Monthly Mechanics: Let’s Design a Roof, Part 2

In case you’re tuning in to Monthly Mechanics for the first time, we are in the process of designing a roof using all the structural principles we’ve learned so far. When we left off, we had calculated the vertical load on a rafter and the reactions at each end. You might want to read (or re-read) Part 1 to review how we got to this point.

Now it’s time to choose a rafter size that can support this load!

Loads and reactions.

Loads and reactions.

Step 5: Determine the internal forces. A rafter is a type of beam, which means most of the external forces on it are applied sideways, rather than along the axis of the rafter. (The opposite of a beam is a column, which sustains mostly axial forces… think of the Parthenon.) This means the rafter will fail in Shear and Flexure if the forces are too big. Remember that a shear failure results from excessive shear force, and a flexure failure results from excessive bending moment. To find out how big these forces are, we need to “read” the rafter from left to right.

The shear diagram jumps up immediately to 2280 pounds because that’s the reaction on the left-side wall. Then the diagram descends by the uniform load of 228 pounds per foot until it reaches the right-side wall, where the other reaction of 2280 pounds brings it back to zero.

Shear diagram.

Shear diagram.

For the moment diagram, add up all the shear forces from left to right. Between the left side and 10 feet, the shear decreases from 2280 pounds to zero, forming a triangle. The bending moment at 10 feet is the area of this triangle, which is half of the rectangle 2280 pounds x 10 feet, or 11,400 foot-pounds. Perform a similar calculation between the left side and 5 feet, and the bending moment at 5 feet is 8550 foot-pounds. Keep repeating this calculation, and the moment diagram turns out to be a curve, peaking at 10 feet.

Moment diagram.

Moment diagram.

Step 6: Select a beam. We need a piece of dimensional lumber that can support a maximum shear force of V = 2280 pounds and a maximum bending moment of M = 11,400 foot-pounds. Time to consult the National Design Specification (NDS) for Wood Construction, which lists the allowable stresses on every type of lumber there is. Let’s use construction-grade spruce-pine-fir because it’s readily available at lumber yards here. The NDS table reproduced below gives a maximum bending stress of 1000 psi and a maximum shear stress of 135 psi.

Reference stresses for all types of lumber, from NDS.

Reference stresses for all types of lumber, from NDS.

Then we get a bonus because our roof has lots of rafters with regular spacing. NDS specifies an adjustment factor of 1.15 for repetitive use, which brings our maximum bending stress up to 1150 psi and our maximum shear stress to 155 psi.

But we previously calculated internal forces, not stresses. To make the conversion, we refer to the article about Stress for two critical formulas: shear stress is σshear = V / A and flexure stress is σflexure = (M*h) / (2*I). Let’s review some symbols here. A is the area of the rafter, which equals base times height: A = b*h. And I is the moment of inertia, which is also a function of base and height: I = b*h3 / 12. So we can rewrite the equations using just shear V, moment M, and the rafter’s dimensions.

σshear = V / (b*h)
σflexure = (M*6) / (b*h2)

Let’s try a 2×12 rafter. The actual dimensions are b = 1.5 in and h = 11.25 in. The required shear stress is σshear = 2280 lb / (1.5 in * 11.25 in) = 136 psi, which is OK because the available shear stress is 155 psi. But we run into trouble when we compute the required flexure stress: σflexure = (11,400 ft-lb * 6) / (1.5 in * [11.25 in]2) = 4324 psi. That’s almost four times the available flexure stress. NO GOOD!

WHAT CAN WE DO? We can reduce the rafter spacing to 16 inches or 12 inches, to lessen the load on each rafter. We can use a different material for the rafters, like glulam lumber or even steel beams. Or we can use roof trusses instead of rafters – a truss can span great distances with an economical amount of material.

A better solution is to refine our load calculations. We made a lot of assumptions to simplify the math, resulting in much higher loads than the roof will actually receive. Actual snow load on a sloping roof is less than the ground snow load of 40 psf, and roof live load is treated differently from regular live load. Also, since the rafters are pitched 12-on-12, vertical loads are not perpendicular to them, which means only a fraction of the calculated loads will actually contribute to shear force and bending moment. (On the other hand, we’d need to account for the non-perpendicular component as axial forces.)

Design is an iterative process, and I hope you appreciate this roof-design exercise not as a failure but as the first stepping stone to success!

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