Steel Joist

Steel Joist Elastic Properties By William Greco z Copyright 2009 William Greco Trident3 [email protected] Normally the moment of inertia is not given for Steel Joists in vendor catalog information, making it difficult to calculate any other cases except uniformly distributed loads. the following procedure can derive this variables formula for any given steel joist. The uniform loads at given spans are usually given as pounds per foot that will cause A 1/360 deflection at center span, this allows us to insert the 1/360 deflection into the equation and simply solve for moment of inertia. After obtaining the moment of inertia the section modulus which is simply the moment of inertia divided by the distance of the most remote fiber from the neutral axis (assumed in this case to be the center of the xx axis of the joist) can be obtained. Deflection equation for uniformly distributed loads…… 4 5. w. l max 384. E. I solving for I = 4 5. L . w max 384. E. I

if a=b then b=a and put terms in order

4 5. L . w max. I . 384 E

max. I

384. E

4 5. L . w I 384. max. E

Example-1:

multiply by I

4 5. L . w if b=a then a=b

divide by Dmax

Determine the section modulus and moment of inertia of an 24 LJ 08 if the standard load table indicates a total load which will cause a 1/360 maximum deflection at center of a 40 foot span is 418 lbs per linear foot. (note: these tables assume open web joist to be a non-compact section of 22,000 psi).

deflection of 40 foot span=

1

= 0.00277778 360 .002777. 480 = 1.333 inches

decimal equivalent of

pounds per inch= 480/12= 34.833

I=

4 5. 480 . 34.833 = 622.826 384. 1.333. 29000000

Steel Joist Elastic Properties z Copyright 2009 William Greco

Trident3 [email protected]

Example-1 (continued) =

S Section modulus =

I

622.826

centroid

24

= 51.902

2 Radius of Gyration =

I AREA AREA= WEIGHT OF STEEL= 490 LBS/CUFT

r

=

WEIGHT OF 24 LJ 08 = 22 LBS / LIN. FT.

622.826 = 9.815 22 490 144