求无限跨连续梁跨中弯矩方程？ 10

The moment equation for an infinite span continuous beam can be derived using the principles of mechanics and calculus. Assuming that the beam is loaded with distributed loads, the moment at any point along the beam can be expressed as the sum of the moments due to the distributed loads and the moments due to any concentrated loads. Let's consider a beam of length L, supported at both ends, and loaded with a distributed load w(x) per unit length. The moment at any point x along the beam can be expressed as: M(x) = -∫(L/2 to x) w(x) * (x - ξ) dξ - ∫(x to L/2) w(x) * (ξ - x) dξ where w(x) is the distributed load at a distance x from one end of the beam, and ξ is a variable representing a distance from the same end of the beam. The first term in the equation represents the moment due to the portion of the distributed load to the left of the point x, and the second term represents the moment due to the portion of the distributed load to the right of the point x. This equation can be simplified by integrating each term and evaluating the constants of integration using the boundary conditions at the ends of the beam. The resulting equation for the moment M(x) at any point x along the beam is: M(x) = -(w0 L^2)/8 - (wx/2)(L - 2x) where w0 is the total distributed load over the entire beam and wx is the distributed load at a distance x from one end of the beam. This equation shows that the moment at any point along an infinite span continuous beam is a function of the distributed load, the distance from the supports, and the length of the beam.

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