What relation holds between stresses in steel and timber?

sigma_s = m sigma_t with m = E_s/E_t = 25

Why does the steel plate govern?

Timber at 6 N/mm^2 would put steel at 150 > 130 N/mm^2, exceeding its limit

Timber has higher permissible stress

How is the permissible load computed?

sigma_s A_s + sigma_t A_t using 130 and 5.2

How is the elongation found?

delta = (sigma_s/E_s) L = 2.476 mm

delta = 6 x 4000 / 8400 mm using the full 6

2078 Kartik · 8 marks

Timber-steel plate composite member: permissible load

Short answer

The steel plate governs (130 N/mm²) with timber at5.2 N/mm². Permissible load =234.0 kN; elongation =2.476 mm.

Two $150\ \text{mm}\times75\ \text{mm}\times4\ \text{m}$ long timber members are reinforced with a steel plate $150\ \text{mm}\times6\ \text{mm}\times4\ \text{m}$ long as shown in figure. The three members are adequately bolted together. The permissible stresses for the timber and the steel members are $6\ \text{N/mm}^2$ and $130\ \text{N/mm}^2$ respectively. E for timber is $8.4\ \text{GN/m}^2$ and for steel is $210\ \text{GN/m}^2$ . Calculate the permissible tensile load for the composite member and the amount of elongation due to this load. [Figure: cross-section 150 mm deep: timber 75 mm, steel plate 6 mm, timber 75 mm side by side.]

The members are bolted together, so they extend equally: $\varepsilon_t = \varepsilon_s$ and $\sigma_s/E_s = \sigma_t/E_t$ .

m = \frac{E_s}{E_t} = \frac{210}{8.4} = 25.0,\qquad \sigma_s = m\,\sigma_t

Areas

A_t = 2\times(150\times75) = 22500\ \text{mm}^2,\qquad A_s = 150\times6 = 900\ \text{mm}^2

Which material governs?

If timber reaches 6 N/mm², steel would be at $25\times6 = 150$ N/mm² $> 130$ N/mm² (not permitted).
If steel reaches 130 N/mm², timber is at $130/25 = 5.2$ N/mm² $< 6$ N/mm² (safe).

Hence the steel plate governs: $\sigma_s = 130$ N/mm² and $\sigma_t = 5.2$ N/mm².

Permissible load

P = \sigma_s A_s + \sigma_t A_t = 130\times900 + 5.2\times22500 = 117000 + 117000 = 234000\ \text{N} = 234.0\ \text{kN}

Elongation (L = 4 m = 4000 mm)

\delta = \frac{\sigma_s}{E_s}L = \frac{130}{210\times10^3}\times4000 = 2.476\ \text{mm}

Answer: permissible tensile load = 234.0 kN; elongation = 2.476 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Strength of Materials chapter 3 questions →

Timber-steel plate composite member: permissible load

Areas

Which material governs?

Permissible load

Elongation (L = 4 m = 4000 mm)

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