Interactive Lecture
Understanding Stresses Due to Different Loadings
by Dr Arnaldo Delli Carri
Mechanical stress represents the internal resistance of a body to external forces. Different loading types leave distinct stress patterns in a component — knowing which is which is how you design against the actual failure mode rather than a worst-case guess.
The fundamental definition of stress:
\[\sigma = \frac{F}{A}\]
where \(\sigma\) is stress (Pa or N/m²), \(F\) is force (N), and \(A\) is the cross-sectional area (m²)
Types of Stress:
Different loading conditions produce different combinations of these fundamental stress types.
Engineers keep the stresses in a structure below its critical values to head off:
Axial stress occurs when forces act along the longitudinal axis of a member, creating uniform tension or compression across the cross-section.
\[\sigma_{\text{axial}} = \frac{N}{A}\]
where \(N\) is the Normal Force (axial load) and \(A\) is the cross-sectional area
Engineering Applications: Axial stress analysis is critical for:
Shear stress develops when forces act parallel to a cross-section, causing layers of material to slide relative to each other. The Jourawski method shows how shear stress varies across different cross-sectional shapes, with parabolic distribution in rectangular sections.
Jourawski's formula: \[\tau = \frac{VQ}{It}\]
where \(V\) = shear force, \(Q\) = first moment of area above/below the cut, \(I\) = second moment of area, \(t\) = width at the cut
Average shear stress: \[\tau_{avg} = \frac{V}{A}\]
Maximum shear stress in rectangular beams: \[\tau_{max} = \frac{3V}{2A}\]
Engineering Applications: Shear stress distribution governs:
Bending creates normal stresses that vary linearly across the beam height - tensile on one side and compressive on the other.
Flexure formula: \[\sigma_{\text{bend}} = \frac{My}{I}\]
Maximum stress: \[\sigma_{max} = \frac{Mc}{I} = \frac{M}{Z}\]
where \(M\) = bending moment, \(y\) = distance from neutral axis, \(I\) = second moment of area,
\(c=y_{max}\) = distance of furthest fibre, \(Z\) = section modulus
Engineering Applications: Bending stress analysis underpins:
Torsion creates shear stresses that act perpendicular to the radius. In circular shafts, these stresses vary linearly from zero at the centre to maximum at the outer surface.
Torsion formula: \[\tau = \frac{Tr}{J}\]
Maximum stress: \[\tau_{max} = \frac{TR}{J} = \frac{T}{Z_p}\]
where \(T\) = torque, \(r\) = radius of fibre of interest, \(J\) = polar moment of inertia,
\(R\) = radius of outermost fibre, \(Z_p\) = Section modulus (polar)
Engineering Applications: Torsional stress governs the design of:
Real members rarely carry just one load type. As long as the material stays linear-elastic and deflections are small, the stresses from each load add point by point — this is superposition. A column carrying an eccentric load, for example, feels an axial stress and a bending stress at the same cross-section:
\[\sigma = \frac{N}{A} \pm \frac{Mc}{I}\]
The uniform axial part \(N/A\) shifts the whole linear bending distribution up or down.
The largest normal stress sits at the extreme fibre where the two contributions share the same sign and add constructively. On the opposite face they partly cancel, so one edge of the section can be the critical one while the other is nowhere near the limit. Normal stresses (axial, bending) only add to other normal stresses; shear stresses (transverse shear, torsion) add to other shear stresses. Combining the two families into a single design stress needs a failure criterion such as Mohr's circle or von Mises — the sim below does exactly that.
Superposition stops being an abstract sum once you can watch it. This circular shaft carries an axial force, a bending moment, and a torque all at once. Move each slider and watch the critical surface point — the top fibre, where bending tension lines up with the axial stress — respond in real time. The two normal stresses stack into a single bar; the torsional shear sits on the very same element, acting at right angles to them.
Read off \(\sigma_{max}\) at that point, then the principal stresses \(\sigma_1, \sigma_2\) and the von Mises stress \(\sigma_{vM}\) that a ductile-yield check compares against. Notice how a modest torque can push \(\sigma_{vM}\) well above the bar stress \(\sigma_{max}\) — the shear term never shows up if you only look at the axial-plus-bending number.
Circular shaft, outer surface:
\[\sigma = \frac{N}{A} + \frac{32M}{\pi d^3}, \qquad \tau = \frac{16T}{\pi d^3}\]
Principal and von Mises stresses at that point:
\[\sigma_{1,2} = \frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2}, \qquad \sigma_{vM} = \sqrt{\sigma^2 + 3\tau^2}\]
The von Mises stress is the number a ductile-yield check compares against the yield strength. Combined shafts are where it earns its keep: gearbox and propeller shafts carry bending from their own weight and gear reactions while transmitting torque, so neither \(\sigma\) nor \(\tau\) alone tells you whether the part is safe.
Most components carry several load types at once, and the combined stresses decide the design across mechanical, structural, and aerospace work:
| Concept | Equation | Description |
|---|---|---|
| Axial | \(\sigma = \frac{N}{A}\) | \(N\) = normal force \(A\) = cross-sectional area |
| Shear (Jourawski) | \(\tau = \frac{VQ}{It}\) | \(V\) = shear force \(Q\) = first moment of area above/below cut \(I\) = second moment of area \(t\) = width at cut |
| Shear (average) | \(\tau = \frac{V}{A}\) | \(V\) = shear force \(A\) = area |
| Shear (max, rectangular) | \(\tau_{max} = \frac{3V}{2A}\) | At neutral axis |
| Bending | \(\sigma = \frac{My}{I}\) | \(M\) = bending moment \(y\) = distance from NA \(I\) = second moment of area |
| Torsion (circular) | \(\tau = \frac{Tr}{J}\) | \(T\) = torque \(r\) = radius \(J\) = polar moment of inertia |
Stress analysis is what connects loading conditions to material limits. Key takeaways:
A steel rod with diameter 25 mm is subjected to a tensile force of 20 kN. Calculate the axial stress.
Show Solution
Given: d = 25 mm = 0.025 m, P = 20 kN = 20,000 N
Area: \(A = \frac{\pi d^2}{4} = \frac{\pi \times 0.025^2}{4} = 4.91 \times 10^{-4} \text{ m}^2\)
Stress: \(\sigma = \frac{P}{A} = \frac{20,000}{4.91 \times 10^{-4}} = 40.7 \text{ MPa}\)
A rectangular beam (100 mm × 200 mm) has a shear force of 80 kN. Calculate the maximum shear stress and the shear stress at 25 mm from the top.
Show Solution
Given: b = 100 mm, h = 200 mm, V = 80 kN = 80,000 N
Area: A = 100 × 200 = 20,000 mm²
I = bh³/12 = 100 × 200³/12 = 66.67 × 10⁶ mm⁴
Maximum shear stress (at neutral axis):
Q = b × (h/2) × (h/4) = 100 × 100 × 50 = 500,000 mm³
τ_max = VQ/(It) = (80,000 × 500,000)/(66.67 × 10⁶ × 100) = 6.0 MPa
Shear stress at 25 mm from top:
Distance from NA = 100 - 25 = 75 mm
Q = 100 × 25 × (100 - 12.5) = 218,750 mm³
τ = VQ/(It) = (80,000 × 218,750)/(66.67 × 10⁶ × 100) = 2.625 MPa
A rectangular beam (80 mm × 150 mm) experiences a bending moment of 10 kN·m. Find the maximum bending stress.
Show Solution
Given: b = 80 mm = 0.08 m, h = 150 mm = 0.15 m, M = 10 kN·m = 10,000 N·m
Second moment of area: \(I = \frac{bh^3}{12} = \frac{0.08 \times 0.15^3}{12} = 2.25 \times 10^{-5} \text{ m}^4\)
Distance to extreme fibre: \(c = \frac{h}{2} = 0.075 \text{ m}\)
Maximum stress: \(\sigma_{max} = \frac{Mc}{I} = \frac{10,000 \times 0.075}{2.25 \times 10^{-5}} = 33.3 \text{ MPa}\)
A solid circular shaft (diameter 50 mm) transmits a torque of 2 kN·m. Calculate the maximum shear stress.
Show Solution
Given: d = 50 mm = 0.05 m, T = 2 kN·m = 2,000 N·m
Polar moment of inertia: \(J = \frac{\pi d^4}{32} = \frac{\pi \times 0.05^4}{32} = 6.14 \times 10^{-7} \text{ m}^4\)
Radius: \(R = \frac{d}{2} = 0.025 \text{ m}\)
Maximum shear stress: \(\tau_{max} = \frac{TR}{J} = \frac{2,000 \times 0.025}{6.14 \times 10^{-7}} = 81.4 \text{ MPa}\)