Question Bank • QB_SWCE_ADV_133 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Surveying and Levelling

Subtopic: Measurement of distance and area

Question Statement:

A $30.0\text{ m}$ steel tape has a cross-sectional area of $A = 4.0\text{ mm}^2$ and Young's modulus of elasticity $E = 2.0 \times 10^5\text{ N/mm}^2$. The tape was standardized under a pull of $P_0 = 60.0\text{ N}$. During a field measurement, a tensile pull of $P = 150.0\text{ N}$ was applied to the tape. The tension (pull) correction $C_p$ in millimeters ($\text{mm}$) is ________ (round off to two decimal places).

Answer Key:

3.38

Step-by-Step Solution & Explanation

1. Pull / tension correction formula: $C_p = \frac{(P - P_0) L}{A E}$ 2. Substituting given values: - $P - P_0 = 150.0 - 60.0 = 90.0\text{ N}$ - $L = 30.0\text{ m} = 30,000\text{ mm}$ - $A = 4.0\text{ mm}^2$ - $E = 2.0 \times 10^5\text{ N/mm}^2$ $C_p = \frac{90.0\text{ N} \times 30,000\text{ mm}}{4.0\text{ mm}^2 \times (2.0 \times 10^5\text{ N/mm}^2)} = \frac{2,700,000}{800,000} = 3.375\text{ mm} \approx 3.38\text{ mm}$ Since applied pull exceeds standardization pull, the correction is positive (+).
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