The maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.

Question

The maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam. Determine how each change affects the beam's maximum load. (Write your answers in terms of M₀,

 the original maximum load.)

(a) tripling the width

The maximum safe load is now

M₀.

(b) doubling the depth

The maximum safe load is now

M₀.

(c) halving the length

The maximum safe load is now

M₀.

(d) halving the width and doubling the length

The maximum safe load is now

M₀.

Answer 1

Max = M = kwd^2/L

triple L and M then = kwd^2/3L = 1/3 the original Max

double d and M then = kw(4)(d^2)/L = 4 times the original Max

half the Length then M = kwd^2/(L/2) = 2kwd^2/L = twice original Max

half width, double Length, then M =k(w/2)(d^2)/2L = kwd^2/4L = 1/4 the original Max