Real-world elasticity problems often combine several ideas: stress, strain, deformation, elastic energy, and safety.
This solver uses simplified first-pass models for axial members, thin pressure vessels, and simply supported beams.
1. Stress and strain
Normal stress is force divided by area:
\[
\boxed{\sigma=\frac{F}{A}}.
\]
In the linear elastic range, Hooke’s law gives
\[
\boxed{\varepsilon=\frac{\sigma}{E}}.
\]
Here \(E\) is Young’s modulus. A larger \(E\) means the material is stiffer.
2. Safety factor
The allowable stress is often estimated from an ultimate strength and a required safety factor:
\[
\boxed{
\sigma_{\mathrm{allow}}
=
\frac{\sigma_{\mathrm{UTS}}}{n}
}.
\]
The actual safety factor is
\[
\boxed{
\mathrm{SF}_{\mathrm{actual}}
=
\frac{\sigma_{\mathrm{UTS}}}{\sigma_{\max}}
}.
\]
A simplified stress check is acceptable only if
\[
\sigma_{\max}\le \sigma_{\mathrm{allow}}.
\]
3. Axial member model
For a straight member under axial force \(F\),
\[
\sigma=\frac{F}{A}.
\]
The axial elongation or shortening is
\[
\boxed{
\Delta L=\frac{FL}{AE}
}.
\]
The approximate stored elastic energy is
\[
\boxed{
U=\frac12F\Delta L
}.
\]
4. Thin cylindrical pressure vessel
For a thin-walled cylindrical pressure vessel, the hoop stress is
\[
\boxed{
\sigma_h=\frac{pr}{t}
}.
\]
The longitudinal stress is
\[
\boxed{
\sigma_l=\frac{pr}{2t}
}.
\]
A common thin-wall requirement is that the wall thickness \(t\) is much smaller than the radius \(r\).
A rough guide is
\[
t<\frac{r}{10}.
\]
Including Poisson’s ratio, the hoop strain is estimated by
\[
\varepsilon_h
=
\frac{\sigma_h-\nu\sigma_l}{E}.
\]
The radial expansion is approximately
\[
\boxed{
\Delta r=\varepsilon_h r
}.
\]
5. Simply supported beam with center load
For a simply supported beam with a center point load \(P\), the maximum bending moment is
\[
\boxed{
M_{\max}=\frac{PL}{4}
}.
\]
The bending stress at the outer surface is
\[
\boxed{
\sigma_{\max}=\frac{M_{\max}c}{I}
}.
\]
Here \(c\) is the distance from the neutral axis to the outer surface, and \(I\) is the second moment of area.
The center deflection is
\[
\boxed{
\delta_{\max}
=
\frac{PL^3}{48EI}
}.
\]
6. Common section formulas
For a rectangular section with width \(b\) and height \(h\),
\[
A=bh,
\qquad
I=\frac{bh^3}{12},
\qquad
c=\frac{h}{2}.
\]
For a circular section with diameter \(d\),
\[
A=\frac{\pi d^2}{4},
\qquad
I=\frac{\pi d^4}{64},
\qquad
c=\frac{d}{2}.
\]
7. Elastic energy
For a gradually applied load, the elastic energy stored is approximately
\[
\boxed{
U=\frac12F\delta
}.
\]
For beams, \(\delta\) is the deflection at the load point. For axial members, \(\delta=\Delta L\).
For pressure vessels, energy can be estimated from stress-strain energy density.
8. What this capstone solver connects
| Problem type |
Main stress |
Main deformation |
Typical use |
| Axial member |
\(\sigma=F/A\) |
\(\Delta L=FL/(AE)\) |
Cables, tie rods, columns in simple axial loading |
| Thin pressure vessel |
\(\sigma_h=pr/t\) |
\(\Delta r=\varepsilon_hr\) |
Pipes, tanks, cylinders with thin walls |
| Simply supported beam |
\(\sigma_{\max}=M_{\max}c/I\) |
\(\delta_{\max}=PL^3/(48EI)\) |
Bridge beams, floor beams, simple structural members |
9. Worked example: center-loaded steel beam
Suppose a simply supported steel beam has
\[
P=25\ \mathrm{kN},
\qquad
L=4\ \mathrm{m},
\qquad
b=120\ \mathrm{mm},
\qquad
h=250\ \mathrm{mm}.
\]
Convert dimensions:
\[
b=0.120\ \mathrm{m},
\qquad
h=0.250\ \mathrm{m}.
\]
The second moment of area is
\[
I=\frac{bh^3}{12}
=
\frac{(0.120)(0.250)^3}{12}
=
1.56\times10^{-4}\ \mathrm{m^4}.
\]
The maximum moment is
\[
M_{\max}=\frac{PL}{4}
=
\frac{(25000)(4)}{4}
=
25000\ \mathrm{N\,m}.
\]
With \(c=h/2=0.125\ \mathrm{m}\),
\[
\sigma_{\max}
=
\frac{M_{\max}c}{I}
\approx
20.0\ \mathrm{MPa}.
\]
If \(E=200\ \mathrm{GPa}\), the midspan deflection is
\[
\delta_{\max}
=
\frac{PL^3}{48EI}
\approx
1.07\ \mathrm{mm}.
\]
Key idea: a real-world elastic design check should compare both strength and deformation, not only force.
