Analysis of Beams -- Complete Revision Notes
Every definition, classification, sign convention, formula, worked example, diagram, mnemonic and exam-style answer you need for this chapter -- built for fast revision and mock practice.
On this page
- Introduction
- What is a Beam
- Classification of Beams (SCFO)
- Types of Supports
- Types of Loads
- Statically Determinate Structures
- Statically Indeterminate Structures + Degree of Indeterminacy (added)
- Shear Force (SF)
- Bending Moment (BM)
- Sign Convention (visual)
- Shear Force Diagram (SFD)
- Bending Moment Diagram (BMD)
- Relationship: Load, SF and BM (+ Area Method, added)
- Point of Contraflexure
- Standard Formulas for Common Beams (added)
- Worked Numerical Examples
- A Note on Frames (added)
- Engineering Applications
- Model Answers by Weightage
- MCQs
- Interview Questions
- Memory Box -- Full Summary
- Full Mock Test (20 Questions)
- One-Page Summary Sheet
Introduction
Every structure carries loads, and those loads must travel down through the structure into the ground. The typical load path is:
Roof -> Beam -> Column -> Foundation -> Soil
Before a beam can be designed, three quantities must be determined first:
- Reactions at the supports
- Shear force at every section
- Bending moment at every section
This chapter builds the tools to find all three -- for any type of beam, support and loading.
What is a Beam?
A beam is a structural member that primarily resists loads applied perpendicular (transverse) to its longitudinal axis. Its main function is to transfer these loads to its supports through internal bending and shear.
Engineering Examples
- Floor beams
- Roof beams (purlins, rafters)
- Bridge girders
- Crane beams (gantry girders)
Classification of Beams (10 Marks -- draw sketches)
(A) Simply Supported Beam
Supports: one pin support + one roller support.
Characteristics: most common beam type; no moment develops at the supports.
(B) Cantilever Beam
One end fixed, the other end completely free.
Examples: balcony slab, traffic signal arm, canopy / sunshade.
(C) Fixed Beam (Built-in Beam)
Both ends are fixed (rigidly built into a wall or column).
Advantages over a simply supported beam of the same span and loading:
- Smaller deflection
- Lower maximum bending moment
(D) Continuous Beam
Supported at more than two supports.
Examples: bridges, multi-span buildings.
(E) Overhanging Beam
One or both ends extend beyond the supports.
(Continuous beam is easy to remember separately -- just "more than 2 supports".)
Types of Supports
(A) Roller Support
Provides: one reaction, normal to the supporting surface.
Allows: horizontal movement and rotation.
Used for: allowing expansion / contraction due to temperature change (e.g. bridge bearings).
(B) Pin (Hinged) Support
Provides: horizontal reaction + vertical reaction.
Allows: rotation.
(C) Fixed Support
Provides: horizontal reaction + vertical reaction + moment reaction.
Allows: no movement, no rotation.
| Support | Horizontal Reaction | Vertical Reaction | Moment Reaction |
|---|---|---|---|
| Roller | No | Yes | No |
| Pin | Yes | Yes | No |
| Fixed | Yes | Yes | Yes |
Types of Loads
(A) Point Load (Concentrated Load)
Acts at a single point on the beam. Unit: kN.
(B) Uniformly Distributed Load (UDL)
Acts uniformly (same intensity) over a length of the beam. Unit: kN/m.
(C) Uniformly Varying Load (UVL)
Load intensity varies linearly along the length -- a triangular (or trapezoidal) distribution. Unit: kN/m, varying from zero (or a base value) up to a maximum.
(D) Moment Load (Applied Couple)
A moment applied directly at a point, not caused by a force acting at a distance. Unit: kN.m.
Statically Determinate Structures
A structure is statically determinate when all reactions and internal forces can be calculated using only the equations of static equilibrium.
Examples: simply supported beam, cantilever beam.
Statically Indeterminate Structures
Requires both:
- Static equilibrium equations, AND
- Compatibility of deformation conditions
Examples: fixed beam, continuous beam, portal frame (in most cases).
| Determinate | Indeterminate |
|---|---|
| Easier analysis | More complex analysis |
| Uses equilibrium equations only | Requires compatibility conditions also |
| Simpler construction | More economical in material usage |
Degree of Static Indeterminacy (DSI) -- for beams
| Beam | Reactions, r | DSI = r - 3 | Remark |
|---|---|---|---|
| Simply supported (pin + roller) | 2 + 1 = 3 | 0 | Determinate |
| Cantilever (fixed - free) | 3 | 0 | Determinate |
| Propped cantilever (fixed + roller) | 3 + 1 = 4 | 1 | Indeterminate to 1st degree |
| Fixed beam (fixed - fixed) | 3 + 3 = 6 | 3 | Indeterminate to 3rd degree |
| Continuous beam on 3 simple supports | 2 + 1 + 1 = 4 | 1 | Indeterminate to 1st degree |
Exam tip: if DSI is negative, the structure is a mechanism (unstable) -- this is a common trap in MCQs.
Shear Force (SF)
Shear force at a section is the algebraic sum of all vertical forces acting on one side of that section. Unit: kN.
Sign Convention
- Positive SF: the left portion of the beam tends to move upward relative to the right portion (common engineering convention).
- Negative SF: the left portion tends to move downward relative to the right portion.
Bending Moment (BM)
Bending moment at a section is the algebraic sum of the moments of all forces acting on one side of that section, taken about that section. Unit: kN.m.
Sign Convention
- Positive BM -- Sagging (beam curves like a smile / cup holding water)
- Negative BM -- Hogging (beam curves like a frown / upside-down cup)
Smile = Sagging = Positive. Sad = Hogging = Negative.
Sign Convention -- Visual
Shear Force Diagram (SFD)
A graph showing the variation of shear force along the length of the beam.
Important Points
- Point load -> sudden jump (vertical step) in the SFD equal to the load.
- UDL -> straight, sloping line (constant slope = -w).
- UVL -> curved (parabolic) variation.
Bending Moment Diagram (BMD)
A graph showing the variation of bending moment along the length of the beam.
Important Points
- Point load -> straight-line (linear) variation between loads, with a slope change (kink) under the load.
- UDL -> parabolic curve.
- UVL -> higher-order (cubic) curve.
Relationship Between Load, Shear Force and Bending Moment
This is one of the most important 10-mark questions in this chapter.
This is one of the most frequently used concepts in beam design -- always locate the section of zero shear first, then compute the moment there.
Because dV/dx = -w and dM/dx = V, the SFD and BMD can also be related through areas:
This "area method" is a fast way to sketch a BMD once the SFD is known, without writing a fresh moment equation for every segment.
Point of Contraflexure
The point where the bending moment changes sign (from positive/sagging to negative/hogging, or vice versa) is called the point of contraflexure.
Common in: continuous beams, fixed beams (any beam/loading that produces both sagging and hogging zones).
Standard Formulas for Common Beams (added)
These standard cases are not explicitly listed in the source notes but are the basis of almost every beam numerical asked in the Loksewa exam.
| Case | Max Shear Force | Max Bending Moment | Location of Mmax |
|---|---|---|---|
| SS beam, central point load P, span L | P/2 | PL/4 | Mid-span |
| SS beam, UDL w over full span L | wL/2 | wL^2/8 | Mid-span |
| SS beam, point load P at distance a from A (b from B, a+b=L) | Larger of Pa/L, Pb/L | Pab/L | Under the load |
| Cantilever, point load P at free end, length L | P (constant) | PL | At fixed end |
| Cantilever, UDL w over full length L | wL | wL^2/2 | At fixed end |
Worked Numerical Examples
Example 1 -- Simply Supported Beam with Central Point Load (from source notes)
Step 1 -- Reactions: by symmetry, Ra = Rb = 20/2 = 10 kN.
Step 2 -- Maximum shear force: Vmax = 10 kN (equal to either reaction).
Step 3 -- Maximum bending moment (central point load): Mmax = PL/4 = (20 x 6)/4 = 30 kN.m.
Example 2 -- Simply Supported Beam with UDL (added)
Step 1 -- Total load: W = w x L = 5 x 4 = 20 kN.
Step 2 -- Reactions: by symmetry, Ra = Rb = W/2 = 10 kN.
Step 3 -- Maximum shear force: Vmax = 10 kN at each support (SF varies linearly, zero at mid-span).
Step 4 -- Maximum bending moment (UDL on SS beam): Mmax = wL^2/8 = (5 x 4^2)/8 = 10 kN.m at mid-span.
Example 3 -- Cantilever Beam with Point Load at Free End (added)
Step 1 -- Reaction: Ra = 15 kN (vertical reaction at the fixed end).
Step 2 -- Shear force: constant V = 15 kN throughout the span (no change until the fixed end).
Step 3 -- Fixed end moment (point load at free end): M = P x L = 15 x 3 = 45 kN.m, hogging (negative), maximum at the fixed end and zero at the free end.
A Note on Frames (added)
The chapter title refers to "Beams and Frames", but the detailed 7th level syllabus for this chapter covers only beam analysis (loads, supports, SF, BM, SFD, BMD, contraflexure). A frame is simply an assembly of beams and columns rigidly (or partly) connected at joints; frame members are analysed using the same SF/BM principles covered here, member by member, after first finding joint reactions and moments -- but frame-specific methods (e.g. moment distribution, slope-deflection) are covered as a separate topic and are not included in this chapter's scope.
Engineering Applications
Structural analysis of beams is essential for:
- RCC beam design
- Steel beam design
- Bridge girders
- Roof trusses
- Industrial structures
Model Answers by Weightage
Answer structure: (1) state the three relationships dV/dx = -w, dM/dx = V, and Mmax at V = 0; (2) explain each in one line with physical meaning; (3) mention the area method (change in V = -area of load diagram, change in M = area of SFD); (4) draw a simple SFD/BMD pair (e.g. Example 1 or 2) to illustrate.
Answer structure: (1) sketch the beam with reactions Ra = Rb = P/2; (2) draw SFD as a horizontal line at +P/2 from A to mid-span, dropping to -P/2 from mid-span to B; (3) draw BMD as two straight lines rising from 0 at A to PL/4 at mid-span, then falling back to 0 at B; (4) state Mmax = PL/4 at mid-span, where V = 0.
Answer: reproduce the comparison table in Section 6/7 as a table, then add the DSI formula (r - 3) with one example of each.
Answer: sketch roller, pin and fixed supports (Fig 2), then reproduce the reaction comparison table in Section 4.
Answer: give both definitions from Sections 8 and 9 with units, then add the sign convention (sagging/hogging) as a bonus point.
Answer: definition (Section 14), condition M = 0, mention it occurs in continuous and fixed beams, and note its importance in reinforcement detailing (curtailment of bars).
MCQs
- A. Horizontal and vertical reactions
- B. Vertical reaction only
- C. Moment only
- D. No reaction
- A. Shear force is maximum
- B. Shear force is zero
- C. Load is zero
- D. Reaction is zero
- A. Two pinned supports
- B. One fixed end and one free end
- C. Two roller supports
- D. No supports
- A. kN
- B. kN/m
- C. kN.m
- D. MPa
- A. Shear force is maximum
- B. Bending moment changes sign
- C. Load is zero
- D. Reaction is zero
- A. A horizontal (constant) line
- B. A straight sloping line
- C. A parabola
- D. A sudden vertical jump
- A. 0
- B. 1
- C. 2
- D. 3
- A. Highly indeterminate
- B. A mechanism (unstable)
- C. Perfectly determinate
- D. A frame
Interview Questions
- Why does the maximum bending moment occur where the shear force is zero?
- Why is a fixed beam more economical than a simply supported beam for the same span and loading?
- What is the practical importance of the point of contraflexure in reinforcement detailing?
- Why are roller supports provided in long bridges?
- How do SFD and BMD help in structural design?
- (Added) How would you quickly check, without a full analysis, whether a given beam is determinate or indeterminate?
Memory Box -- Full Summary
- Beam: member resisting transverse (perpendicular) loads.
- Roller: vertical reaction only. Pin: horizontal + vertical. Fixed: horizontal + vertical + moment.
- SCFO: Simply supported, Cantilever, Fixed, Overhanging (plus Continuous).
- Loads: Point (kN), UDL (kN/m), UVL (triangular, kN/m), Moment (kN.m).
- Determinate: equilibrium only. Indeterminate: equilibrium + compatibility. DSI = r - 3 for beams.
- Shear Force: sum of vertical forces on one side of a section.
- Bending Moment: sum of moments of forces on one side of a section.
- Smile = Sagging = Positive BM. Frown = Hogging = Negative BM.
- dV/dx = -w (Load to Shear). dM/dx = V (Shear to Moment).
- Maximum BM occurs where V = 0.
- Point of Contraflexure: where bending moment changes sign (M = 0).
- SS beam, central point load: Mmax = PL/4. SS beam, UDL: Mmax = wL^2/8.
- Cantilever, point load at free end: Mmax = PL. Cantilever, UDL: Mmax = wL^2/2.
Full Mock Test -- 20 Questions
Attempt all questions first, then expand each answer to check yourself.
1. What is the primary function of a beam?
To resist transverse loads and transfer them to its supports through bending and shear.
2. Name the five types of beams (mnemonic SCFO + one more).
Simply supported, Cantilever, Fixed, Overhanging, and Continuous.
3. What reactions does a pin support provide?
Horizontal reaction and vertical reaction (no moment).
4. What reactions does a fixed support provide?
Horizontal reaction, vertical reaction, and moment reaction.
5. What is the unit of shear force?
kN.
6. What is the unit of bending moment?
kN.m.
7. Define a statically determinate structure.
A structure whose reactions and internal forces can be found using only the equations of static equilibrium.
8. Give the formula for degree of static indeterminacy of a beam.
DSI = r - 3, where r is the total number of reaction components.
9. What shape does a UDL produce on the SFD?
A straight, sloping line (constant slope equal to -w).
10. What shape does a UDL produce on the BMD?
A parabolic curve.
11. Where does maximum bending moment occur?
Where the shear force is zero (V = 0).
12. What is the point of contraflexure?
The point along the beam where the bending moment changes sign (M = 0).
13. Write the formula for maximum bending moment of a simply supported beam with a central point load P and span L.
Mmax = PL/4.
14. Write the formula for maximum bending moment of a simply supported beam with UDL w over span L.
Mmax = wL^2/8.
15. Write the formula for the fixed-end moment of a cantilever with point load P at the free end, length L.
M = PL (hogging), maximum at the fixed end.
16. What does sagging bending moment indicate about the sign convention?
Positive bending moment; beam curves like a smile (concave up).
17. What does hogging bending moment indicate about the sign convention?
Negative bending moment; beam curves like a frown (concave down).
18. State the relationship between slope of SFD and load intensity.
dV/dx = -w.
19. State the relationship between slope of BMD and shear force.
dM/dx = V.
20. Why does a roller support allow horizontal movement?
Because it provides only a vertical reaction, letting the beam expand or contract freely with temperature change without inducing extra stress.
One-Page Summary Sheet (last-minute revision)
| Topic | Key point to remember |
|---|---|
| Beam | Resists transverse loads through bending and shear |
| Beam types | Simply supported, Cantilever, Fixed, Continuous, Overhanging (SCFO) |
| Roller support | Vertical reaction only |
| Pin support | Horizontal + Vertical reactions |
| Fixed support | Horizontal + Vertical + Moment reactions |
| Load types | Point (kN), UDL (kN/m), UVL (triangular), Moment (kN.m) |
| Determinate | DSI = r - 3 = 0 (equilibrium equations suffice) |
| Indeterminate | DSI > 0 (needs compatibility conditions too) |
| Shear Force | Sum of vertical forces on one side of a section (kN) |
| Bending Moment | Sum of moments on one side of a section (kN.m) |
| Sign convention | Sagging = Positive (smile); Hogging = Negative (frown) |
| Load-SF-BM relation | dV/dx = -w; dM/dx = V; Mmax where V = 0 |
| Point of contraflexure | Where bending moment changes sign, M = 0 |
| SS beam + central point load | Mmax = PL/4 at mid-span |
| SS beam + UDL | Mmax = wL^2/8 at mid-span |
| Cantilever + point load at free end | Mmax = PL at fixed end |
| Cantilever + UDL | Mmax = wL^2/2 at fixed end |
Source notes covered beam classification, supports, loads, determinate/indeterminate structures, shear force, bending moment, SFD/BMD rules, the load-SF-BM relationship, point of contraflexure and one worked example. Degree of static indeterminacy (DSI formula), the SFD/BMD area method, a standard-formula table for common beams, two additional worked examples (UDL on a simply supported beam and a cantilever with a point load), and a short clarifying note on frames were added above because they are regularly tested in the Loksewa 7th level paper but were only partially covered or missing from the original notes.
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