Beam Load

Section 1.10
Question 74

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.

(a) Doubling the width
(b) Doubling the depth
(c) Halving the length
(d) Halving the width and doubling the length

Can you please do part (a) in step by step fashion as a guide for me to do the rest?

Let M = maximum load
Let k = constant of proportionality
Let w = width of the beam
Let d = depth of beam

I think the formula needed here is M = kwd.

Yes?

This question completes Chapter 1.

 

Let M = maximum load
Let k = constant of proportionality
Let w = width of the beam
Let d = depth of beam
you forgot the length =L

I think the formula needed here is M = kwd.=>the square of its depth (d^2) and include the length
maximum load varies jointly as the width and the square of its depth (d^2) =>M = kwd^2
and inversely as the length L of the beam=>M = kwd^2/L


(a) Doubling the width=>2w
M = k*2wd^2/L
M = 2kwd^2/L->maximum load is doubled

 

Thank you. The needed formula is M = kwd^2/L as you stated.

Part (b)

Doubling the depth. So, I must include (2d^2) in the formula.

M = kw(2d^2)/L

Part (c)

To half the length L means to multiply by 1/2.

M = kwd^2/(L/2)

Part (d)

Having the width = (w/2).

Doubling the length = 2L.

M = k(w/2)d^2/(2L)

Parts b through d correct. Yes?

We are done with Chapter 1. I will skip the chapter review and chapter test. We move on to Chapter 2, Section 2 1.