Torsion in Strength of Materials | Civil Engineering Notes for JKSSB Aspirants

Introduction

In Strength of Materials (SOM), torsion plays a crucial role in understanding how structural members behave when subjected to twisting loads. Torsion occurs when a shaft or structural member is subjected to a torque (twisting moment), causing it to rotate about its longitudinal axis.

For civil engineering students and JKSSB JE aspirants, mastering the concepts of torsion is essential, as it forms the basis for analyzing shafts, beams, and other load-bearing elements in practical engineering applications.

📘 Definition of Torsion

Torsion is the twisting of a structural member when subjected to a torque (twisting moment) about its longitudinal axis. This produces shear stresses across the cross-section and results in a rotational deformation called the angle of twist.

Key Formulas (in Unicode)

• Torque (T):

Unicode:
T = F × r

where,

• T = Torque or twisting moment
• F = Tangential force applied
• r = Perpendicular distance from axis

• Shear Stress (τ):

Unicode:
τ = (T × r) / J

where,

• τ = Shear stress
• T = Torque
• r = Radius of shaft
• J = Polar moment of inertia

⚙️ Concept of Torque and Twisting Moment

When a force is applied tangentially at a distance from the axis of a shaft or beam, it produces a rotational effect known as torque or twisting moment. This torque is the primary cause of torsional deformation in structural members.

• Torque (T):
  It is the product of the applied force and the perpendicular distance from the axis of rotation.

Formula (Unicode):
T = F × r

where,

• T = Torque or twisting moment (N·m)
• F = Tangential force applied (N)
• r = Distance from axis of shaft (m)

• Twisting Moment:
  The twisting moment is simply another term for torque when it is applied to a shaft or structural member. It causes the shaft to twist and generate shear stresses along its cross-section.
• Work Done by a Torque:
  When a shaft rotates through an angle θ under the action of torque T, the work done (W) is given by:

W = T × θ

where,

• W = Work done (N·m or Joules)
• T = Applied torque (N·m)
• θ = Angle of twist (radians)

📌 Assumptions in Torsion Theory

The theory of torsion is based on certain assumptions to simplify the analysis of shafts under twisting loads. These assumptions are valid for circular shafts (solid or hollow) made of homogeneous, isotropic materials.

Assumptions:

1. The material of the shaft is homogeneous and isotropic (same properties in all directions).
2. The shaft has a circular cross-section (solid or hollow).
3. The applied torque is uniformly distributed along the shaft’s length.
4. Cross-sections that are plane and circular before loading remain plane and circular after loading.
5. The shaft material follows Hooke’s law (stress ∝ strain) within the elastic limit.
6. The radius of the shaft does not change during torsion (no distortion in radial direction).
7. Twist along the shaft is uniform (constant rate of twist).

Important Relation (Torsion Equation – Preview)

The above assumptions lead to the fundamental torsion equation:

τ / r = T / J = Gθ / L

where,

• τ = Shear stress
• r = Radius of shaft
• T = Torque applied
• J = Polar moment of inertia
• G = Modulus of rigidity
• θ = Angle of twist (radians)
• L = Length of shaft

🧮 Torsion Equation (τ / r = T / J = Gθ / L)

The torsion equation is the fundamental relationship in torsion theory. It connects shear stress, torque, shaft dimensions, modulus of rigidity, and angle of twist.

It is written as:

τ / r = T / J = Gθ / L

where,

• τ = Shear stress at radius r (N/m²)
• r = Distance from shaft axis (m)
• T = Applied torque (N·m)
• J = Polar moment of inertia (m⁴)
• G = Modulus of rigidity (N/m²)
• θ = Angle of twist (radians)
• L = Length of shaft (m)

🔹 Derivation Outline of Torsion Equation

1. Shear Strain (γ):
   At radius r, the shear strain is given by:
   γ = rθ / L
2. Shear Stress (τ):
   Using Hooke’s law (τ = Gγ):
   τ = G × (rθ / L)
   ⟹ τ / r = Gθ / L
3. Torque (T):
   Total torque is the summation of elemental shear forces:
   T = ∫ (τ × r × dA)

Since τ / r = constant = k,
T = (τ / r) × ∫ (r² dA)

But ∫ (r² dA) = J (polar moment of inertia),

⟹ T = (τ / r) × J
⟹ τ / r = T / J

🔹 Final Torsion Equation

Thus, combining both relations:

τ / r = T / J = Gθ / L

✅ This formula is the core of torsion analysis and is used to calculate stresses, shaft strength, and angle of twist in civil/mechanical engineering.

📐 Polar Moment of Inertia (J) in Torsion

The polar moment of inertia (J) is a geometric property of a cross-section that measures its resistance to torsional deformation. It plays the same role in torsion as the second moment of area (I) plays in bending.

🔹 Definition

The polar moment of inertia about the axis of the shaft is defined as:

J = ∫ r² dA

where,

• r = radial distance from the axis (m)
• dA = elemental area of the cross-section (m²)

🔹 Polar Moment for Common Sections

1. Solid Circular Shaft (radius R, diameter d = 2R):
   J = (π R⁴) / 2 = (π d⁴) / 32
2. Hollow Circular Shaft (outer radius Rₒ, inner radius Rᵢ, diameters D = 2Rₒ, d = 2Rᵢ):
   J = [π (Rₒ⁴ − Rᵢ⁴)] / 2 = [π (D⁴ − d⁴)] / 32

🔹 Importance in Torsion

• Higher value of J ⇒ shaft is stiffer in torsion and can resist more torque.
• Polar moment depends only on the geometry of the cross-section, not on the material.
• Torsion equation uses J as:

τ / r = T / J = Gθ / L

where J directly controls both shear stress distribution and angle of twist.

✅ Thus, J is the key geometric factor in torsion design.

📊 Shear Stress Distribution in Circular Shafts

When a shaft is subjected to torque (T), shear stresses are induced over its cross-section. These stresses are not uniform but vary linearly with the distance from the center of the shaft.

🔹 General Shear Stress Formula

τ(r) = (T × r) / J

where,

• τ(r) = Shear stress at radius r
• T = Applied torque
• r = Distance from the center
• J = Polar moment of inertia

🔹 Stress Distribution Characteristics

1. Linear Variation
   • At the center (r = 0): τ = 0
   • At the outer surface (r = R): τ = τ_max
   So shear stress increases linearly from center to surface.
2. Maximum Shear Stress
   • At radius R (outermost fiber):
     τ_max = (T × R) / J
3. Nature of Distribution
   • Stress distribution is a straight-line relation between τ and r.
   • If visualized, the shaft cross-section would show zero stress at the core and maximum stress at the boundary.

🔹 Solid Shaft vs Hollow Shaft

• Solid Shaft
  • Stress varies from 0 at center to τ_max at the outer radius.
  • The core region is underutilized (lower stresses).
• Hollow Shaft
  • Material near the center is removed (where stresses are small).
  • Hollow shafts carry more torque per unit weight compared to solid shafts.
  • Widely used in practice (shafts, axles, drill rods).

🔹 Shear Stress Summary

• τ(r) ∝ r (linear relationship).
• Maximum stress always occurs at the outer surface.
• Hollow shafts are more efficient than solid shafts.

✅ This explains how shear stresses are distributed across the shaft under torsion, which is important for design.

🔄 Angle of Twist Formula (θ = T L / G J)

When a shaft is subjected to a torque, it not only develops shear stress but also undergoes rotational deformation. This angular deformation is called the angle of twist (θ).

It measures how much a shaft rotates under torsional loading.

🔹 Formula for Angle of Twist

θ = (T × L) / (G × J)

where,

• θ = Angle of twist (radians)
• T = Applied torque (N·m)
• L = Length of shaft (m)
• G = Modulus of rigidity (N/m²)
• J = Polar moment of inertia (m⁴)

🔹 Derivation (Outline)

From torsion equation:
τ / r = T / J = Gθ / L

Take the relation:
T / J = Gθ / L

Rearranging:
θ = (T × L) / (G × J)

🔹 Observations

1. θ ∝ T → Angle of twist increases directly with torque.
2. θ ∝ L → Longer shafts twist more under the same torque.
3. θ ∝ 1 / G → Stiffer materials (higher G) twist less.
4. θ ∝ 1 / J → Larger cross-sections (higher J) resist twisting better.

🔹 Example (Conceptual)

• A steel shaft (high G) will twist less than an aluminium shaft of the same size under the same torque.
• A hollow shaft of same weight as a solid shaft twists less because it has a higher J.

✅ Thus, the angle of twist formula is vital in designing shafts, axles, and beams to limit excessive rotation under torque.

⚡ Power Transmitted by a Shaft under Torsion

Rotating shafts are commonly used to transmit power in machines, turbines, engines, and civil engineering applications (e.g., pumps). The power transmitted depends on the torque applied and the rotational speed of the shaft.

🔹 Power Transmission Formula

P = 2π N T / 60

where,

• P = Power transmitted (Watts)
• N = Speed of shaft (revolutions per minute, rpm)
• T = Torque applied (N·m)

🔹 Explanation

• Torque T is developed in the shaft due to applied loads.
• Shaft rotates at N rpm, i.e., N/60 revolutions per second.
• Work done per revolution = 2π T (since angle per revolution = 2π radians).
• Power = Work done per second = (2π T × N) / 60.

🔹 Relation with Shear Stress

Using torsion equation (τ_max = (T × R) / J), the maximum torque that can be safely applied is limited by the shaft’s material and size. Therefore,

P ∝ τ_max × shaft dimensions × N

🔹 Key Points

1. Power transmitted increases with both torque (T) and speed (N).
2. Shafts must be designed to avoid excessive shear stress (strength criterion) and excessive twist (stiffness criterion).
3. Hollow shafts are preferred for weight-saving while transmitting the same power.

✅ The power transmission formula is crucial for solving numerical problems in JKSSB exams.

🏗️ Torsional Rigidity

Torsional Rigidity is the measure of a shaft’s resistance to twisting under an applied torque. It shows how stiff the shaft is in torsion.

🔹 Definition

Torsional Rigidity is defined as the torque required to produce unit angle of twist per unit length of shaft.

🔹 Formula

Torsional Rigidity (K) = T / θ = (G × J) / L

where,

• K = Torsional rigidity (N·m/radian)
• T = Applied torque (N·m)
• θ = Angle of twist (radians)
• G = Modulus of rigidity (N/m²)
• J = Polar moment of inertia (m⁴)
• L = Length of shaft (m)

🔹 Explanation

From angle of twist relation:
θ = (T × L) / (G × J)

Rearranging:
T / θ = (G × J) / L

So torsional rigidity depends on:

1. Material property (G) → stiffer materials resist twisting more.
2. Geometry (J) → larger polar moment gives higher rigidity.
3. Length (L) → longer shafts twist more easily (lower rigidity).

🔹 Observations

• Higher G and higher J → greater torsional rigidity.
• Longer L → lower torsional rigidity.
• Hollow shafts with same weight as solid shafts often have greater rigidity.

✅ Torsional rigidity is important in shaft design to ensure safety (no failure) and functionality (limited twist).

🔍 Solid Shaft vs Hollow Shaft under Torsion

Both solid and hollow circular shafts are commonly used in engineering. Their performance under torsional loading depends on geometry, weight, and strength.

🔹 Torque Capacity

From torsion equation:
T = (τ_max × J) / R

• For a solid shaft of diameter d:
  T = (π × τ_max × d³) / 16
• For a hollow shaft with outer diameter D and inner diameter d:
  T = [π × τ_max × (D⁴ − d⁴)] / (16 × D)

🔹 Polar Moment of Inertia

• Solid Shaft (diameter d):
  J = (π d⁴) / 32
• Hollow Shaft (outer D, inner d):
  J = [π (D⁴ − d⁴)] / 32

🔹 Weight Consideration

• Weight is proportional to cross-sectional area.
• Hollow shaft saves material near the center (where shear stress is low), so it can carry the same torque with less weight.

🔹 Efficiency

• For the same material and weight, a hollow shaft is stronger than a solid shaft.
• Hollow shafts are widely used in automobiles, aircraft, and mechanical systems where light weight and high strength are required.

🔹 Comparison Table

✅ Conclusion: Hollow shafts are more efficient than solid shafts because they transmit more torque with less material while maintaining rigidity.

🟦 Torsional Shear Stress in Non-Circular Sections

So far, the torsion equation (τ / r = T / J = Gθ / L) is valid only for circular shafts (solid or hollow).
For non-circular cross-sections (e.g., square, rectangular, I-section), the behavior is different because:

• Cross-sections do not remain plane after twisting.
• There is warping (out-of-plane deformation).
• Stress distribution is non-uniform.

🔹 Key Points

1. In circular shafts:
   • Shear stress varies linearly with radius.
   • Plane sections remain plane.
2. In non-circular shafts:
   • Shear stress distribution is not linear.
   • Sections warp, so the classical torsion equation is not valid.

🔹 Approximate Torsion Formulas

For a rectangular section (breadth b, depth h; h ≥ b):

• Maximum shear stress:
  τ_max = (T × α) / (b × h²)
• Angle of twist:
  θ = (T × L) / (β × G × b × h³)

where α and β are constants depending on h/b ratio (values available in standard tables).

🔹 Example (Thin Rectangular Section, h ≫ b)

• Shear stress is nearly uniform across thickness (b).
• Warping is significant.
• Such sections are less efficient in torsion compared to circular shafts.

🔹 Civil Engineering Application

• Non-circular torsion analysis is important in open thin-walled sections such as I-beams, channels, and thin plates used in bridges and buildings.
• For design, engineers prefer circular or hollow circular shafts whenever torque is dominant, since they distribute shear stress more efficiently.

✅ Hence, for JKSSB exams, remember:

• Circular sections follow τ / r = T / J = Gθ / L.
• Non-circular sections require modified equations due to warping.

⚠️ Failure Theories under Torsional Loading

When shafts are subjected to torsion, shear stresses are developed.
To ensure safe design, engineers use failure theories that predict whether a material will fail under combined stresses.

For pure torsion, the main concern is shear stress failure.

🔹 1. Maximum Shear Stress Theory (Tresca’s Theory)

• Failure occurs when the maximum shear stress (τ_max) in the shaft equals or exceeds the shear stress at yield in a simple tension test.

τ_max = (T × R) / J

Safe condition:
τ_max ≤ τ_allowable

This theory is widely used for ductile materials (e.g., mild steel).

🔹 2. Maximum Distortion Energy Theory (von Mises Theory)

• Failure occurs when the strain energy due to distortion equals that at yield in a uniaxial tension test.
• For pure torsion:

τ_eq = τ_max × √3

Safe condition:
τ_eq ≤ σ_y / √3

where σ_y = yield strength of material.

This theory is often more accurate for ductile materials.

🔹 3. Maximum Normal Stress Theory (Rankine’s Theory)

• States that failure occurs when the maximum principal stress equals the ultimate strength in tension or compression.
• Not suitable for torsional problems since torsion produces pure shear stress (no direct normal stress).

🔹 Summary

• For ductile materials: Tresca or von Mises theories are used.
• For brittle materials: Maximum Normal Stress Theory is sometimes considered.
• In civil engineering exams (JKSSB, SSC JE, RRB), the Torsion Equation combined with Maximum Shear Stress Theory is the most common.

✅ Thus, safe shaft design under torsion requires checking the induced shear stresses against material strength using failure theories.

🏗️ Practical Applications of Torsion in Civil Engineering

Torsion is not just a theoretical concept — it has direct applications in many structural and machine elements. In civil engineering, understanding torsion helps in safe design and stability of structures.

🔹 1. Shafts in Machinery

• Rotating shafts (like in turbines, pumps, and motors) transmit power through torsion.
• Civil engineers encounter this in pumping stations, hydropower plants, and elevators.

🔹 2. Bridge Girders and Beams

• In curved bridges and eccentrically loaded girders, torsion is induced in addition to bending.
• Torsional analysis ensures that beams do not twist excessively under traffic loads.

🔹 3. Structural Columns and Tall Buildings

• High-rise buildings subjected to wind or earthquake forces may experience torsional vibrations.
• Designing for torsion improves stability and safety against seismic forces.

🔹 4. Pavement and Highway Design

• In certain cases, pavement slabs and pre-stressed concrete members undergo torsional stresses due to wheel loads acting away from the center.

🔹 5. Reinforced Concrete (RCC) Design

• RCC members like L-beams, T-beams, and edge beams often experience torsion.
• IS Code provisions provide formulas for torsional reinforcement design.

🔹 6. Hydraulic Structures

• Spillway gates, sluice gates, and hoist shafts are designed considering torsional strength.

✅ Exam Tip (JKSSB/SSC/NTPC):

• Whenever a shaft, beam, or RCC member is loaded eccentrically, torsion comes into play.
• Circular shafts are preferred in torsion because they resist shear stresses more efficiently.

📝 Conclusion & Key Points for JKSSB Preparation

Torsion is one of the most important topics in Strength of Materials (SOM), frequently asked in JKSSB, SSC JE, RRB JE, and other civil engineering exams. It explains how circular shafts and structural members behave when subjected to twisting loads.

🔑 Key Takeaways

1. Torsion Equation:
T∕J = τ∕R = Gθ∕L

2. Power Transmission Formula:
P = (2πNT) ∕ 60

Where:

• P = Power (W)
• N = Speed (r.p.m.)
• T = Torque (N·m)

3. Failure Theories:

• Tresca (Maximum Shear Stress Theory): τmax ≤ τallow
• von Mises (Distortion Energy Theory): σeq ≤ σyield
• Rankine (Maximum Normal Stress Theory): σmax ≤ σallow

✅ Final Tip for JKSSB Aspirants:

• Memorize the torsion relation T∕J = τ∕R = Gθ∕L
• Practice numerical problems on shafts & power transmission
• Revise failure theories – frequently asked in JE-level exams

❓ Frequently Asked Questions (FAQs) on Torsion in SOM

Q1. What is torsion in Strength of Materials?
Torsion is the twisting of a structural member or shaft when subjected to torque, producing shear stresses and angular deformation.

Q2. State the torsion equation.
The torsion equation is: T∕J = τ∕R = Gθ∕L

Q3. What is polar moment of inertia (J)?
It is a geometrical property of a shaft cross-section defined as:
J = ∫ r² dA
For a solid circular shaft: J = πd⁴ ∕ 32
For a hollow shaft: J = (π∕32)(D⁴ − d⁴)

Q4. What is the angle of twist in torsion?
The angle of twist is: θ = TL ∕ (GJ)

Q5. Where is torsion applied in civil engineering?

• Shafts in pumps and turbines
• Bridge girders under eccentric loading
• RCC beams (L-beams, T-beams)
• Tall structures subjected to seismic torsion

📌 Previous Year Questions (PYQs) on Torsion – JKSSB & Other Exams

PYQ 1. The torsion equation is:
(A) T∕I = σ∕y = E∕R
(B) T∕J = τ∕R = Gθ∕L
(C) M∕I = σ∕y = E∕R
(D) P∕A = σ

Answer: (B) T∕J = τ∕R = Gθ∕L

PYQ 2. The polar moment of inertia of a solid circular shaft of diameter d is:
(A) πd⁴ ∕ 64
(B) πd⁴ ∕ 16
(C) πd⁴ ∕ 32
(D) πd² ∕ 4

Answer: (C) πd⁴ ∕ 32

PYQ 3. A solid shaft of diameter d and length L is subjected to torque T. The angle of twist θ is given by:
(A) θ = TL ∕ (GJ)
(B) θ = M∕I
(C) θ = σ∕E
(D) θ = WL∕AE

Answer: (A) θ = TL ∕ (GJ)

PYQ 4. A hollow shaft is preferred over a solid shaft because:
(A) It is lighter and stronger in torsion
(B) It transmits less power
(C) It is easier to manufacture
(D) None of these

Answer: (A) It is lighter and stronger in torsion

PYQ 5. In power transmission by a shaft rotating at N rpm, the power is given by:
(A) P = (2πNT) ∕ 60
(B) P = NT ∕ 60
(C) P = 2πN ∕ T
(D) P = T²N

Answer: (A) P = (2πNT) ∕ 60

📌 Join our Telegram Channel JKSSB CivilsCentral for regular updates, quizzes, PDF notes, and practice sets curated specifically for JKSSB aspirants.

About The Author

Sahil Digra

See author's posts