How much weight can 2 x 4 carry?

[N.G. Reader]

How much load can be suspended on a rope tied to the midpoint of a

*double* 2 x 4 which is supported on 24" centers? Roughly. Assume rope can carry the load.

Thank you.

[Ian Stirling]

Lots. It's probably more important to spread the load evenly across the beam, and avoid twisting the beam.

If the load is evenly spread, I'd guess at least 3000Kg. What are you trying to do, there may be more important issues.

[Wade Lippman]

It has been over 30 years since my ME course, but I vaguely remember that it makes a huge difference if the beam ends are fixed or free.

[N.G. Reader]

Ian:

OK. Situation is this. I have a big garage with an "attic" space on top big

enough to store stuff. There is a pull down staircase to provide access. It is

inconvenient to carry stuff up the stairs, so I popped a floor panel and hung two

eight foot 2 x 4's from the rafters using joist hangers. Because the rafters are

at a 9/12 pitch, I had to use joist hangers intended for 2 x 12's. The joist

hangers (four of them) are screwed into the rafters with 1 5/8" drywall screws, two

on each side of the hanger. I hung a chain operated hoist (no power - I don't know

the proper name of the thing) from the double 2 x 4 between fixation points (24"

o.c.)

Proposed load might include, a lawnmower (push/self propelled), lawn chairs,

sawhorses, leaf blower, things like that. No car motors or anything like that. Only possible critcal load might include me (myself) if my nephew lends me his rock

climbing harness. Possibly.

You've got the idea. Is my beam and fixation method OK in your opinion?

Thanks

Norm

Ian Stirling wrote in news:bj2t5n$t7o$1$ [e-mail]:

[Caliban]

"N.G. Reader" wrote

What are you building?

If you are constructing anything that risks life or property, then you must use the following only as a guideline. Apply a huge factor of safety. For the final result, consult someone you will pay.

Neglecting deflection and neglecting the weight of the beam itself, one needs to consider the stress in bending and the stress in shear. The following is a crude estimate. If you use it and break someone's neck or damage property, you pay, not me. This is Usenet.

Let allowable load, lbs. = P length L, inches = 24 inches base length b = 3.5 inches (nominal base length of 4 inches) height of beam h = 3 inches (nominal height = 1.5 inches per 2 x 4) c = h/2 I = second moment of area = bh^3/12 Max. bending stress allowed in lumber for 2 x 4s is generally given to be

1000 psi Max. shear stress allowed in lumber for 2 x 4s is generally given to be 70 psi

Bending Stress Calcuations

---------------------------- Mc / I < Allowed bending stress

M = PL / 4

(PL/4) c / I < 1000

Solve for P:

P < 1000 (I/c) (4/L)

P < 1000 (bh^3/12 / (h/2) ) (4 / L)

P < 1000 (4 b h^2 / (6L))

P < 1000 4 (3.5) (3^2) / ( 6*24)

P < 875 lbs.

Shear Stress Calcuations

----------------------------

3V / (2A) < Max. allowed shear stress

V = P / 2 A = 2 *1.5 * 3.5 square inches

3 P / 2 / (2*10.5) < 70

P < 980 psi

So ignoring deflection and the weight of the beam, max. P = 874 lbs.

HOWEVER DEFLECTION MAY BE LIMITING AND NEEDS TO BE CHECKED.

Also, the condition of the wood may require further reduction of the allowable load P.

Give more details and maybe we can talk about deflection, and I'll actually do more than an off-the-top-of-my-head recollecting college strength of materials basic techniques calculation.