A head-frame must be strong enough to bear the strain brought on it due to the total load hoisted and the pull of the engine in hoisting this load; it must also be rigid in construction to withstand the severe vibration and shock to which it is subjected on account of the rapid hoisting and the jar due to the landing of the cages.
Fig. 28
The amount and direction of stresses that a head-frame must resist are usually determined by applying the parallelogram of forces as follows: [Fig. 28] is a simple head-frame at a slope; a is the drum of the hoisting engine with the rope coming from its upper side and running over the head-sheave b down to the slope cage c. Assuming that the angles e, f made by the two portions of the rope with the horizontal are equal, and that the pull on each part of the rope is 20,000 pounds, to determine the amount and direction of the resultant of the two rope pulls, proceed as follows: Extend the rope lines to the point of intersection g and from there lay off the two lines g h and g k, to some definite scale, representing the pull of the rope. If a scale of 2,000 pounds to ⅒ inch is taken (⅒ inch = 2,000 pounds), g h and g k will each be 1 inch long. Complete the parallelogram by drawing h l parallel to g k and k l parallel to g h. The diagonal g l represents the direction and amount of the force acting on the head-frame due to the pull of the two portions of the rope. The diagonal, by measurement, is 1½ inches or ¹⁵/₁₀ inches long, and since each tenth inch equals 2,000 pounds, the stress on the head-frame in the line of the diagonal g l is 2,000 × 15 = 30,000 pounds. The figure also shows that the direction of this force is vertical, hence there is no tendency for the frame to be pulled over to either side and, theoretically, side bracing is not needed.
Fig. 29
30. Consider now the case of a vertical shaft, [Fig. 29], in which, as before, a is the drum, b the head-sheave, c the cage, and d the head-frame, and assume the same pull of 20,000 pounds on each part of the rope. As before, extend the lines of the rope, which are the lines of force along which the pulls due to the engine and the load act, until they intersect at g. From this point lay off on these lines distances representing the stresses in the rope to any scale. Using the same scale as before, ⅒ inch = 2,000 pounds, the lines g h and g k representing the two forces will be each 1 inch long. Completing the parallelogram by drawing h l parallel to g k, and k l parallel to g h, and drawing the diagonal g l through g, the resultant, g l = ¹⁹/₁₀ inches, represents a stress of 38,000 pounds. The direction of the resultant is also determined, being in the line of the diagonal g l. If the head-frame shown in [Fig. 28] were used for this case, it would be overturned by this resultant force, unless the leg on the opposite side of the shaft from the engine were securely anchored, so an inclined brace m is added to resist this overturning action. The resultant of all forces acting on the head-frame should generally fall within the structure if the greatest stability is to be secured, but when this cannot be done it is necessary to resist the overturning pull by anchoring the head-frame to its foundations much more securely than is the case where the resultant falls within the structure.
The direction of the resultant force may be obtained by drawing a line through the intersection of the lines of action of the forces at g and the center of the head-sheave b, as may be seen in Figs. [28] and [29].
Fig. 30