16-Civ-A2 Elementary Structural Design · May 2013

Question 7 of 7: C1 — Adequacy Check of a Sawn Timber Floor Beam

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Question 7: C1 — Adequacy Check of a Sawn Timber Floor Beam (10 + 6 + 4 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Sawn timber beam 241 mm × 343 mm, No.1 grade Douglas Fir-Larch, simply supported over 5.0 m, spaced at 2.5 m on centre. Specified loads: dead 2.0 kPa (self-weight included), live 2.5 kPa. Deflection limits $L/180$ under total load and $L/360$ under live load. Dry service, normal-term loading, untreated: $K_D = K_H = K_{Sb} = K_T = 1.0$; $\phi = 0.9$.

Find. Whether the member satisfies bending, shear and both deflection criteria.

Approach. Convert the area loads to a line load on the tributary width, compute the factored moment and shear and the specified-load deflections, then compare each against the O86 factored resistances and the stated deflection limits.

1. Line loads on one beam. The tributary width is the 2.5 m beam spacing: $$w_D=2.0(2.5)=5.0\ \text{kN/m},\qquad w_L=2.5(2.5)=6.25\ \text{kN/m}$$ $$w_f=1.25(5.0)+1.5(6.25)=6.25+9.375=15.6\ \text{kN/m}$$
2. Factored effects on a 5.0 m simple span. $$M_f=\frac{w_fL^2}{8}=\frac{15.625(5.0)^2}{8}=48.8\ \text{kN}\cdot\text{m},\qquad V_f=\frac{w_fL}{2}=\frac{15.625(5.0)}{2}=39.1\ \text{kN}$$
3. Section properties. With $b = 241$ mm and $d = 343$ mm, $$S=\frac{bd^2}{6}=\frac{241(343)^2}{6}=4.73\times10^6\ \text{mm}^3,\qquad A=bd=82\,663\ \text{mm}^2$$ $$I=\frac{bd^3}{12}=\frac{241(343)^3}{12}=810\times10^6\ \text{mm}^4$$
4. Bending resistance. The size factor for a beam-and-stringer section deeper than 305 mm is $K_{Zb} = (305/d)^{1/9} = (305/343)^{1/9} = 0.987$, and with the deck providing continuous lateral restraint $K_L = 1.0$: $$M_r=\phi F_bSK_{Zb}K_L,\qquad F_b=f_b(K_DK_HK_{Sb}K_T)=15.8\ \text{MPa}$$ $$M_r=0.9(15.8)(4.73\times10^6)(0.987)(1.0)=66.3\ \text{kN}\cdot\text{m}$$ $$\boxed{M_f=48.8\ \text{kN}\cdot\text{m}\;<\;M_r=66.3\ \text{kN}\cdot\text{m}\quad\checkmark\ (74\ \%)}$$
5. Shear resistance. For a sawn member without notches, O86 works with two thirds of the gross area: $$V_r=\phi F_v\left(\frac{2A}{3}\right)=0.9(1.5)\left(\frac{2(82\,663)}{3}\right)=74.4\ \text{kN}$$ $$V_f=39.1\ \text{kN}\;<\;74.4\ \text{kN}\quad\checkmark\ (53\ \%)$$ The factor of two-thirds reflects the parabolic distribution of horizontal shear stress across a rectangular section, whose peak is 1.5 times the average.
6. Deflection under total specified load. Deflections are checked at service, so the specified (unfactored) loads are used: $w = 5.0+6.25 = 11.25$ kN/m $= 11.25$ N/mm. $$\Delta_{tot}=\frac{5wL^4}{384EI}=\frac{5(11.25)(5\,000)^4}{384(12\,000)(810\times10^6)}=9.41\ \text{mm}$$ $$\text{limit}=\frac{L}{180}=\frac{5\,000}{180}=27.8\ \text{mm}\quad\checkmark$$
7. Deflection under live load only. Scaling by the load ratio, $$\Delta_{live}=9.41\left(\frac{6.25}{11.25}\right)=5.23\ \text{mm},\qquad \text{limit}=\frac{L}{360}=13.9\ \text{mm}\quad\checkmark$$
8. Verdict. Every criterion is met, with bending at 74 % the governing check: $$\boxed{\text{The 241}\times\text{343 mm No.1 D.Fir-L beam is SATISFACTORY}}$$ The generous deflection margin — barely a third of the permitted total-load value — shows that this member is strength-governed rather than stiffness-governed, which is unusual for a timber floor beam and is a direct consequence of the short 5.0 m span.