8/15/2014 The maximum safe load for a horizontal beam varies jointly as the width of the beam and the square of the thickness of the beam, and inversely with its length. How will the thickness of the beam have changed if the maximum safe load were halved while the width of the beam was doubled, and its length increased by a factor of 3?

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8/15/2014 Solution. Let S denote the maximum safe load for the beam, w, the width of the beam, T, its thickness and l, its length.
Then, S = kwT²/l where k is the constant of proportionality (*)

Initially, let w = S = l = 1unit.

Then, T = 1/√k [Substituting w = S = l = 1 in (*) and solving]

Next, as indicated, let S = ½, w = 2 and l = 3.

Then, T = √(3/4k) [Substituting these values in (*) and solving]

= ½ √(3/k)

T = ≈0.85/√k [√3 ≈ 1.7]

Therefore, the thickness would have dropped by 15%.