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NACA TN 2661 example 1: thin-web beam I-40-4Da in incomplete diagonal tension (Kuhn's engineering theory)
| Tool | Diagonal Tension Web |
| Source | Kuhn, P., Peterson, J. P., and Levin, L. R. (1952). A Summary of Diagonal Tension. Part I: Methods of Analysis. NACA Technical Note 2661, National Advisory Committee for Aeronautics, Washington, DC. Section 7, Example 1 (test beam I-40-4Da of Part II, NACA TN 2662, Tables 1 and 2). |
| Level | Professional |
Problem
A 24S-T3 web 0.0390 in thick and 41.4 in deep between flange centroids (clear depth 37.1 in) is stiffened by double angle uprights 0.750 x 0.625 x 0.125 in (area 0.353 in², radius of gyration 0.351 in, length 38.6 in between rivet centroids) at a 20.0 in pitch (clear 19.37 in), the web sandwiched between the steel flange angles, flange-flexibility factor 1.20. With the test failing shear of 30.3 kips as the design ultimate load, find the web buckling stress, the diagonal-tension factor, the upright stresses, the allowable upright stresses for column failure (rivet strength 4.65 kips against the required value) and forced crippling, and the peak web stress against the allowable of Figure 19.
Inputs to Enter
- h_e 41.4 in, h_U 38.6 in, h_c 37.1 in, d 20.0 in, d_c 19.37 in, t 0.0390 in; double uprights A_U 0.353 in², ρ 0.351 in, t_U 0.125 in, b 0.625 in, Q 0.0388 in³, R 4.65 kips
- Flanges: two angles, t_F 0.3125 in, A_F 3.55 in², E_F 29,000 ksi, ωd 1.20 entered; 24S-T3 curves, E 10,600 ksi, F_tu 62 ksi, F_cy 40 ksi; bolts snug with washers / web sandwiched (basic allowables)
- Web shear S_w 30.3 kips at ultimate
Answers
- k_ss 5.98, R_h = R_d = 1.62 → τ_cr,elastic = 0.416 ksi (book 0.416; its text reads k_ss as 5.92), elastic; τ = 30.3/(41.4 × 0.039) = 18.8 ksi, τ/τ_cr = 45.1 (book 45.2), k = tanh(0.5 log 45.1) = 0.679 (book 0.680)
- Figure 14 iteration: tan α = 0.814 (book 0.81), σ_U/τ = 0.902 (book 0.90), σ_U = 16.9 ksi; Figure 15 ratio 1.143 (book 1.14), σ_U,max = 19.3 ksi (book 19.2)
- Column: L_e = 38.6/√(1 + 0.680²(3 − 2 × 0.519)) = 28.0 in, L_e/ρ = 79.7 (book 79.8), Euler 16.5 ksi, R_R = 8.57 kips (book 8.56), R/R_R = 0.543, Figure 21 factor 0.937 (book 0.94) → allowable 15.4 ksi (book 15.5), M.S. −0.09 (book: 8 % conservative at the failing load, and the beam failed by column action)
- Forced crippling σ_0 = 21 × 0.680^(2/3) × 3.20^(1/3) = 23.9 ksi (book 24.0), 24 % above σ_U,max (book 25 %); web τ′_max = 18.8 × 1.01 × 1.01 = 19.05 ksi (book 19.2) against 21.9 ksi (book 22.0), 15 % reserve (book 15 %); predicted failing loads P_C 27.9 kips (Part II Table 2: 28.5), P_FC 36.9 kips (37.4)
Notes
The tool's default inputs are this beam; press Submit. Every printed value of examples 1 and 2 of TN 2661 (beam V-12-10S: τ_cr 16.1 ksi, k 0.123, σ_U 6.48 ksi, σ_U,max 8.34 ksi, σ_0 7.0 ksi, allowable 25 ksi against 28.56 ksi), thirteen k values and five series V beams of the TN 2662 test tables and the numpy script References/Diagonal-Tension/checks/dtension_check.py are checked in test_diagonal_tension.php. The report's chart readings carry 1 to 3 % reading tolerance; the design curves are digitized from the report.