Notation for Beam Formulas
5.2 Structural Members
Beams
When determining stresses and deflections of wood structural mem-bers, actual—not nominal—sizes must be used.
The proper design of timber beams involves three steps:
1. Compute the member’s section modulus (S= M/f ) and select an appropriate sized beam from Wood Structural Design 1962–
American Forest and Paper Association.
2. Test the beam selected for horizontal shear; increase dimensions if necessary.
3. Test the beam selected for deflection.
Most designers simply perform the first computation (section modu-lus) and then select the proper beam size that they know from experi-ence will meet the requirements for steps 2 and 3. In determining the strength and stiffness of a wood member under transverse loading, it is necessary to consider:
Species Specific gravity Weight (lb/ft3)
Coast Sitka pine 0.39 26.6
Douglas fir—Larch 0.51 34.3
Douglas fir—South 0.48 32.7
Eastern spruce 0.43 28.9
Eastern white pine 0.38 25.5
Hem fir 0.42 28.1
Nor thern pine 0.46 31.2
Oak - red and white 0.67 47.3
Ponderosa pine (Nor th) 0.49 33.0
Southern cypress 0.48 33.6
Southern pine 0.55
Spruce - pine - fir 0.42 26.9
Western cedars 0.35 24.2
Source: American Wood Council
TABLE 5.2 Properties of Selected Structural Lumber Used in Scaffolding and Ladders
1. Bending moment induced by a load 2. Horizontal shear at beam supports 3. Bearing on supporting members
4. Deflection or deformation caused by load
Any of the following four factors may control the design of a struc-tural wood member.
Bending
For structural safety, a beam’s bending moment, induced by the com-bination of live and dead loads on the beam, may not be greater than the resisting moment of the wood member.
For example: Assume a beam having a span L uniformly loaded with w pounds per foot (Fig. 5.2):
FIGURE 5.2 Forces and stresses developed in a uniformly distributed load on a simple beam.
W o o d S t r u c t u r a l M e m b e r s
69
1. Compute the maximum bending moment M, which occurs at the beam’s midpoint L/2, using the formula:
M wL
2. Divide the resisting moment M by the allowable fiber stress in bending Fb for the particular type of wood and grade of lumber, to determine the section modulus S of the structural member.
S M
Fb
=
Since the bending moment may not be greater than the resisting moment, the two formulas can be equated:
F S wL
b = 3 2
2
3. Determine one of the following:
a. Size of beam required—when span and load are known:
S wL
Fb
=3 2
2
b. Allowable span—when size and load per linear foot are known:
L F S
w
= 2 b
3
c. Allowable load per linear foot—when size of beam and span are known:
w F S
L
=2 b
3 2
Alternatively, once the section modulus of a wood structural member is determined, the rigger can select a properly sized beam directly from precomputed tables; however, make sure to select a wood sec-tion that has a secsec-tion modulus equal to or greater than the computed value. (The section modulus of a structural wood member is I/c, which for a rectangular beam is bd2/6.)
Although many beam sizes will meet these requirements, the most practical sizes are those beams whose breadth ranges from one-half to one-third of the beam’s depth.
REMEMBER:
REMEMBER: Beams that are relatively deep compared to their width tend to bend sideways under loading unless properly braced.
Rules-of-thumb for beams having the following ratios of depth to width, or thickness:
2:1 No lateral support is needed 3:1 or 4:1 Ends should be held in position
5:1 One edge should be held in line for entire length of beam 6:1 Lateral supports are required at intervals of 8 ft Although lumber is customarily specified in terms of nominal sizes, always use the actual sizes (net dimensions) for any design computations.
Horizontal Shear
Horizontal or longitudinal shearing forces in a member must be con-sidered in designing a beam subject to vertical shear (Fig. 5.3 ). When a beam is loaded vertically, the upper part of the beam tends to slide along its lower part. To prevent this sliding action, the shear resis-tance of wood member must equal or exceed the horizontal shear that the vertical loading induces.
Maximum horizontal shear occurs in a rectangular beam at the neutral axis of the section and depends on the magnitude of vertical shear in the member:
(Thus, whether the depth or width of a solid beam is doubled, the mass is the same, but the strength/stiffness/weight ratios for a beam are very much better with increased depth rather than increased thickness)
Tension Shear
Load
Breaking strength of a beam:
• Directly proportional to the width, but • Proportional to depth squares, and • Inversely proportional to the span
Stiffness of a beam:
• Directly proportional to width, but • Proportional to depth cubes, and • Inversely proportional to span cubed
FIGURE 5.3 Horizontal or longitudinal shear in a simple beam.
W o o d S t r u c t u r a l M e m b e r s
71
Where fv= Unit Stress
V = Maximum Shear Force
Q = First moment of Area (around neutral axis) I = Moment of Inertia
B = Width of Beam
The unit stress, however, may not be greater than the design value in horizontal shear Fv for the particular type and grade of wood used.
Because notching a beam will affect the shear strength of the member, notching should be avoided whenever possible, especially on a beam’s tension side. Never notch the tension side of a beam that is 4-in. deep or deeper except on its ends. Notches at the ends of a beam, however, should not be greater than one-fourth the depth of the wood member (Fig. 5.4).
To reduce stress concentration, where notching a beam is necessary, use a gradually tapered notch instead of square cornered one. Notches in wood beams should not be deeper than one-sixth the depth of the wood member. And never put a notch in the middle one-third of a span.
Always check the desired bending load of a beam having square cornered notches at its ends against the load obtained by the formula:
V d
When a wooden beam is loaded, its fibers tend to compress at the beam support points. Make sure there is sufficient end-bearing area for the beam to transfer the load without damage to the fibers. Deter-mine the bearing area by dividing the beam’s end reaction by the design value of compression perpendicular to the grain Fcfrom the particular type and grade of wood being used. If the bearing area, from the end of a wood beam, is less than 6 in. long, it is safe to use higher stresses in compression perpendicular to the grain. For such bearing areas, if located 3 in. or more from the ends of a beam, increase the compression values as indicated in Table 5.3.
Deflection
A beam’s deflection Δ is the amount of deformation that results from a load applied to the beam. As long as the induced bending stress does not exceed the design value, deflection does not seriously affect FIGURE 5.4 Notched beam with tapered notches 1/6 the beam’s depth.
the safety of the beam, and deflection can be ignored, with the design of the beam based on strength alone. Deflection, however, becomes a critical factor for the rigger when a specified clearance is required under the beam.
Joists
These wood members are small beams spaced closely together to support floor (or roof) loads. They are designed in the same manner as wood beams. Maximum spans for various sizes of floor loads and stresses are based on the bending strength using the live load indi-cated on each heading, plus a dead load of 10 lb/ft2(span tables for Joists and Rafters–American Wood Council, Washington, DC).
Planks
The span of a scaffold plank is the distance it runs between supports.
The longer the span, the more deflection (bend) it will have, and there-fore, the less its load-bearing capacity will be.
For scaffolding, wood planks 2 in. thick, or thicker, are laid on the flat spanning from beam to beam (similar to the way a floor would be laid). The beams are spaced 5 ft to 12 ft on center, depending on the load to be supported and the kind of wood used (Tables 5.4 and 5.5).
To compute the correct thicknesses of a plank, assume a width of 12 in. and apply beam design methods. Scaffold planks should be unsurfaced, and actual dry dimensions should not be less than spec-ified. For example:
2 × 9 17/8× 83/4
2 × 10 17/8× 9 2 × 12 17/8× 113/4
Length of bearing (in.) Factor
½ 1.75
1 1.38
1½ 1.25
2 1.19
3 1.13
4 1.10
6 or more 1.00
TABLE 5.3 Factors by Which to Increase Compression Values Based on Bearing Length
W o o d S t r u c t u r a l M e m b e r s
73
The longer the span, the more deflection (bend) it will have, and therefore, the less its load-bearing capacity will be (Table 5.6). Because nominal thickness lumber is not cut to its exact dimensions, nominal 2 × 10 is really more like 1½ × 9¼. As a result, it does not have the same load-bearing capacity of full thickness lumber.
For fabricated planks and platforms, maximum spans are to be recommended by the manufacturer. To assure that scaffold planking remains within its safe load-bearing capacity, it may not be allowed to deflect more than 1/60th of its span between supports [OSHA 1926.451(f)(16)]. (rough) Solid Sawn Wood Planks
Rated load capacity Intended load
Light duty 25 lb per ft2 applied uniformly over the entire span area
Medium duty 50 lb per ft2 applied uniformly over the entire span area
Heavy duty 75 lb per ft2 applied uniformly over the entire span area
One person 250 lb placed at the center of the span (total 250 lb)
Two person 250 lb placed 18 in. to the left and right of the center of the span (total 500 lb) Three person 250 lb placed at the center of the span and
250 lb placed 18 in. to the left and right of the center of the span (total 750 lb) Source: OSHA, Washington, DC.
TABLE 5.5 Allowable Spans (for a given load) for Fabricated Planks and Platforms, Maximum Spans to be Recommended by the Manufacturer Based on this Table
Planks should have strength ratios of at least 80 percent of the strength of a theoretically flawless plank.
• Planks, when fully accepted, should be immediately branded on their ends—permanently marked suited for scaffolding.
• Inspected planks should be carefully handled and not dropped in bulk from the delivery truck.
• Use of planks other than those specifically accepted for scaffolding should be forbidden.
REMEMBER:
REMEMBER: Deflection can indicate when a platform is overloaded. To assure that scaffold planking remains within its safe load-bearing capacity, it may not be allowed to deflect more than 1/60th of its span between supports [OSHA 1926.451(f)(16)]. A competent person should not allow employees to occupy a platform that exceeds the 1/60th ratio.
Columns
Most timber columns, or posts, have square or rectangular cross sec-tions and are single lengths of timber. Although columns built up from pieces of timber and joined by nails, bolts, or other mechanical fasteners are often used, they do not have the same strength as a one-piece member of comparable material and dimensions. The strength of a built-up column must be reduced by precomputed per-centages to provide a section equal to the strength of a one-piece col-umn of the same dimension and quality.
The slenderness ratio of a column is a measure of its stiffness and has an important bearing on the load a column will support.
The strength of built-up columns must be reduced by the percent-ages in the table to provide a section equal to the strength of a one-piece column of the same dimension and quality (Table 5.7).
The slenderness ratio is computed by dividing the column’s later-ally unsupported length by the appropriate cross-sectional dimension,
Span of plank
TABLE 5.6 Maximum Permissible Planking Deflection
W o o d S t r u c t u r a l M e m b e r s
75
both in inches (Fig. 5.5). To determine the laterally unsupported length of a column, measure the column’s distance parallel to its longitudinal axis—between the supports that restrain the column against any lat-eral movement.
A short column, one having a slenderness ratio less than 11, will support a load equal to the area of the cross section multiplied by the full allowable compressive strength of the particular type and grade of wood. If the length is increased, however, then the cross-sectional area must also be increased.
Numerous timber column formulas are used in finding the per-missible unit stress. A formula recommended by the National Forest Products Association is
F C I
= ⎛ − d
⎝⎜
⎞ 1 ⎠⎟
80
TABLE 5.7 Adjustment Factors for Built-Up Columns with Cover Plate Slenderness ratio (L/D) of
equivalent solid column
Ratio of built-up column strength to equivalent solid column strength
6 82
10 72
14 71
18 65
22 74
26 82
FIGURE 5.5 Slenderness ratio l/d for simple solid columns.
where F = maximum permissible unit stress for column cross section (lb/in.2)
C = maximum allowable unit stress parallel to grain for short blocks (lb/in.2)
l = unsupported length of column (in.)
d = least width or diameter of cross section (in.)
Timber columns are designed by trial and error, first assuming a cross section and then testing to see whether or not it meets the specified requirements. For example:
1. Assume a cross-sectional area of approximate proper dimensions.
2. Solve for F in the timber column formula, substituting for l the length of the column, in inches; and for d the least width or diameter of the column, in inches.
3. The computed F value is in units of pounds per square inch, and is multiplied by the number of square inches in the assumed column’s cross-sectional area to obtain the allowable safe load for the column.
4. Compare the allowable load found in step 3 with the actual load to be supported.
5. If the actual load is greater, the assumed cross section is too small.
6. Assume a larger cross section, and test again.
While lumber columns may be designed by use of a formula, the rigger will find it more convenient to use precomputed tables to make the proper selection of column sizes (National Forest Products Association).