Theory of Structures: Beams, BMD and SFD, Bending and Deflection Formulas for SSC JE
Rahul Kumar
SSC Exam Expert & Content Editor
In this note
- 1.Types of Beams and Supports
- 2.Shear Force and Bending Moment
- 3.Moment of Inertia and Section Modulus
- 4.Deflection Formulas
- 5.Columns and Buckling
- 6.Trusses and Arches
- 7.Solved Examples
- 8.Quick Tips
- 9.Quick Revision Points
- 10.Practice Questions With Answers
- 11.How to Practise This Chapter
- 12.Frequently Asked Questions
PDF Notes: Theory of Structures: Beams, BMD and SFD, Bending and Deflection Formulas for SSC JE
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Theory of structures deals with how beams, columns and frames carry loads. It is a scoring part of the SSC JE civil paper because most questions are formula-based and can be answered in seconds if you remember the standard results. This note gives the standard formulas for bending moment, deflection, moment of inertia and buckling, with solved examples, practice questions and FAQs.
Types of Beams and Supports
| Beam | Support condition |
|---|---|
| Simply supported | Supported on a hinge at one end and a roller at the other |
| Cantilever | Fixed at one end, free at the other |
| Fixed (built-in) | Both ends fixed |
| Propped cantilever | Fixed at one end and simply supported at the other |
| Continuous | Supported at more than two points |
| Overhanging | The beam extends beyond a support |
- A hinged support gives 2 reactions (horizontal and vertical); a roller gives 1 (vertical); a fixed support gives 3 (horizontal, vertical and moment).
- A structure is statically determinate if the reactions can be found with the equilibrium equations alone (sum of H = 0, V = 0, M = 0). Otherwise it is indeterminate.
- Degree of indeterminacy for a plane frame: r − 3 (for a beam, number of reactions − 3).
Shear Force and Bending Moment
- Shear force (SF) is the algebraic sum of the vertical forces on one side of a section.
- Bending moment (BM) is the algebraic sum of the moments of the forces on one side of the section.
- The rate of change of BM is the shear force (dM/dx = V), and the rate of change of SF is the load intensity (dV/dx = −w).
- The BM is maximum where the SF is zero or changes sign.
- The point of contraflexure is where the BM changes its sign.
| Beam and load | Maximum BM | Maximum SF |
|---|---|---|
| Cantilever, point load W at the free end (length l) | W × l (at the fixed end) | W |
| Cantilever, UDL w per unit length | w l²/2 (at the fixed end) | w l |
| Simply supported, central point load W | W l/4 (at the centre) | W/2 |
| Simply supported, UDL w per unit length | w l²/8 (at the centre) | w l/2 |
| Fixed beam, central point load W | W l/8 at the centre and at the ends | W/2 |
| Fixed beam, UDL w | w l²/24 at the centre; w l²/12 at the ends (fixed end moment) | w l/2 |
Moment of Inertia and Section Modulus
| Section | Moment of inertia about the centroidal axis |
|---|---|
| Rectangle (b × d) | b d³/12 |
| Triangle (base b, height h) about the base | b h³/12 |
| Triangle about the centroid | b h³/36 |
| Circle (diameter d) | π d&sup4;/64 |
| Hollow circle | π (D&sup4; − d&sup4;)/64 |
- Section modulus Z = I / y (where y is the distance of the extreme fibre from the neutral axis). For a rectangle, Z = b d²/6; for a circle, Z = π d³/32.
- The simple bending equation: M/I = σ/y = E/R.
- Bending stress is maximum at the extreme fibres and zero at the neutral axis.
- Shear stress in a rectangular section is maximum at the neutral axis; maximum = 1.5 × average shear stress.
- The torsion equation: T/J = τ/r = Gθ/L.
Deflection Formulas
| Beam and load | Maximum deflection |
|---|---|
| Cantilever, point load W at the free end | W l³ / (3 E I) |
| Cantilever, UDL w | w l&sup4; / (8 E I) |
| Simply supported, central point load W | W l³ / (48 E I) |
| Simply supported, UDL w | 5 w l&sup4; / (384 E I) |
Columns and Buckling
- Euler’s critical load: P = π² E I / (Le)², where Le is the effective length.
- Slenderness ratio = effective length / least radius of gyration.
- Effective length: both ends hinged = L; one end fixed and the other free = 2L; both ends fixed = L/2; one end fixed and the other hinged = L/√2.
- Crippling load is highest when both ends are fixed.
Trusses and Arches
- For a perfect plane truss, the number of members m = 2j − 3, where j is the number of joints.
- The method of joints and the method of sections are used to find member forces.
- A three-hinged arch is statically determinate; two-hinged and fixed arches are indeterminate.
- Influence lines show the effect of a moving load on the reaction, shear or bending moment at a section.
Solved Examples
| Question | Solution |
|---|---|
| Simply supported beam, span 6 m, UDL 10 kN/m. Find the maximum BM. | w l²/8 = 10 × 36 / 8 = 45 kN m; each reaction = 30 kN |
| Cantilever of length 2 m with a 5 kN load at the free end | M = W l = 5 × 2 = 10 kN m |
| Moment of inertia of a 100 mm × 200 mm rectangle (depth 200 mm) | b d³/12 = 100 × 200³ / 12 = 6.67 × 107 mm&sup4; |
| Simply supported beam of span 4 m with a central load of 20 kN. Find the maximum BM. | W l/4 = 20 × 4 / 4 = 20 kN m |
| Section modulus of a rectangle 100 mm wide and 200 mm deep | b d²/6 = 100 × 40000 / 6 = 6.67 × 105 mm³ |
Quick Tips
- Memorise the maximum BM and deflection table; it solves a large number of questions.
- The BM is maximum where the shear force is zero.
- Use consistent units (N and mm, or kN and m) throughout.
- Remember the effective length factors for columns: 1, 2, 0.5 and 0.707.
Quick Revision Points
- Simply supported with UDL: M = wl²/8; central point load: M = Wl/4.
- Cantilever: point load Wl; UDL wl²/2.
- Deflection: cantilever point load Wl³/3EI; simply supported UDL 5wl&sup4;/384EI.
- Euler load P = π²EI/Le².
- Rectangle I = bd³/12.
Practice Questions With Answers
- Maximum BM of a simply supported beam with a UDL w and span l?
Answer: w l²/8 - Maximum BM of a cantilever with a point load W at the free end?
Answer: W l - Moment of inertia of a rectangle about the centroidal axis?
Answer: b d³/12 - Maximum deflection of a cantilever with an end point load?
Answer: W l³/(3 E I) - Effective length of a column with both ends fixed?
Answer: L/2 - Section modulus of a circular section?
Answer: π d³/32 - Where is the bending stress zero in a beam?
Answer: At the neutral axis - The number of reactions given by a fixed support in a plane?
Answer: 3 - What is the point of contraflexure?
Answer: The point where the BM changes sign - For a perfect plane truss m = ?
Answer: 2j − 3
How to Practise This Chapter
Practise 10 questions every day and write the steps on paper. Then attempt timed sets from our practice papers and the previous year papers page. See all topics on the Theory of Structures notes page.
Frequently Asked Questions
What is a simply supported beam?
A beam with a hinge at one end and a roller at the other.
What is a cantilever?
A beam fixed at one end and free at the other.
What is bending moment?
The algebraic sum of the moments of forces on one side of a section.
Where is the bending moment maximum?
At the section where the shear force is zero or changes sign.
What is a point of contraflexure?
A point where the bending moment changes its sign.
What is the maximum BM of a simply supported beam with a UDL?
w l²/8 at the centre.
What is Euler’s buckling load?
P = π²EI/Le².
What is slenderness ratio?
Effective length divided by the least radius of gyration.
What is section modulus?
I divided by the distance of the extreme fibre from the neutral axis.
What is a statically determinate structure?
A structure whose reactions can be found with the equilibrium equations alone.
What is the deflection formula for a simply supported beam with a central load?
W l³/(48 E I).
How many questions come from this in SSC JE?
Strength of materials and structures together give a good number of questions in the civil paper.
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