A timber test specimen of size 50 × 50 mm in cross-section and 150 mm in length was tested for its specific gravity. If the specimen weighs 250 g and has 15% of moisture content, calculate its specific gravity by accounting the moisture content.

⚖️ Specific Gravity and Moisture Content in Timber

The specific gravity of timber is a key indicator of its density and strength. It is defined as the ratio of the oven-dry weight of the wood to the weight of an equal volume of water. However, timber in its natural state contains moisture, which adds to its weight.

To calculate the true specific gravity, we must use the oven-dry weight of the timber. Since the question provides the weight at a specific moisture content (15%), we first need to calculate what the weight would be with 0% moisture before finding the specific gravity.

📝 Step-by-Step Calculation

1. List the Given Data:

• Cross-section = $50 \text{ mm} \times 50 \text{ mm}$
• Length = $150 \text{ mm}$
• Weight at 15% moisture $(W_i)$ = $250 \text{ g}$
• Moisture content $(m)$ = $15\%$

2. Calculate the Volume of the Specimen $(V_i)$:

$$\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}$$ $$\text{Volume} = 150 \text{ mm} \times 50 \text{ mm} \times 50 \text{ mm} = 375,000 \text{ mm}^3$$

To convert $\text{mm}^3$ to $\text{cm}^3$ (since $1 \text{ cm} = 10 \text{ mm}$, therefore $1 \text{ cm}^3 = 1000 \text{ mm}^3$):

$$V_i = \frac{375,000}{1000} = 375 \text{ cm}^3$$

3. Calculate the Specific Gravity:

The standard formula for specific gravity is:

$$\text{Specific Gravity} = \frac{\text{Oven-Dry Weight}}{\text{Volume} \times \text{Density of Water}}$$

Since the density of water is $1 \text{ g/cm}^3$, the formula simplifies to:

$$\text{Specific Gravity} = \frac{W_{\text{dry}}}{V_i}$$

First, we find the oven-dry weight $(W_{\text{dry}})$ from the wet weight $(W_i)$:

$$W_{\text{dry}} = \frac{W_i}{1 + \frac{m}{100}} = \frac{250}{1 + \frac{15}{100}} = \frac{250}{1.15} \approx 217.39 \text{ g}$$

Now, calculate the specific gravity:

$$\text{Specific Gravity} = \frac{217.39 \text{ g}}{375 \text{ cm}^3} \approx \mathbf{0.579}$$

📊 Calculation Summary

Step	Calculation	Result
Calculate Volume	$50 \times 50 \times 150 \text{ mm}^3$	$375 \text{ cm}^3$
Calculate Wet Density	$\frac{250 \text{ g}}{375 \text{ cm}^3}$	$0.667 \text{ g/cm}^3$
Calculate Oven-Dry Weight	$\frac{250 \text{ g}}{(1 + 0.15)}$	$\approx 217.4 \text{ g}$
Calculate Specific Gravity	$\frac{217.4 \text{ g}}{375 \text{ cm}^3}$	$\approx 0.579$

💡 Study Tips

• Specific Gravity uses Dry Weight: The most critical concept to remember is that specific gravity is based on the oven-dry (0% moisture) weight of the timber.
• Don't Use Wet Density: A common mistake is to simply divide the given weight by the volume ($250 / 375 = 0.667$). This gives you the density at 15% moisture, not the specific gravity. Notice that this is one of the incorrect options!
• Master the Moisture Formula: The relationship $W_{\text{wet}} = W_{\text{dry}} \times (1 + m/100)$ is fundamental. You must be able to rearrange it to find the dry weight from the wet weight.
• Unit Conversion: Always pay attention to units. Convert $\text{mm}^3$ to $\text{cm}^3$ when needed by dividing by 1000.
• Double-Check Calculations: Use both the step-by-step method and the combined formula to verify your answer.