Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define and apply the conditions of static equilibrium.
• Construct accurate free body diagrams (FBDs).
• Calculate reactions at supports for determinate beams.
• Determine shear force and bending moment diagrams.
• Analyze the stability of simple structures.

Lesson summary

This week, we delve deeper into applied mechanics and stability in civil structures, focusing specifically on the principles of equilibrium, free body diagrams, and the analysis of simple structures under load. Understanding these concepts is crucial for civil technologists in South Africa. From designing safe and efficient bridges connecting our communities to ensuring the structural integrity of buildings in our bustling cities, the principles of applied mechanics are foundational. Consider the numerous RDP housing projects – proper understanding of load distribution is paramount to prevent collapses and ensure the safety of residents.

Lesson notes

2. 1. Static Equilibrium A body is in static equilibrium when it is at rest and remains at rest. This state is achieved when two conditions are satisfied: Translational Equilibrium: The vector sum of all external forces acting on the body is zero. This prevents the body from accelerating in any direction. Mathematically, this is expressed as: ΣF x = 0 (The sum of all forces in the horizontal direction equals zero). ΣF y = 0 (The sum of all forces in the vertical direction equals zero).

Rotational Equilibrium: The vector sum of all external moments acting on the body is zero. This prevents the body from rotating. Mathematically, this is expressed as: ΣM = 0 (The sum of all moments about any point equals zero).

Why this matters: Buildings, bridges, and other structures must be in static equilibrium to be safe and functional. If the forces and moments are not balanced, the structure will move or collapse. Imagine a poorly constructed shack in a South African township - inadequate consideration of equilibrium can lead to devastating structural failure during heavy rains or strong winds. 2.

2. Free Body Diagrams (FBDs) A free body diagram is a simplified representation of a structure or a part of a structure, showing all the external forces and moments acting on it. The purpose of an FBD is to isolate the body and visualize the forces influencing its equilibrium.

Steps to draw an FBD: Isolate the Body: Draw a sketch of the structure or the specific component you're analyzing.

External Forces: Identify and draw all external forces acting on the body.

This includes: Applied Loads: Forces that are directly applied to the structure (e.g., weight of people on a bridge, wind force on a building).

Reactions: Forces exerted by supports on the structure (e.g., reaction forces at the base of a column). Remember to represent these reactions appropriately: Roller Support:* A roller support provides a reaction force perpendicular to the surface it's resting on.

Hinge Support:* A hinge support provides reaction forces in both the horizontal and vertical directions.

Fixed Support: A fixed support provides reaction forces in both the horizontal and vertical directions, and a moment reaction.

Weight: The force due to gravity acting on the body. It acts vertically downwards at the body's center of gravity.

Dimensions and Angles: Include relevant dimensions and angles on your diagram to facilitate calculations.

Coordinate System: Choose a suitable coordinate system (e.g., x-y plane) and indicate its orientation on the FB

D. Why this matters: FBDs are essential for visualizing and analyzing forces acting on structures. Without a clear FBD, it's difficult to correctly apply the equilibrium equations. 2.

3. Reactions at Supports Support reactions are the forces exerted by the supports on the structure to maintain equilibrium. To determine the support reactions, we apply the equations of static equilibrium to the FBD of the structure.

Example 1: Simply Supported Beam A simply supported beam of length 6m is subjected to a point load of 20kN at a distance of 2m from the left support. Determine the support reactions.

Solution: Draw the FBD: Beam of length 6m. Point load of 20kN acting downwards at 2m from the left support. Vertical reaction forces R A and R B at the left and right supports, respectively.

Apply Equilibrium Equations: ΣF y = 0: R A + R B - 20kN = 0 ΣM A = 0: (20kN 2m) - (R B * 6m) = 0 (Taking moments about support A)

Solve for Reactions: From the moment equation: R B = (20kN 2m) / 6m = 6.67 kN Substitute R B into the force equation: R A + 6.67 kN - 20kN = 0 => R A = 13.33 kN Therefore, the support reactions are R A = 13.33 kN and R B = 6.67 k

N. Example 2: Beam with a UDL A simply supported beam of length 4m is subjected to a uniformly distributed load (UDL) of 5 kN/m. Determine the support reactions.

Solution: Draw the FBD: Beam of length 4m. UDL of 5 kN/m acting downwards along the entire length. We can represent the UDL as a single equivalent point load acting at the center of the beam, with a magnitude of (5 kN/m 4m) = 20 kN. Vertical reaction forces R A and R B at the left and right supports, respectively.

Apply Equilibrium Equations: ΣF y = 0: R A + R B - 20kN = 0 ΣM A = 0: (20kN 2m) - (R B * 4m) = 0 (Taking moments about support A)

Solve for Reactions: From the moment equation: R B = (20kN 2m) / 4m = 10 kN Substitute R B into the force equation: R A + 10 kN - 20kN = 0 => R A = 10 kN Therefore, the support reactions are R A = 10 kN and R B = 10 kN. 2.

4. Shear Force and Bending Moment Diagrams Shear force and bending moment diagrams are graphical representations of the internal shear force and bending moment along the length of a beam. These diagrams are essential for understanding the internal stresses within the beam and are used in structural design to ensure the beam can withstand the applied loads.