Known :
Q = 300 L/s = 0.3 m³/s
D1 = 350 mm = 0.35 m
D2 = 700 mm = 0.7 m
g = 9.81 m/s²
Solution :
A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²
A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²
hL = (kL / 2g) • (U1² - U2²)
hL = (kL / 2g) • Q² (1/A1² - 1/A2²)
hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))
hL = 0.467 m
The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correctly represented by F/A - (My)/(Iz).
<h3>What is the distribution of pressure at some stage in bending?</h3>
Compressive and tensile forces expand withinside the path of the beam axis beneath neath bending loads. These forces set off stresses at the beam. The most compressive pressure is observed on the uppermost fringe of the beam whilst the most tensile pressure is positioned on the decrease fringe of the beam.
The bending pressure is computed for the rail through the equation Sb = Mc/I, wherein Sb is the bending pressure in kilos in keeping with rectangular inch, M is the most bending second in pound-inches, I is the instant of inertia of the rail in (inches)4, and c is the space in inches from the bottom of rail to its impartial axis.
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Answer:
decreases
Explanation:
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