Nội dung text PSAD 16 - Double Integration Method.pdf
To compensate, an additional load continuing the original will be added to the right side of the beam as shown. An equal load in the opposite direction will also be added to not alter the original effects. From this, it can be seen that the figure becomes complicated. In situations like this, it is easier to base x as the distance from the right end. From similar triangles, y1 10 − 4 − 〈x − 4〉 = 4 6 y1 6 − 〈x − 4〉 = 2 3 Express in terms of ⟨x − 4⟩: y1 = 2 3 (6 − ⟨x − 4⟩) = 4 − 2 3 ⟨x − 4⟩ y2 = 4 − y1 y2 = 4 − [4 − 2 3 ⟨x − 4⟩] y2 = 2 3 ⟨x − 4⟩
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