=Tension.= If a weight of 2000 lbs. were hung to the lowest end of a
vertical beam, so that the line of action of the weight and axis of
the beam formed one and the same straight line--the tension on the
beam would be 2000 lbs. But, if the beam were inclined, and the force
acted in a vertical direction, then the strain would be increased in
the ratio of the increase of the diagonal of inclination over the
vertical;--suppose the beam is 20 ft. long and inclined at an angle of
45°--and let 2000 lbs., as before, be suspended from its lower end.
Now the diagonal being 20°,--the vertical will be 14.014 ft.--and the
strain will be found as follows,--
14.014 : 20 :: 2000 : 2854--lbs.
The greater the angle of inclination from the horizontal, the less the
strain from a given load--and when the beam is vertical the weight
causes the least strain.
=Compression.= If we load a vertical post with a weight of 2000 lbs.,
the strain of compression exerted upon the post will be 2000 lbs. Now,
if we incline the post--the strain will be increased, as we have shown
above under the head of tension, and in like manner, dependent upon
the inclination.
But when wood, iron, or any other material is used for a pillar or
strut, it has not only to resist a crushing force, but also a force
tending to bend or bulge it laterally.
A post of circular section with a length of 7 or 8 diameters will not
bulge with any force applied longitudinally, but will split. But if
the length exceeds this limit--it will be destroyed by an action
similar to that of a transverse strain.
A cast iron column of thirty diameters in length, is fractured by
bending; when the length is less than this ratio--by bending and
splitting off of wedge shaped pieces. But by casting the column
hollow, and swelling it in the middle, its strength is greatly
increased.
Barlow's formula for finding the weight that can be sustained by any
beam, acting as a pillar or strut, before bending, is:--
WL² bd³ x 80 E
---- = bd³, whence W = -----------
80 E L²
[TeX: $\frac{WL^2}{80 E} = {bd^3}$, whence $W = \frac{{bd^3} x 80 E}{L^2}$]
now, having the weight given, and assuming the dimensions of
the cross-section--we shall have
-----
/ WL² WL²
d = ³/ -----, and b = ------
\/ 80 Eb 80 Ed³
[TeX: $d = \sqrt[3]{\frac{WL^2}{80 EB}}$, and $b = \frac{WL^2}{80 ED^3}$]
in the above formulæ,
W = weight in pounds.
L = length in feet.
E = a constant.
b = breadth in inches.
d = depth in inches.
=Transverse Strains.= The strain caused by any weight, applied
transversely, to a beam supported at both ends, is directly as the
breadth, and square of the depth, and inversely as the length. It
causes the beam to be depressed towards the middle of its length,
forming a curve, concave to the horizontal and below it. In assuming
this form--the fibres of the upper part of the beam are compressed,
and those of the lower part are extended--consequently there must be
some line situated between the upper and lower surfaces of the beam
where the fibers are subjected to neither of these two forces, this
line is called the _neutral axis_.
These two strains of compression and extension must be equal in
amount--and upon the relative strength of the material to resist these
strains, as well as its form and position, the situation of this axis
depends. If wood resists a compression of 1000 lbs. per square inch of
section, and a tension of 2000 lbs. the axis will be twice as far from
the top as from the bottom in a rectangular beam.
The following table by Mr. G.L. Vose gives, with sufficient accuracy
for practice, the relative resisting powers of wood, wrought, and cast
iron, with the corresponding positions of the axis.
Dist. of axis
Resistance Resistance from top in
to to frac's of
Material. Extension. Compression. Ratio. the depth.
Wrought Iron, 90 66 90/66 90/156, or 0.58.
Cast Iron, 20 111 20/111 20/131, or 0.15.
Wood, 2 1 2/1 2/3, or 0.66.
Thus we see that the resistance of a beam to a cross strain, as well
as to tension and compression, is affected by the incompressibility
and inextensibility of the material.
The formula for the dimensions of any beam to support a strain
transversely is
4 bd²
S = ----
l
[TeX: $S = \frac{4 bd^2}{l}$]
S = the ultimate strength in lbs.
b = the breadth in inches.
d = the depth in inches.
l = the length in inches.
=Detrusion.= Detrusion is the crushing against some fixed point, such
as obtains where a brace abuts against a chord, or where a bridge
rests on a bolster; and the shearing of pins, bolts and rivets, also
comes under this head.
=General Abstract.= The resistance to the above mentioned strains
varies as the area of the cross section; so that by doubling the area
we double the strength. Any material will bear a much greater strain
for a short time than for a long one. The working strength of materials,
or the weight which does not injure them enough, to render them unsafe,
is a mooted point, and varies, according to the authority, from 1-3
to 1-10 of the ultimate strength. The ratio of the ultimate strength
to the working strength is called the _factor of safety_.
The following is a table of ultimate and working strengths of
materials, and factors of safety:
Weight Ult. Ult. Working Strengths Factor Safety.
in lbs. Materials. Ext. Comp. Exten. Comp. Tension Comp.
30 Wood. 14,000 7,000 2,000 1,000 7 7
480 Wrou't Iron. 60,000 64,000 15,000 12,000 4 5.33
450 Cast Iron. 18,000 100,000 4,500 25,000 4 4
=Lateral Adhesion.= Lateral adhesion is the resistance offered by the
fibres to sliding past each other in the direction of the grain, as
when a brace is notched into a chord, or tie beam, at its foot, it is
prevented by the lateral adhesion of the fibres from crowding off the
piece, to the depth of the notch, against which it toes. Barlow's
experiments give the lateral adhesion of fir as 600 lbs. per square
inch, and the factor of safety employed varies in practice from 4 to
6, giving a working strength of from 150 to 100 lbs. per square inch.
=TABLE OF COMPRESSIVE RESISTANCE OF TIMBER.=
Length Safety Length Safety Length Safety
given in Weig't in given in Wt. in given in Wt. in
Diameters. Pounds. Diameters. Pounds. Diameters. Pounds.
6 1000 24 440 42 203
8 960 26 394 44 185
10 910 28 358 46 169
12 860 30 328 48 155
14 810 32 299 50 143
16 760 34 276 52 132
18 710 36 258 54 122
20 660 38 239 56 114
22 570 40 224 58 106
60 99
In tensional strains, the length of the beam does not affect the
strength; but in the beams submitted to compression, the length is a
most important element, and in the table given above, the safety
strains to which beams may be subjected, without crushing or bending,
has been given for lengths, varying from 6 to 60 diameters.