Why are the 'outside' Table 1 stresses used for the column in an open shed?

Because an open shed is exposed to weather, so it is an outside location

Because the column is Group C timber

S/d = 17.8 and K8 = 21.65. How is the column classified and which fc formula is used?

Intermediate (11 < S/d < K8); fc = fcp[1 - (1/3)(S/d / K8)^4]

Intermediate; fc = 0.329E/(S/d)^2

With 225 x 225 mm and fc = 5.85 N/mm^2, what is the conclusion?

P = 296 kN > 250 kN, but S/d exceeds 50 so it fails

P = 350 kN, over-designed and rejected

2076 Baisakh · 6 marks

Square deodar column 4 m, 250 kN, open shed

Short answer

Solid square Deodar column per IS 883:1994. Trial 225×225mm (A=50,625mm²). S/d=17.8, K8=21.65, intermediate column; fc=fcp[1-⅓(S/d/K8)⁴]=5.85 N/mm². Capacity P=fc·A=296 kN > 250 kN. Safe. Grade I, outside location, pin-ended, S/d<50 satisfied.

Design a square Deoder wood column of length 4m to be used in an open shed, to carry an axial load of 250kN.

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 $S$ (cl. 7.6.1.5).

Data and permissible stresses

Length $S = 4.00$ m = 4000 mm. Axial load $N = 250$ kN.

Species: Deodar (Cedrus deodara), Group C. Location: outside. Grade: Grade I (standard) (Table 1 values are for Grade I).

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

Property	Table 1 value (N/mm²)	Used (N/mm²)
Bending / tension along grain $f_b$	8.7	8.70
Compression parallel to grain $f_{cp}$	6.9	6.90
Compression perpendicular to grain $f_{cn}$	2.1	2.10
Horizontal shear $H$	0.7	0.70
Modulus of elasticity $E$	9480	9480

An open shed is an outside location. The grade is not stated, so Grade I (standard) is taken.

Trial section

Try 225 mm × 225 mm (square). Area $A = 50,625$ mm²; least dimension $d = 225$ mm.

Slenderness and permissible axial stress (cl. 7.6.1)

\begin{aligned} S/d &= \frac{4000}{225} = 17.8 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{9480}{6.90}} = 21.65 \end{aligned}

$11 < S/d = 17.8 < K_8 = 21.6$ , so it is an intermediate column:

f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 6.90\left[1 - \frac{1}{3}\left(\frac{17.8}{21.6}\right)^4\right] = 5.85\ \text{N/mm}^2

$S/d = 17.8 < 50$ (cl. 7.6.1.4). OK.

Check

P = f_cA = 5.85\times50,625/10^3 = 296.3\ \text{kN} \ge N = 250\ \text{kN}

Safe.

Answer

Provide a solid Deodar column of 225 mm × 225 mm cross-section (Grade I (standard), outside location). Capacity $296$ kN $> 250$ kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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