Axial loading and thermal stress

Axial bars deform by (elastic). Stepped/composite bars use series/parallel stiffness. Free thermal expansion is; when constrained, thermal stress develops (RK Bansal). Combined mechanical + thermal problems superpose displacements with compatibility.

Write the formula for elongation of a prismatic bar under axial load.

where is length, cross-section, Young’s modulus. Assumes uniform stress, linear elasticity, and load along centroidal axis.

How do you find total elongation of a stepped bar under the same axial load?

Treat segments in series:. If the same passes through all segments,.

What is a composite bar? How is load shared in a parallel composite?

Two or more materials rigidly connected to share axial deformation. Parallel:,, with.

Explain statically indeterminate axial members.

When reactions/internal forces cannot be found from equilibrium alone (e.g. bar fixed at both ends with intermediate load). Compatibility of displacements supplies extra equations.

A rod fixed at both ends is heated. How do you find thermal stress?

Free thermal expansion is fully suppressed. Equivalent mechanical compression gives,.

What is the condition for no thermal stress in a bar?

The bar must be free to expand/contract, or supports must move exactly by. Partial constraint produces stress proportional to the prevented displacement.

How is elongation found when self-weight acts on a vertical bar?

Integrate: stress varies linearly. For uniform bar, where is total weight — equivalent to load acting at mid-length.

Define stiffness of an axially loaded bar.

Axial stiffness. Higher or, or shorter, increases stiffness.

Fixed–fixed bar, intermediate axial load at midspan, uniform. Reactions?

By symmetry (or compatibility), each support takes.

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Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Strength of Materials (SOM)

Axial Loading & Thermal Stress

Axial bars deform by

\delta=PL/(AE)

(elastic). Stepped/composite bars use series/parallel stiffness. Free thermal expansion is

\delta_T=\alpha L\Delta T

; when constrained, thermal stress

\sigma_T=E\alpha\Delta T

develops (RK Bansal). Combined mechanical + thermal problems superpose displacements with compatibility.

Exam tip: keep units consistent — N with mm² gives MPa directly (1 MPa = 1 N/mm²).

Key formulas & points

Skim these first — then read the full notes below.

Elastic elongation: $\delta=\dfrac{PL}{AE}$ ; stiffness $k = \frac{AE}{L}$ .
Series (same $P$ ): $\delta_{\mathrm{total}}=\sum PL_i/(A_i E_i)$ .
Parallel (same $\delta$ ): $P=\sum (A_i E_i/L_i)\delta$ .
Free thermal: $\delta_T=\alpha L\Delta T$ (no stress if free).
Fully fixed bar: $\sigma_T=E\alpha\Delta T$ , $P_T=AE\alpha\Delta T$ .
Statically indeterminate: write equilibrium + compatibility ( $\delta_{\mathrm{mech}}+\delta_T=\delta_{\mathrm{gap}}$ ).
Composite (steel–copper) on rigid supports: shared $\delta$ , load share by $AE$ .

Topic details

Definition and physical meaning

Axial loading means the resultant internal force is along the centroidal axis of a prismatic (or piecewise prismatic) member — pure tension or compression, no bending/torsion in the elementary model.

Deformation (elastic):

\delta=\frac{PL}{AE}

Thermal expansion of an unconstrained bar:

\delta_T=\alpha L\Delta T

where

\alpha

is the linear coefficient of thermal expansion (

\mathrm{1/K}

\mathrm{/^\circ C}

If expansion is fully prevented, an equivalent compressive (heating) or tensile (cooling) force develops so that mechanical shortening cancels

\delta_T

\delta_{\mathrm{mech}}+\delta_T=0\Rightarrow\frac{PL}{AE}+\alpha L\Delta T=0\Rightarrow\sigma=E\alpha\Delta T

(sign: heating + fixed ends → compression).

Physical parameters

Symbol	Meaning	SI unit
$P$	Axial force	$\mathrm{N}$
$A$	Cross-section area	$\mathrm{m^2}$
$L$	Length	$\mathrm{m}$
$E$	Young’s modulus	$\mathrm{Pa}$
$\alpha$	Thermal expansion coeff.	$\mathrm{1/K}$
$\Delta T$	Temperature change	$\mathrm{K}$ (or °C)
$\delta$	Axial deformation	$\mathrm{m}$
$\sigma$	Axial stress	$\mathrm{Pa}$

Typical

\alpha

: steel

\approx 12\times 10^{-6}/\mathrm{K}

, copper

\approx 17\times 10^{-6}

, aluminium

\approx 23\times 10^{-6}

, concrete

\approx 10

–

12\times 10^{-6}

Fig 9.1 — Same P in each section. δ_total = P L₁/(A₁E) + P L₂/(A₂E) + … Compatibility for indeterminate cases.

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Core assumptions (state these in exams)

1. Axial centroidal load — no eccentricity (otherwise bending appears).
2. Linear elastic material; superposition valid.
3. Uniform temperature over the cross-section (or use average

\Delta T

).
4. Plane sections remain plane; stress uniform on section for prismatic bar.
5. Small deformations; support reactions from undeformed geometry.
6. Perfectly rigid supports unless support stiffness is given.
7. Saint-Venant — formulas apply away from local load introduction.

For tapered circular bars (diameters

d_{1}

d_{2}

\delta=\frac{4PL}{\pi d_1 d_2 E}

Stepped and composite bars

Stepped bar in series (same axial force

P

through each segment if no intermediate loads):

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

If intermediate axial loads exist, find

P_{i}

in each segment from equilibrium (cut sections).

Composite / compound bar (two materials rigidly connected to share load, same length change):

\delta_1=\delta_2,\qquad P_1+P_2=P

\frac{P_1 L}{A_1 E_1}=\frac{P_2 L}{A_2 E_2}\Rightarrow\frac{P_1}{P_2}=\frac{A_1 E_1}{A_2 E_2}

Load shares in proportion to axial rigidity

AE

Equivalent stiffness:

k_{\mathrm{eq}}=\sum\frac{A_i E_i}{L}\quad(\text{parallel}),\qquad\frac{1}{k_{\mathrm{eq}}}=\sum\frac{L_i}{A_i E_i}\quad(\text{series})

Thermal stress — free, partial, and fixed

Case A — free bar:

\delta_T=\alpha L\Delta T

\sigma=0

Case B — fully fixed:

\delta_{\mathrm{net}}=0

\sigma_T=E\alpha\Delta T,\qquad P_T=AE\alpha\Delta T

Case C — gap / yielding support: if a gap

g

exists and

\delta_T>g

on heating,

\delta_T-g=\frac{PL}{AE}\Rightarrow P=\frac{AE}{L}(\alpha L\Delta T-g)

Case D — two materials between rigid walls (e.g. copper rod inside steel tube):
Compatibility:

\delta_{\mathrm{st,mech}}+\delta_{\mathrm{st},T}=\delta_{\mathrm{cu,mech}}+\delta_{\mathrm{cu},T}

(both net changes equal if ends remain coplanar), plus

P_{\mathrm{st}}+P_{\mathrm{cu}}=0

for pure thermal (no external load).

External load + temperature: include

P_{\mathrm{ext}}

in equilibrium.

Statically indeterminate axial systems

When reactions exceed equilibrium equations alone:

1. Write equilibrium (

\sum F=0

).
2. Write compatibility of displacements (geometry of supports).
3. Use constitutive relations

\delta=PL/(AE)

and

\delta_T=\alpha L\Delta T

.
4. Solve the linear system for redundant forces.

Example: bar fixed at both ends, intermediate axial load

P

at midspan — reactions

R_{1}+R_{2} = P

and

\delta_{\mathrm{left}}+\delta_{\mathrm{right}}=0

relative to fixed ends.

Step-by-step problem approach

1. Sketch geometry, materials, supports, and temperature change.
2. Classify: determinate vs indeterminate; free vs constrained thermal.
3. Find internal force in each segment (equilibrium / FBD).
4. Write

\delta_i=P_i L_i/(A_i E_i)+\alpha_i L_i\Delta T_i

.
5. Enforce compatibility (sum of

\delta

= support movement or zero).
6. Solve for unknowns; back-substitute for stresses

\sigma_i=P_i/A_i

.
7. Check signs: heating + restraint → compression; cooling → tension.
8. Units: keep

\alpha\Delta T

dimensionless;

E

and

\sigma

in same units.

Common mistakes in exams

• Computing

\sigma=E\alpha\Delta T

for a free bar (stress is zero if free).
• Forgetting

\alpha

differs between materials in a compound bar.
• Using series formula when materials share the same

\delta

(parallel).
• Sign errors on

\Delta T

(rise vs fall) and on tensile/compressive reactions.
• Mixing mm and m in

L

while using

E

in N/mm² inconsistently.
• Ignoring a specified gap between bar and support.

Calculator

Axial deformation

\delta = \frac{P L}{A E}

Load PNLength LmmArea Amm²Modulus EN/mm²

Result

0.9524mm

δ = PL/(AE) = (80000×1000) / (400×2.100e+5) = 0.9524 mm

Worked examples

Try the problem first — open the solution when you are ready to check.

Stepped steel bar elongation

Problem

A stepped steel bar (

E=200\,\mathrm{GPa}

) has segment 1:

A_1=400\,\mathrm{mm^2}

L_1=0.6\,\mathrm{m}

; segment 2:

A_2=200\,\mathrm{mm^2}

L_2=0.4\,\mathrm{m}

. Axial load

P=80\,\mathrm{kN}

(same through both). Find total elongation.

Solution

Formulas (Indian textbook notation)

$\delta_1=\frac{PL_1}{A_1 E}=\frac{(80\times 10^3)(600)}{400\times(200\times 10^3)}=0.60\,\mathrm{mm}$

Formulas (Indian textbook notation)

$\delta_2=\frac{PL_2}{A_2 E}=\frac{(80\times 10^3)(400)}{200\times(200\times 10^3)}=0.80\,\mathrm{mm}$

Formulas (Indian textbook notation)

$\delta=\delta_1+\delta_2=1.40\,\mathrm{mm}$

Fully fixed bar — thermal stress

Problem

A steel rod (

E=210\,\mathrm{GPa}

\alpha=12\times 10^{-6}/\mathrm{K}

) is fixed at both ends. Temperature rises by

\Delta T=40^\circ\mathrm{C}

. Find thermal stress and the force if

A=500\,\mathrm{mm^2}

Compound bar with temperature rise

Problem

A copper rod (

A_c=300\,\mathrm{mm^2}

E_c=100\,\mathrm{GPa}

\alpha_c=18\times 10^{-6}

) and a steel tube (

A_s=600\,\mathrm{mm^2}

E_s=200\,\mathrm{GPa}

\alpha_s=12\times 10^{-6}

) of equal length are connected between two rigid walls.

\Delta T=+50^\circ\mathrm{C}

, no external load. Find stresses.

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
Write the formula for elongation of a prismatic bar under axial load $P$ .
Model answer
$\delta=\dfrac{PL}{AE}$ where $L$ is length, $A$ cross-section, $E$ Young’s modulus. Assumes uniform stress, linear elasticity, and load along centroidal axis.
2
How do you find total elongation of a stepped bar under the same axial load?
Model answer
Treat segments in series: $\delta=\sum P_i L_i/(A_i E_i)$ . If the same $P$ passes through all segments, $\delta=P\sum L_i/(A_i E_i)$ .
3
What is a composite bar? How is load shared in a parallel composite?
Model answer
Two or more materials rigidly connected to share axial deformation. Parallel: $\delta_1=\delta_2$ , $P = P_{1}+P_{2}$ , with $P_i=(\delta A_i E_i)/L_i$ .
4
Explain statically indeterminate axial members.
Model answer
When reactions/internal forces cannot be found from equilibrium alone (e.g. bar fixed at both ends with intermediate load). Compatibility of displacements supplies extra equations.
5
A rod fixed at both ends is heated. How do you find thermal stress?
Model answer
Free thermal expansion $\delta_T=\alpha L\Delta T$ is fully suppressed. Equivalent mechanical compression $\delta=PL/(AE)=\alpha L\Delta T$ gives $P=AE\alpha\Delta T$ , $\sigma=E\alpha\Delta T$ .
6
What is the condition for no thermal stress in a bar?
Model answer
The bar must be free to expand/contract, or supports must move exactly by $\alpha L\Delta T$ . Partial constraint produces stress proportional to the prevented displacement.
7
How is elongation found when self-weight acts on a vertical bar?
Model answer
Integrate: stress varies linearly. For uniform bar, $\delta_{\mathrm{self}}=\dfrac{WL}{2AE}=\dfrac{\gamma L^2}{2E}$ where $W$ is total weight — equivalent to load $W$ acting at mid-length.
8
Define stiffness of an axially loaded bar.
Model answer
Axial stiffness $k=P/\delta=AE/L$ . Higher $A$ or $E$ , or shorter $L$ , increases stiffness.
9
What is Saint-Venant’s principle in axial loading?
Model answer
Localized load application causes nonuniform stress near the ends; far from the load region (roughly one lateral dimension away), stress distribution depends only on the resultant force.
10
How do you analyse a bolt-and-tube assembly tightened by a nut?
Model answer
Initial tightening shares compressive load in tube and tensile load in bolt with equal shortening/elongation magnitudes. External load then redistributes using compatibility and equilibrium.
11
Write the expression for strain energy in an axially loaded bar.
Model answer
$U=\dfrac{P^2 L}{2AE}=\dfrac{AE\delta^2}{2L}=\dfrac{1}{2}P\delta$ for linear elastic behaviour.
12
What is the effect of suddenly applied axial load compared to gradual load?
Model answer
For the same maximum load $P$ applied suddenly (idealized), maximum stress/deflection is twice that of gradual loading: $\delta_{\mathrm{sudden}}=2PL/(AE)$ from energy balance $\tfrac{1}{2}k\delta^2=P\delta$ .
13
How is taper bar elongation under axial load calculated?
Model answer
For circular taper from $d_{1}$ to $d_{2}$ : $\delta=\dfrac{4PL}{\pi E d_1 d_2}$ . Derive by integrating $d\delta=P\,dx/(A(x)E)$ .
14
Differentiate series and parallel axial systems with one example each.
Model answer
Series: stepped shaft — same force, elongations add. Parallel: steel rod inside copper tube with end plates — same elongation, loads add.
15
What assumptions underlie $\delta=PL/AE$ ?
Model answer
Homogeneous isotropic material, constant $A$ and $E$ , centric axial load, stress below proportional limit, plane sections remain plane, neglect of self-weight unless included separately.

Exams & GATE

1
Textbook: RK Bansal Ch. 4–5 (thermal stresses, composite bars).
2
Always draw the bar, mark supports, and write compatibility before solving for redundant reactions.
3
GATE favourites: rigid supports with temperature rise, tapered bars ( $\delta=4PL/(\pi d_1 d_2 E)$ ), and compound bars.

📖 Standard books (India)

Strength of Materials — RK Bansal
Read: Ch. 4–5
SOM — beams, torsion, columns, and deflection

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (state these in exams)

Stepped and composite bars

Thermal stress — free, partial, and fixed

Statically indeterminate axial systems

Step-by-step problem approach

Common mistakes in exams

Axial deformation

Worked examples

Stepped steel bar elongation

Fully fixed bar — thermal stress

Compound bar with temperature rise

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE