Question
Download Solution PDFA symmetrical parabolic arch of span 19 m and rise 4 m is hinged at the springings. It supports a uniformly distributed load of 1.5 tonnes per meter run of the span. The horizontal thrust in tonnes at each of the springing is -
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Horizontal thrust at springings A and B = \(\frac{{w{L^2}}}{{8h}}\)
Where, L = 19 m
h = 4 m
\({H_A} = \frac{{w{L^2}}}{{8h}}\)
\({H_A} = \frac{{1.5 \times {{19}^2}}}{{8 \times 4}}\)
HA = 16.92 tonnes
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