Free HSC Engineering Studies Flashcard Review

The masonry arch carries load almost entirely in compression, which suits stone and brick because they are weak in tension. A steel truss uses members in both tension and compression, so it can span much further with less material.

Concrete could be cast into almost any shape and is strong in compression, and steel reinforcement supplied the tensile strength concrete lacks. Together they allowed fire-resistant frames, slabs and bridges that were cheaper than masonry or iron.

Cement manufacture releases large amounts of CO₂, both from burning fuel in the kiln and from calcining limestone. Quarrying for limestone and aggregate also disturbs land and creates dust.

It can be crushed, with the reinforcing steel removed magnetically for recycling, and reused as recycled aggregate for road bases, fill or new non-structural concrete. This reduces landfill and quarrying.

ΣF_x = 0, ΣF_y = 0 and ΣM = 0 about any point. The body then has no linear or rotational acceleration.

A pin resists movement in any direction, so it provides both horizontal and vertical reaction components. A roller provides only a reaction perpendicular to its supporting surface.

Members are joined by frictionless pins, loads act only at the joints, and members are straight and weightless. Each member therefore carries only an axial force (tension or compression).

Isolate one joint as a free body, then apply ΣF_x = 0 and ΣF_y = 0 to solve for up to two unknown member forces. Work joint by joint, starting at a joint with no more than two unknowns, usually a support.

When only one or a few member forces in the middle of a truss are needed. Cutting through no more than three members and taking moments about a clever point finds a force directly, without solving every joint.

If only two non-collinear members meet at an unloaded joint, both carry zero force. If three members meet and two are collinear, the third carries zero force.

Forces at a joint are drawn to scale head to tail in order. Because the joint is in equilibrium the polygon must close, so the lengths and directions of the closing sides give the unknown member forces and their nature.

Look at the direction of the member force acting on the joint. If it points away from the joint the member is in tension; if it points towards the joint it is in compression.

The algebraic sum of all vertical forces acting on one side of the section. It is the force tending to slide one part of the beam past the other.

The algebraic sum of the moments about that section of all forces on one side of it. It measures how strongly the beam is being bent there.

Where the shear force diagram crosses zero (changes sign). For a single central point load this is at midspan, with M = PL/4.

A series of horizontal steps. It jumps by the value of each point load or reaction and is constant between them.

The SF diagram becomes a sloping straight line under the UDL, and the BM diagram becomes a parabolic curve. For a simply supported beam with a full-span UDL, M_max = wL²/8 at midspan.

M = PL, at the fixed support (hogging). The moment is zero at the free end and increases linearly towards the wall.

The line through the cross-section where bending stress is zero. It passes through the centroid of the section, and fibres on one side are in tension while those on the other are in compression.

A property of a cross-section's shape that measures its resistance to bending (units mm⁴). Placing material far from the neutral axis, as in an I-beam, greatly increases I and reduces bending stress.