Architecture  ·  Level 6
Structural Analysis Principles
Chapter 1: Apply shear force and bending moments structural analysis principles
📚 3 Topics
⚑ What you will be able to do

By the end of this chapter, you will be able to:

  • correctly choose and apply different types of supports according to the needs of structural analysis,
  • accurately calculate support reactions for beams and frames using the proper procedures,
  • prepare clear and accurate shear force and bending moment diagrams that meet industry standards.

Mastering these skills will help you solve real structural problems confidently and contribute to safe, reliable construction projects.

Structural analysis principles such as shear force and bending moment calculations are essential in architectural design to ensure the safety and stability of building elements under load. In Kenya, architects must understand these principles to collaborate effectively with structural engineers and ensure compliance with local building codes. This chapter focuses on applying shear force and bending moment analysis to typical architectural structural elements like beams and frames, emphasizing practical calculations relevant to Kenyan construction contexts.

1.1 Types of supports applied

Structural elements in buildings are subjected to various types of loads and are supported in different ways that influence their behavior under stress. Understanding the types of loads and supports is critical for calculating shear forces and bending moments accurately.

1.1.1 Types of loading

Loads on structural elements can be classified based on their nature and application point. These loads create internal forces such as shear and bending moments that must be analyzed to design safe structures.

1.1.1.1 Point load

A point load is a concentrated force acting at a single point on a beam or frame member. In reality every load is spread over some small area, but when that area is very short compared with the span, the load is treated as acting at one point, which simplifies analysis. It causes abrupt changes in shear force and bending moment diagrams. Point loads are measured in kilonewtons (kN) and are drawn as a single arrow at the point of application. Example: a 2 000 kg steel water tank resting on a short stand over the middle of a roof beam at a primary school is treated as a point load on that beam.

Characteristics and examples of point loads
  • Single line of action: The whole load acts through one point, shown by one arrow.
  • Units: Expressed as a force in newtons (N) or kilonewtons (kN).
  • Column loads: A column or post standing on a transfer beam applies a point load to it.
  • Secondary beams: A beam resting on a main beam applies its reaction as a point load.
  • Equipment loads: Tanks, machines and heavy fittings on small bases act as point loads.
  • Effect on diagrams: The shear force jumps suddenly at the load point, and the bending moment diagram changes slope there.
Worked example: converting a mass to a point load

Force = mass × gravitational acceleration

The tank and its water have a mass of 2 000 kg, with g = 9.81 m/s².

Force = 2 000 × 9.81

Force = 19 620 N

Force = 19.62 kN

1.1.1.2 Uniformly distributed loading

A uniformly distributed load (UDL) is a load spread evenly along the length of a beam, so that every metre carries the same load. It is expressed as a load per unit length, in kilonewtons per metre (kN/m), and is drawn as a row of equal arrows or a rectangle along the loaded length. A UDL produces a linear variation in shear force and a quadratic variation in bending moment. For calculating reactions, a UDL is replaced by a single total load equal to the intensity multiplied by the loaded length, acting at the middle of that length. Example: a reinforced concrete beam supporting the floor slab of a hospital ward carries the weight of the slab, finishes and occupants as a UDL along its span.

Characteristics and examples of UDLs
  • Constant intensity: The load per metre is the same along the loaded length.
  • Units: Expressed in kN/m.
  • Self-weight of beams: The weight of a beam of uniform section acts as a UDL along it.
  • Floor and roof loads: Slabs, screeds and imposed loads transferred to beams act as UDLs.
  • Walls on beams: A wall of uniform height built along a beam applies a UDL.
  • Equivalent total load: Total load = intensity × length, acting at the centre of the loaded length.
Worked example: self-weight of a concrete beam as a UDL

UDL = width × depth × unit weight of concrete

A beam is 0.2 m wide and 0.4 m deep, and reinforced concrete weighs 24 kN/m³.

UDL = 0.2 × 0.4 × 24

UDL = 1.92 kN/m

Worked example: total load from a UDL

Total load = UDL × length

A beam 8 m long carries a UDL of 5 kN/m.

Total load = 5 × 8

Total load = 40 kN, acting at 8 ÷ 2 = 4 m from either end

1.1.1.3 Varying loading

A varying load, also called a non-uniformly distributed load, is a distributed load whose intensity changes along the length of the member. The most common forms are the triangular load, which rises from zero at one end to a maximum at the other, and the trapezoidal load, which changes from one value to another. Load intensity varies over the length, requiring integration, or the use of its resultant, to find shear and moment values. For a triangular load, the total load equals half the maximum intensity multiplied by the length, and it acts at the centroid of the triangle, one third of the length from the maximum end. Example: the wall of a rainwater storage tank on a dairy farm carries water pressure that increases from zero at the surface to a maximum at the base, a triangular varying load.

Characteristics and examples of varying loads
  • Changing intensity: The load per metre increases or decreases along the member.
  • Triangular load: Rises from zero to a maximum intensity, such as water pressure on a tank wall.
  • Trapezoidal load: Changes between two non-zero values, often found where slabs load beams unequally.
  • Soil pressure: Earth pressure on retaining walls increases with depth.
  • Resultant of a triangular load: Total load = ½ × maximum intensity × length, acting at L ÷ 3 from the maximum end.
  • Analysis: A trapezoidal load can be split into a rectangle (UDL) and a triangle and each part treated separately.
Worked example: resultant of a triangular load

Total load = ½ × maximum intensity × length

A beam 10 m long carries a load rising from zero at the left end to 6 kN/m at the right end.

Total load = 0.5 × 6 × 10

Total load = 30 kN

Position from the zero end = 2 × L ÷ 3

Position = 2 × 10 ÷ 3

Position = 6.67 m from the left end (3.33 m from the maximum end)

Moment Load

A moment load is a couple applied at a point creating bending without shear, important in analyzing frames and cantilever beams. It is measured in kilonewton metres (kNm) and is drawn as a curved arrow showing its direction of rotation.

1.1.2 Types of supports

Supports in structural systems restrain movement and rotation in different ways, influencing the internal forces generated in beams and frames. The type of support determines the boundary conditions for analysis.

1.1.2.1 Simply supported

A simply supported beam is a beam carried on two supports, one at each end, usually a pin (hinge) at one end and a roller at the other, so that the beam is free to rotate at both ends and free to expand along its length. This arrangement gives three unknown reactions (two at the pin and one at the roller), which can all be found from the three equations of static equilibrium, so the beam is statically determinate. Under downward loads a simply supported beam sags, with zero bending moment at the supports and maximum moment within the span. Example: the reinforced concrete lintels over windows in a SACCO office block are designed as simply supported beams bearing on the walls at each side.

Features of a simply supported beam
  • Two supports: One support at each end of the span, usually pin and roller.
  • Free rotation at ends: The ends can rotate, so no moment is resisted at the supports.
  • Three reactions: A vertical and a horizontal reaction at the pin and a vertical reaction at the roller.
  • Statically determinate: Reactions are found using ΣV = 0, ΣH = 0 and ΣM = 0.
  • Sagging behaviour: The beam bends downwards, with tension in the bottom fibres.
  • Common uses: Lintels, timber floor joists, precast beams and simple bridge spans.
Worked example: degree of statical indeterminacy

Degree of indeterminacy = number of unknown reactions − number of equilibrium equations

A simply supported beam has 3 unknown reactions and 3 equilibrium equations.

Degree of indeterminacy = 3 − 3

Degree of indeterminacy = 0, so the beam is statically determinate.

1.1.2.2 Fixed support

A fixed support restrains translation and rotation, generating reaction forces and moments; it is common in rigid connections. Because it prevents horizontal movement, vertical movement and rotation, it provides three reactions: a vertical force, a horizontal force and a fixing moment. A beam built into a wall or cast monolithically with a column behaves as if fixed, and a cantilever is a member fixed at one end and free at the other. Example: a reinforced concrete balcony at a hotel is cast as a cantilever fixed into the floor slab and supporting beam, with the outer edge free.

Features of a fixed support
  • No movement or rotation: The member cannot move up, down, sideways or rotate at the support.
  • Three reactions: Vertical reaction, horizontal reaction and a fixing (restraining) moment.
  • Hogging moment: Under downward loads, the fixed end develops a hogging moment, with tension at the top.
  • Cantilevers: One fixed end and one free end, as in balconies, canopies and overhanging slabs.
  • Rigid frames: Beams cast monolithically with columns act as fixed or partly fixed at their ends.
  • Reduced deflection: Fixing the ends makes a beam stiffer than if it were simply supported.
Worked example: indeterminacy of a propped cantilever

A beam fixed at one end and resting on a roller at the other (a propped cantilever) has 3 reactions at the fixed end and 1 at the roller.

Number of unknown reactions = 3 + 1 = 4

Degree of indeterminacy = 4 − 3

Degree of indeterminacy = 1, so equilibrium equations alone cannot find the reactions.

1.1.2.3 Simply supported

A simple support, the most basic form of simply supported end, is one where a member merely rests on its support without being fixed or pinned to it, such as a beam bearing on a wall or a knife-edge. It provides only a vertical reaction pushing up on the member; it cannot hold the member down, so uplift or overturning must be prevented by other means. Horizontal movement is resisted only by friction, which is usually ignored in analysis. Example: the timber floor joists in the gallery of a church are laid directly on the masonry walls, so each end acts as a simple support.

Features of a simple support
  • Bearing only: The member rests on the support with no bolts, dowels or fixings.
  • One reaction: A single vertical reaction acting upwards on the member.
  • No moment resistance: The member is free to rotate at the support.
  • No resistance to uplift: Wind suction or reversed loads can lift the member off the support.
  • Bearing length needed: Enough length must rest on the support to avoid crushing the wall or the member.
  • Common uses: Joists on walls, precast planks on beams and stair flights resting on landings.

1.1.2.4 Hinged support

A hinged support, also called a pinned support, allows rotation but prevents translation in both horizontal and vertical directions. It produces vertical and horizontal reaction forces but no moment, because the pin lets the member turn freely. Hinged supports are used where a member must be held in position but allowed to rotate as it deflects, as in roof trusses and pinned steel frames. Example: the steel roof trusses of a contractor's warehouse are bolted to the column tops with single-bolt connections that act as hinged supports.

Features of a hinged support
  • Free rotation: The member can rotate about the pin.
  • No translation: Movement is prevented both vertically and horizontally.
  • Two reactions: One vertical and one horizontal reaction, or a single inclined reaction resolved into these components.
  • Zero moment: No moment is transmitted at the support.
  • Symbol: Drawn as a triangle with the member attached at its apex, fixed to the ground.
  • Common uses: Truss bearings, pinned column bases and one end of simply supported beams and bridges.

1.1.2.5 Roller support

A roller support allows horizontal movement but prevents vertical translation. It generates a vertical reaction only, perpendicular to the surface on which it rolls, and is used to accommodate expansion or contraction. By allowing the member to move freely along its length as temperature changes, it prevents thermal stresses that would otherwise crack the member or its supports. Example: a pedestrian footbridge built by a county government over a river has a pinned bearing at one end and a roller bearing at the other, so the deck can expand on hot days.

Features of a roller support
  • Free horizontal movement: The member can slide or roll along the support.
  • Free rotation: The member can also rotate at the support.
  • One reaction: A single reaction perpendicular to the rolling surface, usually vertical.
  • Expansion allowance: Lets long members expand and contract with temperature.
  • Symbol: Drawn as a triangle or circle resting on small wheels above a ground line.
  • Common uses: Bridge bearings, long-span roof beams and one end of simply supported beams.

Free End

A free end has no restraint; the element is free to move and rotate, producing no reactions. It occurs at the unsupported tip of a cantilever, where both the shear force and the bending moment are zero unless a load acts exactly at the tip.

Practice Questions

  1. A simply supported beam 7 m long carries a point load of 20 kN at 3 m from the left support. Calculate the support reactions. (6 marks)
  2. A cantilever beam of length 5 m carries a uniformly distributed load of 10 kN/m. Find the reaction forces and bending moment at the fixed support. (6 marks)
  3. Determine the maximum bending moment for a 9 m simply supported beam carrying a uniformly distributed load of 6 kN/m. (5 marks)
  4. A beam fixed at one end and roller-supported at the other, 8 m long, carries a point load of 15 kN at 5 m from the fixed end. Calculate the reactions and moment at the fixed end. (8 marks)
  5. Calculate the bending moment at the midpoint of a simply supported beam 10 m long carrying a triangular load varying from 0 to 8 kN/m. (5 marks)
The rest of this chapter
🔒

Create a free account to open more of this chapter.

Free: practical guides, quick cards, workplace scenarios and more.

Create a free account
🔒1.2 Support reaction on beams and frames calculation

Calculating support reactions is fundamental to structural analysis, as these reactions balance the applied loads and determine internal forces within beams and frames. In architectural practice, accurate reaction calculations ensure that structural elements a…

🔒1.3 Shear force and bending moment drawings preparation

In architectural structural design, understanding shear forces and bending moments in beams and slabs is essential to ensure safety and serviceability. Preparing accurate shear force and bending moment diagrams helps architects coordinate with structural engin…

Chapter Summary

This chapter introduced the fundamental principles of shear force and bending moment analysis in structural engineering. It began by explaining different types of loads including point loads, moments, uniformly distributed loads, and varying loads, which are essential for assessing structural behavior. Various support types were detailed, such as simply supported, fixed, hinged, and roller supports, highlighting their influence on load transfer and structural response. Methods for calculating support reactions on beams and frames were then presented, providing the basis for further analysis. The chapter also covered the preparation of shear force and bending moment diagrams, which visually represent internal forces within structural elements. Finally, the concept of the point of contraflexure was introduced, indicating where bending moments change sign along a beam. Together, these topics equip learners with the skills to analyze and design safe and efficient structural systems.

Self-Assessment

🔒 PDFDownload this self-assessment, with answers

Written Assessment

  1. A simply supported beam of length \(6\,m\) carries a point load of \(12\,kN\) at its mid-span. Calculate the reactions at the supports. (3 marks)

  2. A beam \(8\,m\) long is simply supported and carries a uniformly distributed load of \(5\,kN/m\) over its entire length. Determine the support reactions. (3 marks)

🔒8 more in this section.

Chapter Examination Questions

🔒 PDFDownload these examination questions, with worked answers

SECTION A (40 Marks) - Answer ALL Questions

  1. A simply supported beam in a county government office roof carries a point load of 12 kN at mid-span. The beam length is 6 m. Calculate the support reactions. (4 marks)
  2. Determine the shear force just to the right of a 10 kN point load located 3 m from the left support of a simply supported beam of length 6 m. (4 marks)
🔒18 more in this section.
Flashcards 20 cards Study deck ▾
Question
1

↻ Tap card to reveal answer
🔒

18 more in this section.

Create a free account
Test Yourself 17 questions Start quiz ▾
0%
0 / 2
🔒

15 more in this section.

Create a free account
Am I competent?

At the start of this chapter we promised you would be able to:

  • correctly choose and apply different types of supports according to the needs of structural analysis,
  • accurately calculate support reactions for beams and frames using the proper procedures,
  • prepare clear and accurate shear force and bending moment diagrams that meet industry standards.

Tick each one you can genuinely do.

So, are you there yet?

You're competent when you can confidently do 50% or more of what this chapter promised.

Sign in to record how you're doing.