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Full text of "The steel square, instruction paper"






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COPVRICHT DEPOSm 



THE STEEL SQUARE 



INSTRUCTION PAPER 



PREPARKD BY 

Morris Wili^tams 

Writrr and Expert on Carpentry and 
Building 



AMEiRICAN SCHOOL OF COR RESPONDEJNCE 

v\ 
CHICAGO IIvLINOIS 

U.S.A. 






LIBRARY of CONtiKESS 
I wo Copies rieceivtx: 

JUN 13 iyo8 

OQHY B. / 



Copyright 1908 by 
American Schooi. of Correspondence 



Entered at Stationers' Hall, I^ondon 
All Rights Reserved 




I 



THE STEEL SQUARE 



INTRODUCTORY 

The Standard Steel Square has a blade 24 Inches long and 2 
inches wide, and a tongue from 14 to 18 inches long and 1 J inches wide. 

The blade is at right angles to the tongue. 

The face of the square is shown in Fig. 1. It is always stamped 
with the manufacturer's name and number. 

The reverse is the hack (see Fig. 2). 

The longer arm is the blade; the shorter arm, the tongue. 

In the center of the tongue, on the face side, will be found two 
parallel lines divided into spaces (see Fig. 1); this is the octagon scale. 

The spaces will be found numbered 10, 20, 30, 40, 50, 60, and 70, 
when the tongue is 18 inches long. 

To draw an octagon of 8 inches square, draw a square 8 inches 
each way, and draw a perpendicular and a horizontal line through 
its center. 

To find the length of the octagon side, place one point of a com- 
pass on any of the main divisions of the scale, and the other point of 
the compass on the eighth subdivision ; then step this length off on each 
side of the center lines on the side of the square, which will give the 
points from which to draw the octagon lines. 

The diameter of the octagon must equal in inches the number of 
spaces taken from the square. 

On the opposite side of the tongue, in the center, will be found 
the brace rule (see Fig. 3). The fractions denote the rise and ru7i of 
the brace, and the decimals the length. For example, a brace of 36 
inches run and 36 inches rise, will have a length of 50.91 inches; a 
brace of 42 inches run and 42 inches rise, will have a length of 59.40 
inches; etc. 

On the back of the blade (Fig. 4) will be found the board measure, 
where eight parallel lines running along the length of the blade are 
shown and divided at every inch by cross-lines. Under 12, on the 
outer edge of the blade, will be found the various lengths of the boards, 
as 8, 9, 10, 11, 12, etc. For example, take a board 14 feet long and 9 



THE STEEL SQUARE 



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THE STEEL SQUARE 




THE STEEL SQUARE 




Fig. 5. Use of Steel Square to Find Miter and Side of 
Pentagon. 



inches wide. To 
find the contents, 
look under 12, and 
find 14; then fol- 
low this space along 
to the cross-line un- 
der 9, the width of 
the board ; and here 
is found 10 feet 6 
inches, denoting 
the contents of a 
board 14 feet long 
and 9 inches wide. 
To Find the Mi= 
ter and Length of 
Side for any Poly= 
gon, with the Steel 
Square. In Fig. 5 
is shown a pentagon figure. The miters of the pentagon stand at 
72* degrees with each other, and are found by dividing 360 by 5, the 
number of sides in the pentagon. But the angle when applied to the 
square to obtain the miter, is only one-half of 72, or 36 
degrees, and intersects the blade at 8f |, as shown in Fig. 5. 
By squaring up from 6 on the tongue, intersecting 
the degree line at a, the 
center a is determined 
either for the inscribed 
or the circumscribed di- 
ameter, the radii being 
a h and a c, respec- 
tivelyo 

The length of the 
sides will be 8f | inches 
to the foot. 

If the length of the 
inscribed diameter be 8 
feet, then the sides would 
b- ^ ^ 8|| inches. 



Mill 




Fig. 6. Use of Steel Square to Find Miter and Side of 
Hexagon. 



THE STEEL SQUARE 




The figures to use for other polygons are as follows: 
Triangle 20|f 

Square 12 

Hexagon 7 

Nonagon 4f 

Decagon 3f 

In Fig. 6 the same process is used in finding the 
miter and side of the hexagon polygon. 

To find the degree line, 360 is divided by 6, the num- 
ber of sides, as follows: 
360 -^ 6 = 60; and 
60 -^ 2 = 30 degrees. 

Now, from 12 on 
tongue, draw a line 
making an angle of 30 
degrees with the tongue. 
It wdll cut the blade in 
7 as shown; and from 7 
to m, the heel of the 
square, will be the length 
of the side. From 6 on 
tongue, erect a line to 
cut the degree line in c; and with c as center, describe a circle having 
the radius of c 7; and around the circle, complete the hexagon by 
taking the length 7 m with the compass for each side, as shown. 

In Fig. 7 the same process is shown applied to the octagon. The 
degree line in all the polygons is found by dividing 360 by the number 
of sides in the figure: 

360 -^ 8 = 45; and 45 -^ 2 = 22 J degrees. 
This gives the degree line for the octagon. Complete the process as 
was described for the other polygons. 

By using the following figures for the various polygons, the miter 
lines may be found ; but in these figures no account is taken of the 
relative size of sides to the foot as in the figures preceding: 
Triangle 7 in. and 4 in. 
Pentagon 11 " " 8 " 
Hexagon 4 " " 7" 
Heptagon n\ " " 6" 



Fig. 7. 



Use of Steel Square to Find Miter and Side 
of Octagon. 



THE STEEL SQUARE 




Fig. 8. Use of Square to Find Miter of Equilateral Triangle. 



Octagon 17 in. and 7 in. 
Nonagon 22^ '' "9" 
Decagon 9i " " 3 " 
The miter is to be drawn along the Hne of the first cohimn, as shown 

for the triangle in 
Fig. 8, and for the 
hexagon in Fig. 9. 
In Fig. 10 is 
shown a diagram 
for finding degrees 
on the square. For 
example, if a pitch 
of 35 degrees is re- 
quired, use 8J|- on 
tongue and 12 on 
blade ; if 45 degrees, 
use 12 on tongue 
and 12 on blade; 
etc. 
In Fig. 11 is shown the relative length of run for a rafter and a 
hip, the rafter being 12 inches and the hip 17 inches. The reason, as 
shown in this diagram, why 17 is 
taken for the run of the hip, in- 
stead of 12 as for the common 
rafter, is that the seats of the com- 
mon rafter and hip do not run 
parallel with each other, but di- 
verge in roofs of equal pitch at an 
angle of 45 degrees; therefore, 17 
inches taken on the run of the hip 
is equal to only 12 inches when 
taken on that of the common 
rafter, as shown by the dotted 
line from heel to heel of the two 
squares in Fig. 11. 

In Fig. 12 is shown how 
other figures on the square may be 
found for corners that deviate from the 45 degrees. 




Fig. 9. 



Use of Square to Find Miter of 
Hexagon. 



It is shown that 



THE STEEL SQUARE 




Fig. 10. 



for a pentagon, which makes a 36-degree angle with the plate, the 

figure to be used 

on the square for 

run is 14| inches; 

for a hexagon, \r 

which makes a 

30-degree angle 

with the plate, 

the figure will be 

I3f inches; and 

for an octagon, 

which makes an 

angle of 22J de- 
grees with the 

plate, the figure 

to use on the 

square for run 

of hip to corre- 
spond to the run 

of the common 

rafters, will be 13 inches. It will be observed that the height in each 

case is 9 inches. 

Fig. 13 illustrates a 
method of finding the 
relative height of a hip 
or valley per foot run to 
that of the common raf- 
ter. The square is shown 
placed with 12 on blade 
and 9 on tongue for the 
common rafter; and 
shows that for the hip the 
rise is only 6j\ inches. 

The Steel Square as 
Applied in Roof Fram= 

Pig. 11. Square Applied to Determine Relative inff. Roof framing at 

Length of Run for Rafter and Hip. ** ^ ° ^ 

present is as simple as it 
possibly can be, so that any attempt at a new method would be super- 



Diagram for Finding Pitches of Various Degrees 
by Means of the Steel Square. 




THE STEEL SQUARE 



fluous. There may, however, be a certain way of presenting the sub- 
ject that will carry with it almost the weight assigned to a new theory, 
making what is already simple still more simple. 

The steel square is a mighty factor in roof framing, and without 
doubt the greatest tool in practical potency that ever was invented 





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Pig. 12. Use of Square to Determine Length of Run for Rafters on Corners 
Other than 4.5°. 

for the carpenter. With its use the lengths and bevels of every piece 
of timber that goes into the construction of the most intricate design 
of roof, can easily be obtained, and that with but very little knowledge 
of lines. 

In roofs of equal pitch, as illustrated in Fig. 14, the steel square 
is all that is required if one properly understands how to handle it. 






THE STEEL SQUARE 



What is meant by a pitch of a roof, is the number of inches it 
rises to the foot of run. 

In Fig. 15 is shown the steel square with figures representing 




Fig. 13. Method of Finding Relative Height of Hip or Valley per Foot of Run 
to that of Common Rafter. 

the various pitches to the foot of run. For the J-pitch roof, the figures 
as shown, from 12 on tongue to 12 on blade, are those to be used on 
the steel square for the common rafter; and for f pitch, the figures to 
be used on the square will be 12 and 9, as shown. 




Fig. 14, Diagram to Illustrate Use of Steel Square in Laying Out Timbers 
of Roofs of Equal Pitch. 

To understand this figure, it is necessary only to keep in mind 
that the pitch of a roof is reckoned from the span. Since the run in each 
pitch as shown is 12 inches, the span is two times 12 inches, which 



10 



THE STEEL SQUARE 



equals 24 inches; hence, 12 on blade to represent the foot run, and 12 
on tongue to represent the rise over J the span, will be the figures on 
the square for a ^-pitch roof. 

For the | pitch, the figures are shown to be 12 on tongue and 9 
on blade, 9 being J of the span, 24 inches. 

The same rule applies to all the pitches. The 4 pitch is shown 
to rise 4 inches to the foot of run, because 4 inches is J of the span, 24 
inches, the ^ pitch is shown to rise 8 inches to the foot of run, because 

8 inches is J of the span, 24 inches; etc. 

The roof referred to in Figs. 16 and 17 is to 
rise 9 inches to the foot of run; it is therefore a 
|-pitch roof. For all the common rafters, the fig- 
ures to be used on the square will be 12 on blade 
to represent the run, and 9 on tongue to represent 
the rise to the foot of run; and for all the hips 
and valleys, 17 on blade to represent the run, and 

9 on tongue to represent the rise of the roof to the 
foot of run. 

Why 17 represents the run for all the hips 
and valleys, will be understood by examining 
Fig. 19, in which 17 is shown to be the diag- 
onal of a foot square. 

In equal-pitch roofs the 
corners are square, and the 
plan of the hip or valley will 
always be a diagonal of a 
square corner as shown at 1, 2, 
3, and 5 in Fig. 14. 

In Fig. 18 
are shown ^ 
pitch, I pitch 
and J pitch over 
a square corner. 






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Fig. 15. Steel Square Giving Various Pitches to Foot of Run. 



The figures to be used on the square for the hip, will be 17 for 
run in each case. For the ^ pitch, the figures to be used would be 
17 inches run and 4 inches rise, to correspond with the 12 inches run 
and 4 inches rise of the common rafter. For the f pitch, the figures 
to be used for hip would be 17 inches run and 9 inches rise, to corre- 



I 



i 



THE STEEL SQUARE 



11 



spond with the 12 inches run and 9 inches rise of the common rafter; 
and for the ^1 pitch, the figures to be used on the square will be 17 
inches run and 12 inches rise, to correspond with the 12 inches run 
and 12 inches rise of the common rafter. 

It will be observed from above, that in all cases where the plan 
of the hip or valley is a diagonal of a square, the figures to be used on 



Top cut for i3ft.6in. 




Plurr>b Cut 



Top cut for i3ft/ \ \ 



\ 

Fig. 16. Method of Laying Out Common Rafters of a %-Pitcli Roof. 

the square for run will be 17 inches; and for the rise, whatever the roof 
rises to the foot of run. It should also be remembered that this is the 
condition in all roofs of equal pitch, where the angle of the hip or 
valley is a 45-degree angle, or, in other words, where w^e have the 
diagonal of a square. 

It has been shown in Fig. 12 how other figures for other plan 
angles may be found; and that in each case the figures for run vary 



Heel cut of hip 




T op cut for laft.ein. run of hip 




Top cut for 13 ft. run' of hip 

Fig. 17. Method of Laying Out Hips and Valleys of a -/s-Pitch Roof. 

according to the plan angle of the hip or valley, while the figure for the 
height in each case is similar. 

In Fig. 14 are show^n a variety of runs for common rafters, but 
all have the same pitch; they rise 9 inches to the foot of run. The main 



12 



THE STEEL SQUARE 



roof is shown to have a span of 27 feet, which makes the run of the 
common rafter 13 feet 6 inches. The run of the front wing is shown 
to be 10 feet 4 inches; and the run of the small gable at the left corner 
of the front, is shown to be 8 feet. 

The diversity exhibited in the runs, and especially the fractional 
part of a foot shown in two of them, will afford an opportunity to treat 
of the main difficulties in laying out roof timbers in roofs of equal 
pitch. Let it be determined to have a rise of 9 inches to the foot of 

run; and in this connec- 
tion it may be well to re- 
member that the propor- 
tional rise to the foot run 
for roofs of equal pitch 
makes not the least dif- 
ference in the method of 
treatment. 

To lay out the common 
rafters for the main roof, 
which has a run of 13 feet 
6 inches, proceed as shown 
in Fig. 16. 

Take 12 on the blade 
and 9 on the tongue, and 
step 13 times along the 
rafter timber. This will 
give the length of rafter 
for 13 feet of run. In 
this example, however, 
there is another 6 inches 
of run to cover. For this additional length, take 6 inches on the blade 
(it being J a foot run) for run, and take J of 9 on the tongue (which is 
4 J inches), and step one time. This, in addition to what has already 
been found by stepping 13 times with 12 and 9, will give the full length 

of the rafter. 

The square with 12 on blade and 9 on tongue will give the heel 

and plumb cuts. 

Another method of finding the length of rafter for the 6 inches 
is shown in Fig. 16, where the square is shown applied to the rafter 




Fig. 18. Method of Laying Out Hips and Rafters for 
Roofs of Various Pitches over Square Corner. 



THE STEEL SQUARE 13 



timber for the plumb cut. Square No. 1 is shown applied with 12 on 
blade and 9 on tongue for the length of the 13 feet. Square from this 
cut, measure 6 inches, the additional inches in the run; and to this 
point move the square, holding it on the side of the rafter timber 
with 12 on blade and 9 on tongue, as for a full foot run. 

It will be observed that this method is easily adapted to find any 
fractional part of a foot in the length of rafters. 

In the front gable, Fig. 14, the fractional part of a foot is 4 inches 
to be added to 10 feet of run; therefore, in that case, the line shown 
measured to 6 inches in Fig. 16 would measure only 4 inches for the 
front gable. 

Heel Cut of Common Rafter. In Fig. 16 is also shown a method 
to lay out the heel cut of a common rafter. The square is shown 
applied with 12 on blade and 9 on tongue; and from where the 12 on 
the square intersects the edge of the rafter timber, a line is drawn 
square to the blade as shown by the dotted line from 12 to a. Then 
the thickness of the part of the rafter that is to project beyond the 
plate to hold the cornice, is gauged to intersect the dotted line at a; 
and from a, the heel cut is drawn v/ith the square having 12 en blade 
and 9 on tongue, marking along the blade for the cut. 

The common rafter for the front wing, which is shown to have 
a run of 10 feet 4 inches, is laid out precisely the same, except that 
for this rafter the square with 12 on blade and 9 on tongue will have 
to be stepped along the rafter timber only 10 times for the 10 feet of 
run; and for the fractional part of a foot (4 inches) which is in the run, 
either of the two methods already shown for the main rafter may 
be used. 

The proportional figures to be used on the square for the 4 inches 
will be 4 on blade and 2^ on tongue ; and if the second method is used, 
make the addition to the length of rafter for 10 feet, by drawing a 
line 4 inches square from the tongue of square No. 1 (see Fig. 16), 
instead of 6 inches as there shown for the main rafter. 

Hips. Three of the hips are shown in Fig. 14 to extend from 
the plate to the ridge-pole; they are marked in the figure as 1, 2, and 
3 respectively, and are shown in plan to be diagonals of a square 
measuring 13 feet 6 inches by 13 feet 6 inches; they make an angle, 
therefore, of 45 degrees with the plate. 



14 



THE STEEL SQUARE 



In Fig. 18 it has been shown that a hip standing at an angle of 
45 degrees with the plate will have a run of 17 inches for every foot 
run of the common rafter. Therefore, to lay out the hips, the figures 
on the square will be 17 for run and 9 for rise; and by stepping 13 
times along the hip rafter timber, the length of hip for 13 feet of run 
is obtained. The length for the additional 6 inches in the run may 
be found by squaring a distance of 8 J inches, as shown in Fig. 17, 

from the tongue of the square, and 
moving square No; 1 along the edge 
of the timber, holding the blade on 
17 and tongue on 9, and marking 
the plumb cut where the dotted line 
is shown. 

In Fig. 18 is shown how to find the 
relative run length of a portion of a 
hip to correspond to that of a frac- 
tional part of a foot in the length 
of the common rafter. From 12 
inches, measure along the run of 
the common rafter 6 inches, and 
drop a line to cut the diagonal line 
in m. From m to a, along the diagonal line, will be the relative run 
length of the part of hip to correspond with 6 inches run of the common 
rafter, and it measures 8J inches. 

The same results may be obtained by the following method of 
figuring: 

As 12 : 17 : : 6 
6 




19. Diagram Showing Relative 
Lengths of Run for Hips and 
Common Rafters in Equal- 
Pitch Roofs. 



12)102 



8-6 




In Fig. 19 is shown a 12-inch square, 
the diagonal m being 17 inches. By 
drawing lines from the base a 6 to cut the 
diagonal line, the part of the hip to corre- 
spond to that of the common rafter will be 
indicated on the line 17. In this figure 

it is shown that a 6-inch run on ah, which represents the run of a 
foot of a common rafter, will have a corresponding length of 8i 



a ^ 



Fig. 20. Method of Determining 

Run of Valley for Additional 

Run in Common Rafter. 



THE STEEL SQUARE 



15 



inches run on the hne 17, which represents the plan Hne of the hip or 
valley in all equal-pitch roofs. 

In the front gable, Fig. 14, it is shown that the run of the common 
rafter is 10 feet 4 inches. To find the length of the common rafter, 




Fig. 21. Corner of Square Building, Show- 
ing Plan Lines of Plates and Hip. 



Fig. 23. Corner of Square Building, Show- 
ing Plan Lines of Plates and Valley. 



take 12 on blade and 9 on tongue, and step 10 times along the rafter 
timber; and for the fractional part of a foot (4 inches), proceed as was 
shown in Fig. 16 for the rafter of the main roof; but in this case measure 
out square to the tongue of square No. 1, 4 inches instead of 6 inches. 
The additional length for the fractional 4 inches run can also be 
found by taking 4 inches on blade and 3 inches on tongue of square, 
and stepping one time; this, in addition to the length obtained by 



Heel cut of Valley 




Fig. 23. Use of Square to Determine Heel Cut of Valley. 



stepping 10 times along the rafter timber with 12 on blade and 9 on 
tongue, will give the full length of the rafter for a run of 10 feet 4 inches. 
In the intersection of this roof with the main roof, there are shown 
to be two valleys of different lengths. The long one extends from the 
plate at n (Fig. 14) to the ridge of the main roof at m; it has therefore 



16 



THE STEEL SQUARE 





a run of 13 feet 6 inches. For the length, proceed as for the hips, by 

taking 17 on blade of the square and 9 on tongue, and stepping 13 

times for the length of the 13 feet; and for the fractional 6 inches, 

proceed precisely as shown in Fig. 17 for the hip, by squaring out from 

the tongue of square No. 1, 8 J inches; this, in addition to the length 

obtained for the 13 feet, will give the full length of the long valley n m. 

The length of the short valley a c, as shown, extends over the 

run of 10 feet 4 inches, and butts against the side of the long valley at e. 

By taking 17 on blade and 9 on tongue, and stepping along the rafter 

timber 10 times, the length for the 10 feet is found; and for the 4 

inches, measure 5f 

inches square from 

the tongue of 

square No. 1, in 

the manner shown 

-ID 1 * V* c.- in Fier. 17, where 

- z' Bevel to fit hips . 

Z^ against a deep the 8 J inches is 
^ roof or Tid^eboard shown added for 
the 6 inches addi- 
tional run of the 
main roof for the 
hips. 

The length 5f is 
found as shown in 
Fig. 20, by meas- 
uring 4 inches from 
a to m along the run 
of common rafter for one foot. Upon m erect a line to cut the seat of 
the valley at c; from c to a will be the run of the valley to correspond 
with 4 inches run of the common rafter, and it will measure 5f inches. 
How to Treat the Heel Cut of Hips and Valleys. Having found 
the lengths of the hips and valleys to correspond to the common rafters, 
it will be necessary to find also the thickness of each above the plate 
to correspond to the thickness the common rafter will be above the 
plate. 

In Fig. 21 is shown a corner of a square building, showing the 
plates and the plan lines of a hip. The length of the hip, as already 
found, will cover the span from the ridge to the corner 2; but the sides 



Fig. 24. 



Steel Square Applied to Finding Bevel for Fitting 
Top of mp or Valley to Ridge. 



THE STEEL SQUARE 



17 





of the hip intersect the plates at 3 and 3 respectively; therefore the 
distance from 2 to 1 , as shown in this diagram, is measured backwards 
from a to 1 in the manner shown in Fig. 17; then a plumb line is drawn 
through 1 to m, parallel to the plumb cut a-17. From m to o on this 
line, measure the same thickness as that of the common rafter; and 
through draw the heel cut to a as shown. 

In like manner the thickness of the valley above the plate is found ; 
but as the valley as shown in the plan figure. Fig. 22, projects beyond 
point 2 before it intersects the outside of the plates, the distance from 
2 to 1 in the case of the valley will have to be measured outwards from 
2, as shown from 
2tol in Fig. 23; 
and at the point 
thus found the 
thickness of the 
valley is to be 
measured to cor- 
respond with 
that of the com- 
mon rafter as 
shown at m n. 

In Fig. 24 is 
shown the steel 
square applied to 
a hip or valley 
timber to cut the 
bevel that will 

fit the top end against the ridge. The figures on the square are 17 
and 19 J. The 17 represents the length of the plan line of the hip 
or valley for a foot of run, which, as was shown in previous figures,, 
will always be 17 inches in roofs of equal pitch, where the plan lines 
stand at 45 degrees to the plates and square to each other. 

The 19i taken on the blade represents the actual length of a hip 
o-r valley that will span over a run of 17 inches. The bevel is marked 
along the blade. 

The cut across the back of the short valley to fit it against the 
side of the long valley, will be a square cut owing to the two plan lines 
being at right angles to each other. 



^jj /"Bevel tofitbacK 

^ -1/ of jacks against 

hip or valley 



Fig. 25. Steel Square Applied to Jack Rafter to Find Bevel for 
Fitting against Side of Hip or Valley. 



18 



THE STEEL SQUARE 



In Fig. 25 is shown the steel square applied to a jack rafter to 
cut the back bevel, to fit it against the side of a hip or valley. The 
figures on the square are 12 on tongue and 15 on blade, the 12 repre- 
senting a foot run of a common rafter, and the 15 the length of a 
rafter that will span over a foot run; marking along the 
blade will give the bevel. 

The rule in every case to find the back bevel for jacks in 
roofs of equal pitch, is to take 12 on the tongue to represent 
the foot run^ and the length of the rafter for a foot of run on 
the blade, marking along the blade in each case for the 
bevel. 

In a J-pitch roof, which is the 
most common in ah parts of the 




Run of Rafter 



Fig. 26. Finding Length to Shorten 

Rafters for Jacks per Foot 

of Run. 



country, the length of rafter for a \ 

foot of run will be 1 7 inches ; hence 

it will be well to remember that 12 

on tongue and 17 on blade, marking 

along the blade, will give the bevel to fit a jack against a hip or a 

valley in a J-pitch roof. 

In a roof having a rise of 9 inches to the foot of run, such as the 
one under consideration, the length of rafter for one foot of run will 
be 15 inches. The square as shown in Fig. 25, with 12 on tongue and 
15 on blade, will give the bevel by marking along the blade. 

To find the length of a rafter for a foot of run for any other pitch, 
place the two-foot rule diagonally from 12 on the blade of the square 
to the figure on tongue representing the rise of the roof to the foot of 

run; the rule will give the length of the 
rafter that will span over one foot of 
run. 

The length of rafter for a foot of 

run will also determine the difference 

in lengths of jacks. For example, if a 

roof rises 12 inches to one foot of run, 

the rafter over this span has been found 

to be 17 inches; this, therefore, is the 

number of inches each jack is shortened in one foot of run. If the 

rise of the roof is 8 inches to the foot of run, the length of the rafter is 

found for one foot of run, by placing the rule diagonally from 12 on 




Fig. 27. Finding Length of Jack 
Rafter in >^-Pitch Roof. 



THE STEEL SQUARE 



19 



tongue to 8 on blade, which gives 14J inches, as shown in Fig. 26. 
Tliis, therefore, will be the number of inches the jacks are to be 
shortened in a roof rising 8 inches to the foot of run. If the jacks are 
placed 24 inches from center to center, then multiply 14^ by 2 = 29 
inches. 

In Fig. 27 is shown how to find ^Q ^ 

the length with the steel square. The 
square is placed on the jack timber 
rafter with the figures that have been ^ 
used to cut the common rafter. In 
Fig. 27, 12 on blade and 12 on tongue 
were the figures used to cut the com- 
mon rafter, the roof being ^ pitch, 

rising 12 inches to the foot of run. In the diagram it is shown how 
to find the length of a jack rafter if placed 16 inches from center to 
center. The method is to move the square as shown along the line of 
the blade until the blade measures 16 inches; the tongue then would be 
as shown from w to m, and the length of the jack would be from 12 on 
blade to m on tongue, on the edge of the jack rafter timber as shown. 

This latter method becomes convenient w^hen the space between 
jacks is less than 18 inches; but if used when the space is more than 



Fig. 28. Finding Length of Jack 
Rafter in %-Pitch Roof. 




Ridq 



/Valley 



V 



k 



Plate' 



Fig= 29. Method of Determining Length of Jacks Between Hips and Valleys; 
also Bevels for Jacks, Hips, and Valleys. 



18 inches it will become necessary to use two squares; otherwise the 
tongue as shown at m would not reach the edge of the timber. 

In Fig. 28 the same method is shown for finding the length of a 
jack rafter for a roof rising 9 inches to the foot of run, with the jacks 
placed 18 inches center to center. The square in this diagram is 
shown placed on the jack rafter timber with 12 on blade and 9 on 



20 



THE STEEL SQUARE 



tongue; then it is moved forward along the Hne of the blade to w. 
The blade, when in this latter position, will measure 18 inches. The 
tongue will meet the edge of the timber at m, and the distance from 
m on tongue to 12 on blade will indicate the length of a jack, or, in 
other words, will show the length each jack is shortened when placed 




Miter Bevel for Boards 



Bevel to cut the Rn^rH Back Bdvel for Jacks *, 




Fig. 30. Method of Finding Bevels for All Timbers in Roofs of Equal Pitch. 

18 inches between centers in a roof having a pitch of 9 inches to the 
foot of run. 

When jacks are placed between hips and valleys as shown at 
1, 2, 3, 4, etc., in Fig. 14, a better method of treatment is shown in 
Fig. 29, where the slope of the roof is projected into the horizontal 
plane. The distance from the plate in this figure to the ridge m, equals 
the length of the common rafter for the main roof. On the plate ann 
is made equal to a ti n in Fig. 14. By drawing a figure like this to a 
scale of one inch to one foot, the length of all the jacks can be measured 



THE STEEL SQUARE 



21 



Side cut of hip 
aqainst t'he 
"*" ridqe board 




and also the lengths of the hip and the two valleys. It also gives the 
bevels for the jacks, as well as the bevel to fit the hip and valley against 
the ridge; but this last bevel must be applied to the hip and valley 
when backed. 

It has been shown before, thiit the figures to be used on the 
square for this bevel when the timber is left square on back as is the 
custom in construction, are the 
length of a foot run of a hip or val- 
ley, which is 17, on tongue, and the 
length of a hip or valley that will j 
span over 17 inches run, on blade — 
the blade giving the bevel. 

Fig. 30 contains all the bevels or 
cuts that have been treated upon so 
far, and, if correctly understood, will enable any one to frame any 
roof of equal pitch. In this figure it is shown that 12 inches run and 
9 inches rise will give bevels 1 and 2, which are the plumb and heel 
cuts of rafters of a roof rising 9 inches to the foot of run. By taking 
these figures, therefore, on the square, 9 inches on the tongue and 12 
inches on the blade, marking along the tongue will give the plumb cut, 
and marking along the blade will give the heel cut. 

Bevels 3 and 4 are the plumb and heel cuts for the hip, and are 
shown to have the length of the seat of hip for one foot run, which is 
17 inches. By taking 17 inches, therefore, on the blade, and 9 inches 
on the tongue, marking along the tongue for the plumb cut, and along 

Miter cut for roo< board 



Fig. 31, Method of Finding Bevel 5, Fig. 
30, for Fitting Hip or Valley Against 
Ridge when not Backed. 





Fig. 33. Method of Finding Baclc Bevel 6, 

Fig. 30, for Jack Rafters, and Bevel 

7, for Roof-Board. 



Fig. 33. 



Determining Miter Cut for Roof- 
Board. 



the blade for the heel cut, the plumb and heel cuts are found. Bevel 
5, which is to fit the hip or valley against the ridge when not backed, 
is shown from o w, the length of the hip for one foot of run, which is 
19i inches, and from o s, which always in roofs of equal pitch will 
be 17 inches and equal in length to the seat of a hip or valley for one 
foot of run. 



22 



THE STEEL SQUARE 



These figures, therefore, takei 



the 



19^ 



the blade, 




Laying Out. Timbers of One-half Gable of %-Pitch Roof. 



square, 

and 17 on the tongue, will give the bevel by marking along the blade 

as shown in Fig. 31, where the square is shown applied to the hip 

timber with 19 J on blade and 17 on tongue, 

the blade showing the cut. 

Bevels 6 and 7 in Fig. 30 are shown 

formed of the length of the rafter for one foot 

of run, which is 15 inches, and the run of the 

rafter, which is 12 inches. These figures are 

applied on the 

square, as shown 

in Fig. 32, to a 

jack rafter tim- 
ber; taking 15 on 

the blade and 12 

on the tongue, 

marking along 

the blade will 

give the back bevel for the jack rafters, and marking along the tongue 

will give the face cut of roof -boards to fit along the hip or valley. 

It is shown in Fig. 30, also, that by taking the length of rafter 

15 inches on blade, and rise of roof 9 inches on tongue, bevel 8 will 

give the miter cut for the roof-boards. 

In Fig. 33 the square is shown applied to a roof-board with 15 

on blade, which is the length of the rafter to one foot of run, and 

with 9 on tongue, which is the rise of the roof to the foot run; marking 

along the tongue will give the miter for the boards. 

Other uses may be made of these 
figures, as shown in Fig. 34, which 
is one-half of a gable of a roof ris- 
ing 9 inches to the foot run. The 
squares at the bottom and the top 
will give the plumb and heel cuts of 
the common rafter. The same 

figures on the square applied to the studding, marking along the 

tongue for the cut, will give the bevel to fit the studding against the 

rafter; and by marking along the blade we obtain the cut for the 

boards that run across the gable. By taking 19 J on blade, which is 




Fig. 35. 



Finding Backing of Hip in 
Gable Roof. 



THE STEEL SQUARE 



23 



the lengith of the hip for one foot of run, and taking on the tongue the 
rise of tlio roof to the foot of run, which is 9 inches, and applying 
these as shown in Fig. 35, we obtain the backing of the hip by 
marking along the tongue of the two squares, as shown. 

It will be observed from what has been said, that in roofs of 
equal pitch the figure 12 on the blade, and whatever number of inches 
the roof rises to the foot run on the t')ngue, will give the plumb and 
heel cuts for the common rafter; and that by taking ;.7 on the blade 
instead of 12, and taking on the tongue the figure representing the 
rise of the roof to the foot run, the plumb and heel cuts are found for 
the hips and valleys. 

By taking the length of the common rafter for one foot of run 
on blade, and the run 12 on tongue, marking along the blade will give 




6 e 

Fig. 36. Laying Out Timbers of Roof with Two Unequal Pitches. 



the back bevel for the jack to fit the hip or valley, and marking along 
the tongue will give the bevel to cut the roof-boards to fit the line of 
hip or valley upon the roof. 

With this knowledge of what figures to use, and why they are 
used, it w^ill be an easy matter for anyone to lay out all rafters for 
equal-pitch roofs. 

In Fig. 36 is shown a plan of a roof with two unequal pitches. 
The main roof is shown to have a rise of 12 inches to the foot run. The 
front wing is shown to have a run of 6 feet and to rise 12 feet; it has 
thus a pitch of 24 inches to the foot run. Therefore 12 on blade of the 
square and 12 on tongue will give the plumb and heel cuts for the 
main roof, and by stepping 12 times along the rafter timber the length 
of the rafter is found. The figures on the square to find the heel and 



24 



THE STEEL SQUARE 



plumb cuts for the rafter in the front wing, will be 12 run and 24 rise, 
and by stepping 6 times (the number of feet in the run of the rafter), 
the length will be found over the run of 6 feet, and it will measure 13 
feet 6 inches. 

If, in place of stepping along the timber, the diagonal of 12 and 
24 is multiplied by 6, the number of feet in the run, 
the length may be found even to a greater exactitude. 

Many carpenters use this method of framing; and 
to those who have confidence in their ability to figure 
correctly, it is a saving of time, and, as before said, 
will result in a more accurate measurement; but the 
better and more scientific method of framing is to work 
to a scale of one inch, as has already been explained. 

According to that method, the 
diagonal of a foot of run, and the 
number of inches to the foot run the 
roof is rising, measured to a scale, 
will give the exact length. For 
example, the main roof in Fig. 36 is 
rising 12 inches to a foot of run. The diagonal of 12 and 12 is 17 




i I I I I I I 



Fig. 37. Finding Length of Rafter for 

Front Wing in Roof Shown in 

Fig. 36. 



inches, which, considered as a scale of one inch to a foot, will give 




Fig. 38. Laying Out Timbers of Roof Shown in Fig. 36, by Projecting Slope of 
Roof into Horizontal Plane. 

17 feet, and this will be the exact length of the rafter for a roof rising 
12 inches to the foot run and having a run of 12 feet. 

The length of the rafter for the front wing, which has a run of 6 
feet and a rise of 12 feet, may be obtained by placing the rule as shown 



THE STEEL SQUARE 



25 



Elevation 



in Fig. 37, from 6 on blade to 12 on tongue, which will give a length of 
13i inches. If the scale be considered as one inch to a foot, this will 
equal 13 feet 6 inches, which will be the exact length of a common 
rafter rising 24 inches to the foot run and having a run of 6 feet. 

It will be observed that the plan lines of the valleys in this figure 
in respect to one another deviate from forming a right angle. In 
equal-pitch roofs the plan lines are always at right angles to each other, 
and therefore the diagonal of 12 and 12, which is 17 inches, will be 
the relative foot run of valleys and hips in equal-pitch roofs. 

In Fig. 36 is shown how to find the figures to use on the square 
for valleys and hips when deviating 
from the right angle. A line is 
drawn at a distance of 12 inches 
from the plate and parallel to it, 
cutting the valley in m as shown. 
The part of the valley from m to 
the plate will measure 13J inches, 
which is the figure that is to be 
used on the square to obtain the 
length and cuts of the valleys. 

It will be observed that this 
equals the length of the common 
rafter as found by the square and 
rule in Fig. 37. In that figure is 
shown 12 on tongue and 6 on blade. 
The 12 here represents the rise, and 
the 6 the run of the front roof. If 
the 12 be taken to represent the 
run of the main roof , and the 6 to 

represent the run of the front roof, then, the diagonal 13 J will indi- 
cate the length of the seat of the valley for 12 feet of run, and there- 
fore for one foot it will be 13 J inches. Now, by taking 13^ on the 
blade for run, and 12 inches on the tongue for rise, and stepping 
along the valley rafter timber 12 times, the length of the valley 
will be found. The blade will give the heel cut, and the tongue the 
plumb cut. 

In Fig. 38 is shov/n the slope of the roof projected into the hori- 
zontal plane. By drawing a figure based on a scale of one inch to one 




Fig. 39. Method of Finding Length and 
Cuts of Octagon Hips Intersect- 
ing a Roof. 



26 



THE STEEL SQUARE 



foot, all the timbers on the slope of the roof can be measured. Bevel 
2, shown in this figure, is to fit the valleys against the ridge. By 
drawing a line from w square to the seat of the valley to m, making 



^ Ridge in vSecon d POvSition 



r~^^^ 




c Cornice 



Fig. 40, Showing How Cornice Aflects Valleys and Plates in Roof with Unequal Pitches. 

10 2 equal in length to the length of the valley, as shown, and by con- 
necting 2 and m, the bevel at 2 is found, which will fit the valleys 
against the ridge, as shown at 3 and 3 in Fig. 36. 

In Fig. 39, is shown how to find the length and cuts of octagon 
hips intersecting a roof. In Fig. 36, half the plan of the octagon is 
shown to be inside of the plate, and the hips o, 2, o intersect the slope 
of the roof. In Fig. 39, the lines below x y are the plan lines; and those 

above, the elevation. From 2, o. 
o, in the plan, draw lines to x y, 
as shown from o to m and from z 
to m; from m and m, draw the ele- 
vation lines to the apex o", inter- 
secting the line of the roof in d" 
and c''. From c^'' and c", draw 
the lines d" v" and c" a" parallel 
io X y\ from c", drop a line to in- 
tersect the plan line ao in c. 
Make a w equal in length to a" o" 
of the elevation, and connect \o c; 

Fig. 41. Showing Relative Position of i p n i • i 

Plates in Roof with Two un- measure irom w to 71 the lull neignt 

equal Pitches. " 

of the octagon as shown from xy 
to the apex o"; and connect c n. The length from lo to c is that of 




THE STEEL SQUARE 



27 



the two hips shown at o o in Fig. 36, both being equal hips intersect- 
ing the roof at an equal distance from the plate. The bevel at'w;is the 
top bevel, and the bevel at c will fit the roof. 

Again, drop a line from d'^ to intersect the plan line azind. 
Make a 2 equal to v'^ o" in the elevation, and connect 2 d. Measure 
from 2 to 6 the full height of the tower as shown from xyio the apex 
o" in the elevation, and connect d b. 



Seat ofVa-lley 




Fig. 43. 



The length 2 d represents the 
length of the hip z shown in 
Fig. 36; the bevel at 2 is that of 
the top ; and the bevel at d, the 
one that will fit the foot of the 
hip to the intersecting roof. 

When a cornice of any con- 
siderable width runs around a 
roof of this kind, it affects the 
plates and the angle of the val- 
leys as shown in Fig. 40. In 
this figure are shown the same 
valleys as in Fig. 36 ; but, owing 
to the width of the cornice, the 
foot of each has been moved the 
distance a b along the plate of the 
main roof. Why this is done is 

shown in the drawing to be caused by the necessity for the valleys 
to intersect the corners c c of the cornice. 

The plates are also affected as shown in Fig. 41, where the plate 
of the narrow roof is shown to be much higher than the plate of the 
main roof. 

' The bevels shown at 3, Fig. 40, are to fit the valleys against the 
ridge. 

In Fig. 42 is shown a very simple method of finding the bevels for 
purlins in equal-pitch roofs. Draw the plan of the corner as shown, 
and a line from m to o; measure from o the length x y, representing 
the common rafter, to w; from w draw a line to m; the bevel shown 
at 2 will fit the top face of the purlin. Again, from o, describe an 
arc to cut the seat of the valley, and continue same around to S; con- 
nect S m; the bevel at 3 will be the side bevel. 



Method of Finding Bevels for Pur- 
lins in Equal-Pitch Roofs. 



EXAMINATION PAPER 



THE STEEL SQUARE 



Read Carefully: Place your name aud full address at the head of the 
paper. Anj^ cheap, light paper like the sample previously seutyou may be 
used. Do uot crowd your work, but arrange it neatly and legibly. Do not 
copy the ansivers from the Instruction Paper; use your oivn words, so that 
we may be sure that you understand the subject. 

1. On what part of the square will you find the octagon scalef 
Describe its use. 

2. On what part of the square will you find the brace rule? 
Describe its use. 

3. On what part of the square would you look for the hoard 
measure? Describe its use. 

4. What is meant by the pitch of a roof? 

5. What is meant by a bevel in roof framing? 

6. Draw a diagram of a roof, indicating thereon the ridge, 
common rafters, jack rafters, hips, valleys, plate. 

7. How would you lay out a pentagon by means of the square? 
A hexagon? Illustrate with diagrams. 

8. How, with a square, would you find the miter of an equi- 
lateral triangle? Of a hexagon? Illustrate with diagrams. 

9. What are meant by plumb cut and heel cut? Draw a 
diagram to illustrate. 

10. Show how to lay out the heel cut of a common rafter. 

11. Draw diagrams illustrating use of the square in finding the 
relative length of run for rafters and hips. Explain. 

12. Show how to apply the square to a hip or valley timber to 
cut the bevel that will fit the top end against the ridge. 

13. Describe the application of the square in finding the 
relative height of a hip or valley per foot of run, to that of the common 
rafter. Draw a diagram. 

14. Describe a method of finding the bevels for purlins in 
equal-pitch roofs. 

15. Describe, with diagram, the use of the square in cutting the 
back bevel to fit a jack rafter against the side of a hip or valley. 



THE STEEL SQUARE 



16. How would you use the square to find the length to cut 
jack rafters in roofs of J-pitch? |-pitch? 

17. Show how to find the length and cuts of octagon hips 
intersecting a roof. 

After completing the work, add and sign the following statement : 

I hereby certify that the above work is entirely my own. 

(Signed) 



JUN 13 1908 



^/ 



LIBRARY OF CONGRESS 



028 145 854 1