_THE STRENGTH OF A SCAFFOLD_
The strength of a scaffold equals the resistance that its members
and their connections can offer to the strains that act upon it. The
timbers used may fail in various ways.
First.—As beams, they may fail to resist a cross strain.
Secondly.—As pillars and struts, they may fail to resist compression in
the direction of their length.
Thirdly.—As ties and braces, they may fail to resist tension; that is,
a strain tending to pull a beam asunder by stretching.
The loads that act upon a scaffold may be dead, that is to say, they do
not create a shock; or live, which means they are not stationary, and
may cause shock and vibration.
A live load causes nearly double the strain that a dead load produces.
If, therefore, both are of equal weight, the timber under strain
will carry as a live load only one half of what it would carry as a
dead load. The breaking weight is the load that will cause fracture
in the material. The safe load is the greatest weight that should be
allowed in practice. It is in proportion to the breaking load, and that
proportion is termed the factor of safety.
Experiments have been made to determine the resistance of timber to
fracture under the various forces that act upon it.
The result of these experiments is expressed by a given number, termed
a constant, which varies with the different growths of timber and the
different strains to which they are subjected.
The constants (C) for the strength of different kinds of timber under a
cross strain are as follows:
TABLE I.
+-----------------+-------------------+-----------+
| | C in lbs. for | |
| Material | rectangular beams | Authority |
| | supported at both | |
| | ends | |
+-----------------+-------------------+-----------+
| Spruce | 403 |} |
| Larch | 392 |} |
| Fir, Northern } | |} |
| Fir, Dantzic } | 448 |} |
| Fir, Memel } | |} |
| Fir, Riga | 392 |} Hurst |
| Elm | 336 |} |
| Birch | 448 |} |
| Ash, English | 672 |} |
| Oak, English | 560 |} |
| Oak, Baltic | 481 |} |
+-----------------+-------------------+-----------+
The factor of safety for beams under a cross strain is one-fifth.
TABLE II.—_Constants_ (D) _for compression_.
+-----------------+-----------------+-------------+-----------------+
| |D[3] in lbs. for | | D[4] in lbs. for|
| Material | crushing only | Material | flexure for long|
| | | | pillars |
+-----------------+-----------------+-------------+-----------------+
| | Wet Dry | | |
|Spruce |6,499 to 6,819 | Spruce | -- |
|Larch (fallen two| | Larch | 1,645 |
| months) |3,201 to 5,568 | Fir, Riga | 2,035 |
|Fir (white deal) |6,781 to 7,293 | Fir, Memel | 2,361 |
|Elm | 10,331 | Elm | 1,620 |
|Birch, English |3,297 to 6,402 | Birch | -- |
|Birch, American | 11,663 | | |
|Ash |8,683 to 9,363 | Ash | 1,840 |
|Oak, English |6,484 to 10,058 | Oak, English| 2,068 |
|Oak, Dantzic | | Oak, Dantzic| 2,410 |
| (very dry) | 7,731 | | |
| | | | |
|Pillars of medium length—the constant for bending is taken at 2·9 |
| for all timbers. |
+-----------------+-----------------+-------------+-----------------+
The working load on pillars should not be greater than one-tenth of the
breaking weight; but if the pillars are used for temporary purposes,
and are over 15 and under 30 diameters one-eighth, and under 15
diameters one-fifth may be taken as the factor of safety.
TABLE III.—_The Constants_ (E) _for Breaking Weight
under tensional stress_.
+----------+----------------------------------+-----------+
| Material | E in lbs. per unit of 1 sq. inch | Authority |
+----------+----------------------------------+-----------+
| Spruce | 3,360 | } |
| Larch | 3,360 | } |
| Fir | 3,360 | } Hurst |
| Elm | 4,480 | } |
| Ash | 4,480 | } |
| Oak | 6,720 | } |
+----------+----------------------------------+-----------+
The working load should not exceed one-fifth of the weight that would
cause rupture.
=Beams subject to a Transverse Strain.=—The loads that act upon a
beam may be concentrated, that is, acting at one point, or distributed,
which means that the load is evenly placed over the entire length of
the beam or a portion of the beam.
An evenly distributed load is considered to act at a point immediately
below its centre of gravity.
If a beam carries several loads they are considered to act at a point
immediately below the resultant centre of gravity of the whole.
Beams may be fixed, loose, or continuous. When fixed they are built
into the structure; if loose they are supported only; and are
continuous when they have more than two points of support in their
length.
For practical purposes, a continuous beam may be considered one-half
stronger than when supported at two points only.
The strength of a solid rectangular beam varies directly as its
breadth, the square of its depth, and inversely as its length.
The formula for finding the breaking weight (B.W.) of a solid
rectangular beam supported at each end and loaded at its centre is as
follows:
B.W. = (_bd_^2)/_l_ × _c_
where _b_ = breadth in inches;
_d_^2 = depth multiplied by itself in inches;
_l_ = length of beam between supports in feet;
_c_ = the constant found by experiment on the timber in use.
_Example._—Find the B.W. of a solid rectangular beam of Northern
fir, 9 in. by 9 in., and 10 feet between supports, loaded at its
centre.
The constant for Northern fir (Table I.) is 448.
B.W. = (9 × 9^2)/10 × 448
= (9 × 9 x 9 x 448)/10
= 14 tons 11cwt. 2qrs. 11-1/5 lbs.
One-fifth of this must be taken as a safe load, say 2 tons 18 cwt.
* * * * *
The strength of a solid cylindrical beam varies directly as its
diameter cubed, and inversely as its length.
The formula for finding the breaking weight (B.W.) of a solid
cylindrical beam supported at each end and loaded at its centre is as
follows:
B.W. = _d_^3/_l_ × _c_ × 10/17
where _d_^3 = least diameter in inches cubed;
_l_ = length of beam between supports in feet;
_c_ = the constant found by experiment on rectangular beams;
10/17 = the proportion that cylindrical beams bear to square beams,
_Example._—Find the B.W. of a solid cylindrical beam of spruce
fir 6 inches in diameter and 4 feet between supports.
The constant C for spruce fir is (Table I.) 403.
B.W. = (6 × 6 × 6 × 403 × 10)/(4 × 17)
= 12,801-3/17 lbs.
= 5 tons 14 cwt. 1qr. 5-3/17lbs.
One-fifth of this must be taken as a safe load, say 1 ton 2 cwt. 3
qrs.
[Illustration: FIG. 132]
The above diagram (fig. 132) illustrates the supports and loading of
the beams just considered. The following diagrams show beams variously
supported and loaded, and their relative strength to the first.
By adding to the formulæ already given, the proportion that these
following bear to the first, the same formulæ can be used for all.
[Illustration: FIG. 133]
Beam supported at both ends and load equally distributed bears double
of fig. 132.
[Illustration: FIG. 134
Beam fixed, load in centre, bears half more than fig. 132.]
[Illustration: FIG. 135
Beam fixed, load evenly distributed, bears triple of fig. 132.]
[Illustration: FIG. 136
Beam fixed one end only, load on free end, bears a quarter of fig. 132.]
[Illustration: FIG. 137
Beam fixed one end only, load evenly distributed, bears half of fig.
132.]
To find the effective length of a beam supported at both ends when the
position of the load is varied:
_Rule._—Divide four times the product of the distances of the load
from both supports by the whole span, in feet.
_Example._—Find the effective length of a beam the supports of
which are 6 feet apart, the load acting at a point 1 foot from its
centre.
The distances of the load from each support in this case are 2 feet and
4 feet.
Therefore (2 × 4 × 4)/6 = 5-1/3 feet, effective length of beam.
=Posts and Struts subject to Compression.=[5]—Posts and struts,
when above 30 diameters in length, tend to fail by bending and
subsequent cross breaking. When below 30 and above 5 diameters in
length, their weakness may partly be in their bending and partly by
crushing, and when below 5 diameters in length by crushing alone; but,
as these latter are rarely met with in scaffolding, they need not be
considered.
_On the resistance of long posts._—To find the greatest weight, W,
that a square post of 30 diameters and upwards will carry:
Multiply the fourth power of the side of the post in inches by the
value of D (Table II.), and divide by the square of the length in feet.
The quotient will be the weight required in pounds.
The formula is as follows:
W = (_d_^4 × D)/L^2 = weight in lbs. for square posts to resist bending,
where _d_ = depth in inches,
D = constant in pounds, for flexure,
L = length in feet.
_Example._—Find the weight that may be placed upon a post of elm
12 feet long and 4 inches square.
The value of D (Table II.) being 1,620,
(4^4 × 1,620)/12^2 = (256 × 1,620)/144 = 2,880 lbs.
One-tenth of this must be taken as a safe working load = 288 lbs.
* * * * *
The strength of cylindrical posts is to square ones as 10 is to 17, the
rest of the formula remaining the same, except that _d_ = diameter
of post in inches.
* * * * *
_When the pillar is rectangular._—Multiply the greater side by the
cube of the lesser in inches, and divide by the square of the length in
feet. The quotient multiplied by the value of D (Table II.) will give
the weight which the post will carry in pounds.
The formula is as follows:
W = (BT^3 × D)/L^2,
where B = breadth of post in inches;
T = least thickness in inches, and the rest of notation as before.
_Example._—Find the weight that may be placed on a rectangular
post of Dantzic oak 8 in. by 6 in. and 18 feet long.
The value of D (Table II.) being 2,410 lbs.,
(8 × 6^3 × 2,410)/18^2 = (8 × 216 × 2,410)/324 = 12,853 lbs.
The safe working load = say, 1,285 lbs.
_On the strength of posts of medium length._—When a post is less
than 30, but over 5, diameters in length, and therefore tending to fail
by crushing as well as by bending, the resistance to crushing is a
considerable proportion of the strength. This must, in consequence, be
allowed for in any calculation for the breaking weight.
To find the weight that would break a square or rectangular post of
between 5 and 30 diameters in length:
Multiply the area of the cross section of the post in square inches by
the weight in pounds that would crush a short prism of 1 inch square
(Table II.), and divide the product by 1·1 added to the square of the
length in feet, divided by 2·9 times the square of the least thickness
in inches.
The formula is as follows:
W = DS/(1·1 + L^2/T^2 2·9)
where D = the constant for the resistance to crushing.
S = the area of the cross section of the post in square inches.
1·1 is a modification introduced in order that the result
may be in accord with the result of experiments.
L = the length of the post in feet.
T = the least thickness in inches.
2·9 = the constant for the resistance to bending, and which
is taken at that figure for all timbers (Table II.).
_Example._—Find the breaking weight of a post of Dantzic oak 10
feet long and 6 inches square.
The constant for Dantzic oak (Table II.) is 7,731.
W = (7,731 × 36)/(1·1 + 100/(2·9 × 36)) = 278,316/2·05 = 135,763 lbs.
The factor safety is 1/8. Therefore the safe load will be 16,970 lbs.
The strength of cylindrical pillars is to square ones as 10 is to 17.
=Braces and Ties subject to a Tensional Strain.=—The weight that
will produce fracture in a beam strained in the direction of its
length, is in proportion to the area of the cross section of the beam
multiplied by the weight that would fracture a unit of that area.
The formula is as follows:
B.W. = E S
where E = the cohesive force in lbs. per unit of 1 square inch, as
in Table III.
S = the sectional area of the beam in square inches.
_Example._—Find the B.W. of a rectangular elm brace 9 in. by 3 in.
under a tensional strain.
The value of E (Table III.) is 4,480.
B.W. = 4,480 × 9 × 3 = 120,960 lbs.
One-fifth of this should be taken as a safe load,
= 24,192 lbs. = 10 tons 16 cwt.
which is the safe load required.
To find the sectional area of a cylindrical beam square its diameter
and multiply by ·7854.
The effect of fracture of a member of a scaffold depends upon its cause
and upon the importance of the member destroyed.
If the fracture is caused by a live load, say a heavy stone being
suddenly placed over a putlog, it is probable that the suspending
rope, if still attached, would prevent more damage being done. If the
fracture arose from an increasing dead load, say a stack of bricks
being gradually built up by labourers, the mass would probably tear its
way through all obstructions. Nevertheless, the entire scaffold, if
well braced and strutted, should not come down, the damage remaining
local.
The result of fracture of a standard under direct crushing would be
somewhat different, as, providing that the scaffold is rigid, the
greater strain thrown upon the ledgers, due to the increased distance
between supports, would probably cause them to fracture. In this
case the damage would probably still remain local. If, owing to the
fracture, the effect of the bracing were lost, the whole scaffold would
probably fail, as shown in the chapter on Stability.
It should be noted that the ledgers, together with the putlogs when
fixed at both ends, apart from carrying the loads, have an important
effect upon the standards, as, when securely connected, they divide the
uprights into a series of short posts, thus dispelling any likelihood
of failing by flexure.