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Qwestrum Engineering360 · Mechanical Engineering · Machine Design

Design of Shafts

A shaft is sized so the combined torsional shear from τ = 16T/(πd³) and bending stress σ_b = 32M/(πd³) stay within the allowable value fixed by the ASME code. Diameter follows from d = (16/π · √(M² + T²)/τ_allow)^(1/3), the equivalent-torque route in VB Bhandari.
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.

  • ASME:τallow=0.3Sutor0.18SypforductileshaftsASME: \tau_{allow} = 0.3 S_{ut} or 0.18 S_{yp} for ductile shafts
  • Hollowshaft:J=π(do4−di4)32Hollow shaft: J = \pi\frac{(d_{o}^{4} - d_{i}^{4})}{32}
  • Keyway reduces torsional strength — use kb ≈ 0.75

Topic details

Introduction

Shaft design is a guaranteed question in Indian machine-design and GATE papers because it couples torsion and bending in one member. The ASME code approach caps the allowable shear at τ_allow = 0.30 σ_ut or 0.18 σ_yp (whichever is smaller), further reduced for keyways.

Scope in B.Tech and GATE syllabus

The standard exam workflow is: draw the bending-moment and torque diagrams along the shaft, locate the section where M and T are jointly largest, then compute the equivalent twisting moment T_e = √(M² + T²) and equivalent bending moment M_e = ½(M + √(M² + T²)). VB Bhandari and SS Rattan both present these as the working formulae.

Why this topic matters in practice

For hollow shafts the polar modulus uses J = π(d_o⁴ − d_i⁴)/32, and the candidate is expected to show the weight saving over a solid shaft of equal strength — a favourite comparison in viva examinations.

Key relations & formulas

τ=T⋅rJ=16T(πd3)\tau = T\cdot \frac{r}{J} = \frac{16T}{(\pi d^{3})}
(solid circular shaft, torsion)
σb=32M(πd3)\sigma_{b} = \frac{32M}{(\pi d^{3})}
(bending stress)
τeq=σb2+4τ2\tau_{eq} = \sqrt{\sigma_{b}^{2} + 4\tau^{2}}
(combined bending and torsion, ASME code)
d=(16π⋅M2+T2/τallow)(13)d = (\frac{16}{\pi} \cdot \sqrt{M^{2} + T^{2}}/\tau_{allow})^(\frac{1}{3})
(shaft diameter)

Notation and sign conventions

Relation 1 —
τ=T⋅rJ=16T/\tau = T\cdot \frac{r}{J} = 16T/
τ=T⋅rJ=16T(πd3)\tau = T\cdot \frac{r}{J} = \frac{16T}{(\pi d^{3})}
(solid circular shaft, torsion)
Write this relation with symbols exactly as in Design of Machine Elements — VB Bhandari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
σb=32M/\sigma_{b} = 32M/
σb=32M(πd3)\sigma_{b} = \frac{32M}{(\pi d^{3})}
(bending stress)
Write this relation with symbols exactly as in Design of Machine Elements — VB Bhandari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
τeq=√\tau_{eq} = √
τeq=σb2+4τ2\tau_{eq} = \sqrt{\sigma_{b}^{2} + 4\tau^{2}}
(combined bending and torsion, ASME code)
Write this relation with symbols exactly as in Design of Machine Elements — VB Bhandari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
d=d =
d=(16π⋅M2+T2/τallow)(13)d = (\frac{16}{\pi} \cdot \sqrt{M^{2} + T^{2}}/\tau_{allow})^(\frac{1}{3})
(shaft diameter)
Write this relation with symbols exactly as in Design of Machine Elements — VB Bhandari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

The governing physics is torsion of a circular section: τ = T·r/J with J = πd⁴/32 gives τ = 16T/(πd³) at the surface. Superposed bending adds σ_b = 32M/(πd³) at the same outer fibre.

Governing relations in practice

Because both act at the surface, the design combines them through a failure theory. The ASME code, based on maximum-shear-stress, defines the equivalent torque T_e = √(M² + T²) so that d = (16 T_e/(π τ_allow))^(1/3). When shock and fatigue matter, combined factors K_b and K_t multiply M and T respectively.

Design and analysis considerations

Allowable stress is code-driven: τ_allow = min(0.30 σ_ut, 0.18 σ_yp), and a keyway cuts this by about 25 % (factor ≈ 0.75) because it removes material and raises stress concentration.

Advanced theory and extensions

Hollow shafts exploit that material near the axis carries little torque; for the same outer stress a hollow shaft of J = π(d_o⁴ − d_i⁴)/32 is markedly lighter. Bhandari's design tables let students pick the next standard diameter after the calculated value.

Assumptions and validity limits

State assumptions explicitly before using any relation for design of shafts — 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 Machine Design 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 Machine Design 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 design of shafts.
4. Use equation 1:
τ=T⋅rJ=16T/\tau = T\cdot \frac{r}{J} = 16T/
.
5. Use equation 2:
σb=32M/\sigma_{b} = 32M/
.
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

Design of Shafts appears in shafts, keys, bearings, springs, and fasteners. In Indian mechanical curricula this topic is tested because it connects theory to safe sizing of mechanical components.
GATE and semester exams often combine design of shafts with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use design of shafts?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Using T alone and ignoring bending M when both act at the critical section
• Forgetting the keyway reduction factor (≈0.75) on allowable shear stress
• Taking allowable stress from yield only, not the smaller of 0.30σ_ut and 0.18σ_yp
• Leaving diameter as the raw cube-root value instead of rounding up to a standard size

Quick revision checklist

Before attempting design of shafts problems, confirm you can:
1.
ASME:τallow=0.3Sutor0.18SypforductileshaftsASME: \tau_{allow} = 0.3 S_{ut} or 0.18 S_{yp} for ductile shafts

2.
Hollowshaft:J=π(do4−di4)32Hollow shaft: J = \pi\frac{(d_{o}^{4} - d_{i}^{4})}{32}

3. Keyway reduces torsional strength — use kb ≈ 0.75
Revise the solved examples in Design of Machine Elements — VB Bhandari 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.

Solid shaft diameter (ASME)

Problem

A shaft transmits a bending moment M = 300 N·m and a torque T = 400 N·m. Allowable shear stress τ_allow = 40 MPa. Find the required solid diameter.

Solution

T_e = √(M² + T²) = √(300² + 400²) = √(90000 + 160000) = 500 N·m = 500000 N·mm
d = (16 T_e/(π τ_allow))^(1/3) = (16×500000/(π×40))^(1/3) = (63661.9)^(1/3) ≈ 39.9 mm → use 40 mm.

Conceptual check — Design of Shafts

Problem

In a Machine Design semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of design of shafts." What should a complete answer include?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is Design of Shafts, and why does it appear in B.Tech / GATE syllabi?

    Model answer

    A shaft is sized so the combined torsional shear from τ = 16T/(πd³) and bending stress σ_b = 32M/(πd³) stay within the allowable value fixed by the ASME code. Diameter follows from d = (16/π · √(M² + T²)/τ_allow)^(1/3), the equivalent-torque route in VB Bhandari.
  2. 2
    State the relation τ = T·r/J = 16T/ and name each symbol.

    Model answer

    The governing relation is τ=T⋅rJ=16T/\tau = T\cdot \frac{r}{J} = 16T/. Write every symbol with SI units before substituting numbers.
  3. 3
    State the relation σ_b = 32M/ and name each symbol.

    Model answer

    The governing relation is σb=32M/\sigma_{b} = 32M/. Write every symbol with SI units before substituting numbers.
  4. 4
    State the relation τ_eq = √ and name each symbol.

    Model answer

    The governing relation is τeq=√\tau_{eq} = √. Write every symbol with SI units before substituting numbers.
  5. 5
    State the relation d = and name each symbol.

    Model answer

    The governing relation is d=d =. Write every symbol with SI units before substituting numbers.
  6. 6
    Explain: ASME: τ_allow = 0.3 S_ut or 0.18 S_yp for ductile shafts

    Model answer

    ASME:τallow=0.3Sutor0.18SypforductileshaftsASME: \tau_{allow} = 0.3 S_{ut} or 0.18 S_{yp} for ductile shafts — state the assumption range and one exam trap linked to this point.
  7. 7
    Explain: Hollow shaft: J = π(d_o⁴ − d_i⁴)/32

    Model answer

    Hollowshaft:J=π(do4−di4)32Hollow shaft: J = \pi\frac{(d_{o}^{4} - d_{i}^{4})}{32} — state the assumption range and one exam trap linked to this point.
  8. 8
    Explain: Keyway reduces torsional strength — use kb ≈ 0.75

    Model answer

    Keyway reduces torsional strength — use kb ≈ 0.75 — state the assumption range and one exam trap linked to this point.
  9. 9
    How would you correct this error in a viva: Using T alone and ignoring bending M when both act at the critical section?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  10. 10
    How would you correct this error in a viva: Forgetting the keyway reduction factor (≈0.75) on allowable shear stress?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  11. 11
    How would you correct this error in a viva: Taking allowable stress from yield only, not the smaller of 0.30σ_ut and 0.18σ_yp?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  12. 12
    How would you correct this error in a viva: Leaving diameter as the raw cube-root value instead of rounding up to a standard size?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.

Exams & GATE

  • 1
    GATE favourite: combined M and T on stepped shafts — use equivalent torque Te = √(M² + T²).
  • 2
    Avoid: Using T alone and ignoring bending M when both act at the critical section
  • 3
    Avoid: Forgetting the keyway reduction factor (≈0.75) on allowable shear stress
  • 4
    Avoid: Taking allowable stress from yield only, not the smaller of 0.30σ_ut and 0.18σ_yp

📖 Standard books (India)

  • Design of Machine Elements — VB Bhandari

    Read: Syllabus unit

    Machine design, shafts, bearings, springs, and joints

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