Roof Load on a Single Beam
FBD of Left Half of Beam


When designing structures, many times a dimension needs to be determined. In this problem, the load and beam cross section is given, but its length is not. This is the main concern since the length also determines the moment. This problem can be solved by first finding the moment in terms of the unknown length L. Then determine the moment of inertia. Finally, use the bending stress equation and solve for the length.


Since the bending stress depends on the moment, it needs to be determined. The
center beams carry the roof load over a two foot width of the roof as shown
in the diagram. The total load on the beam in terms of inches will be,
w = (60 lb/ft^{2})(2 ft)(1 ft/12 in) =
10 lb/in
This constant distributed load will cause a maximum moment at the center (this
can be determined by constructing a
moment diagram, but has been omitted here). Cutting the beam at the center and
summing moments gives the moment equation,
ΣM_{right edge}=
0
M  5L(L/2) + [10 (L/2)] (L/4) = 0
M = 1.25 L^{2} lbin
As expected, it is a function of the unknown length L.

Beam Cross Section


The moment of inertia needs to be found for the complete beam cross section,
but this shape is not in any handbook. To simplify finding the moment of inertia,
the beam can be separated into two
rectangular shapes, giving,
I_{top} = h^{3}b/12 = 1.5^{3} (7.25)
/ 12 = 2.0391 in^{4}
I_{bottom} = h^{3}b/12 = 7.25^{3} (1.5)
/ 12 = 47.635 in^{4}
Using the
parallel axis theorem, the moment of inertia for both of these can be found
at the centroid of the actual "T" beam. But first, the center needs to
be located. The vertical centroid is
= 5.8125 in
Now that the center is known, the moment about the NA can be determined as
I_{NA} = I_{t} + A_{t} y_{t}^{2}
+ I_{b} + A_{b} y_{b}^{2}
= 2.039 + 7.25(1.5)(7.25
+ 0.75  5.8125)^{2} +
47.635
+ 7.25(1.5)(5.8125  7.25/2)^{2}
= 2.039 + 52.04 + 47.63 + 52.04
= 153.7 in^{4}

Beam Bending Stresses at Section aa:
Positive Moment Causes Compression
at top and Tension at Bottom. 

The material bending stress cannot exceed 7,500 psi. and there is also a factor
of safety of 2.0. Using the bending stress equation gives,
σ = My/I
7,500/2.0 = (1.25 L^{2})(y) / 153.7
The y distance is the furthest from the neutral axis. For this T shaped beam,
it would be in the tension zone, with y = 5.8125 in.
576,400 = 1.25 L^{2} (5.8125)
Solving for L gives
L = 281.7 in = 23.47 ft
It is interesting to note, that if just a single 2x8 is used in the vertical
position (I = 47.63, y = 3.25), then the span would be limited to 16.55 ft. Thus, even with the moment of inertia over three times larger for the Tbeam, the span is not even double. This is due to the squared length term in the moment.
