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Select Sal column, 200 kN + 30 kNm

Short answer

Provide solid Sal column 200 mm × 275 mm (Select grade, inside). Permissible axial stress 10.12 N/mm², bending 19.60 N/mm²; interaction ratio 0.97 < 1, safe.

Design a Select grade Sal wood rectangular column having height 3.2m and subjected to axial compressive load of 200 kN and bending moment 30 kNm.

Approach. Working stress design to IS 883:1994 (cl. 7.6 for columns, cl. 7.7 for combined axial compression and bending). Ends are taken as pin-ended, so the effective length equals the unsupported length SS (cl. 7.6.1.5).

Data and permissible stresses

Length S=3.20S = 3.20 m = 3200 mm. Axial load N=200N = 200 kN, bending moment M=30M = 30 kN·m.

Species: Sal (Shorea robusta), Group A. Location: inside. Grade: Select grade (Grade I values multiplied by 1.16, IS 883:1994 cl. 6.3).

Permissible stresses (IS 883:1994 Table 1, inside location):

PropertyTable 1 value (N/mm²)Used (N/mm²)
Bending / tension along grain fbf_b16.919.60
Compression parallel to grain fcpf_{cp}10.612.30
Compression perpendicular to grain fcnf_{cn}4.65.34
Horizontal shear HH0.941.09
Modulus of elasticity EE1267012670

Location is not stated, so inside is taken. The bending moment acts about the major axis (depth dd); buckling is checked about the minor axis (least dimension bb).

Trial section

Try 200 mm × 275 mm. Area A=55,000A = 55,000 mm²; least dimension d=200d = 200 mm.

Slenderness and permissible axial stress (cl. 7.6.1)

S/d=3200200=16.0K8=0.584Efcp=0.5841267012.30=18.75\begin{aligned} S/d &= \frac{3200}{200} = 16.0 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{12.30}} = 18.75 \end{aligned}

11<S/d=16.0<K8=18.711 < S/d = 16.0 < K_8 = 18.7, so it is an intermediate column:

fc=fcp[1−13(S/dK8)4]=12.30[1−13(16.018.7)4]=10.12 N/mm2f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 12.30\left[1 - \frac{1}{3}\left(\frac{16.0}{18.7}\right)^4\right] = 10.12\ \text{N/mm}^2

S/d=16.0<50S/d = 16.0 < 50 (cl. 7.6.1.4). OK.

Combined axial compression and bending (cl. 7.7.1)

fac=NA=20000055,000=3.636 N/mm2fab=6Mbd2 (bending about the strong axis)=6×30000000200×2752=11.901 N/mm2\begin{aligned} f_{ac} &= \frac{N}{A} = \frac{200000}{55,000} = 3.636\ \text{N/mm}^2 \\ f_{ab} &= \frac{6M}{bd^2}\ \text{(bending about the strong axis)} = \frac{6\times30000000}{200\times275^2} = 11.901\ \text{N/mm}^2 \end{aligned}

Permissible bending stress fb=19.60f_b = 19.60 N/mm²; the form factor is not applied for depths up to 300 mm (cl. 7.5.4).

facfc+fabfb=3.63610.12+11.90119.60=0.966≤1.0\frac{f_{ac}}{f_c} + \frac{f_{ab}}{f_b} = \frac{3.636}{10.12} + \frac{11.901}{19.60} = 0.966 \le 1.0

Safe.

Answer

Provide a solid Sal column of 200 mm × 275 mm cross-section (Select grade, inside location). Combined stress ratio =0.97<1= 0.97 < 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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