# Sheet metal drafting/Chapter 6

CHAPTER VI INTERSECTING RECTANGULAR PRISMS
Problem 24 THREE-PIECE RECTANGULAR ELBOW
45. The Three-piece Rectangular Elbow.—This problem deals with a three-piece, 90° elbow, Fig. 131, having a throat radius of 4½ in. Looking at the elevation of this elbow, Fig. 132, one would be unable to tell whether the fitting was round or rectangular piping. The profile shows that the elevation is of a rectangular pipe elbow.
The Elevation.—First, an angle of 90°, one side of which is to be used as the base line of the elbow, is drawn. From the vertex of the angle, distances of 4½ in. and 2½ in., as shown in Fig. 132, are set off. The arc of the throat and the arc of the back are drawn, using the vertex of the angle as the center of the elbow. The arc of the back is divided into four equal parts. Miter lines are drawn through the first and third divisions of the arc, above the base line. The elevation is completed by drawing straight lines tangent to the arcs. The detailed description of an elbow elevation is given in Problem 10, Chapter III.
The Profile.—A profile, Fig. 133, is drawn, using extension lines to locate the view properly. Each corner of the profile is numbered. It should be remembered that the seam always occurs at number 1 in the profile. In this case the seam comes at one corner of the elbow. Many prefer to have the seam at the center of one of the sides, or faces, of the elbow.
The Patterns.—Three lines of stretchout, one at right angles to each piece of the elbow, must be drawn. The spacing of the profile is transferred to each line of stretchout and is numbered to correspond. Measuring lines are drawn through each point in the lines of stretchout. Starting at point 1 of the profile an extension line should be traced upwards to the miter lines, and from there an extension line to lines 1 of the stretchouts should be drawn. Notice that two stretchouts are served by each intersection of the miter line. Extension lines from the elevation into any stretchout must always be drawn at right angles to the sides of the pipe. In like manner intersections of the stretchouts can be located and the patterns completed by drawing straight lines between these points as shown in Figs. 134, 135, and 136. Three-sixteenths inch single edges and ⅜-inch double edges, as shown, join the pieces of the elbow by double seaming.
​46. Related Mathematics on Elbows.—Solids of Revolution.—All elbows may be treated mathematically as solids of revolution.
Any surface moving about a fixed point will generate a solid of revolution. Suppose a piece of round rod is formed in the rolls to a true circular profile. A solid of revolution would be created ​because a ring slipped over the rod and caused to move around it to the right or to the left would always be the same distance from the center of the profile to which the rod was formed. A piece of bar iron formed to a circular profile would also be a solid of revolution. This solid could be regarded as being generated by a rectangle revolving about a center point. Figure 132 is drawn around two arcs whose center is the center of the elbow. It can also be seen that the more pieces "put in" the elbow, the nearer the straight lines come to the arcs about which they are drawn. If the elbow were made of a very great number of pieces, these would become so much like arcs that they could hardly be distinguished from them. This or any other elbow can be treated as a solid of revolution.
Solids of Revolution Have Three Diameters.—Every solid of revolution may be considered as having three diameters. The radius of the throat of Fig. 132 is 4½″. If four of these elbows were joined so as to make a complete ring, it would have a diameter of 4½″×2, or 9″. This would be the inside diameter of the solid of revolution. The radius of the back, Fig. 132, is 4½″+2½″, or 7″. The corresponding diameter for the whole ring would be 7″×2 or 14″. This would be the outside diameter of the solid of revolution. The third diameter is twice the center line radius of the elbow. In Fig. 132 the center line radius is 5¾″, and for the whole ring this would be 5¾″×2 = 11½″. This is called the neutral diameter of the solid of revolution, because when any rod or bar is formed into a circular profile the metal near this line stands still, that outside of the line stretches, and that inside of the line shrinks a like amount. This can be proved by drawing straight lines on a pencil eraser and bending the eraser, at the same time noting the distances between the lines.
Rule for Surface Area.—The surface area of a solid of revolution is equal to the circumference of its right section (profile) multiplied by the length of its neutral zone (diameter of the neutral×3.1416).
Sample Problem.— What is the surface area of the elbow shown in Fig. 132?
| Outside | diameter | of | ring | =14″ |
| Inside | " | " | " | =9″. |
| Neutral | " | | " | =11½″. |
| Length of neutral zone (11½″×π)=36.128″. ||||||
| Perimeter of right section (length of line of stretchout)=20″. ​ ||||||
| Surface area of entire ring 36.128″×20″=722.56 sq. in. ||||||
| Surface area of elbow (90° or ¼ of entire ring) 722.56 sq. in.÷4=180.64 sq. in. ||||||
| Ans. 180.64 sq. in. ||||||
Problem 24A.—A 90° elbow to fit a rectangular pipe 7″×12″ has a 10½″ throat radius. What is its surface area?
Problem 24B.—A 60° elbow ($\scriptstyle {\frac {1}{6}}$ of the solid of revolution) to fit a rectangular pipe 30″×61″ has a throat radius of 17″. What is its surface area?
Problem 25 RECTANGULAR PIPE OFFSET
47. The Rectangular Pipe Offset.—As was pointed out in the preceding problem, the elevation of a fitting for rectangular piping presents much the same appearance as an elevation for a round pipe fitting. For this reason the description given below will answer equally as well for an offset in round piping, the only difference being the shape of the profile.
The Profile.—A profile, as shown in Fig. 138, using the dimensions given, should be drawn. Each vertex (corner) of the profile should be numbered. Extension lines are carried upwards from points 2 and 3.
The Elevation.—A base line AH in Fig. 137 should be drawn equal in length to line 2–3 of the profile. A perpendicular 3 inches high should be erected at point A. The upper point of this fine should be lettered B. The line BC is drawn according to the dimensions given in Fig. 137. The 3-inch dimension gives the fitting its name, it being the amount that the third piece sets off to one side of the first piece of the elbow. The line CD is drawn at right angles to the base line. This completes the outline of one side of the elevation. The miter line BG must next be drawn. Every miter line of an elbow bisects the angle formed by the adjacent sides of the pipe. Therefore, in order to get the miter line BG the angle ABC must be bisected.
The procedure for bisecting an angle is as follows: With B as a center and any radius, set off equal distances on each side of the point B, on lines AB and BC. Letter these points P and R. With P and R as centers and any radius greater than RB, draw intersecting arcs (to the right of the figure). Letter this intersection J.
The straight line BJ will bisect the angle and can be used as the first miter line of the fitting. A perpendicular is erected at the point H until it cuts the miter line at the point G, Fig. 137. The next line to be drawn is FG.
In order to draw the line FG the angle HGB must be copied. This is done as follows: With G as a center and any radius, draw an arc cutting line HG at the point K and also cutting the miter line at the point M. With M as a center and a radius equal to ​MK, set off a distance equal to MK on the other side of the miter line. Letter this point N. The straight line GN will form the angle NGB, which will exactly equal the angle HGB. Make the line GF equal in length to line BC. Draw the line CF, which is the second miter line of the elevation.
The elevation is completed by drawing lines DE and EF.
The Pattern.—There are two methods of construction in general use in shop practice. One method calls for each piece to be developed separately, Fig. 139, the other calls for the body of the fitting to be made as shown in Fig. 140, the ends being "double seamed in." Figure 139 shows the patterns for the offset laid out in ​such a manner that no stock will be wasted. The measurements for the several pieces are taken from the elevation and are plainly marked. The pattern for the body piece, Fig. 140, is obtained by adding 3-inch double edges to the front elevation, and notching for a 1⅜-inch cleat as shown. The stretchout for the end piece is determined from the front elevation, and the width from the profile. Single edges ¼ in. wide are added to each side. The top is notched for a 1½-inch cleat.
48. Related Mathematics on Rectangular Pipe Offset.—Problem 25A.—How much would the stock cost for four rectangular pipe offsets, Fig. 137, made from No. 24 galvanized steel (1.156 lb. per square foot) if the steel cost $8.75 per 100 lb.?
Problem 25B.—How much more would the stock cost for the patterns shown in Fig. 140 than those shown in Fig, 139? Stock cut from body pattern corners would be regarded as waste.
Problem 26 DIAGONAL OFFSET
49. The Diagonal Offset.—This type of fitting is used in ventilating and heating ducts. It is also frequently encountered in running rectangular copper conductor pipes. The elevation shows a section of a lintel cornice such as is frequently seen above the first floor windows of a building. A conductor pipe running down an inside corner of a building having a lintel cornice would have to be offset diagonally in order to clear the obstruction.
The Plan.—Figure 142 shows the outline of a conductor pipe (lines AB, BC, and CD). The entire elevation is not shown because it plays no part in the development. The plan, Fig. 143, is drawn making the angle equal to 45°. If the required angle is other than 45°, it presents an entirely different problem and cannot be drawn by this method. The profiles are numbered as shown. Profile 1, 2, 3, and 4 is that of the lower part of the fitting and profile 5, 6, 7, 8, that of the upper part. While both are the same size their numbering must be different.
Diagonal Elevation.—A base line for the diagonal elevation must be drawn parallel to the line 4–8 of the plan. Extension lines are carried from each point in both profiles at right angles to this base line. Extension lines from points 1 and 3 will locate points A and E on the base line. A perpendicular is erected at point A and the distance AB set off equal to AB of the elevation, Fig. 142. The extension line from point 5, of Fig. 143, will locate points C and D in Fig. 144, heights being taken from Fig. 142. Drawing the line BC will complete the outline of that part of the fitting that rests directly upon the lintel cornice. A perpendicular is now erected at point E of Fig. 144. With B as a center and a radius equal to AE, an arc is drawn as shown in Fig. 144. Any other point F on the line CB is selected and another arc of the same radius drawn. The fine HG must be drawn tangent to both arcs. The point G occurs at the intersection of this tangent and the extension line from point 3 of Fig. 143. Line GH should be equal in length to BC. At the point D a line is, drawn at right angles to line CD. This line will intersect the extension line from point 7 of Fig. 143. The intersection should be lettered K. A straight line KH will complete the outline of the diagonal ​elevation. The miter lines BG and CH should be drawn in. The solid and dotted lines representing the other edges of the pipe are located by extension lines running from points 2, 4, 6, and 8 of
Fig. 143. The patterns are drawn in the manner described in Problem 25.
Locks are purposely left off of the patterns in order to avoid ​a confusion of lines. It is recommended to the student that he make paper or metal models of this problem in order to understand thoroughly the basic principles involved.
50. Related Mathematics on Diagonal Offsets.—Problem 26A.—Sixteen-ounce cold-rolled copper costs 25¼ cents per pound. How much would twenty-four diagonal offsets, Fig. 144, cost? In finding the area of this fitting multiply the distance (perimeter) around the profile by the combined length of lines AB, BC, and CD of Fig. 142. Sixteen-ounce copper weighs 16 ounces to the square foot. Allow 5 per cent for locks.
Problem 27 CURVED ELBOW IN RECTANGULAR PIPE
51. The Curved Elbow in Rectangular Pipe.—This type of fitting is extensively used in ducts for heating and ventilating systems because it offers a minimum of friction to the moving air. The elbow discussed here has an angle of 90°.
The Profile.—A plan or profile, Fig. 148, is drawn according to the dimensions given. The corners are lettered, and extension lines are carried upwards from points A and D to locate the elevation properly.
The Elevation.—After a base line is drawn, points 1 and 12 are located, and a distance of 3¾ in. (to scale) set off to the left of point 1 to serve as the center point of the elbow. The limits of the elbow are defined by erecting a perpendicular at this point. The arcs of the throat and the back are drawn. These arcs are divided into equal spaces and are numbered as shown.
The Pattern.—The pattern for the body, Fig. 150, is a copy of the elevation. To this is added a ¼-inch single edge on the throat and back. A 1½-inch edge for "shipping" the pipe is added as shown. Figure 151 shows the pattern of the throat piece which is a rectangle whose width is equal to line AB of plan, and whose length is equal to the stretchout of the spacing of the arc of the throat. Fig. 149. To this rectangle must be added 1½ in. for "shipping" and to each long side a $\scriptstyle {\frac {9}{16}}$-inch edge for a hammer lock. Figure 152 differs from Fig. 151 only with regard to its length, which is taken from the stretchout of the arc of the back.
The Hammer Lock.—The hammer lock is so called because it can be made up on the job, the only tool required being a hammer. Straight strips of metal are formed in the brake, to act as the sides of the fitting as shown in Fig. 153. The ¼-inch single edges of the body are worked up to a right angle and slipped into the slot of the hammer lock. The protruding edge is then closed down with a hammer as shown in Fig. 154. This gives the job an appearance of being double seamed, and requires much less time and effort than the double seamed job.
52. Related Mathematics on Curved Elbows.—Problem 27A.—How much would three curved elbows, Fig. 149, weigh, made from No. 28 galvanized iron (.7812 lb. per square foot), allowing 20 per cent for waste?