Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Calculate reactions at supports for determinate beams.
• Determine shear force and bending moment in beams.
• Explain centroid and second moment of area (moment of inertia).
• Calculate maximum bending stress in a beam.
• Evaluate the stability of simple structures.

Lesson summary

This week, we delve into the crucial area of applied mechanics and stability within civil structures. Understanding these principles is paramount for ensuring the safety, durability, and functionality of all types of structures, from houses and bridges to dams and skyscrapers. In South Africa, with its diverse climate conditions (ranging from arid deserts to subtropical regions) and varying soil types, the application of these principles is even more critical. Consider the impact of unstable structures on communities: a poorly designed bridge collapsing, a building suffering from subsidence due to inadequate foundations, or a dam failing and causing devastating floods.

Lesson notes

2.1 Reactions at Supports Supports are crucial for transferring loads from the structure to the ground. Statically determinate beams have support conditions that allow us to calculate the reactions using the equations of static equilibrium: ΣFx = 0 (Sum of horizontal forces equals zero) ΣFy = 0 (Sum of vertical forces equals zero) ΣM = 0 (Sum of moments equals zero)

Common support types: Roller Support: Provides a vertical reaction force only.

Hinge Support: Provides both vertical and horizontal reaction forces.

Fixed Support: Provides vertical, horizontal, and moment reactions.

Example 1: A simply supported beam, 6m long, carries a point load of 10kN at 2m from the left support and a UDL of 2kN/m over the entire span. Calculate the reactions at the supports.

Step 1: Draw a Free Body Diagram (FBD): This is essential for visualizing the forces and distances. Label the supports A (left) and B (right) and the reactions Ra and Rb. The UDL acts at the midpoint of the beam.

Step 2: Apply the Equations of Equilibrium: ΣFy = 0: Ra + Rb - 10kN - (2kN/m 6m) = 0 => Ra + Rb = 22kN ΣMA = 0 (Taking moments about point A): (10kN 2m) + (2kN/m 6m 3m) - (Rb * 6m) = 0 => 20 + 36 - 6Rb = 0 => Rb = 9.33 kN Substitute Rb back into the first equation: Ra + 9.33 = 22 => Ra = 12.67 kN Therefore, the reaction at support A (Ra) is 12.67 kN and the reaction at support B (Rb) is 9.33 kN. 2.2 Shear Force and Bending Moment Shear Force (SF): The algebraic sum of all vertical forces acting to the left or right of a section.

Bending Moment (BM): The algebraic sum of the moments of all forces acting to the left or right of a section, about that section.

Example 2: For the same beam in Example 1, determine the shear force and bending moment at a point 4m from the left support.

Step 1: Calculate the reactions (already done in Example 1): Ra = 12.67 kN, Rb = 9.33 kN Step 2: Consider the section 4m from the left support: Shear Force (SF): SF = Ra - 10kN - (2kN/m 4m) = 12.67 kN - 10kN - 8kN = -5.33 kN (The negative sign indicates the shear force is acting downwards).

Bending Moment (BM): BM = (Ra 4m) - (10kN 2m) - (2kN/m 4m 2m) = (12.67 kN 4m) - (10kN * 2m) - (16kN) = 50.68 - 20 - 16 = 14.68 kNm Therefore, the shear force at 4m from the left support is -5.33 kN, and the bending moment is 14.68 kNm. 2.3 Centroid and Second Moment of Area (Moment of Inertia)

Centroid: The geometric center of a shape.

Second Moment of Area (I): A measure of a shape's resistance to bending. It depends on the shape and dimensions of the cross-section. For a rectangle with width 'b' and height 'h', I = (b*h^3)/12 (about its centroidal axis). For a circle with radius 'r', I = (π*r^4)/

4. For I-sections, you need to divide the section into rectangles and apply the parallel axis theorem (not covered in detail this week, but good to research). 2.4 Bending Stress The bending stress (σ) at a point in a beam is given by: σ = (M*y)/I Where: M is the bending moment at the section y is the distance from the neutral axis (centroid) to the point where the stress is being calculated I is the second moment of area.

Example 3: A rectangular beam (100mm wide and 200mm deep) is subjected to a bending moment of 20 kNm. Calculate the maximum bending stress.

Step 1: Calculate the second moment of area (I): I = (bh^3)/12 = (0.1m * (0.2m)^3)/12 = 6.67 x 10^-5 m^4 Step 2: Determine the distance 'y' to the outermost fiber: y = h/2 = 0.2m/2 = 0.1m Step 3: Calculate the bending stress (σ): σ = (My)/I = (20000 Nm * 0.1m) / (6.67 x 10^-5 m^4) = 30 x 10^6 N/m^2 = 30 MPa Therefore, the maximum bending stress in the beam is 30 MPa. 2.5 Stability (Buckling and Overturning)

Buckling: The sudden failure of a structural member under compressive stress, where the member bends or bows sideways. Slender columns are more prone to buckling.

Overturning: The rotation of a structure about a pivot point due to applied loads, often caused by wind or seismic forces. The restoring moment (due to the structure's weight) must be greater than the overturning moment for stability. Qualitative

Example: Consider a tall, narrow wall. It's more susceptible to overturning due to wind pressure than a short, wide wall with the same weight per unit area. The short wide wall has a greater restoring moment due to a longer lever arm for the weight. Reinforcing such walls with buttresses adds stability. Guided Practice (With Solutions)

Question 1: A cantilever beam, 4m long, carries a point load of 5kN at its free end. Calculate the reaction at the fixed support.

Solution: ΣFy = 0: Ra - 5kN = 0 => Ra = 5kN (The reaction at the support is equal and opposite to the applied load). ΣMA = 0: Ma - (5kN 4m) = 0 => Ma = 20 kNm (The moment reaction is equal to the load multiplied by the distance).

Question 2: A simply supported beam, 8m long, carries a UDL of 3kN/m over its entire span. Calculate the maximum bending moment.

Solution: Step 1: Calculate the reactions.