Structural analysis of two-dimensional (2D) framed structures is fundamental in ensuring the safety and stability of buildings in Kenya’s construction industry. These frames, consisting of beams and columns arranged in a plane, must be carefully analyzed to determine the internal forces and moments generated under various loading conditions. Mastery of 2D frame analysis enables building technologists to design structures that resist failure and comply with Kenyan building codes and standards.

1.1 Introduction to 2D frame analysis

2D frame analysis involves calculating the forces and moments acting on planar structural systems composed of interconnected beams and columns. These frames typically support floors, roofs, and walls in buildings, transferring loads safely to the foundations. The analysis assumes all members lie in a single plane and connections are usually rigid or semi-rigid, affecting how forces distribute.

1.1.1 Concept of 2D frames

2D frames are assemblies of linear elements, beams and columns, joined at nodes, forming a stable structural skeleton. The members resist bending moments, shear forces, and axial forces, unlike simple trusses which carry only axial forces. This complexity requires methods accounting for bending stiffness and joint moments.

The general equilibrium equations for 2D frames include:

$$ \sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0 $$

These equations ensure the frame is statically determinate or indeterminate, influencing the analysis approach.

Worked Examples

Example 1:
A simple 2D frame has a horizontal beam of length 6 m supported by two columns. Calculate the reactions assuming a vertical load of 12 kN at mid-span.
Given: Load \( P = 12 \, \text{kN} \), Beam length \( L = 6 \, \text{m} \)$$ \sum F_y = 0 $$$$ R_A + R_B = 12 \, \text{kN} $$$$ \sum M_A = 0 $$$$ R_B \times 6 = 12 \times 3 $$$$ R_B = \frac{36}{6} = 6 \, \text{kN} $$$$ R_A = 12 - 6 = 6 \, \text{kN} $$ Answer: Reaction at A = 6 kN, Reaction at B = 6 kN

Example 2:
Calculate the axial force in a vertical column supporting a beam with a 10 kN point load at 2 m from the left support. Beam length is 5 m.
Given: Load \( P = 10 \, \text{kN} \), Load distance \( a = 2 \, \text{m} \), Beam length \( L = 5 \, \text{m} \)$$ R_A + R_B = 10 \, \text{kN} $$$$ \sum M_A = 0 $$$$ R_B \times 5 = 10 \times 2 $$$$ R_B = \frac{20}{5} = 4 \, \text{kN} $$$$ R_A = 10 - 4 = 6 \, \text{kN} $$ Axial force in left column = \( R_A = 6 \, \text{kN} \)
Answer: Axial force = 6 kN

Example 3:
Determine the bending moment at mid-span of a simply supported beam with a uniform load of 4 kN/m over a 6 m span.
Given: Uniform load \( w = 4 \, \text{kN/m} \), Span \( L = 6 \, \text{m} \)$$ M_{max} = \frac{wL^2}{8} $$$$ M_{max} = \frac{4 \times 6^2}{8} $$$$ M_{max} = \frac{4 \times 36}{8} = 18 \, \text{kNm} $$ Answer: Maximum bending moment = 18 kNm

1.1.2 Static determinate and indeterminate frames

Frames can be statically determinate, where equilibrium equations alone solve all reactions and internal forces, or statically indeterminate, requiring compatibility conditions and material properties. Most building frames are indeterminate due to rigid connections, demanding advanced methods like moment distribution or matrix analysis.

The degree of static indeterminacy \( d \) is:

$$ d = r, e $$

where \( r \) is the number of reaction components and \( e \) is the number of equilibrium equations (usually 3 per joint in 2D).

Worked Examples

Example 1:
Calculate the degree of static indeterminacy for a frame with 3 supports each providing 2 reactions and 4 unknown internal forces.
Given: \( r = 3 \times 2 = 6 \), \( e = 3 \)$$ d = 6 - 3 = 3 $$ Answer: Degree of indeterminacy = 3

Example 2:
Determine if a frame with 2 fixed supports (each 3 reactions) and 4 unknown member forces is statically determinate.
Given: \( r = 2 \times 3 = 6 \), \( e = 3 \)$$ d = 6 - 3 = 3 $$ Since \( d > 0 \), frame is statically indeterminate.
Answer: Frame is statically indeterminate by 3

Example 3:
A simply supported beam has 2 supports and 3 unknown reactions. Find its degree of static indeterminacy.
Given: \( r = 3 \), \( e = 3 \)$$ d = 3 - 3 = 0 $$ Answer: Frame is statically determinate

1.1.3 Assumptions in 2D frame analysis

Analysis assumes members behave as linear elastic materials, joints are either rigid or pinned, and loads act in the plane of the frame. Deflections are small, and shear deformations are often neglected. These assumptions simplify calculations but must be validated for each project.

Worked Examples

Example 1:
Calculate the deflection at mid-span of a beam with length 4 m, modulus of elasticity \( E = 200 \times 10^9 \, \text{Pa} \), moment of inertia \( I = 8 \times 10^{-6} \, \text{m}^4 \), under a central point load of 5 kN.
Given: \( P = 5 \, \text{kN} \), \( L = 4\, \text{m} \), \( E = 200 \times 10^9 \, \text{Pa} \), \( I = 8 \times 10^{-6} \, \text{m}^4 \)$$ \delta = \frac{PL^3}{48EI} $$$$ \delta = \frac{5 \times 10^3 \times 4^3}{48 \times 200 \times 10^9 \times 8 \times 10^{-6}} $$$$ \delta = \frac{5 \times 10^3 \times 64}{48 \times 200 \times 10^9 \times 8 \times 10^{-6}} $$$$ \delta = \frac{320000}{76800 \times 10^3} = 4.17 \times 10^{-3} \, \text{m} = 4.17 \, \text{mm} $$ Answer: Deflection = 4.17 mm

Example 2:
Assuming small deflections, calculate the bending stress at the extreme fiber of a beam with bending moment 10 kNm, section modulus \( Z = 0.0002 \, \text{m}^3 \).
Given: \( M = 10 \times 10^3 \, \text{Nm} \), \( Z = 0.0002\, \text{m}^3 \)$$ \sigma = \frac{M}{Z} $$$$ \sigma = \frac{10 \times 10^3}{0.0002} = 50 \times 10^6 \, \text{Pa} = 50 \, \text{MPa} $$ Answer: Bending stress = 50 MPa

Example 3:
Calculate the shear force at a support of a simply supported beam of length 5 m carrying a uniform load of 3 kN/m.
Given: \( w = 3 \, \text{kN/m} \), \( L = 5 \, \text{m} \)$$ V = \frac{wL}{2} $$$$ V = \frac{3 \times 5}{2} = 7.5 \, \text{kN} $$ Answer: Shear force at support = 7.5 kN

1.1.4 Methods of analysis for 2D frames

Common methods include:

• Moment distribution method: Iterative approach balancing moments at joints.
• Slope-deflection equations: Use member rotations and displacements.
• Matrix stiffness method: Computer-based, using system matrices.
• Approximate methods: Simplify for quick estimates.

Each method suits different frame complexity and available computation resources.

Worked Examples

Example 1:
Use moment distribution for a fixed-ended beam of 6 m with a central load of 9 kN. Calculate fixed-end moments.
Given: \( P = 9 \, \text{kN} \), \( L = 6 \, \text{m} \)$$ M_{AB} = M_{BA} = -\frac{PL}{8} = -\frac{9 \times 6}{8} = -6.75 \, \text{kNm} $$ Answer: Fixed-end moments = -6.75 kNm

Example 2:
Calculate slope-deflection moment at joint A of a beam AB with length 5 m, modulus \( E = 200 \times 10^9 \, \text{Pa} \), moment of inertia \( I = 10^{-5} \, \text{m}^4 \), fixed at A and free at B, subjected to a moment \( M = 5 \, \text{kNm} \).
Given:$$ M_{AB} = \frac{2EI}{L} (2\theta_A + \theta_B) + M_{fixed} $$ Assuming \( \theta_B = 0 \), \( \theta_A \) unknown, solve iteratively.
Answer: Requires iterative calculation beyond scope here.

Example 3:
Using matrix stiffness, calculate member stiffness \( k \) for a beam of length 4 m, \( E = 210 \times 10^9 \, \text{Pa} \), \( I = 12 \times 10^{-6} \, \text{m}^4 \).
Given:$$ k = \frac{4EI}{L} $$$$ k = \frac{4 \times 210 \times 10^9 \times 12 \times 10^{-6}}{4} $$$$ k = 252000 \, \text{Nm/rad} $$ Answer: Member stiffness = 252000 Nm/rad

Chapter Summary

This chapter introduced the analysis of forces in two-dimensional framed structures, emphasizing the importance of understanding load types, support conditions, and internal forces for structural stability. Various loads acting on 2D frames were examined, including point loads, distributed loads, and moments, each affecting the frame differently. The chapter detailed the characteristics of common supports such as pinned, fixed, and roller supports, explaining their influence on the structure’s response. The fundamental equilibrium equations were presented as the basis for analyzing forces within the frame. Internal forces in members, including axial forces, shear forces, and bending moments, were discussed to reveal how members carry loads. Different methods for force analysis were explored, focusing on the methods of joints and sections for determining member forces and the moment distribution method for analyzing indeterminate frames. Together, these concepts provide a comprehensive framework for evaluating and designing safe and efficient 2D framed structures.