Discussion
In a three-point bending mode for flexural test of a polymer sample with the following data:
load applied in the mid-span = 80 kg
width of the specimen = 3 cm
depth of the specimen = 2 cm
length of the specimen = 30 cm
the flexural strength will be
Answer: B
The flexural strength is about 29.4 MPa.
For three-point bending of a rectangular bar, the maximum stress at mid-span is \(\sigma = 3FL/(2bd^{2})\), with the 30 cm length taken as the span and the 80 kg load as a force of 784.8 N:
\[ \begin{aligned} \sigma &= \frac{3(784.8)(0.30)}{2(0.03)(0.02)^{2}} \\ &= \frac{706.3}{2.4 \times 10^{-5}} = 2.94 \times 10^{7}\ \text{Pa} \end{aligned} \]
that is 29.4 MPa. Using a rectangular section with width and depth swapped gives about 19.6 MPa, and forgetting to convert kilograms to newtons gives 3 MPa; neither is offered, and the other options do not follow from the formula.
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