Question

EXAMPLE 8.3 Each simply supported floor joist shown in the photo is subjected to a uniform design loading of 4 kN/m, Fig. 8-12a. Determine the maximum deflection of the joist. El is constant. 4 kN/m 10 20 kN 20 kN (4 x) N Elastic Curve. Due to symmetry, the joist's maximum deflection will occur at its center. Only a single x coordinate is needed to determine the internal moment. Moment Function. From the free-body diagram, Fig. 8-12b, we have 20 kN Slope and Elastic Curve. Applying Eq. 8-4 and integrating twice gives Fig. 8-12 E1_ = 20x-2x2 dx2 El-= 10x2-06667x3 + Cl dx Here u = 0 at x = 0 so that C2 = 0, and u-0 at x = 10, so that C1166.7. The equation of the elastic curve is therefore Elv-3.333x30.1667x4- 166.7x At x = 5 m. note that du/dr =0. The maximum deflection is therefore 521 EI Vmax = Ans

Question: How to find boundaries? (the stuff in a red box) like: why is x=10?

Answer 1

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