By the end of this chapter, you will be able to:
Mastering these skills will help you confidently analyze and work with composite materials, a key part of modern structural design and construction.
Compiled by: Beatrice Aoko
Composite materials are increasingly important in modern building technology, offering enhanced performance by combining two or more distinct materials. In the Kenyan construction industry, the use of composites can improve structural efficiency, reduce costs, and provide durability under diverse environmental conditions. Understanding the principles and calculations related to composite materials is essential for building technologists aiming to design and analyze composite structural elements accurately.
Composite materials consist of two or more constituent materials with different physical or chemical properties that, when combined, produce a material with characteristics distinct from the individual components. The analysis of composite materials in building technology involves understanding their behavior under load, their stiffness, and strength characteristics.
Composite materials exhibit combined stress-strain behavior influenced by the properties of their constituents. The overall stress in a composite is the sum of stresses in each material, weighted by their respective areas or volumes. The fundamental formula governing the axial stress in a composite member under load is:
$$ \sigma_c = \frac{P}{A_c} $$
where \( \sigma_c \) is the composite stress, \( P \) is the axial load, and \( A_c \) is the equivalent composite cross-sectional area.
When two materials are bonded and subjected to axial load, the strain \( \varepsilon \) is the same in both materials, assuming perfect bonding and no slip. The stresses in each material are related to their moduli of elasticity as:
$$ \sigma_1 = E_1 \varepsilon, \quad \sigma_2 = E_2 \varepsilon $$
The load carried by each material is:
$$ P_1 = \sigma_1 A_1 = E_1 \varepsilon A_1, \quad P_2 = \sigma_2 A_2 = E_2 \varepsilon A_2 $$
The total load is:
$$ P = P_1 + P_2 = \varepsilon (E_1 A_1 + E_2 A_2) $$
The composite modulus \( E_c \) is:
$$ E_c = \frac{P}{\varepsilon (A_1 + A_2)} = \frac{E_1 A_1 + E_2 A_2}{A_1 + A_2} $$
Example 1: A composite column consists of a steel section and a concrete section bonded together. The steel area \( A_1 = 5000 \, \text{mm}^2 \), modulus of steel \( E_1 = 200 \times 10^3 \, \text{MPa} \), concrete area \( A_2 = 10000 \, \text{mm}^2 \), modulus of concrete \( E_2 = 25 \times 10^3 \, \text{MPa} \). Calculate the composite modulus \( E_c \).
Given:
\( A_1 = 5000 \, \text{mm}^2 \), \( E_1 = 200000 \, \text{MPa} \)
\( A_2 = 10000 \, \text{mm}^2 \), \( E_2 = 25000 \, \text{MPa} \)
Formula:
\( E_c = \frac{E_1 A_1 + E_2 A_2}{A_1 + A_2} \)
Substitution:
\( E_c = \frac{200000 \times 5000 + 25000 \times 10000}{5000 + 10000} \)
Calculate numerator:
\( 200000 \times 5000 = 1,000,000,000 \)
\( 25000 \times 10000 = 250,000,000 \)
Sum = \( 1,250,000,000 \)
Calculate denominator:
\( 5000 + 10000 = 15000 \)
Calculate \( E_c \):
\( E_c = \frac{1,250,000,000}{15000} = 83333.33 \, \text{MPa} \)
Answer: \( E_c = 83,333.33 \, \text{MPa} \)
Example 2: A composite beam has a steel area of \( 3000 \, \text{mm}^2 \) with \( E_1 = 210 \times 10^3 \, \text{MPa} \), and a timber area of \( 7000 \, \text{mm}^2 \) with \( E_2 = 12 \times 10^3 \, \text{MPa} \). Find the composite modulus.
Given:
\( A_1 = 3000 \, \text{mm}^2 \), \( E_1 = 210000 \, \text{MPa} \)
\( A_2 = 7000 \, \text{mm}^2 \), \( E_2 = 12000 \, \text{MPa} \)
Formula:
\( E_c = \frac{E_1 A_1 + E_2 A_2}{A_1 + A_2} \)
Substitution:
\( E_c = \frac{210000 \times 3000 + 12000 \times 7000}{3000 + 7000} \)
Calculate numerator:
\( 210000 \times 3000 = 630,000,000 \)
\( 12000 \times 7000 = 84,000,000 \)
Sum = \( 714,000,000 \)
Calculate denominator:
\( 3000 + 7000 = 10000 \)
Calculate \( E_c \):
\( E_c = \frac{714,000,000}{10000} = 71400 \, \text{MPa} \)
Answer: \( E_c = 71,400 \, \text{MPa} \)
Example 3: A reinforced concrete beam has steel reinforcement area \( A_1 = 1500 \, \text{mm}^2 \) with \( E_1 = 200 \times 10^3 \, \text{MPa} \) and concrete area \( A_2 = 30000 \, \text{mm}^2 \) with \( E_2 = 30 \times 10^3 \, \text{MPa} \). Calculate the composite modulus.
Given:
\( A_1 = 1500 \, \text{mm}^2 \), \( E_1 = 200000 \, \text{MPa} \)
\( A_2 = 30000 \, \text{mm}^2 \), \( E_2 = 30000 \, \text{MPa} \)
Formula:
\( E_c = \frac{E_1 A_1 + E_2 A_2}{A_1 + A_2} \)
Substitution:
\( E_c = \frac{200000 \times 1500 + 30000 \times 30000}{1500 + 30000} \)
Calculate numerator:
\( 200000 \times 1500 = 300,000,000 \)
\( 30000 \times 30000 = 900,000,000 \)
Sum = \( 1,200,000,000 \)
Calculate denominator:
\( 1500 + 30000 = 31500 \)
Calculate \( E_c \):
\( E_c = \frac{1,200,000,000}{31500} = 38095.24 \, \text{MPa} \)
Answer: \( E_c = 38,095.24 \, \text{MPa} \)
The modular ratio \( n \) is a key concept used to transform composite sections into an equivalent section of one material for analysis. It is defined as the ratio of the modulus of elasticity of one material to another, typically steel to concrete in building structures:
$$ n = \frac{E_s}{E_c} $$
where \( E_s \) is the modulus of elasticity of steel and \( E_c \) is that of concrete.
The transformed section method replaces the area of one material with an equivalent area of the other material by multiplying by \( n \), enabling easier stress and strain calculations.
Example 1: Calculate the modular ratio for steel with \( E_s = 200 \times 10^3 \, \text{MPa} \) and concrete with \( E_c = 25 \times 10^3 \, \text{MPa} \).
Given:
\( E_s = 200000 \, \text{MPa} \), \( E_c = 25000 \, \text{MPa} \)
Formula:
\( n = \frac{E_s}{E_c} \)
Substitution:
\( n = \frac{200000}{25000} \)
Calculate:
\( n = 8 \)
Answer: \( n = 8 \)
Example 2: For steel \( E_s = 210 \times 10^3 \, \text{MPa} \) and timber \( E_c = 12 \times 10^3 \, \text{MPa} \), calculate the modular ratio.
Given:
\( E_s = 210000 \, \text{MPa} \), \( E_c = 12000 \, \text{MPa} \)
Formula:
\( n = \frac{E_s}{E_c} \)
Substitution:
\( n = \frac{210000}{12000} \)
Calculate:
\( n = 17.5 \)
Answer: \( n = 17.5 \)
Example 3: A composite beam has steel \( E_s = 195 \times 10^3 \, \text{MPa} \) and concrete \( E_c = 30 \times 10^3 \, \text{MPa} \). Find the modular ratio.
Given:
\( E_s = 195000 \, \text{MPa} \), \( E_c = 30000 \, \text{MPa} \)
Formula:
\( n = \frac{E_s}{E_c} \)
Substitution:
\( n = \frac{195000}{30000} \)
Calculate:
\( n = 6.5 \)
Answer: \( n = 6.5 \)
A composite beam has a steel area \(A_s = 1800\,\text{mm}^2\) and concrete area \(A_c = 9000\,\text{mm}^2\). The modular ratio is \(n = 8\). Find the equivalent transformed steel area in terms of concrete.
Given:
Formula:
Substitution:
Answer: Equivalent transformed steel area is 14,400 mm² (in terms of concrete)
A composite beam has a concrete area \(A_c = 12000\,\text{mm}^2\) and steel area \(A_s = 2000\,\text{mm}^2\). The modular ratio is \(n = 10\). Find the equivalent transformed concrete area in terms of steel.
Given:
Formula:
Substitution:
Answer: Equivalent transformed concrete area is 1200 mm² (in terms of steel)
Determining the neutral axis (NA) location and moment of inertia (I) for composite sections is essential for designing beams and columns. Using the transformed section method, the area of one material is converted using the modular ratio to an equivalent area, and then the NA and I are calculated as for a homogeneous section.
The neutral axis location \( \bar{y} \) from a reference axis is given by:
$$ \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} $$
where \( A_i \) are the areas (transformed where necessary) and \( y_i \) are the distances from the reference axis to the centroid of each area.
The moment of inertia \( I \) is:
$$ I = \sum ( I_i + A_i d_i^2 ) $$
where \( I_i \) is the moment of inertia of each area about its own centroid, and \( d_i \) is the distance from the area centroid to the neutral axis.
Example 1: A composite beam has a concrete rectangular section 300 mm wide and 500 mm deep, with steel reinforcement at the bottom. Steel area \( A_s = 2000 \, \text{mm}^2 \), modular ratio \( n = 8 \). Find the neutral axis from the top.
Given:
Concrete section: width \( b = 300 \, \text{mm} \), depth \( h = 500 \, \text{mm} \)
Steel area \( A_s = 2000 \, \text{mm}^2 \) at 480 mm from top (20 mm cover)
Modular ratio \( n = 8 \)
Step 1: Transform steel area:
\( A_s' = n \times A_s = 8 \times 2000 = 16000 \, \text{mm}^2 \)
Step 2: Calculate concrete area:
\( A_c = b \times h = 300 \times 500 = 150000 \, \text{mm}^2 \)
Step 3: Calculate centroid of concrete area from top:
\( y_c = \frac{h}{2} = \frac{500}{2} = 250 \, \text{mm} \)
Step 4: Calculate neutral axis:
$$ \bar{y} = \frac{A_c y_c + A_s' y_s}{A_c + A_s'} $$
Substitute:
$$ \bar{y} = \frac{150000 \times 250 + 16000 \times 480}{150000 + 16000} $$
Calculate numerator:
\( 150000 \times 250 = 37,500,000 \)
\( 16000 \times 480 = 7,680,000 \)
Sum = \( 45,180,000 \)
Calculate denominator:
\( 150000 + 16000 = 166000 \)
Calculate \( \bar{y} \):
\( \bar{y} = \frac{45,180,000}{166000} = 272.05 \, \text{mm} \)
Answer: Neutral axis is 272.05 mm from the top
Example 2: For a composite beam with timber width 200 mm and depth 400 mm, steel reinforcement area 1500 mm² at 380 mm from top, modular ratio \( n = 15 \), find the neutral axis.
Given:
Timber area \( A_t = 200 \times 400 = 80000 \, \text{mm}^2 \)
Steel area \( A_s = 1500 \, \text{mm}^2 \)
Steel centroid \( y_s = 380 \, \text{mm} \)
Modular ratio \( n = 15 \)
Timber centroid \( y_t = 200 \, \text{mm} \)
Step 1: Transform steel area:
\( A_s' = n \times A_s = 15 \times 1500 = 22500 \, \text{mm}^2 \)
Step 2: Calculate neutral axis:
$$ \bar{y} = \frac{A_t y_t + A_s' y_s}{A_t + A_s'} $$
Substitute:
$$ \bar{y} = \frac{80000 \times 200 + 22500 \times 380}{80000 + 22500} $$
Calculate numerator:
\( 80000 \times 200 = 16,000,000 \)
\( 22500 \times 380 = 8,550,000 \)
Sum = \( 24,550,000 \)
Calculate denominator:
\( 80000 + 22500 = 102500 \)
Calculate \( \bar{y} \):
\( \bar{y} = \frac{24,550,000}{102500} = 239.51 \, \text{mm} \)
Answer: Neutral axis is 239.51 mm from the top
Example 3: A reinforced concrete beam has concrete width 250 mm, depth 450 mm, steel area 1800 mm² located 430 mm from the top. Modular ratio \( n = 7 \). Find the neutral axis location.
Given:
Concrete area \( A_c = 250 \times 450 = 112500 \, \text{mm}^2 \)
Steel area \( A_s = 1800 \, \text{mm}^2 \)
Steel centroid \( y_s = 430 \, \text{mm} \)
Concrete centroid \( y_c = 225 \, \text{mm} \)
Modular ratio \( n = 7 \)
Step 1: Transform steel area:
\( A_s' = 7 \times 1800 = 12600 \, \text{mm}^2 \)
Step 2: Calculate neutral axis:
$$ \bar{y} = \frac{112500 \times 225 + 12600 \times 430}{112500 + 12600} $$
Calculate numerator:
\( 112500 \times 225 = 25,312,500 \)
\( 12600 \times 430 = 5,418,000 \)
Sum = \( 30,730,500 \)
Calculate denominator:
\( 112500 + 12600 = 125100 \)
Calculate \( \bar{y} \):
\( \bar{y} = \frac{30,730,500}{125100} = 245.74 \, \text{mm} \)
Answer: Neutral axis is 245.74 mm from the top
In composite beams, bending stresses vary according to the transformed section properties and the position relative to the neutral axis. The bending stress \( \sigma \) at a distance \( y \) from the neutral axis is given by:
$$ \sigma = \frac{M y}{I} $$
where \( M \) is the bending moment, \( I \) is the moment of inertia of the transformed section, and \( y \) is the distance from the neutral axis to the point of interest.
For composite materials, stresses in the actual materials are found by adjusting the transformed stress by the modular ratio:
$$ \sigma_s = n \times \sigma_{transformed} $$
Example 1: A composite beam has a bending moment of 50 kNm, moment of inertia \( I = 2.5 \times 10^{10} \, \text{mm}^4 \), and neutral axis at 200 mm from the top. Calculate the bending stress at the top fiber (distance \( y = 200 \, \text{mm} \)).
Given:
\( M = 50 \, \text{kNm} = 50 \times 10^6 \, \text{Nmm} \)
\( I = 2.5 \times 10^{10} \, \text{mm}^4 \)
\( y = 200 \, \text{mm} \)
Formula:
\( \sigma = \frac{M y}{I} \)
Substitution:
\( \sigma = \frac{50 \times 10^6 \times 200}{2.5 \times 10^{10}} \)
Calculate numerator:
\( 50 \times 10^6 \times 200 = 1 \times 10^{10} \)
Calculate stress:
\( \sigma = \frac{1 \times 10^{10}}{2.5 \times 10^{10}} = 0.4 \, \text{N/mm}^2 = 0.4 \, \text{MPa} \)
Answer: Bending stress at top fiber is 0.4 MPa (compression or tension depending on loading)
Example 2: For the same beam, calculate bending stress at the bottom fiber 300 mm from neutral axis.
Given:
\( y = 300 \, \text{mm} \)
Formula:
\( \sigma = \frac{M y}{I} \)
Substitution:
\( \sigma = \frac{50 \times 10^6 \times 300}{2.5 \times 10^{10}} \)
Calculate numerator:
\( 50 \times 10^6 \times 300 = 1.5 \times 10^{10} \)
Calculate stress:
\( \sigma = \frac{1.5 \times 10^{10}}{2.5 \times 10^{10}} = 0.6 \, \text{MPa} \)
Answer: Bending stress at bottom fiber is 0.6 MPa
Example 3: A composite beam with modular ratio \( n = 7 \) has a transformed bending stress in steel of 20 MPa. Find the actual steel stress.
Given:
\( n = 7 \), \( \sigma_{transformed} = 20 \, \text{MPa} \)
Formula:
\( \sigma_s = n \times \sigma_{transformed} \)
Substitution:
\( \sigma_s = 7 \times 20 \)
Calculate:
\( \sigma_s = 140 \, \text{MPa} \)
Answer: Actual steel stress is 140 MPa
Example 4: A beam section has \( I = 3 \times 10^{10} \, \text{mm}^4 \), bending moment \( M = 80 \, \text{kNm} \), and distance to steel reinforcement \( y = 450 \, \text{mm} \). Calculate bending stress in concrete at the steel level.
Given:
\( M = 80 \times 10^6 \, \text{Nmm} \), \( I = 3 \times 10^{10} \, \text{mm}^4 \), \( y = 450 \, \text{mm} \)
Formula:
\( \sigma = \frac{M y}{I} \)
Substitution:
\( \sigma = \frac{80 \times 10^6 \times 450}{3 \times 10^{10}} \)
Calculate numerator:
\( 80 \times 10^6 \times 450 = 36 \times 10^{9} \)
Calculate stress:
\( \sigma = \frac{36 \times 10^{9}}{3 \times 10^{10}} = 1.2 \, \text{MPa} \)
Answer: Bending stress in concrete at steel level is 1.2 MPa
Example 5: For the above, if modular ratio \( n = 8 \), calculate the actual steel stress when transformed steel stress is 15 MPa.
Given:
\( n = 8 \), \( \sigma_{transformed} = 15 \, \text{MPa} \)
Formula:
\( \sigma_s = n \times \sigma_{transformed} \)
Substitution:
\( \sigma_s = 8 \times 15 \)
Calculate:
\( \sigma_s = 120 \, \text{MPa} \)
Answer: Actual steel stress is 120 MPa
A composite column has steel area 4000 mm² with modulus 210,000 MPa and concrete area 12,000 mm² with modulus 30,000 MPa. Calculate the composite modulus of elasticity. (5 marks)
Calculate the modular ratio for steel modulus 200,000 MPa and concrete modulus 28,000 MPa. (3 marks)
A composite beam has concrete section 250 mm wide and 450 mm deep, steel reinforcement area 1800 mm² placed 430 mm from top. If modular ratio is 7, find the neutral axis location from the top. (7 marks)
For a composite beam with bending moment 60 kNm and moment of inertia \( 2.8 \times 10^{10} \, \text{mm}^4 \), calculate the bending stress at a fiber 300 mm from the neutral axis. (5 marks)
Given a transformed steel bending stress of 18 MPa and modular ratio 9, find the actual steel stress. (3 marks)
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Create a free accountThis chapter introduces composite materials as engineered combinations of two or more constituent materials with distinct physical or chemical properties, designed to achieve superior performance. It then develops the fundamental compatibility and equilibrium equations necessary to analyze the behavior of composite structures under load, ensuring deformation compatibility and force balance between components. The chapter proceeds to solve composite materials problems by applying these equations to determine stresses and strains in layered or bonded materials subjected to various loading conditions. Finally, it examines the calculation of composite material forces, detailing how the internal forces are distributed among the different constituents based on their mechanical properties and geometric arrangements. Through this progression, learners gain a comprehensive understanding of how composite materials respond structurally and how to analyze them effectively for engineering applications.
A composite beam consists of a 200 mm wide steel plate and a 400 mm wide concrete slab connected rigidly. If the modulus of elasticity for steel is \(E_s = 200 \times 10^9\) Pa and for concrete is \(E_c = 25 \times 10^9\) Pa, calculate the modular ratio \(n\) of steel to concrete. (2 marks)
A composite beam has a steel flange 150 mm wide and 20 mm thick and a concrete web 300 mm wide and 200 mm thick. Using the modular ratio \(n = 8\), find the equivalent transformed width of the concrete web in terms of steel. (3 marks)
Type: Individual
| Tools & Equipment | Materials |
|---|---|
| Magnifying glass | Fiber Reinforced Polymer (FRP) sample panels |
| Digital scale | Metal Matrix Composite (MMC) blocks |
| Micrometer screw gauge | Ceramic Matrix Composite (CMC) tiles |
| Hardness testing kit | Polymer Matrix Composite (PMC) sheets |
| Notebook and pencil | Reference classification chart for composite materials |
| Personal Protective Equipment (PPE) |
| S/N | Item | Quantity |
|---|---|---|
| 1 | Fiber Reinforced Polymer (FRP) sample panels | 3 pcs per Candidate |
| 2 | Metal Matrix Composite (MMC) blocks | 2 pcs per Candidate |
| 3 | Ceramic Matrix Composite (CMC) tiles | 2 pcs per Candidate |
| 4 | Polymer Matrix Composite (PMC) sheets | 3 pcs per Candidate |
| 5 | Magnifying glass | 1 pc per Candidate |
| 6 | Digital scale (up to 500 g capacity) | 1 pc per 5 Candidates |
| 7 | Micrometer screw gauge | 1 pc per 5 Candidates |
| 8 | Hardness testing kit | 1 kit per 5 Candidates |
| 9 | Reference classification chart for composite materials | 1 chart per Candidate |
| 10 | Notebook and pencil | 1 set per Candidate |
| 11 | Personal Protective Equipment (PPE): dustcoat, safety boots, gloves | 1 set per Candidate |
| Items to be Evaluated | Marks Available | Marks Obtained | Comments |
|---|---|---|---|
| TASK 1: Identification and Classification Process | |||
| Donning the required PPE (dustcoat, safety boots, gloves) (Award 1 mark for each correctly donned PPE or zero) | 3 | ||
| Carefully inspecting each composite sample using magnifying glass (Award 1 mark for inspecting each type of composite sample: FRP, MMC, CMC, PMC) | 4 | ||
| Measuring sample dimensions accurately with micrometer screw gauge (Award 1 mark for measuring thickness of each composite type correctly) | 4 | ||
| Weighing samples using digital scale and recording weights (Award 1 mark for weighing each composite type and recording values) | 4 | ||
| Performing hardness test on samples using hardness testing kit (Award 1 mark for hardness test on each composite type and recording results) | 4 | ||
| Using reference classification chart to categorize each sample correctly (Award 1 mark for correct classification of each composite sample) | 4 | ||
| Recording all observations and classifications clearly in the notebook (Award 1 mark for clarity and completeness of notes per composite type) | 3 | ||
| Cleaning and returning tools and samples to designated area (Award 2 marks for proper cleaning and organization) | 2 | ||
| Sub-Total | 28 | ||
| PRODUCT CHECKLIST | |||
| Accurate classification of all four composite material samples (Award 2 marks for each correctly classified composite: FRP, MMC, CMC, PMC) | 8 | ||
| Correct recording of sample dimensions (thickness) matching measured values (±0.1 mm tolerance) (Award 1 mark for each composite sample's dimension accuracy) | 4 | ||
| Correct recording of sample weight matching measured values (±1 g tolerance) (Award 1 mark for each composite sample's weight accuracy) | 4 | ||
| Complete and clear report with observations, test results, and classification summary (Award up to 6 marks for completeness, clarity, and neatness) | 6 | ||
| Sub-Total | 22 | ||
| GRAND TOTAL | 50 | ||
Type: Individual
| Tools & Equipment | Materials |
|---|---|
| Pencil | Engineering graph paper |
| Ruler (300 mm) | Sample composite beam section model (steel and concrete layers) |
| Set square | |
| Steel scale (300 mm) | |
| Calculator |
| S/N | Item | Quantity |
|---|---|---|
| 1 | Engineering graph paper | 1 Sheet per Candidate |
| 2 | Pencil | 1 Pc per Candidate |
| 3 | Ruler (300 mm) | 1 Pc per Candidate |
| 4 | Calculator | 1 Pc per Candidate |
| 5 | Set square | 1 Pc per Candidate |
| 6 | Steel scale (300 mm) | 1 Pc per Candidate |
| 7 | Sample composite beam section model (steel and concrete layers) | 1 Pc per Candidate |
| 8 | Personal Protective Equipment (dust coat, helmet, safety boots) | 1 Set per Candidate |
| Items to be Evaluated | Marks Available | Marks Obtained | Comments |
|---|---|---|---|
| TASK 1: Preparation and Setup | |||
| Donned Personal Protective Equipment (dust coat, helmet, safety boots) (Award 1 mark for each PPE worn) | 3 | ||
| Examined the composite beam section model for layer identification (Award 2 marks for correctly identifying steel and concrete layers) | 2 | ||
| Set up drawing tools and materials on the workspace (Award 1 mark for proper setup) | 1 | ||
| Marked the beam section dimensions on graph paper accurately (Award 1 mark for each correct dimension line: width, depth, layer thickness) | 3 | ||
| Labeled all layers and interfaces clearly on the drawing (Award 2 marks for clear and correct labeling) | 2 | ||
| Prepared coordinate axes and neutral axis location for section (Award 3 marks for correct axes and neutral axis setup) | 3 | ||
| Sub-Total | 14 | ||
| TASK 2: Derivation of Compatibility Equations | |||
| Applied strain compatibility concept between steel and concrete layers (Award 4 marks for correct conceptual application) | 4 | ||
| Expressed strain relationships at the interface of steel and concrete layers (Award 5 marks for accurate strain relationship expressions) | 5 | ||
| Formulated the compatibility equations relating strains and displacements (Award 6 marks for complete and correct equations) | 6 | ||
| Used appropriate assumptions for linear strain variation across the section (Award 3 marks for correct assumptions) | 3 | ||
| Presented final compatibility equations neatly and logically (Award 4 marks for clear and logical presentation) | 4 | ||
| Sub-Total | 22 | ||
| TASK 3: Verification and Conclusion | |||
| Verified equilibrium conditions for the composite section (Award 3 marks for correct equilibrium verification) | 3 | ||
| Explained physical meaning of compatibility equations in context (Award 3 marks for clear explanation) | 3 | ||
| Cleaned and organized workspace after completion (Award 2 marks for cleanliness and organization) | 2 | ||
| Sub-Total | 8 | ||
| PRODUCT CHECKLIST | |||
| Correctly drawn composite beam section with dimensions 600mm width and 300mm depth (Award 5 marks for accurate dimensions and neat drawing) | 5 | ||
| Accurate and complete compatibility equations derived for the composite layers (Award 10 marks for correctness and completeness) | 10 | ||
| Clear labeling of layers, interfaces, neutral axis, and strain variables (Award 5 marks for clarity and completeness) | 5 | ||
| Logical presentation and neatness of final work (Award 5 marks for overall presentation and neatness) | 5 | ||
| Sub-Total | 25 | ||
| GRAND TOTAL | 69 | ||
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