Question:
I am adding a second story to my house, and need to know if the
existing steel beam under the first floor, which is supporting the one
load bearing wall, can bear the weight of a second story. The roof is
bearing on the exterior walls, so it is not a factor. I am looking at
two floors with live loads of 30 psf and dead loads of 20 psf, plus
one 2nd floor ceiling with no live load and a dead load of 10 psf.
(The ceiling joists are not roofing members, and thus are bearing on
the wall.) The existing steel beam is an old-style S-flange, namely
an S 7×20 (Ixx=42.4, Iyy=3.17). It joists are 22′ long, so that’s a
tributary span of 11′. The beam is 24′ and is supported at 12′ with a
column on a concrete slab.
I need to know how to calculate (a) the load that beam is bearing with
one floor, (b) what it bears with the second floor and extra ceiling,
and (c) whether the beam is strong enough to handle that.
I’d love to know the actual equations for these things, so I can do
this calculation next time on my own, and so when I present this to
the the building inspector I know what I’m talking about.
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Answer:
Hello cpopetz, you seem to have done a very good job of describing
your problem. I think we can get some answers.
(a) the load that beam is bearing with
one floor
NOTE: What you have here is a continuous beam with two equal spans.
However, to make things a little easier we can consider it as a simple
beam. The answer will be a little conservative, but that is probably
good and these formulas are not 100% accurate anyway.
The beam formulas for this loading are:
M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
NOTE: Maximum deflection is limited to D = l/360 = 12 x 12 / 360 = 0.40 in
w (load per foot) = (30 + 20 + 10)psf x 11′ = 660 lb per ft
l (beam span) = 12 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia
Solving for M:
M = 660 x 12^2 / 8 = 11,880 ft lb = 142,560 in lb
The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi
= 19,800 psi
The section modulus of the required beam (S) = M/s = 142,560/19,800
= 7.2 in^3
Now we must calculate the required I (moment of inertia):
Solving for I in the above formula for deflection we get:
I = 5wl^4/384 ED = (5 x 660 x 12^4 / 384 x 30,000,000 x 0.40) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 25.66 in^4
This tells us what we already know, that the existing beam was more than adequate.
(b) what it bears with the second floor and extra ceiling
The only difference in these calcs is that we now add in another 30 +
20 psf. So, w now becomes: 110 x 11′ = 1210 lb per ft
Solving for M:
M = 1210 x 12^2 / 8 = 21,780 ft lb = 261,360 in lb
The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi
= 19,800 psi
The section modulus of the required beam (S) = M/s = 261,360/19,800
= 13.2 in^3
Now we must calculate the required I (moment of inertia):
Solving for I in the above formula for deflection we get:
I = 5wl^4/384 ED = (5 x 1210 x 12^4 / 384 x 30,000,000 x 0.40) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 47.04 in^4
(c) whether the beam is strong enough to handle that
I checked your numbers for the existing S 7×20 and you are exactly
right. So, we are comparing our calculated required value of 47.04
in^4 to 42.4 in^4. I would say that your existing beam is adequate. I
have a beam program that will handle the actual end conditions that
you have. I will check out the beam using that program and let you
know what results I get. If there is any of what I have posted that
you don’t understand, please ask for a clarification.
Back with you soon, Redhoss
