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A Study to Establish Equivalence of Thermal and Mechanical Loads

  • Research Article - Mechanical Engineering
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Abstract

A proper combination of thermal and mechanical loads is generally maintained to provide the necessary complicated shape and profile to the sheet metal in shipbuilding, aerospace, and others, for which an experienced operator should have the knowledge of the equivalence of these two loads. A study was conducted to establish the equivalence of thermal and mechanical loads in terms of their capability to produce the same deflection. The maximum deflection of a simply supported beam-like structure with small thickness subjected to only thermal load had been determined using finite element analysis. Moreover, the temperature profile followed during the heating and cooling was plotted. A test rig was fabricated for determination of the same and necessary comparison to establish this equivalence.

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Abbreviations

\( B \) :

Breadth of beam

\( C_{p} \) :

Specific heat

\( E \) :

Young’s modulus of elasticity

\( h \) :

Thickness of the plate

\( h_{f} \) :

Coefficient of convective heat transfer

\( H \) :

Thickness of beam

I :

Current

\( I' \) :

Moment of inertia

\( k \) :

Thermal conductivity

\( K^{ - 1} \) :

Expansion coefficient

\( L \) :

Length of the beam

P :

Load

\( q_{\text{c}} \) :

Convection heat loss

\( q_{\text{r}} \) :

Radiation heat loss

\( Q \) :

Input power

\( Q_{\text{s}} \) :

Surface heat flux

\( Q_{\text{v}} \) :

Volume heat flux

\( r_{\text{b}} \) :

Beam radius

t :

Time

T :

Temperature

\( T_{0} \) :

Ambient temperature

V :

Voltage

\( \gamma_{\text{s}} \) :

Fraction of input power falling on the surface

\( \gamma_{\text{v}} \) :

Fraction of input power falling on the volume

δ :

Deflection

η :

Efficiency

ρ :

Density

σ :

Stefan–Boltzmann constant

ε :

Emissivity

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Acknowledgement

The help and cooperation of the staff members of the Workshop and Mr. Abhijit Basak of Electrical Engineering Laboratory of NIT Rourkela, India, are gratefully acknowledged.

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Authors and Affiliations

  1. Department of Chemical Engineering, National Institute of Technology Rourkela, Rourkela, India

    Bitan Pratihar

  2. Department of Mechanical Engineering, Indian Institute of Technology Kharagpur, Kharagpur, India

    Debasish Das & Dilip Kumar Pratihar

Authors
  1. Bitan Pratihar
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  2. Debasish Das
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  3. Dilip Kumar Pratihar
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Corresponding author

Correspondence to Dilip Kumar Pratihar.

Appendix

Appendix

1.1 Numerical Analysis Conducted to Establish the Equivalence of Thermal and Mechanical Loads

Load–deflection equation for the simply supported beam with concentrated point load acting at its mid-span is as follows [38]:

$$ \delta = \frac{{PL^{3} }}{{48EI^{{\prime }} }}, $$

where \( \delta \) represents the maximum deflection (m), \( P \) denotes the load (N), \( L \) indicates the length of the beam (m), \( E \) represents Young’s modulus of elasticity of the beam material (N/m2), and \( I^{'} \) denotes the area moment of inertia (m4), which is calculated as \( I^{{\prime }} = \frac{1}{12}BH^{3} \), where \( B \) represents breadth of the beam (m) and \( H \) denotes thickness of the beam (m).

1.2 Test Material: Steel

$$ \begin{array}{*{20}l} {\delta = 0. 0 1 4\;{\text{m}}} \hfill \\ {L = 0. 2 2\;{\text{m}}} \hfill \\ {B = 0. 0 0 5\;{\text{m}}} \hfill \\ {H = 0. 0 0 2\;{\text{m}}} \hfill \\ {E = 190, 0 0 0 , 0 0 0 , 0 0 0\;{\text{Pa}}} \hfill \\ {I^{'} = \frac{1}{12}0.005 \times (0.002)^{3} = 3.33 \times 10^{ - 12} \;{\text{m}}^{4} } \hfill \\ {P = \frac{{48\delta EI^{'} }}{{L^{3} }} = \frac{{48 \times 0. 0 1 4\times 190, 0 0 0 , 0 0 0 , 0 0 0\times 3.33 \times 10^{ - 12} }}{{ ( 0. 2 2)^{3} }} = 39.937\;{\text{N}}} \hfill \\ \end{array} $$

  • Maximum temperature = 1290 °C

  • Mechanical equivalent load = 39.937 N

  • Total heat input = 1.6 × 1200 kJ, as 1600 W heat source was used for 1200 s.

Therefore, 1.0 N of mechanical load corresponds to 48.07 kJ of thermal load to produce the same amount of deflection in steel specimen.

1.3 Test Material: Aluminum

$$ \begin{array}{*{20}l} {\delta = 0.009\;{\text{m}}} \hfill \\ {L = 0.22\;{\text{m}}} \hfill \\ {B = 0.005\;{\text{m}}} \hfill \\ {H = 0.002\;{\text{m}}} \hfill \\ {E = 81,5 0 0 , 0 0 0 , 0 0 0\;{\text{Pa}}} \hfill \\ {I^{'} = \frac{1}{12}0.005 \times (0.002)^{3} = 3.33 \times 10^{ - 12} \;{\text{m}}^{4} } \hfill \\ {P = \frac{{48\delta EI^{'} }}{{L^{3} }} = \frac{{48 \times 0.009 \times 815 0 0 0 0 0 0 0 0\times 3.33 \times 10^{ - 12} }}{{(0.22)^{3} }} = 11.01\;{\text{N}}} \hfill \\ \end{array} $$

  • Maximum temperature = 610 °C

  • Mechanical equivalent load = 11.0 N

  • Total heat input = 0.9 × 1620 kJ, as 900 W heat source was utilized for 1620 s.

Therefore, 1.0 N of the mechanical load was found to be equivalent to 132.42 kJ of thermal load in order to produce the same amount of deflection in the aluminum specimen.

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Pratihar, B., Das, D. & Pratihar, D.K. A Study to Establish Equivalence of Thermal and Mechanical Loads. Arab J Sci Eng 45, 631–639 (2020). https://doi.org/10.1007/s13369-019-04116-x

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