 100 lbs. suspended from the end of the boom. Now, at the point x, three stresses are applied, viz., the load a, tensile stress in C, and a compressive stress in B. To find the stresses in B and C draw a vertical line. 1 - 2 equal to 100 lbs. (the load) at a scale of, say, 100 lbs. to the inch, and from the upper end of this line draw a line parallel to B and from the lower end a line parallel to C, the lines intersecting at 3. Then the line b gives the stress in B.and the line c in the stress C. To find the stress in the guy rope D, when it is in the same plane as the boom, draw a line from 3 parallel with the guy until it meets the load line at 5. Then the triangle of forces for the point y must be 3 - 2, 2 - 5, 5 - 3, and the arrow heads should point in the direction indicated by the notation. The distance from 2 to 5 must represent the compression in the mast. If we draw a horizontal line from 3 to the load line at 4, then the distance 2 - 4 represents the vertical component of the stress in C and the distance 4 - 5, the vertical component of the stress in the guy, and it is evident that the sum of these components must be the stress in the mast, or in other words, the stress in the mast, is the downward pull of C and D. The reaction at P must be the stress in the mast, plus the vertical component of b (1 - 4) or the amount of the load, plus 4 - 5 the vertical component of D. The distance 4 - 5 in this example measures 22 lbs., consequently the reaction P must be 122 lbs.

127 Stresses In A Derrick 300278

Fig. 274.

In § 117 it was stated that the sum of the parallel forces acting in one direction must equal the sum of those actions in the opposite direction, and the reader might at first think that the reaction at P should equal the load of 100 lbs., but there is a third vertical force not shown in the figure and that is the anchorage of the guy D. The guy must pull up on its anchorage to the same amount that it pulls down on the mast, which is the distance 4 - 5 or 22 lbs. To counteract this uplift at the end of the guy there must be a weight or force of 22 lbs. acting downwards, consequently the sum of the weights or forces acting downwards is 122 lbs., which must be resisted by the single reaction at P. . 128. - Points to be observed in connection with the foregoing examples. - If we stop to consider the process by which the stresses in the foregoing examples, Figs. 271-274, were obtained, we will notice, first, that the amount or magnitude of the stress in the different members of the frame depends only upon the load and the inclination of the piece, and is not affected by the length of the member. Second, that the stresses are proportional to the load. Thus, in any one of the foregoing examples, if the load is doubled, the magnitude of each line in the stress diagram will be doubled, for doubling the load will be the same as increasing the scale in the same ratio, i. e., a line one inch long represents 100 lbs. at a scale of 100 lbs. to the inch, or 200 lbs. at a scale of 200 lbs. to the inch.