Tools › Civil and Geotechnical › Timber and Masonry › Timber Beam and Column › Example

4 × 12 timber beam on a 14 ft span and a 6 × 6 column: NDS adjustment factors, Ylinen stability factors C_L and C_P (Wood Handbook chapter 9)

Tool	Timber Beam and Column
Source	Forest Products Laboratory (2010). Wood Handbook: Wood as an Engineering Material. General Technical Report FPL-GTR-190, U.S. Department of Agriculture, Forest Service, Madison, WI. Chapter 9 (eqs 9-25, 9-28, 9-38, Table 9-1) and chapter 7 (derivation of design values); Ritter, M. A. (1990). Timber Bridges: Design, Construction, Inspection, and Maintenance. EM 7700-8, USDA Forest Service, Washington, DC. Chapter 5 (effective-length rules, Table 5-13).
Level	Undergraduate

Problem

A nominal 4 × 12 sawn-lumber beam (dressed 3.5 × 11.25 in) spans 14 ft on 3 in bearings under 200 lb/ft with its compression edge unbraced over the whole span, and a nominal 6 × 6 column (5.5 × 5.5 in) 10 ft long with pinned ends carries 15 kip concentrically. With illustrative reference design values F_b 1,000, F_v 180, F_c 1,500, F_c⊥ 625, E 1.7 × 10⁶ and E_min 0.62 × 10⁶ lb/in² and every adjustment factor 1.0, find the beam's bending, shear, bearing and deflection ratios and the column's allowable load.

Inputs to Enter

Beam 4 × 12 nominal (PS 20 dressed 3.5 × 11.25 in), span 14 ft, w 200 lb/ft, l_u 14 ft, NDS uniform-load effective-length rule, bearing 3 in at the ends, deflection limit L/240
Column 6 × 6 nominal (5.5 × 5.5 in), length 10 ft, K_e 1.0 (pinned), P 15 kip concentric; sawn lumber c = 0.8
F_b 1,000, F_v 180, F_c 1,500, F_c⊥ 625, E 1,700,000, E_min 620,000 psi (illustrative); C_D 1.0 (ten years), C_M, C_t, C_F, C_fu, C_i 1.0, no repetitive-member factor

Open the Timber Beam and Column tool (loaded with this example)

Answers

Beam: M = wL²/8 = 4,900 lb-ft, V = 1,400 lb, Δ = 5wL⁴/(384EI) = 0.245 in (L/240 = 0.700 in, ratio 0.350); S = 73.83 in³, f_b = 796 psi
l_e = 1.63 l_u + 3d = 307.6 in (l_u/d = 14.9 ≥ 7), R_B = √(l_e d/b²) = 16.81, F_bE = 1.20 E_min/R_B² = 2,634 psi, C_L = 0.9716 (Ylinen form with c = 0.95), F_b′ = 972 psi, bending ratio 0.820
f_v = 3V/(2bd) = 53 psi against 180 (ratio 0.296); f_c⊥ = R/(b l_b) = 133 psi against 625 (ratio 0.213)
Column: l_e/d = 21.82, F_cE = 0.822 E_min/(l_e/d)² = 1,071 psi, C_P = 0.5661 (eq. 9-28 with c = 0.8), F_c′ = 849 psi, P_allow = 25.68 kip; f_c = 496 psi, ratio 0.584; Euler on the mean E 2,937 psi, FPL fourth-power parabola 1,442 psi

Notes

The tool's default inputs are this case; press Submit. The reference design values are illustrative (the NDS Supplement tables are not shipped) and the case is the tool's own default checked in test_timber_beam_column.php (431 checks) against an independent numpy/scipy script (Ylinen's equation recovered from the stress-strain law 9-27, the closed forms against double integration) and the Timber Bridges manual's printed Examples 5-2, 5-9 and 5-10, reproduced to their digits with the pre-1991 NDS forms the manual prints.