Unit 2: Mensuration Study Guide

Introduction to Mensuration and Module Objectives

• Definition: Mensuration is the application of mathematical formulae in the setting out and calculation of dimensions.
• Objective: To demonstrate the relationship between the application of formulae and the units of measurement as described in the latest edition of the Standard System of Measuring Builders' Work (SSMBW).
• Outcomes: By the end of this module, the student should be able to:     - Identify areas and volumes from a technical drawing.     - Successfully book dimensions of various formulae on standard dimension paper.
• Significance in Building Work: Mensuration is essential for determining:     - Dimensions: length, breadth, depth, thickness, and height.     - Descriptions of building elements.     - Units of measurement.     - Physical quantities.     - Methodology and systematic approach to measurement.

Method and Components of Measurements

• Setting Out the Work: This involves utilizing descriptions and dimensions to calculate quantities.
• Sidecasts: These are supporting calculations used when dimensions must be amplified, amended, explained, or clarified.
• References: Documenting the sources of dimensions or information used in descriptions.
• Units of Measurement: Standardized units used for different spatial qualities:     - Perimeter ($P$), Length, or Circumference: Measured in metres ($m$).     - Area ($A$): Measured in square metres ($m^2$).     - Volume ($V$): Measured in cubic metres ($m^3$).

Measurement of 2D Shapes and Areas

• Square:     - Example: Side length of $3000\,mm$ ($3.00\,m$).     - Perimeter ($P$): $3.00\,m \times 4 = 12.00\,m$.     - Area ($A$): $3.00\,m \times 3.00\,m = 9.00\,m^2$.
• Rectangle:     - Example: Dimensions of $7000\,mm \times 3000\,mm$ ($7.00\,m \times 3.00\,m$).     - Perimeter ($P$): $(2 \times 7.00\,m) + (2 \times 3.00\,m) = 20.00\,m$.     - Area ($A$): $7.00\,m \times 3.00\,m = 21.00\,m^2$.
• Trapezium:     - Formula for Area ($A$): $\frac{\text{Sum of top and bottom parallel sides}}{2} \times \text{height}$.     - Example: Top ($3100\,mm$), Bottom ($4750\,mm$), height ($1900\,mm$).     - Calculation Step: $\frac{3.10\,m + 4.75\,m}{2} = 3.925\,m$.     - Result Area: $3.10\,m + 4.75\,m = 7.85\,m$ total of bases; $\frac{7.85\,m}{2} = 3.925\,m$; final calculation shown as $3.93 \times 1.89 = 7.43\,m^2$ (subject to rounding).
• Triangle:     - Formula for Area ($A$): $\frac{1}{2} \times \text{base} \times \text{height}$.     - Example: Base ($7.15\,m$), Sides ($8.98\,m$), Height ($8.24\,m$).     - Perimeter ($P$): $(2 \times 8.98\,m) + 7.15\,m = 25.11\,m$.     - Area ($A$): $\frac{1}{2} \times 7.15\,m \times 8.24\,m = 29.46\,m^2$.
• Circle:     - Constants: $\pi = \frac{22}{7}$.     - Example: Radius ($r$) = $1500\,mm$ ($1.50\,m$), Diameter ($d$) = $3000\,mm$ ($3.00\,m$).     - Circumference ($P$): $2 \times \pi \times r = 9.43\,m$.     - Area ($A$): $\pi \times r^2 = 7.07\,m^2$.
• Sector of a Circle:     - Length ($L$): $2 \times \pi \times r \times \frac{n}{360}$, where $n$ is the angle in degrees.     - Example Sector: Angle $80^{\circ}$, Radius $1.50\,m$.     - Perimeter: $(2 \times 1.50\,m) + (\frac{80}{360} \times 2 \times \pi \times 1.50\,m) = 5.10\,m$.     - Area ($A$): $\pi \times r^2 \times \frac{n}{360} = 1.57\,m^2$.
• Fillet:     - Formula for Area: $R^2 - \frac{\pi \times R^2}{4}$.     - Example: Radius ($R$) = $2500\,mm$ ($2.50\,m$).     - Area calculation: $(2.50 \times 2.50) - \frac{\pi \times 2.50 \times 2.50}{4} = 1.34\,m^2$.

Measurement of 3D Shapes and Solids

• Pyramid (Square Base):     - Volume ($V$): $\frac{1}{3} \times \text{Area of base} \times \text{height}$.     - Area ($A$): $2 \times S \times L$, where $S$ is side length and $L$ is sloping length (Note: This specific formula applies only to square pyramids).     - Alternative Area: (Area of 4 triangles) = $( \frac{1}{2} \times \text{base} \times \text{height} ) \times 4$.
• Pyramid (Non-Square Base):     - Volume calculation based on the specific base area provided.
• Cone:     - Volume ($V$): $\frac{1}{3} \times \text{Area of base} \times \text{height} = \frac{1}{3} \pi r^2 h$.     - Area of sloping side: $\pi \times r \times \text{sloping height}$.
• Partial Pyramid or Frustum (Square Base):     - Volume ($V$): $\frac{H}{3} \times [A + a + \sqrt{A \times a}]$, where $A$ is the area of the large base and $a$ is the area of the small base.     - Area: Area of top + Area of bottom + Area of 4 sloping sides.
• Partial Pyramid (Non-Square Base):     - Volume ($V$): $\frac{H}{3} \times [A + a + \sqrt{A \times a}]$.     - Area: Area of top + Area of bottom + Area of 4 sloping sides.
• Frustum of a Cone:     - Volume ($V$): $\frac{H}{3} \times [A + a + \sqrt{A \times a}]$.     - Area: Sum of top surface area, bottom surface area, and $\frac{1}{2} \times \text{Slant Height} \times (\text{Circumference A} + \text{Circumference B})$.
• Prismoid:     - Definition: A solid with parallel ends and planar sides (like a prism).     - Volume ($V$): $\frac{L}{6} \times [A + a + 4M]$, where $A$ and $a$ are areas of planar ends and $M$ is the area of the middle section.
• Spiral:     - Length ($L$): $\sqrt{C^2 + P^2} \times P_s$.     - Where $C = \pi \times D$ (Circumference), $P$ is pitch (e.g., $3000\,mm$), and $D$ is diameter (e.g., $900\,mm$).

Calculations for Regular Polygons and Irregular Shapes

• Regular Polygons (Equal Sided Figures):     - Formula for Area: $A = S^2 \times \text{constant}$, where $S$ is the length of one side.     - Table of Constants:         - Triangle (3 sides): $0.433013$         - Square (4 sides): $1$         - Pentagon (5 sides): $1.720477$         - Hexagon (6 sides): $2.59876$         - Heptagon (7 sides): $3.633912$         - Octagon (8 sides): $4.828427$         - Nonagon (9 sides): $6.181824$         - Decagon (10 sides): $7.694209$         - Undecagon (11 sides): $9.36564$         - Dodecagon (12 sides): $11.19615$     - Volume determination: $\text{Area} \times \text{height}$.
• Irregular Shapes (Simpson's Rule):     - Used to determine the area of irregular shapes.     - Formula: $\frac{D}{3} \times [\text{first ordinate} + 4(\text{even ordinates}) + 2(\text{remaining ordinates})]$.     - Note: To apply Simpson's rule, the shape must be divided into an even number of strips.

Mensuration Exercises and Practical Problems

• Exercise 1: Octagon Building:     - Internal dimensions: $3436\,mm$ per side.     - External dimensions: $3784\,mm$ per side.     - Wall thickness: $220\,mm$. Height: $3000\,mm$.     - Building Details: 4 standard doors ($813 \times 2032\,mm$), 1 double door with fanlight ($600\,mm$) and sidelight ($400\,mm$), 4 windows ($1295 \times 996\,mm$).     - Material specs: $54\text{ bricks}/m^2$ for $110\,mm$ wall; $15\%$ waste allowance on bricks.     - Tasks: Calculate (a) vinyl floor area, (b) net external plaster area, (c) net number of bricks for superstructure and foundations ($900\,mm$ deep).
• Exercise 2: Circular Column with Frustum Head:     - Column: Diameter $300\,mm$, Height $3\,m$.     - Head (Upside down conical frustum): Top diameter $1\,m$, Head height $350\,mm$.     - Concrete mix: $1:2:5$; Bulking: $30\%$; Waste: $7\%$; Constant: $30.3\text{ pockets}/m^3$.     - Costs: Cement R $85.00/pocket$, Sand R $350.00/m^3$, Stone R $400.00/m^3$.
• Exercise 3: Circular Dam Construction:     - Requirements: Water for $1000$ cattle ($10\,L/day$), $5000$ sheep ($3\,L/day$), and $5000$ goats ($4\,L/day$) for $100$ days.     - Specifications: Wall height $2.5\,m$, wall/floor thickness $500\,mm$.     - Constants: $1000\,litres = 1\,m^3$; $\text{Radius} = \sqrt{\text{Area} \div \frac{22}{7}}$.
• Exercise 4: Steel Springs for Manufacturing:     - Volume: $36,900$ bakkies/year for $6$ years; $4$ springs/bakkie.     - Spring specs: $250\,mm$ diameter, $7$ full circles, each circle $65\,mm$ apart.     - Raw material: High tensile steel rods, each $6000\,mm$ long.     - Tasks: Determine the number of rods required and the percentage waste.
• Exercise 5: Site Excavation Logistics:     - Site ABCD over sloping ground. Sides excavated to angle of repose ($45^{\circ}$).     - Dimensions: $AB = 180\,m$, $CD = 300\,m$, $AD = 200\,m$.     - Depths: $A = 2389\,mm$, $B = 0$, $C = 0$, $D = 3217\,mm$.     - Logistics: $20\%$ bulking factor. Trucks are $7\,m^3$. Working day is $8$ hours. Truck turnaround time (load/travel) is $45$ minutes. Total available time: $20$ days.