Axial Deformation

Axial Deformation

Learning Objectives

Define axial deformation and understand its relation to stress and strain.
Calculate axial deformation using the fundamental equation.
Determine deformations in members with varying loads or cross-sections, and members subjected to their own self-weight.
Calculate thermal deformation and thermal stress in constrained and unconstrained members.
Calculate strain energy and modulus of resilience.
Analyze impact loading and its effect on axial deformation.
Solve statically indeterminate axial members using equilibrium and compatibility equations.

Axial deformation refers to the change in length of a member subjected to axial loads (tension or compression). It represents how much a material elongates when pulled or shortens when compressed. This deformation is a direct consequence of the stress-strain relationship of the material and is critical for ensuring serviceability in design.

Fundamental Equations

Axial Rigidity

The product $AE$ (Cross-sectional Area $\times$ Modulus of Elasticity) is often referred to as the Axial Rigidity of the member. A higher axial rigidity means the member will deform less under a given load.

Deformation Formula

For a homogeneous, prismatic bar subjected to a uniform axial load, the deformation ( $\delta$ ) is given by the formula derived from Hooke's Law ( $\sigma = E\epsilon$ ). By substituting $\sigma = P/A$ and $\epsilon = \delta/L$ into Hooke's law, we can derive the fundamental deformation formula.

Fundamental Deformation Formula

Calculates the axial deformation of a prismatic member.

\delta = \frac{PL}{AE}

Variables

Symbol	Description	Unit
$\delta$	Total axial deformation	mm
$P$	Internal axial force at the section	N
$L$	Length of the member	mm
$A$	Cross-sectional area	$mm^2$
$E$	Modulus of Elasticity	MPa

Varying Loads or Cross-Sections

If the load, area, or material varies along the length, the total deformation is the sum of the deformations of individual segments.

Deformation for Discrete Segments

Sum of individual deformations.

\delta = \sum \frac{P_i L_i}{A_i E_i}

Variables

Symbol	Description	Unit
$\delta$	Total axial deformation	mm
$P_i$	Internal axial force of segment i	N
$L_i$	Length of segment i	mm
$A_i$	Cross-sectional area of segment i	$mm^2$
$E_i$	Modulus of Elasticity of segment i	MPa

For continuous variation (e.g., a tapered bar or distributed axial load):

Deformation for Continuous Variation

Integration over the member length.

\delta = \int_0^L \frac{P(x)}{A(x)E(x)} dx

Variables

Symbol	Description	Unit
$\delta$	Total axial deformation	mm
$P(x)$	Internal axial force at distance x	N
$L$	Total length of the member	mm
$A(x)$	Cross-sectional area at distance x	$mm^2$
$E(x)$	Modulus of Elasticity at distance x	MPa

Sign Convention

Tension (+): Causes elongation (positive $\delta$ ).
Compression (-): Causes contraction (negative $\delta$ ).

Deformation Due to Self-Weight

When a vertical bar is suspended from one end, it deforms not only from applied external loads but also under its own weight. The internal force $P(x)$ varies linearly from zero at the free end to a maximum (the total weight) at the support.

For a prismatic bar with constant area $A$ , length $L$ , modulus $E$ , and specific weight $\gamma$ (weight per unit volume):

Deformation Due to Self-Weight

Elongation of a vertically suspended bar due to its own weight.

\delta_{\text{weight}} = \frac{\gamma L^2}{2E} = \frac{W L}{2AE}

Variables

Symbol	Description	Unit
$\delta_{\text{weight}}$	Elongation due to self-weight	mm
$\gamma$	Specific weight (weight per unit volume)	$N/mm^3$
$W$	Total weight of the bar	N
$L$	Length of the member	mm
$A$	Cross-sectional area	$mm^2$
$E$	Modulus of Elasticity	MPa

This indicates that a bar elongates half as much under its own self-weight as it would if the total weight $W$ were applied entirely as a concentrated load at its free end.

Thermal Effects and Energy

Thermal Deformation

Changes in temperature cause materials to expand or contract. If this deformation is free to occur, no stress is induced. This pure change in length is known as thermal deformation.

Thermal Stress

If the thermal deformation is restrained (prevented by supports), internal stresses develop. The thermal stress is equivalent to the force required to push the member back to its original constrained length.

Thermal Deformation Formula

Change in length due to temperature.

\delta_T = \alpha L (\Delta T)

Variables

Symbol	Description	Unit
$\delta_T$	Thermal deformation	mm
$\alpha$	Coefficient of Linear Thermal Expansion	$/^\circ C$
$L$	Original length of the member	mm
$\Delta T$	Change in temperature ( $T_{\text{final}} - T_{\text{initial}}$ )	$^\circ C$

Thermal Stress Formula

Stress developed in a restrained member under temperature change.

\sigma_T = E \alpha (\Delta T)

Variables

Symbol	Description	Unit
$\sigma_T$	Thermal stress	MPa
$E$	Modulus of Elasticity	MPa
$\alpha$	Coefficient of Linear Thermal Expansion	$/^\circ C$
$\Delta T$	Change in temperature	$^\circ C$

Strain Energy ( $U$ )

When an axial load is applied, the member deforms and absorbs strain energy. For a gradually applied axial load, the strain energy within the proportional limit is given by the area under the load-deformation diagram.

Strain Energy from Axial Loads

Potential energy stored due to axial deformation.

U = \frac{1}{2} P \delta = \frac{P^2 L}{2 A E}

Variables

Symbol	Description	Unit
$U$	Strain energy	J
$P$	Applied axial force	N
$\delta$	Axial deformation	m
$L$	Length of the member	m
$A$	Cross-sectional area	$m^2$
$E$	Modulus of Elasticity	Pa

Modulus of Resilience ( $U_{\text{r}}$ )

The maximum strain energy density (strain energy per unit volume) a material can absorb without undergoing permanent plastic deformation.

Modulus of Resilience

Maximum elastic strain energy per unit volume.

U_{\text{r}} = \frac{\sigma_{\text{yield}}^2}{2E}

Variables

Symbol	Description	Unit
$U_{\text{r}}$	Modulus of resilience	$J/m^3$
$\sigma_{\text{yield}}$	Yield stress	Pa
$E$	Modulus of Elasticity	Pa

Impact Loading

When an axial load is applied suddenly (dynamic load), the strain energy must absorb the kinetic energy of the impact, resulting in an internal force and deformation that can be significantly larger than the static effect.

Deformation from Impact

If a weight $W$ is dropped from a height $h$ onto an axial member, the maximum deformation $\delta_{\text{max}}$ is derived from the conservation of energy, equating the potential energy lost by the weight to the strain energy gained by the member.

Impact Deformation Formula

Maximum dynamic deformation from a falling weight.

\delta_{\text{max}} = \delta_{\text{st}} \left[ 1 + \sqrt{1 + \frac{2h}{\delta_{\text{st}}}} \right]

Variables

Symbol	Description	Unit
$\delta_{\text{max}}$	Maximum dynamic deformation	m
$\delta_{\text{st}}$	Static deformation if weight W was applied gradually ( $WL/AE$ )	m
$h$	Height from which the weight is dropped	m

Note that if $h = 0$ (load applied suddenly but without a drop height), the dynamic deformation $\delta_{\text{max}}$ is exactly $2\delta_{\text{st}}$ , meaning a suddenly applied load produces twice the deformation (and twice the stress) of a gradually applied static load.

Statically Indeterminate Members

Statically Indeterminate Members

A structural member is considered statically indeterminate when the equations of static equilibrium ( $\sum F = 0$ , $\sum M = 0$ ) alone are insufficient to determine all the internal forces and support reactions. This typically happens when there are redundant supports—more constraints than necessary to maintain stability.

Compatibility Equations

To solve statically indeterminate problems, we must supplement the equilibrium equations with Compatibility Equations. These equations relate the deformations of various parts of the structure to ensure geometric consistency (e.g., ensuring a continuous beam doesn't break apart at its joints).

Solving Statically Indeterminate Members

STEP-BY-STEP

Equilibrium Equations: Draw a Free Body Diagram (FBD) and write the static equilibrium equations (e.g., $\sum F_x = 0$ ).
Compatibility Equations: Formulate a relationship between the deformations of the members based on geometric constraints (e.g., $\delta_{\text{total}} = 0$ for fixed supports).
Force-Deformation Relations: Substitute $\delta = PL/AE$ into the compatibility equation.
Solve: Solve the system of simultaneous equations for the unknown forces.

Initial Gaps and Misfits

Sometimes, structural members are fabricated slightly too short or too long. If a member is too short by a gap distance $\Delta_{\text{gap}}$ , and forces are applied to close the gap and connect it, internal stresses will develop.

The compatibility equation in such a scenario becomes:

Compatibility for Initial Gaps

Total deformation equates to the gap distance.

\delta_{\text{total}} = \Delta_{\text{gap}}

Variables

Symbol	Description	Unit
$\delta_{\text{total}}$	Total deformation of the structure	m
$\Delta_{\text{gap}}$	Initial gap distance	m

This means the total elongation of the structure must equal the initial gap distance in order for the connection to be made.

Interactive Lab: Axial Deformation

Interactive Simulation

Use the simulator below to explore how changes in Force, Length, Area, and Material stiffness affect the total axial deformation.