Discussion
The deflection of a structure under loading is measured at five different values of the force applied X in kilo Newtons (kN). The deflection Y is in centimetres. Data are given in the following table. Employing Lagrange's interpolation with quadratic polynomial, compute the extrapolated value of the deflection Y for zero load (X = 0 kN). How many points are used ?
| X (kN) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |
| Y (cm) | 3.0 | 3.9 | 5.2 | 7.3 | 10.5 |
Answer: B
The extrapolated deflection is 2.50 cm, using 3 points.
A quadratic needs three points. Using the three nearest X = 0, namely (0.5, 3.0), (1.0, 3.9) and (1.5, 5.2), the Lagrange coefficients at X = 0 are:
\[ \begin{aligned} L_{0} &= \frac{(0 - 1)(0 - 1.5)}{(0.5 - 1)(0.5 - 1.5)} = 3 \\ L_{1} &= \frac{(0 - 0.5)(0 - 1.5)}{(1 - 0.5)(1 - 1.5)} = -3 \\ L_{2} &= \frac{(0 - 0.5)(0 - 1)}{(1.5 - 0.5)(1.5 - 1)} = 1 \end{aligned} \]
\[ \begin{aligned} Y(0) &= 3(3.0) - 3(3.9) + 1(5.2) \\ &= 9.0 - 11.7 + 5.2 = 2.50 \end{aligned} \]
Two points give only a straight line, and four or five points would give a cubic or quartic rather than the quadratic asked for.
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