The deformation of an open-section bar subjected to pure torsion can be solved by
choosing an appropriate Prandtl stress function. Which of the following statements
is/are true about the Prandtl stress function?
Options:
(A) It satisfies the equilibrium equation
(B) It is zero on the lateral surfaces of the bar
(C) It satisfies the compatibility equation
(D) It does not satisfy the equilibrium equation
Official Answer Key:
A, B, C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.45)**
**1. Intuitive Physical Setup & Given Parameters**:
The problem concerns the Prandtl stress function $\Phi(x, y)$ used in the formulation of the Saint-Venant torsion of prismatic bars with open or arbitrary cross-sections.
We are asked to select all true statements about $\Phi(x, y)$ (MSQ).
**2. Step-by-Step Theory of Elasticity Derivation**:
1. **Definition of Prandtl Stress Function**:
- For pure torsion of a cylinder with axis along $z$, the only non-zero stress components are shear stresses $\tau_{zx}$ and $\tau_{zy}$.
- The Prandtl stress function $\Phi(x, y)$ is defined such that:
$\tau_{zx} = \frac{\partial \Phi}{\partial y}, \quad \tau_{zy} = -\frac{\partial \Phi}{\partial x}$
2. **Equilibrium Equations**:
- In the absence of body forces, the 3D equilibrium equations reduce to:
$\frac{\partial \tau_{zx}}{\partial z} = 0, \quad \frac{\partial \tau_{zy}}{\partial z} = 0$
$\frac{\partial \tau_{zx}}{\partial x} + \frac{\partial \tau_{zy}}{\partial y} = \frac{\partial}{\partial x}\left(\frac{\partial \Phi}{\partial y}\right) + \frac{\partial}{\partial y}\left(-\frac{\partial \Phi}{\partial x}\right) = \frac{\partial^2 \Phi}{\partial x \partial y} - \frac{\partial^2 \Phi}{\partial y \partial x} = 0$
- The equilibrium equation is **identically satisfied** by the definition of $\Phi$. Statement A is **TRUE**, Statement D is **FALSE**.
3. **Compatibility Equation**:
- Compatibility of the shear strains gives the governing Poisson equation:
$\nabla^2 \Phi = \frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial y^2} = -2G\theta$
where $G$ is shear modulus and $\theta$ is angle of twist per unit length. Statement C is **TRUE**.
4. **Boundary Condition on Lateral Surface**:
- The lateral surface of the cylinder is traction-free ($T_i = 0$):
$\tau_{zx} n_x + \tau_{zy} n_y = 0 \implies \frac{\partial \Phi}{\partial y}\frac{dy}{ds} + \left(-\frac{\partial \Phi}{\partial x}\right)\left(-\frac{dx}{ds}\right) = \frac{d\Phi}{ds} = 0$
- Thus, $\Phi$ is constant along the boundary curve. For simply connected cross-sections or open sections, this constant is arbitrarily set to zero: $\Phi = 0$ on the boundary. Statement B is **TRUE**.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) It satisfies the equilibrium equation**: ✅ **CORRECT**. By definition of stress components via partial derivatives.
- **(B) It is zero on the lateral surfaces of the bar**: ✅ **CORRECT**. Required boundary condition for traction-free lateral surface.
- **(C) It satisfies the compatibility equation**: ✅ **CORRECT**. Leads to $\nabla^2 \Phi = -2G\theta$.
- **(D) It does not satisfy the equilibrium equation**: ❌ **INCORRECT**. Directly contradicts Option A.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **A, B, C**
- **Exam Takeaway**: Prandtl's stress function satisfies equilibrium identically, compatibility via $\nabla^2\Phi = -2G\theta$, and traction-free lateral boundaries via $\Phi_{\text{boundary}} = 0$.
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