Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Identify different types of structural members.
• Calculate forces using free body diagrams and equilibrium.
• Determine stress (tensile, compressive, shear).
• Explain tension, compression, shear, and bending.
• Solve for unknown forces in simple truss structures.

Lesson summary

This week, we delve into the fundamental principles of structural members and the forces they experience within simple structures. Understanding these principles is crucial for any aspiring civil technologist. From the houses we live in to the bridges we cross, every structure relies on the careful design and execution of these core concepts. In South Africa, with its diverse infrastructure needs – from rural housing projects to urban development – a solid grasp of structural mechanics is vital for building safe, durable, and sustainable structures.

Lesson notes

2.1 Types of Structural Members Structural members are the individual components of a structure that are designed to resist loads. The primary types of structural members are: Tension Members: These members are designed to resist pulling forces or tension. Examples include cables in suspension bridges, tie rods, and the bottom chord of a truss. In South Africa, you'll see tension members in the supporting cables of overhead power lines and in the steel reinforcement within concrete structures. Think of the steel cables used in the Bloukrans Bridge bungee jump – they are primarily tension members.

Compression Members: These members are designed to resist pushing forces or compression. Examples include columns, struts, and the top chord of a truss. Consider the reinforced concrete pillars supporting many buildings in Johannesburg's CBD; these are primarily compression members. In housing, wooden or steel studs within walls act as compression members supporting the roof load.

Beams: Beams are structural members that are primarily designed to resist bending moments caused by loads applied perpendicular to their longitudinal axis. Examples include roof beams, floor joists, and bridge girders. Look at the support beams under a shack in an informal settlement - even if not designed by an engineer, they are acting as beams and resisting bending. 2.2 Forces and Equilibrium Forces: A force is a vector quantity that tends to cause motion or change the motion of an object. Forces are measured in Newtons (N).

External Forces: These are forces applied to the structure from external sources, such as gravity (dead load), wind, or people (live load). Dead load refers to the weight of the structure itself, while live load refers to the weight of occupancy and usage. Think of the weight of roof tiles as dead load and the weight of people walking on a floor as live load.

Internal Forces: These are forces that develop within the structural members in response to external forces. These forces are responsible for holding the structure together and maintaining equilibrium. Examples include tensile force, compressive force, and shear force.

Equilibrium: A structure is in equilibrium when the sum of all forces and moments acting on it is equal to zero. This means that the structure is not moving or rotating. The three equations of static equilibrium are: ΣFx = 0 (Sum of horizontal forces is zero) ΣFy = 0 (Sum of vertical forces is zero) ΣM = 0 (Sum of moments about any point is zero) 2.3 Stress Stress is the internal force acting per unit area within a material. It is measured in Pascals (Pa) or N/m². Tensile Stress (σt): This is the stress caused by a tensile force pulling on the material. σt = Force / Area. Compressive Stress (σc): This is the stress caused by a compressive force pushing on the material. σc = Force / Area. Shear Stress (τ): This is the stress caused by a force acting parallel to the surface of the material, causing it to slide or shear. τ = Force / Area. 2.4 Free Body Diagrams (FBD) A free body diagram is a simplified representation of a structure or a structural member, showing all the external forces acting on it. Drawing an FBD is crucial for analyzing forces and ensuring equilibrium.

Example 1: Tension in a Cable A concrete block weighing 500 N is suspended from a cable attached to the ceiling. Calculate the tension in the cable.

Solution: Draw a Free Body Diagram: Draw the concrete block as a point. Show the weight of the block (500 N) acting downwards and the tension in the cable (T) acting upwards.

Apply Equilibrium Equations: Since the block is in equilibrium, the sum of the vertical forces must be zero. ΣFy = 0 T - 500 N = 0 T = 500 N Therefore, the tension in the cable is 500

N. Example 2: Compression in a Column A column supports a load of 10 kN. The cross-sectional area of the column is 0.01 m². Calculate the compressive stress in the column.

Solution: Calculate Compressive Stress: σc = Force / Area σc = 10,000 N / 0.01 m² σc = 1,000,000 Pa = 1 MPa Therefore, the compressive stress in the column is 1 MPa.

Example 3: Forces in a Simple Truss Consider a simple truss with two members, AB and BC, pinned at joint B. A vertical load of 1000 N is applied at joint B. Member AB is horizontal, and member BC is inclined at 45 degrees to the horizontal. Determine the forces in members AB and B

C. Solution: Draw a Free Body Diagram of joint B: Show the 1000 N load acting downwards, the force in member AB (Fab) acting horizontally to the left (assuming compression), and the force in member BC (Fbc) acting at 45 degrees.