## Trusses – Example 4, Pratt roof truss with snow

This problem, like the Howe truss, comes from J.L. Meriam, “Statics”, John Wiley & Sons, 1966.

This diagram actually shows concentrated forces at the joints – which is where they must be for what we’re doing. Nevertheless, the problem as posed says that we have a total weight of 4000 pounds distributed uniformly across the top of the truss. We take the 1000 pounds on each of these four beams, and assign 500 pounds to each of its endpoints. Points A and E have 500 pounds, but the other three joints connect two loaded beams and get 1000 pounds.

I need labels…

I am – effectively – given coordinates of the joints:

Let’s make our own picture:

A quick count: we have 13 beams (members, m = 13) and 8 joints (j = 8), and we want

2j = 16 = 13 + 3

so we want 3 possible reaction forces. but we’ve got 2: all the external forces are downward, so the reaction forces at A and E are upward. Still, it will work out, even though it’s 16 equations in 15 unknowns.

Alternatively, we could specify that the bridge is fixed at A and rolling at E – so there are two possible reaction force components at A, and only one at E. This would give us one more unknown, 16 in total. The fact remains, however, that there is no horizontal reaction force at A.

Despite that, having solved the Howe truss with one fewer unknown than equations, I will do this one using an unnecessary horizontal reaction at A.

Off we go….

Get the vectors between every possible pair of points:

Make unit vectors of them:

Let me clear my force arrays:

As ever, there is a force at the tail of every nonzero vector. let f denote the magnitude, F the vector… and do it so that fij = =fji, by sorting on the indices on f. That is,

$F_{ij} = f_{ij}\ u_{ij}\$

expresses the forces F in terms of the magnitudes f_ij and the unit vectors u_ij, and that the force at one end of the beam has the same magnitude as the force at the other, namely

$f_{ji} = f_{ij}\$,

is taken care of by keeping only half of the f_ij. The unit vectors take care of the signs, because the two forces at the ends of a beam are “equal and opposite”.

I need to outright declare that the following magnitudes are zero because there are no beams with these indices:

Lay out the table of forces:

Collect the components for each joint:

I have allowed for a reaction force (R1, R2) at the fixed point A, but only a vertical reaction (0, R3) at the rolling point E.

The key to dealing with a continuous load, at this level of modeling, is to divide the load on each beam evenly between its two endpoints. Not all the relevant joints receive loading from two beams… our 4000 lb. load on four beams gets distributed as 500-1000-1000-1000-500 on the 5 joints of the top chord.

Add the external forces to the internal beam forces…

Set every total force to zero:

Now solve the system. (The inside Flatten gets me one set of 16 equations instead of two sets of eight.)

NOTE that R1 = 0, as it must be.

I can check a few answers against the book. The force in BH is

Change the signs, as I did for the Howe truss, for the drawings. In principle I suppose I should be able to handle the other convention, but in practice, I am used to interpreting arrows pointing into a beam as external compressive forces, rather than arrows pointing out of the beam to represent the internal resistance to the external compression.

Whoa! Is CG really unstressed? Yes:

That’s not to say I would remove it. In the real world, beams are deformed, and CG may play a role.

The verticals BH and DF in compression? Yes:

The diagonals CH and CF in tension? Yes:

And that’s it for this Pratt truss, and a distributed load.