Given. Beam B1 spans 9.6 m total (2.4 m cantilever + 4.8 m centre span between posts P1 + 2.4 m cantilever), carrying nine factored 25 kN point loads at 1.2 m centres from end to end (Figure 3). The two loads that land exactly over the posts (at $x=2.4$ m and $x=7.2$ m) bypass the beam and go straight into the post below.
all 1.0 (part a); $K_{zc}=1.14$, $K_c=0.70$ (part b)
Find. (a) the number of 38 × 286 plies required for lintel B1; (b) whether post P1 is adequate given a 44% wind-bending stress ratio.
Figure 3 — B1/P1 elevation: nine 25 kN loads at 1.2 m o.c.; the two loads over the posts bypass the beam.
Approach. (a) Remove the two post-line loads from the beam, find the reaction at each post by symmetry, sketch the moment diagram to find the governing (hogging) moment and shear, then size the number of plies for bending and check shear. (b) Add the direct post-line load to the beam reaction for the post's total axial load, compute its factored compressive resistance, and combine the axial and given bending ratios.
Beam reactions. Removing the two loads that land on the posts leaves seven 25 kN loads on the beam ($7\times25=175$ kN), symmetric about mid-span. Each post reaction is
$$R_p=\frac{175}{2}=87.5\text{ kN}$$
Moment diagram. Cutting the beam just inside each post face, only the overhang loads (at $x=0$ and $x=1.2$ m, arms 2.4 m and 1.2 m from the post) contribute:
$$M(2.4)=-\big[25(2.4)+25(1.2)\big]=-90.0\text{ kN}\cdot\text{m}$$
By symmetry $M(7.2)=-90.0$ kN·m as well, and the interior span never returns to positive moment (mid-span $M(4.8)=-30.0$ kN·m) — the whole beam stays in hogging, so the two post sections govern.
$$\boxed{M_f=90.0\text{ kN}\cdot\text{m}}$$
The largest shear occurs immediately each side of a post, jumping from $-50.0$ kN (sum of the two overhang loads) to $+37.5$ kN as the reaction is picked up:
$$V_f=50.0\text{ kN}$$
Size the lintel for bending. A single 38 × 286 ply has section modulus and resistance-per-ply
$$S_1=\frac{38(286)^2}{6}=518{,}041\text{ mm}^3, \qquad M_{r,1}=\phi_b f_b S_1=0.9(10.0)(518{,}041)\times10^{-6}=4.66\text{ kN}\cdot\text{m}$$
For $n$ plies, $M_r=n\,M_{r,1}$ must reach $M_f=90.0$ kN·m, i.e. $n\ge90.0/4.66=19.3$, so $n=20$ (19 plies gives $M_r=88.6<90.0$, insufficient):
$$M_{r,20}=20(4.66)=93.25\text{ kN}\cdot\text{m} \ge 90.0\text{ kN}\cdot\text{m}\quad\checkmark$$
Check shear at n = 20.
$$V_{r,20}=\phi_v f_v\left(\tfrac{2}{3}\right)(20\,A_1)=0.9(1.9)\left(\tfrac23\right)\big(20\times38\times286\big)\times10^{-3}=99.1\text{ kN} \ge 50.0\text{ kN}\quad\checkmark$$
Bending governs; shear has ample reserve.
$$\boxed{\text{B1 = 20-ply built-up } 38\times286 \text{ SPF No.1}}$$
Post P1 axial load. Each post carries its beam reaction plus the load applied directly over it:
$$P_f=R_p+P=87.5+25.0=112.5\text{ kN}$$
Post P1 compressive resistance. With full cross-section $A=140(185)=25{,}900\text{ mm}^2$ and the given $K_{zc}=1.14$, $K_c=0.70$:
$$P_r=\phi_c f_c A\,K_{zc}K_c=0.8(8.7)(25{,}900)(1.14)(0.70)\times10^{-3}=143.85\text{ kN}$$
Combined axial-bending interaction. The axial utilization ratio is
$$\frac{P_f}{P_r}=\frac{112.5}{143.85}=0.782$$
Adding the given wind-bending stress ratio of 0.44:
$$\boxed{\frac{P_f}{P_r}+\left(\frac{M_f}{M_r}\right)_{\text{wind}}=0.782+0.44=1.222 > 1.0}$$
The combined ratio exceeds unity, so post P1 is not structurally adequate for the combined axial and wind-bending demand at this size — a larger post (e.g. 140×235) or a reduced unbraced/effective length to raise $K_{zc}$ would be needed.
