Applying the first law, if we have two columns of liquid iron connected at the bottom, they would just balance each other. For convenience we shall leave out of our calculations the upward pressure on the gates in the following examples, for in practical work they need seldom be taken into account.
Fig. 28. Illustrating Pressure of Liquids in Molds.
In A, Fig. 28, suppose these columns stand 6 inches above the joint bd, and that the column cd, has an area of 1 square inch. In B, suppose the area of the right-hand column cdef is 5 times the area of column cd. In both cases the level with top of runner a will be maintained. The depth of the cavity below the joint bdf makes no difference in maintaining these levels. The weight of one cubic inch of iron, .26 pound, is taken as the basis of all calculations.
Now if we close the column cd at d, as in C, it is clear that it would require the actual weight of that column to balance the lifting pressure of the surface d, or 6X.26Xl = 1.56 pounds. And if the larger area df is closed over, as in D, it takes 5 times this weight to resist the pressure exerted upon it by the runner, or 6X.26X5 = 7.8 pounds. If the pattern projected 2 inches into the cope, the height of the runner above the surface acting against the cope would be 4 inches, and the pressure to be overcome would be equal to the weight of cghe, or 4 X .26 X 5 = 5.2 pounds.
The important factors, then, are: height of runner; and area of mold which presses against the cope. We can therefore state a rule: To calculate the upward pressure of molten iron, multiply the depth in inches by the weight of one cubic inch of iron (.26), and multiply this product by the area in square inches upon which the pressure acts.
Applying the second law cited, the strains on sides and bottom of molds and upon cores is explained.
Fig. 29. Diagram Showing Analysis of Liquid Pressure.
By the rule just stated we first find the pressure per square inch at any given level by multiplying the depth by .26, and it is obvious that this pressure increases, the lower in the mold a point is taken.
In Fig. 29, the pressure at a equals hx.26. This also acts against the sides at ee. The pressure at b is h' X .26, and is exerted sidewise and downward. The pressure at c is h"x.26. This point, being half way between the levels a and b, represents the average sidewise or lateral pressure on all of the sides.
If this mold, then, is 11 inches square and 9 inches deep, with the pouring basin 6 inches above the joint, we have the following conditions:
Area of a = 121 sq. in.
Area of b = 121 sq. in.
Area of c (one side) = 99 sq. in.
Area of four sides =396 sq. in.
Height of h = 6 in.; pressure head = 1.56 lb. per sq. in. Height of V =15 in.; pressure head = 3.90 lb. per sq. in. Height of h" = 10 1/2 in.; pressure head = 2.73 lb. per sq. in.
Multiplying these together, we have the pressures on the various faces as follows:
Upward pressure on a = 188.76 lb.
Total pressure on side c = 270.27 lb.
Total pressure on four sides = 1081.08 lb. Total downward pressure on b = 471.90 lb.
A study of these figures shows the necessity of well-made flasks and bottom boards, for these must resist a greater pressure even than that required to keep the cope from lifting. They also show clearly why the lower parts of the casting resist the pressure of the gases more and require firmer ramming then the upper portions.
A difference in the way a pattern is molded may make a great difference in the weight required on the cope. Compare A and B, Fig. 30. Supposing this pattern is cylindrical in shape and with the dimensions as indicated, we would have the following basis:
Area of circle a =113.10 sq. in.
Ares of circle 6 = 78.54 sq. in.
Area of ring c'c' (b subtracted from a) = 34.56 sq. in. Then:
Total lift on cope A is 8X.26X 113.10 =235.24 lb.
The lift on cope B is 8X.26X34.56= 71.88 + (8+5)X.26x78.54=265.46 Total lift on B = 337.34 lb.
Fig. 30. Diagram Showing Difference in Pressure on Cope Due to Placing of Pattern.
Variation of Pressure Distribution. Fig. 31 is an example of a core 5 inches square surrounded by 1 inch of metal, with a runner 6 inches high. We have here:
Pressure per square inch on a is 7X.26 = 1.82 lb.
Pressure per square inch on b is 12x.26 = 3.121b.
The difference in these pressures is 1.30 pounds per square inch. Then for every foot of length in the core we must balance a lifting pressure on the bottom of the core of 5X12X3.12 = 187.2 pounds, until the metal covers surface a, when it will exert a counteracting downward pressure, and the strain on the chaplets will be only 5X12X1.30 = 78 pounds.
Fig. 31. Diagram Showing Difference in Pressure on Top and Bottom of a Cube.
Some of the ordinary defects which the beginner will find on his castings are as follows:
The amount of metal in the ladle is misjudged with the result that the mold is not completely filled.
These come from gases becoming pocketed in the metal instead of passing off through the sand. This is due to hard ramming, wet sand, etc.
These form when two streams of metal chill so much before they meet, that their surfaces will not fuse when forced against each other, as illustrated in Fig. 32.
These come from the washing of loose sand or excess of facing into the mold cavity when pouring. They are usually bedded in the cope side of the casting.
Scabs show like small warts or projections on the surface of the casting. They result from small patches of the mold face washing off. They may be caused from too much slicking, which draws the moisture to the surface of the mold, making the skin flake under the drying effect of the incoming metal.
Swells are bulged places on a casting and are due to soft ramming which leaves the walls of the mold too soft to withstand the pressure of the liquid metal.
These are due to unequal cooling in the casting. They are sometimes caused by the mold being so firm that it resists the natural shrinkage of the iron, causing the metal to pull apart when only partially cold.
This occurs when these strains cause the casting to bend or twist, but are not sufficient to actually crack the metal.