Reduction of a simple distributed loading problem

Renan · Jun 11, 2018

Two textbook diagrams of a horizontal shaft with a distributed load: (a) a parabolic load w = (60 x^2) N/m over 2 m, total 240 N/m at the end; (b) a resultant force F_R = 160 N at x-bar = 1.5 m, with worked integration solution for Example 4.21.

Handwritten solution to a distributed load problem: load w = (0.5x^3) N/m over a 10 m platform, integrated to give resultant force F_R = 1.25 kN and centroid x̄ = 8.00 m, with a 500 N/m peak load shown at the right end.

Diagram of a parabolic area under a curve, shaded region bounded by a curve rising to the right, with height b on the left, base a, A = (1/3)ab labeled, centroid C, and dimensions 3/4 a and 3/10 b marked.

Why can't we use the NOTE method (exercise 4.21) on the exercise 4-156 ?

A = ab/3 = (10.500)/3 = 1666,66 N --> this not right

kuruman · Jun 11, 2018

Renan said:

View attachment 226820
Why can't we use the NOTE method (exercise 4.21) on the exercise 4-156 ?

Because the location function is different. In one case is goes as ##x^2## and in the other as ##x^3##. If you do the integral in each case, you will get a different answer.

Reduction of a simple distributed loading problem

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