Basis of Structural Design Course 3 Structural action: trusses and beams Course notes are available for download at http://www.ct.upt.ro/users/AurelStratan/ Arch → Truss rafter tie Linear arch supporting a concentrated force: large spreading reactions at supports Relieving of support spreading: adding a tie between the supports 1 Truss forces Truss members connected by pins: axial forces (direct stresses) only Supports: – one pinned, allowing free rotations due to slight change of truss shape due to loading – one roller bearing support ("simple support") - allowing free rotations and lateral movement due to loading and change in temperature - (C) - (C) + (T) Forces in the truss: – tie is in tension (+) – rafters are in compression (-) Truss forces If more forces are present within the length of the rafter ⇒ bending stresses To avoid bending stresses, diagonal members and vertical posts can be added - + + - + More diagonals and posts can be added for larger spans in order to avoid bending stresses 2 Alternative shape of a truss For a given loading find out the shape of a linear arch (parabolic shape) Add a tie to relieve spreading of supports Highly unstable shape Alternative shape of a truss Add web bracing (diagonals and struts) in order to provide stability for the pinned upper chord members If the shape of the truss corresponds to a linear arch web members are unstressed, but they are essential for stability of the truss Reverse bowstring arches: – advantage: longer members are in tension – disadvantage: limited headroom underneath 3 Truss shapes Curved shape of the arch: difficult to fabricate ⇒ trusses with parallel chords Trusses with parallel chords: web members (diagonals and struts) carry forces whatever the loads Pratt truss: – top chord in compression – bottom chord and diagonals in tension – economical design as longer members (diagonals) are in tension Truss shapes Howe truss: – top chord in compression – bottom chord in tension – diagonals in compression Warren truss: – top chord in compression – bottom chord in tension – diagonals in tension and compression – economy of fabrication: all members are of the same length and joints have the same configuration 4 Truss joints Pinned joints ⇒ statically determinate structures ⇒ member forces can be determined from equilibrium only Rigid joints ⇒ small bending stresses will be present, but which are negligible due to the triangular shape Traditionally trusses are designed with pinned joints, even if members are connected rigidly between them Space trusses The most common plane truss consists of a series of triangles The corresponding shape in three dimensions: tetrahedron (a) The truss at (b) is a true space truss – theoretically economical in material – joints difficult to realise and expensive Two plane trusses braced with cross members are usually preferred 5 Statically indeterminate trusses Indeterminate trusses: large variety Example (a): cross diagonals in the middle panel, so that one of the diagonals will always be in tension Example (b): Sydney Harbour Bridge, Australia - both supports pinned Beams Beam: a structure that supports loads through its ability to resist bending stresses Leonardo da Vinci (1452-1519): the strength of a timber beam is proportional to the square of its depth Leonhard Euler and Daniel Bernoulli were the first to put together a useful theory around 1750 6 Beams: analogy with trusses Forces in a Pratt truss loaded by a unit central force Forces in a Howe truss Forces in a truss with double diagonals (reasonable estimate) Beams: analogy with trusses Chords: – The forces in the top and bottom chord members in any panel are equal, but of opposite signs, and they increase with the distance from the nearest support – Chords have to resist the bending moment, proportional to the distance from the nearest support Diagonals: – The forces in the diagonal members are equal, but opposite in sign, and have the same values in all panels – Diagonals have to resist the shear forces, the same in all panels 7 Beams: analogy with trusses Bending and shear deformations in a truss Steel plate girder Steel plate girder: heavy flanges and thin web welded together, and reinforced by transversal stiffeners Unit vertical force at the midspan Top flange: in compression Bottom flange: tension Web: shear, with principal tension and compression stresses similar to those in a truss After web buckling, only tensile loads are resisted by the web, plate girder acting as a Pratt truss 8 Beams: bending action Top flange in compression linear variation of normal stress Bottom flange in tension Normal stress proportional to distance from the neutral plane Simplifications: – Thin web, thick flanges ⇒ web has a small contribution to the bending resistance (ignore it) – Normal stress can be considered uniform on flanges Beams: bending action Moment resistance – Idealised double T beam: M = σ⋅A⋅⋅d/2 – Rectangular beam of the same area and depth: M = σ⋅b⋅⋅d2/6 = σ⋅A⋅⋅d/6 The best arrangement of material for bending resistance: away from the neutral axis A/2 F = σ ·(A/2) d d −σ +σ F = σ ·(A/2) A/2 F = σ ·(0.5d·b/2) d 2d/3 −σ A +σ M = σ ·A·d/2 M = σ ·A·d/6 F = σ ·(0.5d·b/2) b 9 Beams: bending action Examples of efficient location of material for bending resistance – light roof beams (trusses) – hot-rolled and welded girder Beams: bending action Examples of efficient location of material for bending resistance – panel construction 10 Beams: bending action Examples of efficient location of material for bending resistance – corrugated steel sheet Beams: bending action Examples of efficient location of material for bending resistance – castellated joist 11 Beams: bending action Examples of efficient location of material for bending resistance – columns requiring bending resistance in any direction: tubular sections Beams: shear stresses Simply supported beam of uniform rectangular crosssection loaded by a concentrated central force W: – can carry a moment M = σ⋅b⋅⋅d2/6 – has a deflection δ If the beam is cut in two parts along the neutral plane: – sliding takes place between the two overlapped beams – the two overlapped beams can carry a moment M = 2⋅⋅[σ⋅b⋅⋅(d/2)2/6] = σ⋅b⋅⋅d2/12, half of the uncut beam – the deflection of the two overlapped beams is 4δ 12 Beams: shear stresses In the uncut beam stresses should be present along the neutral plane to prevent sliding of the lower and upper halves of the beam: shear stresses Smaller stresses would be required to keep the unity of action if the beam were cut above the neutral plane Shear stresses – parabolic variation in a rectangular cross-section – carried mainly by the web, on which they can be considered to be constant for a steel double T beam Structural shapes Simply supported beam subjected to a uniformly distributed load The "perfect" use of material for bending resistance in a beam with idealised double T crosssection (M = σ⋅A⋅⋅d/2): parabolic variation of height A/2 A/2 13 Structural shapes Simply supported truss subjected to a uniformly distributed load The "perfect" use of material for "bending" action: parabolic variation of height Structural shapes Bridge with a simply supported central span and two cantilevered sides The shape of the truss must resemble the bending moment diagram in order to make efficient use of material in upper and bottom chords Quebec railway bridge 14 Structural shapes Forth bridge, Scotland Angel Saligny bridge, Romania 15

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