Question
Download Solution PDFA symmetrical two-hinged parabolic arch when subjected to a uniformly distributed load on the entire horizontal span, is subjected to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFIn particular, if the 2-hinge arch has a parabolic shape and it is subjected to a uniform horizontally distributed vertical load, then only compressive forces will be resisted by the arch. Under these conditions, the arch shape is called a funicular arch because no bending or shear forces occur within the arch.
Bending moment, Mx = 0
Radial shear, Sx = 0
Normal thrust, Nx ≠ 0
Important Point:
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Arches |
Horizontal Thrust |
|
Semicircular arch subjected to concentrated load (W) at Crown |
\(H = \left( {\frac{W}{\pi}} \right)\) |
|
Semicircular arch subjected to UDL (w/per unit length) over the entire span |
\(H = \frac{4}{3}\left( {\frac{{wR}}{\pi }} \right)\) |
|
Two hinged semi-circular arches subjected to uniformly varying load |
\(\rm{H = \frac{2}{3}\times \frac{wR}{\pi}}\) |
|
Parabolic arch subjected to concentrated load (W) at Crown |
\(H = \frac{{25}}{{128}}\left( {\frac{{WL}}{h}} \right)\) |
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Parabolic arch subjected to UDL (w/per unit length) over entire span |
\(\rm{H = \left( {\frac{{w{L^2}}}{{8h}}} \right)}\) |
|
Two hinged parabolic arch subjected to uniformly distributed load on the half span |
\(\rm{H = \left( {\frac{{w{L^2}}}{{16h}}} \right)}\) |
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Two hinged parabolic arch subjected to varying distributed load on the full span |
\(\rm{H = \left( {\frac{{w{L^2}}}{{16h}}} \right)}\) |
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