Structural Engineering Dynamics

CHAPTER 3: ELASTICITY IN SOLIDS

3.1 EQUILIBRIUM EQUATIONS & 3.2 STRAIN-DISPLACEMENT

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

Derive the differential equation of equilibrium for three dimensional problems. [5-7] (2073 Magh, Q7; 2070 Magh, Q3a; 2073 Bhadra, Q7b)
Derive Strain-Displacement relationship for 3-D solid element. [6] (2080 Chaitra, Q7)
Derive strain-displacement and constitutive relationships that exist in plane stress problem for isotropic material. [6] (2072 Ashwin, Q6)

Differential Equations of Equilibrium for 3D Solids

Consider an infinitesimal rectangular parallelepiped (a very small 3D solid element) extracted from a stressed elastic body, with side lengths $ dx $, $ dy $, and $ dz $ parallel to the coordinate axes $ x $, $ y $, and $ z $. Let the body forces acting per unit volume be $ X $, $ Y $, and $ Z $.

Figure 3.1: Normal and shear stress components acting on a 3D infinitesimal differential element.

The normal stress $ \sigma_x $ acts on the left face. Due to stress gradients across the body, the normal stress on the right face (at a distance $ dx $) becomes $ \sigma_x + \frac{\partial \sigma_x}{\partial x} dx $. Similarly, shear stresses $ \tau_{yx} $ and $ \tau_{zx} $ vary across the element.

For the element to be in static equilibrium, the sum of all forces in the $ x $-direction must equal zero ($ \Sigma F_x = 0 $):

$$ \left( \sigma_x + \frac{\partial \sigma_x}{\partial x} dx - \sigma_x \right) dy dz + \left( \tau_{yx} + \frac{\partial \tau_{yx}}{\partial y} dy - \tau_{yx} \right) dx dz + \left( \tau_{zx} + \frac{\partial \tau_{zx}}{\partial z} dz - \tau_{zx} \right) dx dy + X (dx dy dz) = 0 $$

Dividing the entire equation by the element volume ($ dx dy dz $) yields the first differential equation of equilibrium. Repeating this for the $ y $ and $ z $ directions yields the full set of 3D equilibrium equations:

$$ \frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} + X = 0 $$

$$ \frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{zy}}{\partial z} + Y = 0 $$

$$ \frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_z}{\partial z} + Z = 0 $$

Note: From moment equilibrium ($ \Sigma M = 0 $), it can be proven that shear stresses are complementary, i.e., $ \tau_{xy} = \tau_{yx} $, $ \tau_{xz} = \tau_{zx} $, and $ \tau_{yz} = \tau_{zy} $.

Strain-Displacement Relationships

Let $ u $, $ v $, and $ w $ be the continuous displacement functions of a point in the $ x $, $ y $, and $ z $ directions respectively. The normal strains ($ \epsilon $) are defined as the change in length per unit original length. Taking the limit as the length approaches zero, normal strains are the first partial derivatives of displacements:

$$ \epsilon_x = \frac{\partial u}{\partial x}, \quad \epsilon_y = \frac{\partial v}{\partial y}, \quad \epsilon_z = \frac{\partial w}{\partial z} $$

The shear strains ($ \gamma $) are defined as the change in the right angle between two originally orthogonal line segments. It depends on the relative displacement gradients:

$$ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}, \quad \gamma_{yz} = \frac{\partial v}{\partial z} + \frac{\partial w}{\partial y}, \quad \gamma_{zx} = \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z} $$

In matrix notation, the strain vector $ \{\epsilon\} $ is related to the displacement vector $ \{u\} $ through a differential operator matrix $ [B] $, such that $ \{\epsilon\} = [B]\{u\} $.

3.3 CONSTITUTIVE RELATIONS & LAMÉ'S CONSTANTS

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

Derive the constitutive relation ({σ}=[D]{E}) for an elastic isotropic material. [5-6] (2071 Bhadra, Q6a; 2079 Chaitra, Q7; 2078 Chaitra, Q3a)
Explain how strain can be expressed in terms of stresses for a 3-D object. [5] (2076 Bhadra, Q6a)
Derive the expression for Lamé's constants. Explain their necessity. Differentiate between isotropic and anisotropic material. [2-5] (2077 Chaitra, Q3a; 2070 Bhadra, Q2a; 2080 Chaitra, Q7; 2074 Bhadra, Q7b)

Isotropic vs. Anisotropic Materials

An isotropic material has physical and mechanical properties (like Young's Modulus $ E $ and Poisson's ratio $ \nu $) that are identical in all directions at a given point (e.g., steel, amorphous plastics). An anisotropic material has properties that vary depending on the direction of measurement (e.g., wood, fiber-reinforced composites).

Strain Expressed in Terms of Stresses (3D Generalized Hooke's Law)

For a 3D isotropic material, the total normal strain in any direction is the sum of the direct strain due to the stress in that direction, minus the Poisson's effect (lateral contraction) from the orthogonal stresses:

$$ \epsilon_x = \frac{1}{E} [ \sigma_x - \nu(\sigma_y + \sigma_z) ] $$

$$ \epsilon_y = \frac{1}{E} [ \sigma_y - \nu(\sigma_x + \sigma_z) ] $$

$$ \epsilon_z = \frac{1}{E} [ \sigma_z - \nu(\sigma_x + \sigma_y) ] $$

The shear strains are related to shear stresses by the shear modulus $ G $:

$$ \gamma_{xy} = \frac{\tau_{xy}}{G}, \quad \gamma_{yz} = \frac{\tau_{yz}}{G}, \quad \gamma_{zx} = \frac{\tau_{zx}}{G} \quad \text{where} \quad G = \frac{E}{2(1+\nu)} $$

Derivation of Lamé's Constants and the Constitutive Matrix [D]

In Finite Element Method, we need to express stresses in terms of strains: $ \{\sigma\} = [D]\{\epsilon\} $. To invert the generalized Hooke's equations above, we define the volumetric strain $ \epsilon_v = \epsilon_x + \epsilon_y + \epsilon_z $.

Adding the three normal strain equations yields:

$$ \epsilon_v = \frac{1 - 2\nu}{E} (\sigma_x + \sigma_y + \sigma_z) \implies (\sigma_x + \sigma_y + \sigma_z) = \frac{E}{1 - 2\nu}\epsilon_v $$

Substituting this back into the equation for $ \epsilon_x $:

$$ \epsilon_x = \frac{\sigma_x}{E} - \frac{\nu}{E}\left( \frac{E}{1 - 2\nu}\epsilon_v - \sigma_x \right) $$

Solving explicitly for $ \sigma_x $, we obtain the form:

$$ \sigma_x = \frac{\nu E}{(1+\nu)(1-2\nu)}\epsilon_v + \frac{E}{1+\nu}\epsilon_x $$

To greatly simplify the complex algebra in 3D constitutive equations, mathematicians introduced Lamé's Constants ($ \lambda $ and $ \mu $):

$$ \lambda = \frac{\nu E}{(1+\nu)(1-2\nu)} \quad \text{and} \quad \mu = G = \frac{E}{2(1+\nu)} $$

Necessity of Lamé's Constants: Lamé's constants are necessary because they compress the mathematically bulky generalized Hooke's Law into a highly compact, mathematically elegant tensor form: $ \sigma_{ij} = \lambda \epsilon_{kk} \delta_{ij} + 2\mu \epsilon_{ij} $. This significantly reduces computational coding effort and theoretical derivation length.

Using Lamé's constants, the 3D constitutive matrix relation $ \{\sigma\} = [D]\{\epsilon\} $ is strictly written as:

$$ \begin{Bmatrix} \sigma_x \\ \sigma_y \\ \sigma_z \\ \tau_{xy} \\ \tau_{yz} \\ \tau_{zx} \end{Bmatrix} = \begin{bmatrix} \lambda+2\mu & \lambda & \lambda & 0 & 0 & 0 \\ \lambda & \lambda+2\mu & \lambda & 0 & 0 & 0 \\ \lambda & \lambda & \lambda+2\mu & 0 & 0 & 0 \\ 0 & 0 & 0 & \mu & 0 & 0 \\ 0 & 0 & 0 & 0 & \mu & 0 \\ 0 & 0 & 0 & 0 & 0 & \mu \end{bmatrix} \begin{Bmatrix} \epsilon_x \\ \epsilon_y \\ \epsilon_z \\ \gamma_{xy} \\ \gamma_{yz} \\ \gamma_{zx} \end{Bmatrix} $$

3.4 PLANE STRESS & 3.5 PLANE STRAIN ELEMENTS

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

Define / Differentiate plane stress and plane strain problems with necessary conditions and suitable examples. [3-8] (2073 Magh, 2070 Bhadra, 2074 Magh, 2071 Magh, 2079 Jestha, 2078 Chaitra, 2078 Kartik, 2073 Bhadra, 2079 Shrawan)
Derive constitutive law for a two-dimensional plane stress problem. [4-6] (2078 Kartik, 2079 Shrawan, 2075 Bhadra)
Derive constitutive laws for a two dimensional plane strain problems. [6] (2071 Magh, Q3)

1. Plane Stress Problem

Definition: A state of plane stress exists in a solid where one dimension (usually the thickness in the z-direction) is extremely small compared to the other two dimensions (x and y directions). The loads are applied strictly in the x-y plane.

Conditions: Because the body is very thin and unconstrained in the z-direction, the normal stress and shear stresses acting on the z-face are strictly zero.

$$ \sigma_z = 0, \quad \tau_{xz} = 0, \quad \tau_{yz} = 0 $$

(Note: While stress in the z-direction is zero, the normal strain $ \epsilon_z $ is NOT zero due to the Poisson effect, allowing the plate to thin out).

Examples: Thin plates with central holes under tension, thin gears, membrane shells, and thin retaining walls.

Derivation of Constitutive Matrix [D] for Plane Stress:
Substituting $ \sigma_z = 0 $ into the generalized Hooke's equations: $$ \epsilon_x = \frac{1}{E}(\sigma_x - \nu\sigma_y) \quad \text{and} \quad \epsilon_y = \frac{1}{E}(\sigma_y - \nu\sigma_x) $$ Solving these two simultaneously for $ \sigma_x $ and $ \sigma_y $ yields:

$$ \begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix} \begin{Bmatrix} \epsilon_x \\ \epsilon_y \\ \gamma_{xy} \end{Bmatrix} $$

2. Plane Strain Problem

Definition: A state of plane strain exists in a solid where one dimension (the z-direction) is infinitely long compared to the cross-section (x-y plane). The loads act uniformly along the entire length of the body.

Conditions: Because the body is infinitely long and constrained by its own massive continuity, there is absolute zero displacement in the z-direction ($ w=0 $). Therefore, all strains involving the z-direction are strictly zero.

$$ \epsilon_z = 0, \quad \gamma_{xz} = 0, \quad \gamma_{yz} = 0 $$

(Note: While strain in the z-direction is zero, the normal stress $ \sigma_z $ is NOT zero. $ \sigma_z = \nu(\sigma_x + \sigma_y) $).

Examples: Long retaining walls, gravity dams, continuous strip footings, and long pressurized pipes.

Derivation of Constitutive Matrix [D] for Plane Strain:
Substituting $ \epsilon_z = 0 $ into the 3D relation $ \sigma_z = \nu(\sigma_x + \sigma_y) $ and substituting this back into the equations for $ \epsilon_x $ and $ \epsilon_y $ yields:

$$ \begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = \frac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & 0 \\ \nu & 1-\nu & 0 \\ 0 & 0 & \frac{1-2\nu}{2} \end{bmatrix} \begin{Bmatrix} \epsilon_x \\ \epsilon_y \\ \gamma_{xy} \end{Bmatrix} $$

Figure 3.2: 3D Visualization differentiating Plane Stress (unconstrained thickness) and Plane Strain (fully constrained length) conditions.

3.6 AXI-SYMMETRIC ELEMENTS

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

Explain plain stress, plain strain and axi-symmetric problems with examples. [6] (2072 Magh, Q3; 2079 Shrawan, Q6a; 2074 Bhadra, Q7a)
Explain about axisymmetric problems with examples. Write down the constitutive relations and strain displacement relation for axisymmetric problems. [5] (2077 Chaitra, Q3b; 2076 Bhadra, Q6b; 2071 Bhadra, Q6b; 2070 Magh, Q3b)

Definition and Examples

An axisymmetric problem arises when a 3D structural solid, its material properties, its boundary conditions, and the applied external loads are completely symmetrical about a central axis of revolution (typically the z-axis).

Because of this perfect rotational symmetry, any variable (stress, strain, displacement) depends solely on the radial coordinate $ r $ and the axial coordinate $ z $. There is absolutely no variation with respect to the circumferential/tangential coordinate $ \theta $. This effectively reduces a complex 3D problem into a 2D problem modeled in the $ r-z $ plane.

Examples: Cylindrical storage tanks, pressure vessels, cooling towers, rocket nozzles, circular footings, and flywheels.

Figure 3.3: An axisymmetric cylindrical wedge showing coordinates (r, z, θ) and primary stresses. The 3D problem is solved entirely on the 2D r-z cross-section plane.

Strain-Displacement Relations for Axisymmetric Condition

In a cylindrical coordinate system $ (r, \theta, z) $, the displacements are $ u $ (radial) and $ w $ (axial). Due to symmetry, tangential displacement $ v = 0 $ and all derivatives $ \frac{\partial}{\partial \theta} = 0 $.

Crucially, as a point displaces radially by $ u $, the circumference increases from $ 2\pi r $ to $ 2\pi(r+u) $. This geometric expansion creates a circumferential (hoop) strain $ \epsilon_\theta $. The strain-displacement relations are:

$$ \epsilon_r = \frac{\partial u}{\partial r} \quad \text{(Radial Strain)} $$

$$ \epsilon_\theta = \frac{u}{r} \quad \text{(Circumferential/Hoop Strain)} $$

$$ \epsilon_z = \frac{\partial w}{\partial z} \quad \text{(Axial Strain)} $$

$$ \gamma_{rz} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial r} \quad \text{(Shear Strain)} $$

Constitutive Relations for Axisymmetric Condition

Applying the 3D generalized Hooke's law in cylindrical coordinates, and removing the zero shear stresses ($ \tau_{r\theta} = \tau_{z\theta} = 0 $), the stress-strain relation maps four non-zero strains to four non-zero stresses:

$$ \begin{Bmatrix} \sigma_r \\ \sigma_z \\ \sigma_\theta \\ \tau_{rz} \end{Bmatrix} = \frac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & \nu & 0 \\ \nu & 1-\nu & \nu & 0 \\ \nu & \nu & 1-\nu & 0 \\ 0 & 0 & 0 & \frac{1-2\nu}{2} \end{bmatrix} \begin{Bmatrix} \epsilon_r \\ \epsilon_z \\ \epsilon_\theta \\ \gamma_{rz} \end{Bmatrix} $$

This constitutive relation shows that even though the mathematical model is mapped onto a 2D $r-z$ plane, the physics must strictly account for the out-of-plane hoop stress ($ \sigma_\theta $) that intrinsically develops due to the 3D volume constraints of the cylinder.