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9.2.1 Describe what is meant by an external load acting on a structure.
This involves loads where physical contact is made.
9.2.2 Describe what is meant by body load.
This is a load without physical contact, for example, a structure’s own weight.
The easiest and simplest form of load is the load produced by physical contact. When someone leans against the wall he produces a load (external load from separate source), when someone pushes your against your chest you feel a load. Even standing on the floor you impose a body load by standing on it.
9.2.3 Describe the difference between weight and mass.
Refer to the effect of gravity and how commonly people refer to the weight of an object when they should refer to its mass.
9.2.4 State the units of weight and mass.
9.2.5 Explain the relationship of external loads to internal forces and the concept of the balance of equilibrium of forces within a structure.
9.2.6 Explain how a structure “works” by interpreting how external loads give rise to internal forces within the structural members.
A static structure is in equilibrium, otherwise it would move (the forces acting upon it are equal in size and opposite in direction).
Another example is a bridge or building that is standing. If it wasn't would collapse.
9.2.7 Explain the differences between tensile and compressive forces and how they affect equilibrium within a structure.
Tensile loads tend to extend or stretch a structural member. Compressive loads tend to compress or shorten a structural member. Tensile and compressive forces must balance if the structure is to maintain equilibrium. Only forces that are parallel or perpendicular need to be considered. Knowledge of trigonometry or quantitative resolution of vectors into components is not required.
The link below explores multiple kinds of forces and how they are utilized in certain structures such as bridges.
9.2.8 Calculate a tensile or compressive stress, given values of force and area.
Stress = force/area
9.2.9 Calculate a tensile or compressive strain, given values of the original dimension and the change in dimension.
Strain = change of length/original length
Find the strain in a tug-o-war rope which resists a load of 700N and is 15mm in diameter
Area: 15mm --> calculate into meters: 0.015
Note: always work in the units that are equivalent to the resultant answer, in this case: N/m2
Tensile stress: Ft
Ft = force / area
Ft = 700N / π(pi) r2
Ft = 39611896.95
Ft = 39.6 N/m2
9.2.10 Evaluate the importance of forces in a design context.
The main problem that arises is if the failure to support specific force will fail. If we apply forces, will the failure to support the tension, compression, shear, bending, torsion or elastic instability occur?
As designers, we must remember that the forces have to be distributed evenly over the whole area. The limitations of the bending and torsion equations must also be taken into account. In deriving the bending equation it was assumed that each cross-section of the beam is symmetrical about the plane in which the loads are applied.
Further complications arise when a component carries two or more types of loading simultaneously. This would result in the fracture in the materials.
Among other things, the designer has to specify the material from which each component is made. Many materials are not equally strong in tension and compression and the designer must therefore consider both tensile and compressive stresses.
Concrete, for example, has little strength in tension and steel rods offer little resistance to compression. But if the steel rods are embedded in a concrete beam near the tension face, the bending moments are resisted by tension in the steel and compression in the concrete.
There is a body load and an external load that acts upon buildings thus reinforced concrete is used.
"Force is action or agency that causes a body of mass to accelerate. It may be experienced as a lift, a push, or a pull. The acceleration of the body is proportional to the vector sum of all forces acting on it (known as net force or resultant force). In an extended body, force may also cause rotation, deformation, or an increase in pressure for the body. Rotational effects are determined by the torques, while deformation and pressure are determined by the stresses that the forces create." wikipedia.com
The designer must consider other factors that affect the strength of the components. If a bar is subjected to a load many times it may fail at a lower stress than if the load ware applied only once , and effect known as fatigue. It if operates at a high temperature strain may become progressive at a constant load, a phenomenon called creep.
Bulleted list and italicised paragraphs are excerpted from Design Technology: guide. Cardiff Wales, UK: International Baccalaureate Organization, 2007.
Images are clickable links to its location.