 TABLE VII  SHEARING STRENGTH ALONG THE GRAIN OF SMALL CLEAR PIECES OF 41 WOODS IN GREEN CONDITION (Forest Service Cir. 213)  When When COMMON NAME surface of surface of OF SPECIES failure is failure is radial tangential ++ Lbs. per Lbs. per sq. inch sq. inch Hardwoods Ash, black 876 832 white 1,360 1,312 Basswood 560 617 Beech 1,154 1,375 Birch, yellow 1,103 1,188 Elm, slippery 1,197 1,174 white 778 872 Hackberry 1,095 1,161 Hickory, big shellbark 1,134 1,191 bitternut 1,134 1,348 mockernut 1,251 1,313 nutmeg 1,010 1,053 pignut 1,334 1,457 shagbark 1,230 1,297 water 1,390 1,490 Locust, honey 1,885 2,096 Maple, red 1,130 1,330 sugar 1,193 1,455 Oak, post 1,196 1,402 red 1,132 1,195 swamp white 1,198 1,394 white 1,096 1,292 yellow 1,162 1,196 Sycamore 900 1,102 Tupelo 978 1,084 Conifers Arborvitae 617 614 Cedar, incense 613 662 Cypress, bald 836 800 Fir, alpine 573 654 amabilis 517 639 Douglas 853 858 white 742 723 Hemlock 790 813 Pine, lodgepole 672 747 longleaf 1,060 953 red 812 741 sugar 702 714 western yellow 686 706 white 649 639 Spruce, Engelmann 607 624 Tamarack 883 843 
Both shearing stresses may act at the same time. Thus the weight carried by a beam tends to shear it off at right angles to the axis; this stress is equal to the resultant force acting perpendicularly at any point, and in a beam uniformly loaded and supported at either end is maximum at the points of support and zero at the centre. In addition there is a shearing force tending to move the fibres of the beam past each other in a longitudinal direction. (See Fig. 12.) This longitudinal shear is maximum at the neutral plane and decreases toward the upper and lower surfaces.
[Illustration: FIG. 12.Horizontal shear in a beam.]
Shearing across the grain is so closely related to compression at right angles to the grain and to hardness that there is little to be gained by making separate tests upon it. Knowledge of shear parallel to the grain is important, since wood frequently fails in that way. The value of shearing stress parallel to the grain is found by dividing the maximum load in pounds (P) by the area of the cross section in inches (A).
{ P } { Shear =  } { A }
Oblique shearing stresses are developed in a bar when it is subjected to direct tension or compression. The maximum shearing stress occurs along a plane when it makes an angle of 45 degrees P with the axis of the specimen. In this case, shear = . When 2 A the value of the angle [Greek: theta] is less than 45 degrees, P the shear along the plane =  sin [Greek: theta] cos [Greek: A theta]. (See Fig. 13.) The effect of oblique shear is often visible in the failures of short columns. (See Fig. 14.)
[Illustration: FIG. 13.Oblique shear in a short column.]
[Illustration: FIG. 14.Failure of short column by oblique shear.]
 TABLE VIII  SHEARING STRENGTH ACROSS THE GRAIN OF VARIOUS AMERICAN WOODS (J.C. Trautwine. Jour. Franklin Institute. Vol. 109, 1880, pp. 105106)  KIND OF WOOD Lbs. per KIND OF WOOD Lbs. per sq. inch sq. inch +++ Ash 6,280 Hickory 7,285 Beech 5,223 Locust 7,176 Birch 5,595 Maple 6,355 Cedar (white) 1,372 Oak 4,425 Cedar (white) 1,519 Oak (live) 8,480 Cedar (Central Amer.) 3,410 Pine (white) 2,480 Cherry 2,945 Pine (northern yellow) 4,340 Chestnut 1,536 Pine (southernyellow) 5,735 Dogwood 6,510 Pine (very resinous yellow) 5,053 Ebony 7,750 Poplar 4,418 Gum 5,890 Spruce 3,255 Hemlock 2,750 Walnut (black) 4,728 Hickory 6,045 Walnut (common) 2,830  NOTE.Two specimens of each were tested. All were fairly seasoned and without defects. The piece sheared off was 5/8 in. The single circular area of each pin was 0.322 sq. in. 
TRANSVERSE OR BENDING STRENGTH: BEAMS
When external forces acting in the same plane are applied at right angles to the axis of a bar so as to cause it to bend, they occasion a shortening of the longitudinal fibres on the concave side and an elongation of those on the convex side.
Within the elastic limit the relative stretching and contraction of the fibres is directly[9] proportional to their distances from a plane intermediate between themthe ~neutral plane~.
(N_{1} P in Fig. 15.) Thus the fibres halfway between the neutral plane and the outer surface experience only half as much shortening or elongation as the outermost or extreme fibres.
Similarly for other distances. The elements along the neutral plane experience no tension or compression in an axial direction. The line of intersection of this plane and the plane of section is known as the ~neutral axis~ (N A in Fig. 15) of the section.
[Footnote 9: While in reality this relationship does not exactly hold, the formulae for beams are based on its assumption.]
[Illustration: FIG. 15.Diagram of a simple beam. N_{1} P = neutral plane, N A = neutral axis of section R S.]
If the bar is symmetrical and homogeneous the neutral plane is located halfway between the upper and lower surfaces, so long as the deflection does not exceed the elastic limit of the material. Owing to the fact that the tensile strength of wood is from two to nearly four times the compressive strength, it follows that at rupture the neutral plane is much nearer the convex than the concave side of the bar or beam, since the sum of all the compressive stresses on the concave portion must always equal the sum of the tensile stresses on the convex portion. The neutral plane begins to change from its central position as soon as the elastic limit has been passed. Its location at any time is very uncertain.
The external forces acting to bend the bar also tend to rupture it at right angles to the neutral plane by causing one transverse section to slip past another. This stress at any point is equal to the resultant perpendicular to the axis of the forces acting at this point, and is termed the ~transverse shear~ (or in the case of beams, ~vertical shear~).
In addition to this there is a shearing stress, tending to move the fibres past one another in an axial direction, which is called ~longitudinal shear~ (or in the case of beams, ~horizontal shear~). This stress must be taken into consideration in the design of timber structures. It is maximum at the neutral plane and decreases to zero at the outer elements of the section. The shorter the span of a beam in proportion to its height, the greater is the liability of failure in horizontal shear before the ultimate strength of the beam is reached.
_Beams_
There are three common forms of beams, as follows:
(1) ~Simple beam~a bar resting upon two supports, one near each end. (See Fig. 16, No. 1.)
(2) ~Cantilever beam~a bar resting upon one support or fulcrum, or that portion of any beam projecting out of a wall or beyond a support. (See Fig. 16, No. 2.)
(3) ~Continuous beam~a bar resting upon more than two supports. (See Fig. 16, No. 3.)
[Illustration: FIG. 16.Three common forms of beams. 1. Simple.
2. Cantilever. 3. Continuous.]
_Stiffness of Beams_
The two main requirements of a beam are stiffness and strength.
The formulae for the _modulus of elasticity (E)_ or measure of stiffness of a rectangular prismatic simple beam loaded at the centre and resting freely on supports at either end is:[10]
[Footnote 10: Only this form of beam is considered since it is the simplest. For cantilever and continuous beams, and beams rigidly fixed at one or both ends, as well as for different methods of loading, different forms of cross section, etc., other formulae are required. See any book on mechanics.]
P' l^{3} E =  4 D b h^{3}
b = breadth or width of beam, inches.
h = height or depth of beam, inches.
l = span (length between points of supports) of beam, inches.
D = deflection produced by load P', inches.
P' = load at or below elastic limit, pounds.
From this formulae it is evident that for rectangular beams of the same material, mode of support, and loading, the deflection is affected as follows:
(1) It is inversely proportional to the width for beams of the same length and depth. If the width is tripled the deflection is onethird as great.
(2) It is inversely proportional to the cube of the depth for beams of the same length and breadth. If the depth is tripled the deflection is one twentyseventh as great.
(3) It is directly proportional to the cube of the span for beams of the same breadth and depth. Tripling the span gives twentyseven times the deflection.
The number of pounds which concentrated at the centre will deflect a rectangular prismatic simple beam one inch may be found from the preceding formulae by substituting D = 1" and solving for P'. The formulae then becomes:
4 E b h^{3} Necessary weight (P') =  l^{3}
In this case the values for E are read from tables prepared from data obtained by experimentation on the given material.
_Strength of Beams_
The measure of the breaking strength of a beam is expressed in terms of unit stress by a _modulus of rupture_, which is a purely hypothetical expression for points beyond the elastic limit. The formulae used in computing this modulus is as follows:
1.5 P l R =  b h{^2}
b, h, l = breadth, height, and span, respectively, as in preceding formulae.
R = modulus of rupture, pounds per square inch.
P = maximum load, pounds.
In calculating the fibre stress at the elastic limit the same formulae is used except that the load at elastic limit (P_{1}) is substituted for the maximum load (P).
From this formulae it is evident that for rectangular prismatic beams of the same material, mode of support, and loading, the load which a given beam can support varies as follows:
(1) It is directly proportional to the breadth for beams of the same length and depth, as is the case with stiffness.
(2) It is directly proportional to the square of the height for beams of the same length and breadth, instead of as the cube of this dimension as in stiffness.
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