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EDEXCEL NATIONAL CERTIFICATE/DIPLOMA
FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS
UNIT 11 - NQF LEVEL 3
OUTCOME 1 - FRAMES AND BEAMS
TUTORIAL 1 - PIN JOINTED FRAMES
Be able to determine the forces acting in pin-jointed framed structures and simply supported beams
Pin-jointed framed structures: solution e.g. graphical (such as use of Bow’s notation, space and
force diagram), analytical (such as resolution of joints, method of sections, resolution of forces in
perpendicular directions (Fx = F Cosθ, Fy = F Sinθ), vector addition of forces, application of
conditions for static equilibrium (ΣFx= 0, ΣFy = 0, ΣM = 0 ))
Forces: active forces e.g. concentrated loads; uniformly distributed loads; reactive forces e.g.
support reactions, primary tensile and compressive force in structural members
Simply supported beams: distribution of shear force and bending moment for a loaded beam e.g.
concentrated loads, uniformly distributed load (UDL); types of beam arrangement e.g. beam
without overhang, beam with overhang and point of contraflexure.
It is assumed that the student has studied Mechanical Principles and Applications Unit 6
All static structures such beams and frames are in a state of equilibrium. This must mean that:
• all forces in any given direction add up to zero.
• all the turning moments about a given point must add up to zero.
If we only use Cartesian coordinates it follows that:
• all the vertical forces upwards (+ve) must equal all the vertical forces downwards (-ve). In
other words Σ Fx = 0
• all the horizontal forces to the right (+ve) must equal all the horizontal forces to the left (-
ve) In other words Σ Fy= 0
• all the clockwise turning moments (+ve) must equal all the anticlockwise turning
moments (-ve). In other words Σ M = 0
This is known as D'Alambert's Principle.
It is of interest to note that in maths anti-clockwise is normally positive and the use of the oppo