# Sheet metal drafting/Chapter 3

CHAPTER III CYLINDERS CUT BY PLANES
Problem 9 THE SCOOP
25. The Scoop.—Figure 66 shows an ordinary flour or sugar scoop. Briefly described, any scoop is a cylinder cut off at an angle. A head is soldered in, and a handle is attached to the head. Figure 72 shows another type of scoop, the body being cut by a curved plane and a cylindrical handle being attached to the head.
The elevation should be drawn, using the dimensions given in Fig. 66. It is not necessary to show the handle in the elevation. After the profile. Fig. 67, has been drawn, it should be divided into twelve equal spaces. Extension lines from each division of the profile should be carried through the elevation, Fig. 68, until they meet the miter line.
Definition of a Miter Line.—The miter line is the line of junction between two shapes; these shapes may be alike or unlike. The miter line of the scoop is the line of junction between the body of the scoop and an imaginary cutting plane.
The line of stretchout is drawn at right angles to the elevation. The spacing of the profile must then be transferred to the line of stretchout and numbered to correspond. The measuring lines are now drawn in. The extension lines from the profile meet the miter line at seven points as shown. From each of the seven points of intersection on the miter line a dotted extension line is carried over into the stretchout. These extension lines must be drawn parallel to the line of stretchout. Starting from point 1 of the profile, follow the extension line until it meets the miter line, and from there follow the dotted line until it meets lines 1 and 1 of the stretchout. Small circles are placed where the dotted line crosses the measuring lines No. 1 of the stretchout in order to mark them definitely. In like manner every point of the profile can be located in its proper position in the stretchout. A curved line drawn through these points will give the miter cut of the pattern. A standard tin lock is added to each side as shown. Over-all dimensions should be placed on the pattern as shown. The pattern for the head can be obtained by reproducing the profile and adding a ⅛-inch burr. It is not necessary to allow for the dish of the head because it is so slight.
Up to this point the discussion applies to both types of scoop, ​Figs. 66 and 72. It should be noticed in drawing Fig. 68, that points 2 and 12, 3 and 11, 4 and 10, 5 and 9, and 6 and 8 fall on
the same extension lines. In view of this fact, many draftsmen save time by drawing a half-profile as shown in the elevation of Fig. 73.
​The pattern of the handle, Fig. 69, is a straight piece of metal ¾ in. wide and 3.1416 in. long. Hems are turned on the long side, and a ⅜-inch lap added for joining the ends when the piece is "formed up." The handle for the scoop shown in Fig. 73 is developed by the same method that was used for the body. This is not an exact pattern, due to the double curvature, but is near enough for practical purposes on small work. The cap for the handle is made by a 1⅛-inch hollow punch on a lead piece. By driving the punch and "punching" into the lead piece, a burr is formed on the cap.
Problem 10 TWO-PIECE ELBOW
26. The Two-piece Elbow.—A model of a two-piece elbow, Fig. 76, can be constructed from a cylindrical piece of wood such as a broom handle. The handle should be cut through at an angle and the two pieces put together so that they will form an angle similar to that shown in the elevation of Fig. 77. It should be noticed that the cut portions are not circles but that the section is longer in one direction than in the other. The two pieces fit together perfectly to form an elbow.
The following facts concerning all elbows are illustrated in Fig. 77. They should be memorized by the student.
The Base Line.—Every elbow is represented as starting from a horizontal line. This line is called the base line of the elbow.
Arcs of the Elbow.—Every elbow is made around the arcs of two circles. These arcs have the same center.
Center of the Elbow.—The center of the arcs around which the elbow is made is also the center of the elbow.
Throat of the Elbow.—That part of the elbow drawn around the smaller arc is the throat of the elbow, and the arc is the arc of the throat.
Back of the Elbow.—That part of the elbow made around the larger arc is the back of the elbow, and the arc is the arc of the back.
Throat Radius.—The throat radius is the distance measured along the base line, from the center of the elbow to the throat.
Center Line Radius.—The center line radius is the distance, measured along the base line, from the center of the elbow to the center line of the big end.
The Backset of an Elbow.—The backset is the amount the back rises (sets) above the throat. This vertical distance is indicated by the dash line drawn horizontally from the highest point of the throat of the big end.
Number of Backsets.—The first piece of an elbow has one back-set, the last piece has one, and every other piece in the elbow has two backsets.
Rule for the Number of Backsets.—The number of backsets is equal to the number of pieces in the elbow less one, multiplied by two. A four-piece elbow would have (4-1)×2=6 backsets.
​Big End of an Elbow.—The big end is the piece that starts at the base line. Its "cut" is equal to the diameter of the elbow×π, plus the necessary allowance for locks or laps. The big end cut of a 7-inch elbow would be (7×π)+1=22.991 in. (or 23 in.)
Small End of an Elbow.—The small end is the last piece of an elbow. Its "cut" is found by subtracting seven times the thickness of the metal used from the big end cut. Thus the small end ​cut of a 7-inch elbow made from No. 24 U. S. S. Gage would be 22.991-(.025×7)=22.991-.175=22.816 in. ($\scriptstyle {22{\frac {13}{16}}}$ in.)
Angle of an Elbow.—The angle of an elbow is a measure of the opening formed by two straight lines drawn from the center of the elbow to the extremities (ends) of the arc of the back.
Miter Lines of an Elbow.—The miter lines are the lines of junction between the pieces of the elbow. All miter lines must meet at the center of the elbow.
The student is required to make a drawing similar to Fig. 77 and to name all the parts defined above. The profile should be made about four inches in diameter but the size of the drawing is left to the student's discretion.
Problem 11 TWO-PIECE 60° ELBOW
27. The Two-piece 60° Elbow.—Figure 79 shows the elevation of a two-piece 60° elbow having a throat radius of 3 in. to fit over a pipe 4½ in. in diameter, and to be made of No. 24 galvanized steel. The elevation is started by drawing a base line 7½ in. long. The base line of an elbow is always equal in length to the sum of the diameter of the elbow and the throat radius. Using this base line as one side, an angle of 60° must be laid off. A distance equal to the throat radius (3 in.) is set off from the vertex (point) of the angle. The arcs of the throat and back are drawn, using the vertex of the angle as a center. The number of backsets in the elbow is equal to (No. of pieces-1)×2. For this elbow it will be (2-l)×2=2 backsets. The arc of the back is divided into as many equal spaces as there are backsets in the elbow; in this case, two equal parts. The miter line is drawn from the center of the elbow through the first division of the arc, above the base line. Perpendiculars (lines drawn at right angles) to the base line are erected from each end of the diameter of the big end. These perpendiculars must stop at the miter line. Straight lines drawn from these intersections to the extremities of the arcs complete the elevation of the small end.
Problem 12 FOUR-PIECE 90° ELBOW
28. The Four-piece 90° Elbow.—In laying out this elbow, Fig. 82, an angle of 90° should first be drawn. A distance equal to the sum of the throat radius and the diameter of the elbow should be laid off upon the horizontal side of this angle. The arcs of the throat and the back are then drawn in. A four-piece elbow has six backsets. Consequently, the arc of the back should be divided into six equal parts. Miter lines are next drawn through the first, third, and fifth divisions of the arc of the back, above the base line. This gives the first piece of the elbow one backset, the second piece two, the third piece two, and the fourth, or last piece, one.
Perpendiculars from the starting point of each arc are erected until they meet the first miter line. From these points straight lines are drawn so that they just touch the arc at one point and continue on until they meet the next miter line. In like manner straight lines representing the third piece of the elbow are drawn. The elevation is completed by straight lines drawn from the intersection of the third miter line to the ends of the arcs. The length of each miter line thus established can be tested with the dividers. If all are not of equal length, the drawing is incorrect. The elevation of an elbow is always drawn around the outside of the arcs. The straight lines of the throat and back are never drawn inside of the arcs. Many students make this mistake in the elevation and thereby produce an elbow wholly different from the one intended. The profile, Fig. 83, should be drawn and divided into sixteen equal parts. Extension lines are carried upwards from each division until they meet the miter line. The line of stretchout and the measuring lines of the stretchout. Fig. 84, are drawn. The spacings of the profile are transferred to the line of stretchout with numbers to correspond. Extension lines from each intersection of the miter line are carried over into the stretchout. Each division of the profile should be traced by means of the extension lines, first to the miter line, and thence to the correspondingly numbered line in the stretchout. Each point thus located in the stretchout should be marked with small circles. A curve drawn through these points will give the miter cut of the big end.
​It has already been shown that all miter cuts in the same elbow are exactly alike as to shape and size. Therefore, it is only necessary to reverse the pattern of the big end to get the miter cuts for the other pieces. Figure 84 shows all four pieces as they would appear when laid out on the metal in the shop. The man in the shop cuts a rectangular piece of iron the proper size, sets off the
dimensions of the backs and throats, and by reversing the pattern for the big end, gets the entire layout. The section in the lower right-hand corner. Fig. 85, shows the method used to join the pieces of the elbow together. An elbow made in this manner is known to the trade as a "peened elbow." The single and double edges are prepared in the turning machine (thick edge) and, after being slipped together, the double edge is clinched over the single edge with the peen of the hammer.
Problem 13 LONG RADIUS ELBOWS
29. The Long Radius Elbows.—The principles of pattern cutting that apply to long radius elbows such as used in conveyor systems are the same as in the preceding problems. There are, however, certain rules that apply to "Blow Pipe Elbows" that should be thoroughly understood. Figure 86 shows a partial elevation of a five-piece elbow. Since the pattern for all pieces can be laid out from the pattern of the first piece, it follows that all necessary information can be obtained from an elevation of the first two pieces of an elbow. A draftsman rarely draws more than two pieces of the elevation and divides the arc of the throat instead of the arc of the back. He uses the arc of the throat because it requires less room than the arc of the back and produces the same result.
Center Line Radius.—It has been determined by careful experiment that an elbow having a center line radius equal to twice the diameter of the pipe to which the elbow is to be joined, offers the least resistance to the flow of air, or other material, through the pipe. According to this rule an elbow for 12-inch pipe would have a throat radius of 18 in. and a center line radius of 24 in. A blow pipe elbow should never be "peened." All laps and edges should be closely riveted and soldered air-tight, the inside to be made as smooth as possible. All laps should be made in the direction of flow of air or other material through the pipe.
Laps for Riveting.—In the case of the "peened" elbow no allowance is made for joining the pieces. This alters the throat radius somewhat but this fact is generally neglected. In blow pipe systems the work must be exactly to measurements. Laps for riveting are, therefore, added as shown in Fig. 89. It should be noted that the rivet holes for the longitudinal seams are on the circumference lines of the pattern, while those for the transverse seams are in the center of the lap. The rivet holes are equally spaced and as the lap is ¾ in. wide, the centers of the holes are ⅜ in. in from the edge of the lap, and ⅜ in. in from the miter cut of the adjoining piece of the elbow. Laps are added to one miter cut only (of each piece) and start with the lap on the big end.
Thickness of Metal Used.—Another rule always to be observed is to make the elbow at least two gages heavier than the pipe to ​which the elbow is to be joined. The patterns shown in Fig. 88 should be separated sufficiently to allow for a lap between each piece and must be so drawn by the student. The laps should be ¾ in. wide. Rivet holes on all sides of each piece should be shown. Figure 90 shows the first and second pieces of an elbow after being "fitted." The throat is "laid off" with the stretching hammer, and the back is "drawn in" with a mallet or raising hammer. By
this method the miter cuts of each piece are "butted" and the true curvature of the elbow preserved.
30. Related Mathematics on Elbows.—Problem 13A.—Back­sets of an Elbow.—(See description of Fig. 77 for rule.)
(a) How many backsets has a four-piece, 90° elbow? (b) How many backsets has a six-piece, 75° elbow? (c) How many pieces has an elbow having fourteen back­sets? (d) How many miter lines has a six-piece elbow?
Problems on the Rise of the Miter Line.—An elbow is always ​measured by the degrees of the angle formed by straight lines drawn from its extremities to the center of the elbow, Fig. 77. Since all backsets in the same elbow are equal, the value of the backset can be expressed in degrees. A five-piece elbow has 8 backsets. A five-piece 90° elbow would have each backset equal to 90°÷8 or 11¼°. The first piece of any elbow contains one backset, the last piece one, and every other piece contains two. Therefore, the rise of the miter lines for a five-piece 90° elbow would be:
| Rise of | 1st | miter | line would be | 11¼°×1 | =11¼°, having but one backset. |
| " | 2nd | " | " | 11¼×(1+2) | =33¾°, by adding two backsets. |
| " | 3rd | " | " | 11¼×(3+2) | =56¼°, by adding two backsets. |
| " | 4th | " | " | 11¼×(5+2) | =78¾°, by adding two backsets. |
Problem 13B.—Give the rise of each miter line in a four-inch, four-piece, 75° elbow.
Problem 13C.—Give the rise of each miter line in a three-piece, 90° elbow.
Problem 13D.—Give the value in degrees of the backset of a two-piece, 12° elbow.
Problem 13E.—Give the value in degrees of the backset of a three-piece, 24° elbow.
Problem 13F.—Give the value in degrees of the backset of a four-piece, 36° elbow.
Problem 13G.—Give the value in degrees of the backset of a five-piece, 48° elbow.
Problem 13H.—Give the value in degrees of the backset of a six-piece, 60° elbow.
Problem 13I.—Give the value in degrees of the backset of a seven-piece, 72° elbow.
Problem 13J.—Give the value in degrees of the backset of an eight-piece, 84° elbow.
Problem 13K.—For the same big end diameter, why would one pattern answer for all of the elbows mentioned in problems 13D to 13J inclusive?
PROBLEMS ON THE "CUTS OF AN ELBOW"
Problem 13L.—What would be the "big end cuts" of the following sizes of elbows? Add the standard stovepipe lock of one inch.
(a) An elbow for 12" pipe? (b) An elbow for 14" pipe? (c) An elbow for 18" pipe? (d) An elbow for 24" pipe?
Problem 13M.—The "small end cut" of an elbow, or pipe, is always found by deducting seven times the thickness of the metal used from the cut of the big end. What would be the "small end cut" of the elbows in Problem 13L, if No. 20 U.S.S. Gage steel was used? Number 20 gage is .037".
Problem 13N.—Fill in the columns in the table of deductions given below. The figures for the third column are obtained by multiplying those of the second column by 7. The figures for the fourth column are obtained by dividing those of the third column by .0156. This will give answers in 64ths of an inch since .0156 is the decimal for $\scriptstyle {\frac {1}{64}}$.
Example of columns filled in:
| Gage | Decimal Thickness | Decimal Deduction | Fractional Deduction |
| No. 23 | .028125" | .196875" | $\scriptstyle {\frac {13}{64}}$" (nearly) |
Table of Deductions for Small End Cuts
| U.S.S. Gage | Decimal Equivalent, Thickness in Inches. |
| No. | |
| 28 | .015625 |
| 26 | .01875 |
| 24 | .025 |
| 22 | .03125 |
| 20 | .037 |
| 18 | .05 |
| 16 | .0625 |
| 14 | .078125 |
| 12 | .109375 |
STANDARD CUTS OF PIPE
Manufacturers of pipe and elbows have adopted the following standards for big end cuts for stove and conductor pipe.
Table of Standard Big End Cuts for Pipe and Elbows :
| Pipe Size | Stovepipe (1" lock) | Conductor Pipe (½" lock) | Conductor Pipe (½" lock) |
| | | Size | Cut |
| 4" | 14" | 2" | 6⅞" |
| 4½" | 15½" | 2½" | 8⅝" |
| 5" | 17" | 3" | 9⅞" |
| 6" | 20" | |
| 7" | 23" | |
| 8" | 26" | |
| 9" | 29¼" | |
| 10" | 32" | |
CENTER LINE RADIUS
As explained in Fig. 77 the center line radius is the distance measured along the base line from the center of the elbow to the center point of the diameter of the big end. The arc of the center line is also shown in Fig. 77.
Problem 13 O.—What will be the center line radius for the following elbows? (a) A 14" diameter elbow? (b) A 7" diameter elbow? (c) A 9" diameter elbow?
WEIGHT OF AN ELBOW
To get the weight of an elbow multiply the length of the center line arc by the "cut" of the big end, and this quantity by the weight per square foot of the material used.
Example.—What will be the weight of a 4", four-piece, 60° elbow made from No. 24 galvanized iron? Diameter of the elbow is 4"; Center Line Radius is 8".
Since the center line radius is 8" the center line arc must be a part of the circumference of a circle whose diameter is 16". The circumference of a 16" circle=16×π or 50.625". Since the elbow has an angle of 60° the center line arc can be but $\scriptstyle {\frac {60}{360}}$ or $\scriptstyle {\frac {1}{6}}$ of the whole circle. Therefore, the length of the center line arc would be $\scriptstyle {\frac {1}{6}}$ of 50.625" or 8.377". The big end cut for a 4" elbow ​is 14" and the surface area is 14×8.377=117 sq. in. (nearly). One square foot or 144 sq. in. of No. 24 galvanized iron weighs 1.156 lb. Therefore, 117 sq. in. would equal $\scriptstyle {\frac {117}{144}}$ of 1.156 or 1.136 lb. (Ans.)
The table given below shows the weight in pounds per square foot of the gages of metal in common use in the shop.
Table of Weights Per Square Foot of Galvanized and Black Sheets :
| Gage U. S. S. Standard | Galvanized Steel Wt. per sq. ft. in lb. | Black Steel Wt. per sq. ft. in lb. |
| 28 | 0.7812 | 0.6375 |
| 26 | 0.9062 | 0.765 |
| 24 | 1.156 | 1.02 |
| 22 | 1.406 | 1.275 |
| 20 | 1.656 | 1.53 |
| 18 | 1.156 | 2.04 |
| 16 | 2.656 | 2.55 |
| 14 | 3.281 | 3.187 |
| 12 | 4.531 | 4.462 |
Problem 13P.—How much would a 90°, 7" elbow weigh if made from No. 22 galvanized iron? Elbow to have standard radius but to be "peened."
Problem 13Q.—What would be the weight of a 90°, 8" elbow having a throat radius of 8" and made of No. 20 black iron?
Note.—When the throat radius is less than standard, add one-half of the diameter of the big end to get the center line radius.
Problem 13R.—How large a piece of iron would be required to make a 75°, 10" diameter, standard blow pipe elbow of eight pieces? Add ¾", for a lap on each piece, to the length of the center line radius.
Problem 14 THE BACKSET METHOD
31. The Backset Method.—The Backset Method is a short, but accurate, method of developing an elbow pattern. Figures 91 and 92 show the elevation and profile of an elbow, the pattern of which is developed in the manner previously described. It should be noticed that the pattern has been moved over to the right to allow a half circle to be drawn between it and the elevation. The extension lines cut this half circle at points A, B, C, D, E, F, G, H, and J. These views should be carefully drawn and placed in the position shown.
The distances between A and B, B and C, etc., should be exactly equal if the drawing is carefully made. The diameter of this half circle is equal to the height of the backset. Because of the foregoing, the elevation and profile are not necessary if the backset height in inches is known. The half circle divided into equal spaces can be used instead.
Figure 94 shows a five-piece, 90° long radius elbow laid out by this Backset Method. A piece of metal is cut of sufficient size to make the whole elbow. A horizontal line is drawn to represent the height in the throat of the first piece. Above this, another line is drawn to represent the height of the backset of the elbow. Lock lines are next drawn ½ in. in from the right and left-hand edges of the blank. The half circle is next drawn in the backset as shown. This half circle is divided into eight equal parts. The girth (distance between the lock lines) is next divided into sixteen equal parts. The girth in Fig. 94 is 28 in. and each space would equal 28÷16 or 1¾ in. The dividers can be set 1¾ in. and the spacing performed without repeated trials. Measuring lines are then drawn through each division at right angles to the bottom of the blank. A rectangular piece of metal with one edge turned to a right angle is a convenient tool for drawing these lines. Extension lines are carried over from each division of the half circle, and points of intersection determined. A curved line through these intersections will complete the pattern of the first piece. Experienced men can locate the intersections with four settings of the compass because point B is the same distance from the top as point H is from the bottom line. This ​is also true of points E and G, and D and F, while point E is in the center. The student should try and see if he can learn the
method of doing this, thereby saving time by doing away with the extension lines drawn from the half circle. The laps necessary for riveting the pieces together are shown in Fig. 94.