GATE AE 2026 • Q26 MSQ • 1 Mark

Propulsion

Thermodynamic Cycles of Gas Turbines

Subtopic: Brayton Cycle with Real Component Efficiencies

Question Statement:

The state of stress at a point in a 2-D body, in the x-y Cartesian coordinate system, is represented in matrix form as [σ]. The transformation matrix [Q] rotates the coordinate system to a new x'-y' Cartesian coordinate system. Select the CORRECT option(s) that represent(s) the state of stress in the new coordinate system.

Options:

  • (A) [Q] [σ] [Q]T
  • (B) [Q] [σ] [Q]-1
  • (C) ([Q]-1)T [σ] [Q]T
  • (D) [Q]-1 [σ] [Q]

Official Answer Key:

A, B, C

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2026 Q.26)** **1. Intuitive Physical Setup & Given Parameters**: Let $[\sigma]$ denote the Cauchy stress tensor in the original Cartesian coordinate system $x$-$y$. Let $[Q]$ denote the coordinate transformation (direction cosine) matrix that transforms vectors as: $\{x'\} = [Q]\{x\}$ Since coordinate transformations between Cartesian frames are orthogonal, $[Q]$ satisfies: $[Q]^{-1} = [Q]^T$ We need to determine the correct matrix representation of the transformed stress tensor $[\sigma']$ in the new coordinate system $x'$-$y'$ (MSQ). **2. Step-by-Step Tensor Transformation Derivation**: 1. **Tensor Transformation Rule**: - Stress is a second-order Cartesian tensor. Under a coordinate transformation with direction cosines $Q_{ij}$, the transformation law in indicial notation is: $\sigma'_{ij} = Q_{ik} Q_{jl} \sigma_{kl} = Q_{ik} \sigma_{kl} (Q^T)_{lj}$ - In matrix notation, this corresponds to: $[\sigma'] = [Q] [\sigma] [Q]^T$ - Thus, **Option A** is correct. 2. **Utilizing Orthogonality of $[Q]$**: - Because $[Q]$ represents a pure Cartesian coordinate rotation, it is an orthogonal matrix: $[Q]^T = [Q]^{-1}$ - Substituting $[Q]^T = [Q]^{-1}$ into Option A yields: $[\sigma'] = [Q] [\sigma] [Q]^{-1}$ - Thus, **Option B** is also correct. 3. **Checking Inverse Transpose**: - Since $[Q]^{-1} = [Q]^T$, taking transpose on both sides gives $([Q]^{-1})^T = Q$. - Therefore: $([Q]^{-1})^T [\sigma] [Q]^T = [Q] [\sigma] [Q]^T = [\sigma']$ - Thus, **Option C** is also mathematically equivalent and correct. 4. **Checking Option D**: - Option D is $[Q]^{-1} [\sigma] [Q] = [Q]^T [\sigma] [Q]$, which represents the inverse transformation from $x'$ back to $x$, not from $x$ to $x'$. Thus, Option D is incorrect. **3. Comprehensive Options & Distractor Analysis**: - **(A) [Q] [σ] [Q]T**: ✅ **CORRECT**. Standard second-order tensor transformation formula. - **(B) [Q] [σ] [Q]-1**: ✅ **CORRECT**. Equivalent to A because $[Q]^T = [Q]^{-1}$ for orthogonal Cartesian rotations. - **(C) ([Q]-1)T [σ] [Q]T**: ✅ **CORRECT**. Since $([Q]^{-1})^T = ([Q]^T)^T = [Q]$, this evaluates identically to $[Q][\sigma][Q]^T$. - **(D) [Q]-1 [σ] [Q]**: ❌ **INCORRECT**. This represents the backwards transformation from the primed system to the unprimed system. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **A, B, C** - **Exam Takeaway**: For orthogonal Cartesian coordinate rotations, always remember $[Q]^T = [Q]^{-1}$. Both $[Q][\sigma][Q]^T$ and $[Q][\sigma][Q]^{-1}$ represent the exact same forward tensor rotation.

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