Non-uniform distributed loads - Structure Bench

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Non-uniform distributed loads - Structure Bench

Three loads, three statics checks

Case	Total load	Eaves reaction	Ridge reaction	Peak moment
Snow, tapering	43.5 kN	24.8 kN	18.6 kN	32.8 kN·m at 2.89 m
Plant, over part of the span	16.0 kN	8.3 kN	7.7 kN	20.1 kN·m at 3.03 m
A pull along the slope	20.0 kN	19.3 kN along, 5.2 kN up	0	0
All three		38.3 kN up	26.4 kN	52.8 kN·m at 2.97 m

Hand check: the trapezoid and the part-length load

The rafter. The member runs 6 m across and 1.61 m up, at 15°, so its length is,

L = \frac{6}{\cos 15°} = \frac{6}{0.9659} = 6.212\ \text{m}

Every load here is given per metre of that length, so it's the 6.212 m we multiply by, not the 6 m on plan.

The snow. The snow load falls from 10 kN/m at the eaves to 4 kN/m at the ridge. That's a trapezoid, and the total load is its area,

W = \frac{w_1 + w_2}{2} L = \frac{10 + 4}{2} \times 6.212 = 43.48\ \text{kN}

The resultant sits at the trapezoid's centroid, nearer the heavy end. Measured along the member from the eaves, that's,

\bar{s} = \frac{L(w_1 + 2w_2)}{3(w_1 + w_2)} = \frac{6.212 \times (10 + 2 \times 4)}{3 \times (10 + 4)} = 2.662\ \text{m}

Now we take moments about the eaves pin. The load and the ridge reaction are both vertical, so their lever arms are horizontal distances, $2.662 \cos 15°$ and $6.212 \cos 15°$ . The cosines cancel, and the ridge reaction is,

R_\text{ridge} = \frac{W \bar{s}}{L} = \frac{43.48 \times 2.662}{6.212} = 18.6\ \text{kN}

Taking moments about the ridge instead, the eaves reaction is $43.48 \times (6.212 - 2.662) / 6.212 = 24.8$ kN.

The plant. The plant load is 8 kN/m over the 2 m from 2 m to 4 m along the member: 16 kN, centred 3 m along. Moments about the eaves give the ridge $16 \times 3 / 6.212 = 7.73$ kN, and the eaves takes the other 8.27 kN.

The peak moment is where the shear is zero, and that's where the load from the start of the patch has added up to the eaves reaction, $8(s - 2) = 8.27$ , at $s = 3.03$ m along. To find the moment there, we need to remember that a vertical force pushes on a sloping member partly across it and partly along it, and only the part across it bends it. That part is the force times $\cos 15°$ , so,

M = \cos 15° \left[ 8.27 \times 3.03 - \frac{8 \times (3.03 - 2)^2}{2} \right] = 0.9659 \times (25.06 - 4.24) = 20.1\ \text{kN m}

The pull. The 20 kN pull acts along the slope, down towards the eaves. Resolved, it's $20 \cos 15° = 19.3$ kN horizontally and $20 \sin 15° = 5.2$ kN vertically. Its line of action runs along the member and straight through the eaves pin, so it has no moment about the pin. Taking moments about the pin, the ridge roller has nothing to resist, so it carries nothing, and the pin takes the whole pull, 19.3 kN horizontally and 5.2 kN vertically. With no force across the member there's no bending at all. The pull simply compresses the member between the pin and the load, -20 kN, and leaves the rest of it unloaded.