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Contents | 1. Introduction to structural design | 2. Loads |

Introduction to wood | Material properties | Sectional properties | Design approaches | Construction systems |

Basic tabular values of allowable stresses in tension are shown in Appendix Table A-3.1 for some
common species and grades of visually-graded lumber. The allowable stress in tension (parallel to
the grain) for timber elements needs to be modified, or adjusted, to account for the variations in
material properties discussed earlier in this chapter. The three most important adjustment factors,
corresponding to these material properties, are as follows: *C _{M}* for wood structural elements exposed to wet service conditions;

The actual tension stress within the structural element is computed by dividing the internal tension
force by the cross-sectional area available to resist that force. Where bolt holes are present, the
gross area, *A _{g}*, of the cross section is reduced by the nominal hole area, as shown in Figure 1.53

(3.1)

where *D _{h}* is the bolt hole diameter, taken somewhat larger than the bolt diameter; and

(3.2)

where no bolt holes are used; and

(3.3)

where bolt holes are present. These equations can be rewritten to solve for the capacity (allowable load), using the adjusted allowable stress, *F _{t}*', instead of the actual stress. Where tension elements are nailed rather than bolted, no reduction for nail holes is made; the full gross area is assumed to be available to resist the internal forces:

(3.4)

Where bolt holes are used:

(3.5)

Where wood elements are loaded in tension parallel to grain, another potential mode of
failure must be checked where closely-spaced groups of bolts (or lag screws that fully penetrate the
main member) are used as fasteners — this phenomenon does not apply to small-diameter nailed connections. As shown in Figure 3.13, the forces transmitted through a fastener group could cause
entire "slots" of wood — either within each row of fasteners or for the entire fastener group taken
as a whole — to tear out under the load. To prevent failure in the first case ("row tear-out," Figure
3.13*b*), and in spite of the fact that the forces acting on the element itself are *tensile*, the external force acting on the connection must be no greater than the total allowable *shear* force that can be developed on all the potential failure planes along the rows of fasteners. In the second case ("group tear-out," Figure 3.13*c*), the external force must be no greater than the allowable shear and tension forces that can be safely resisted by the three surfaces forming the boundary failure planes for the
group of fasteners as a whole. Of these three surfaces, the resistance of the top and bottom parallel
planes, stressed in shear, is equivalent to a single row subjected to row tear-out; the third surface,
labeled At in Figure 3.13*c*, is stressed in tension.

Figure 3.13: Forces on a wood member *(a)* loaded parallel to grain can cause *(b)* row tear-out or (c) group tear-out

In calculating row and group tear-out, adjusted allowable stresses for shear and tension are used that correspond to the species and grade of the wood elements being checked. The total length of the surface assumed to be resisting shear stress along a given row of fasteners is taken as the smallest distance between fasteners (or between the end of the wood member and the first fastener) multiplied by the number of fasteners in that row. This accounts for the fact that shear stress along the potential failure planes defined by fastener rows is not uniform, but is higher where the area between fasteners along the shear plane is smallest. Additionally, this shear stress is not uniformly distributed between fasteners, but is assumed to step up and down in a triangular pattern, from maximum to zero, so that the average value of shear stress is actually half of its maximum value (corresponding to the average height of such a triangle).

Taking all these caveats into consideration, the allowable maximum force at a connection using bolts or lag screws — where the load is parallel to grain — is limited by the smaller of the following values for row and group tear-out:

a. *For row tear-out:* multiply the force safely resisted by the two shear planes at each row of fasteners by the number of rows (or, if not all rows of fasteners are the same, add the values for each row computed separately). The force resisted by a single row (arbitrarily called row 1) is equal to:

(3.6)

where *n*_{1} = the number of fasteners in row 1; *F _{v}*' = the adjusted allowable shear stress for the wood element;

(3.7)

b. *For group tear-out:* add the shear force resisted at the parallel planes defined by the top and bottom fastener rows (in typical cases where the top and bottom fastener rows have the same geometry, this is equal to the value of *Z*'_{RT}_{-1} computed above for single row tear-out; otherwise, add *Z*'_{RT}_{-1} /2 for the top and bottom rows) plus the tension force resisted by the plane surface joining, and perpendicular to, these shear planes. For *A _{t}* representing the area subjected to tension stress between the top and bottom rows of fasteners (see Figure 3.13

(3.8)

These limitations based on row and group tear-out are summarized in Appendix Table A-3.14.

** Problem definition.** Find the maximum load that
can be applied to a 2 × 8 tension element connected with six 1/2 in. diameter bolts. The wood
used is Hem-Fir No. 1. Assume live, dead, and wind loads only, dry conditions, and spacing as shown in Figure 3.14.

Figure 3.14: Bolted 2 × 8 cross sections for Example 3.1

** Solution overview.** Find gross area and net area;
compute adjusted allowable stress; find capacity,

Figure 3.15: Possible failure modes for Example 3.1 include *(a)* tension failure on net area; *(b)* row tear-out; and *(c)* group tear-out

*Problem solution*

**1.** From Appendix Table A-3.12 the cross-sectional
area, *A _{g}* = 10.875 in

**2.** Use Equation 3.1 to find the net area, *A _{n}* :
Notice that even though there are six bolt holes, only two of them are subtracted from the gross area in calculating the net area, since the "failure plane" passes through only two holes. The hole diameter is taken as 1/16 in. larger than the bolt diameter, so

**3.** From Appendix Table A-3.1 find the tabular value of allowable tension stress, *F _{t}* = 625 psi.

**4.** Compute adjusted allowable tension stress:

a. From Appendix Table A-3.2, find adjustments to tabular value: *C _{D}* = 1.6;

b. *F _{t}*' =

**5.** Using Equation 3.5, find capacity (allowable load) based on failure through the net area: P =
*F _{t}*'

**6.** From Appendix Table A-3.14, check capacity based on row and group tear-out, since the orientation
of the load is parallel to grain and the member is in tension. The adjusted allowable stress
in shear is found from Appendix Tables A-3.7 and A-3.8: *F _{v}* = 150 psi and the relevant adjustments
are for wet service (

a. *Z*'* _{RT}* =

b. *Z*'* _{GT}* =

Because the smaller of these two values (4320 lb) is smaller than the capacity found in step 4, row tear-out governs the connection design, and the total adjusted connection capacity, *P* = 4320 lb.

**7.** *Conclusion:* the capacity of the 2 × 8 tension element, *P* = 4320 lb. The capacity of the bolts themselves has not been checked; the design and analysis of such fasteners is discussed later in this chapter.

** Problem definition.** Find an appropriate 2 × cross
section (Hem-Fir No. 2) to support a tensile load
of 5 kips consisting only of live and dead loads
in normal proportions. Use two lines of three
3/8 in. diameter bolts, as shown in Figure 3.16,
to connect the wood element to another part
of the structure. The bolt hole diameter = bolt
diameter + 1/16 in. = 7/16 in. = 0.4375 in.

Figure 3.16: Bolted 2× cross section for Example 3.2

** Solution overview.** Compute provisional adjusted
allowable stress; find required net area; find required
gross area; select provisional cross section;
check cross section by finding adjusted allowable
stress, required net area, and required
gross area. Check row and group tear-out.

*Problem solution*

**1.** From Appendix Table A-3.1, find the tabular value of the allowable tension stress, *F _{t}* = 525 psi (use "dimension lumber" for 2× element).

**2.** Compute provisional adjusted allowable tension stress:

a. From Appendix Table A-3.2 find adjustments to tabular value: *C _{D}* = 1.0;

b. *F _{t}*' =

**3.** Find required net area, *A _{n}* = load/stress = 5000/525 = 9.52 in

**4.** Using Equation 3.1 (but solving for *A _{g}*), and referring to Figure 3.16, find the required gross area,

**5.** We need a provisional 2× cross section with *A _{g}* ≥ 10.83 in

*Trial 1: 2×8*

Because the actual value of the size factor for a 2 × 8, *C _{F}* = 1.2, is larger than the value initially assumed, the adjusted allowable stress will be higher than assumed, and therefore a 2 × 8 cross section will certainly work. However, it is necessary to analyze (check) the next smaller cross section, since this cross section has an even larger size factor than does the 2 × 8, and so has an even higher adjusted allowable stress.

*Trial 2: 2×6*

**1.** From Appendix Table A-3.12 the cross-sectional area of a 2 × 6, *A _{g}* = 8.25 in

**2.** Use Equation 3.1 to find the net area, *A _{n}* =

**3.** As before, find the tabular value of allowable tension stress, *F _{t}* = 525 psi.

**4.** Compute adjusted allowable tension stress:

a. From Appendix Table A-3.2, find adjustments to tabular value: *C _{D}* = 1.0;

b. *F _{t}*' =

**5.** Using Equation 3.5, find capacity (allowable load), P = *F _{t}*'

*Check row and group tear-out*

**1.** From Appendix Table A-3.14, check capacity based on row and group tear-out, since the orientation of the load is parallel to grain and the member is in tension. The adjusted allowable stress in shear is found from Appendix Tables A-3.7 and A-3.8: *F _{v}* = 150 psi and the relevant adjustments are for wet service (

a. *Z*'* _{RT}* =

b. *Z*'* _{GT}* =

**2.** Conclusion: Because the smaller of the capacities for row and group tear-out (5400 lb) is larger than the actual load of 5000 lb, neither row nor group tear-out governs the connection design, and the 2 × 8 provisionally selected above may be used.

© 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.