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Qwestrum Engineering360 · Civil Engineering · Strength of Materials

Bending and Shear Stress

Use σ = M/Z for the extreme-fibre bending stress and τ = VQ/(Ib) for the transverse shear; the section modulus Z = bh²/6 turns most rectangular-beam numericals into a single division.
Exam tip: keep units consistent — N with mm² gives MPa directly (1 MPa = 1 N/mm²).

Key formulas & points

Skim these first — then read the full notes below.

  • Neutral axis: bending stress = 0; max tension and compression at extreme fibres
  • Shear stress parabolic in rectangular section; zero at top and bottom
  • I=bh312forrectangle;Z=bh26I = \frac{bh^{3}}{12} for rectangle; Z = \frac{bh^{2}}{6}

Topic details

Introduction

The flexure formula M/I = σ/y = E/R is one of the most heavily examined relations in civil SOM because it feeds directly into RCC and steel design. Bending stress varies linearly across the depth, zero at the neutral axis and maximum at the top and bottom fibres.

Scope in B.Tech and GATE syllabus

Shear stress from τ = VQ/(Ib) is distributed parabolically over a rectangular section, being zero at the extreme fibres and maximum (1.5 V/A) at the neutral axis — the opposite pattern to bending. Confusing where each stress peaks is the most common conceptual error.

Why this topic matters in practice

Exam problems usually give a loaded beam, ask you to locate maximum bending moment and shear force from the SFD/BMD, then compute the corresponding stresses. So the numerical always starts with statics before the stress formula.

Key relations & formulas

MI=σy=ER\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}
(bending equation, flexure formula)
τ=(VQ)(Ib)\tau = \frac{(V Q)}{(I b)}
(shear stress in rectangular beam, RK Bansal)
σmax=MymaxI=MZ\sigma_{max} = M \frac{y_{max}}{I} = \frac{M}{Z}
(Z = section modulus)

Notation and sign conventions

Relation 1 —
MI=σy=ER\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}
MI=σy=ER\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}
(bending equation, flexure formula)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
τ=\tau =
τ=(VQ)(Ib)\tau = \frac{(V Q)}{(I b)}
(shear stress in rectangular beam, RK Bansal)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
σmax=MymaxI=MZ\sigma_{max} = M \frac{y_{max}}{I} = \frac{M}{Z}
σmax=MymaxI=MZ\sigma_{max} = M \frac{y_{max}}{I} = \frac{M}{Z}
(Z = section modulus)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Bending stress arises because fibres above the neutral axis shorten and those below lengthen (for sagging), producing a linear stress distribution σ = My/I. The second moment of area I measures how the material is distributed about the neutral axis; placing area far from the axis (as in an I-section) raises I and lowers stress for the same moment.

Governing relations in practice

The section modulus Z = I/y_max packages this into a single design parameter, so σ_max = M/Z. For a rectangle Z = bh²/6, which shows stress falls with the square of depth — doubling depth quarters the bending stress, the reason beams are deep rather than wide.

Design and analysis considerations

Transverse shear stress τ = VQ/(Ib) depends on the first moment Q of the area beyond the fibre considered; Q is maximum at the neutral axis and zero at the top and bottom, giving the parabolic profile. In rectangular beams τ_max = 1.5 times the average shear V/A.

Advanced theory and extensions

Because bending governs at mid-span (where M peaks) and shear governs near supports (where V peaks), a complete design check evaluates both at their respective critical sections rather than at a single location.

Assumptions and validity limits

State assumptions explicitly before using any relation for bending and shear stress — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Mechanics of Materials (Civil) viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Mechanics of Materials (Civil) papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to bending and shear stress.
4. Use equation 1:
MI=σy=ER\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}
.
5. Use equation 2:
τ=\tau =
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Bending and Shear Stress appears in beams, slabs, and columns. In Indian civil curricula this topic is tested because it connects theory to stress and deformation in civil structures.
GATE and semester exams often combine bending and shear stress with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use bending and shear stress?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Placing maximum shear stress at the top fibre instead of the neutral axis.
• Using I = bh³/12 but forgetting to convert to Z when the question asks for stress.
• Reading maximum bending moment off the shear diagram instead of the moment diagram.
• Taking τ_avg = V/A as the maximum shear for a rectangle instead of 1.5 V/A.

Quick revision checklist

Before attempting bending and shear stress problems, confirm you can:
1. Neutral axis: bending stress = 0; max tension and compression at extreme fibres
2. Shear stress parabolic in rectangular section; zero at top and bottom
3.
I=bh312forrectangle;Z=bh26I = \frac{bh^{3}}{12} for rectangle; Z = \frac{bh^{2}}{6}
Revise the solved examples in Strength of Materials — RK Bansal and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Maximum bending stress in a rectangular beam

Problem

A simply supported rectangular beam 230 mm wide and 450 mm deep carries a maximum bending moment of 50 kNm. Find the maximum bending stress.

Solution

Section modulus Z = bh²/6 = 230 × 450² / 6 = 7.76 × 10⁶ mm³. Maximum bending stress σ_max = M/Z = 50 × 10⁶ N·mm / 7.76 × 10⁶ mm³ = 6.44 N/mm² (MPa). This modest value shows the section is comfortably within the flexural capacity of M20 concrete or structural steel.

Conceptual check — Bending and Shear Stress

Problem

In a Mechanics of Materials (Civil) semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of bending and shear stress." What should a complete answer include?

Exams & GATE

Draw SFD/BMD first — bending moment governs flexure, shear governs near supports.

📖 Standard books (India)

  • Strength of Materials — RK Bansal

    Read: Syllabus unit

    SOM — beams, torsion, columns, and deflection

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