Structural Engineering & Statics
where the load goes
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Drawing as calculation — page 2

Essays 25 to 29 of 29 on this thread, in the same order.
w = 20 kN/mcorner 65.2 kNmmidspan 94.8 kNm94.8 + 65.2 = 160.0 = wL²/8 Internal forces

The moment that goes round the corner

At a rigid knee the bending moment does not stop at the end of the beam. It turns and runs down the column, and in the same instant the beam's shear becomes the column's axial force — while the block of steel that has to carry the turn appears on no member diagram anywhere.

 +67.0 N/mm²−67.0 N/mm²+67.0 N/mm²−67.0 N/mm²I-section, ω = ±12169 mm² at a tiptop flangebottom flangeh = 295V_f = 1.69 kN each way · V_f × h = 0.500 kN·m = the torque at the built-in endM_f = 3.04 kN·m each way · M_f × h = 0.897 kN·m² = the bimomentflange plan, sideways movement ×20 against the length Sections and stress

The section that cannot stay flat

Twist an I-section and its cross-section dishes out of its own plane. Stop that happening at one end and the member finds a second way to resist — the flanges bend in opposite directions — and the stress resultant that describes it has units nothing else in statics has.

 the cutcentre of curvatureN = 50 kNR = 79.70 mmr_i = 50r_o = 120My/I248.7 N/mm²r = 75.04centroid 79.70free body: everything below the cut, so the face carries N = 50 kN and M = N·R = 3.985 kN·m248.7 N/mm² of tension inside against 140.6 of compression outside, on a neutral axis 4.65 mm off the centroid Sections and stress

The bar that was bent before it was loaded

In a curved bar plane sections still stay plane, and the bending formula is wrong anyway. The fibres were different lengths before anything was applied, so an equal rotation of two plane faces produces unequal strain — the stress is a hyperbola, the neutral axis has moved inward, and a crane hook carries half as much again as My/I reports.

 100 kNthe gap is the shear: 16.3% of the totalbending alone, and what the beam really doesthe shear part alone, magnified 3 times furthertwo straight lines meeting under the loadshearγ = V/GAs is a slope the section is racked through, not a curvature —so this diagram is integrated once, where the moment diagram above it is integrated twice Deflection

The deflection that is not bending

Every deflection on this site so far has been the second integral of a moment, and that calculation silently drops a term. The beam also shears, and the shear deflection is not a correction to the curve — it is a different shape, and for a deep member it is most of the answer.

 δ = 631.06 read here10 kN at every top node — each member drawn at the width of its own sharetop chord (four members)47.2%bottom chord (six members)28.5%diagonal (six members)22.3%vertical (five members)2.0%the members that moved the roof, ranked — a symmetric pair is two members and appears twicetop chord, at mid-span14.80%F -52.9 × f -1.765top chord, at mid-span14.80%F -52.9 × f -1.765bottom chord, at mid-span8.77%F 47.1 × f 1.176top chord, near the left support8.77%F -47.1 × f -1.176bottom chord, at mid-span8.77%F 47.1 × f 1.176top chord, near the right support8.77%F -47.1 × f -1.176diagonal, near the left support6.20%F -38.6 × f -0.772diagonal, near the right support6.20%F -38.6 × f -0.772virtual work and a stiffness solution agree to 3.6e-15 — two methods sharing no arithmetic Deflection

Which member moved the roof

A beam sags because it curves. A truss has no curvature anywhere — it comes down because every one of its members changes length, and the sum of those changes, weighted member by member, is a ranking that names which ones are worth stiffening. Usually not the ones a designer worries about.

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