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Mechanics of Materials · Chapter 8 of 10 · Advanced

Combined Loadings

Real parts rarely see one load at a time. Superposition lets you add the stresses from axial force, bending, torsion, and pressure, then transform them to find the worst case at a point.

Readiness check

This chapter combines the single-load formulas of earlier chapters. Tick only what you can do closed-notes.

Compute axial stress P/A and bending stress Mc/I.
Compute torsion stress Tr/J.
Transform a stress state (principal stresses).
Track tension and compression signs.
Recall thin-wall geometry (r, t).

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview bending (Chapter 4) and transformation (Chapter 7).

3 or more weak itemsRevisit axial, torsion, and bending stresses first.

The core idea

When several loads act together, add the stress each would cause on its own, then transform the combined state to find the principal stresses at the critical point.

σ = P/A ± Mc/Ihoop σ = pr/t, long σ = pr/2tthen σ_1,2 = σ_avg ± R

Because stresses are linear in load (within the elastic range), superposition holds: compute the axial, bending, torsional, and pressure stresses separately and sum the like components at a point. Then, since these may be normal and shear together, transform to principal stresses and maximum shear (Chapter 7). The art is choosing the critical point, usually where the contributions add rather than cancel.

The skill works when: you superpose like stress components at the worst point and then transform the combined state.

The skill breaks down when: normal and shear stresses are added directly, or a point where loads cancel is mistaken for the critical one.

The concept. Superposition: a uniform axial stress plus a linear bending stress add, point by point, to a combined distribution. One edge may end up in tension even under a compressive load.

The skills, taught in order

Combined loading is superposition plus transformation. Five skills cover the principle, pressure vessels, eccentric loading, shafts, and the general procedure.

8.1 The superposition principle

In the linear-elastic range, the stress from several loads equals the sum of the stresses from each load alone. Add only like components: normal stresses to normal, shear to shear. This reduces any combined problem to the single-load formulas already learned.

8.2 Thin-walled pressure vessels

A thin cylinder under internal pressure has a hoop (circumferential) stress σ = pr/t and a longitudinal stress σ = pr/2t, so the hoop stress is twice the longitudinal, which is why pipes split along their length. A sphere has σ = pr/2t in every direction.

Vessel	Hoop / circumferential	Longitudinal / axial
Cylinder	pr/t	pr/2t
Sphere	pr/2t	pr/2t

8.3 Eccentric loading (axial plus bending)

A load applied off the centroid is equivalent to a central load plus a moment M = Pe. The stress superposes: σ = P/A ± Mc/I. The eccentricity can push one face into tension even when the load is compressive, the reason masonry and short columns are kept within the middle third.

8.4 Combined bending and torsion

A shaft carrying a transverse load and a torque has, at its surface, a bending normal stress and a torsional shear stress at the same point. These form a plane-stress state that must be transformed (Chapter 7) into principal stresses and maximum shear for a failure check.

8.5 The general procedure

Identify the critical point (where contributions add), compute each load's stress there, superpose like components into σ_x, σ_y, τ_xy, then transform to σ₁, σ₂, and τ_max. Compare the governing value with the material's strength.

Engineering connection: pressure vessels, crane hooks and C-frames (eccentric), and power shafts (bending plus torsion) are all combined-loading problems; this chapter unifies the whole course.

Worked example 1: a cylindrical pressure vessel

A cylindrical pressure vessel 1.5 m in diameter has a 12 mm wall and holds an internal gauge pressure of 2.5 MPa. Find the hoop and longitudinal stresses.

Figure 1. Internal pressure stretches the wall two ways. The hoop stress (around the circumference) is twice the longitudinal, so cylinders fail by splitting along a line, not around a ring.

ProblemFind the hoop and longitudinal stresses in the vessel in Figure 1.
Given / findd = 1.5 m so r = 750 mm, t = 12 mm, p = 2.5 MPa. Find σ_h and σ_l.
AssumptionsThin wall (t ≪ r), uniform stress through the thickness, closed ends.
ModelApply the thin-walled cylinder formulas for hoop and longitudinal stress.
Equationsσ_h = pr/t σ_l = pr/2t
Solveσ_h = 2.5 × 750/12 = 156 MPa. σ_l = 2.5 × 750/(2 × 12) = 78 MPa.
CheckThe hoop stress is exactly twice the longitudinal, as the formulas demand. The thin-wall ratio t/r = 12/750 = 0.016 ≪ 0.1, so the thin-wall assumption is valid.
ConclusionThe hoop stress governs, which is why welded seams run lengthwise and why pipes burst with a longitudinal crack. The wall element is in biaxial tension, ready for a transformation check if needed.

Result. Hoop stress 156 MPa; longitudinal stress 78 MPa (hoop is twice).

Worked example 2: an eccentric load

A short rectangular post (100 mm × 150 mm) carries a 120 kN compressive load applied 40 mm off-center along the 150 mm direction. Find the stress on each face.

Figure 2. The off-center load is a central load plus a moment Pe. Adding the uniform compression to the linear bending leaves one face more compressed and the other in tension.

ProblemFind the stress on each face of the post in Figure 2.
Given / findP = 120 kN (compression), e = 40 mm, section 100 × 150 mm (c = 75). Find σ on both faces.
AssumptionsShort post (no buckling), linear-elastic, eccentricity about one axis.
ModelReplace the eccentric load by a central load plus M = Pe, then superpose axial and bending stresses.
EquationsM = Pe σ = −P/A ± Mc/I I = bh³/12
SolveA = 15 000 mm², so P/A = 8.0 MPa (compression). I = 100(150)³/12 = 2.81×10⁷ mm⁴; M = 120 000 × 40 = 4.8×10⁶ N·mm; Mc/I = ±12.8 MPa. Combined: −8.0 − 12.8 = −20.8 MPa on the near face, −8.0 + 12.8 = +4.8 MPa (tension) on the far face.
CheckThe bending term (12.8) exceeds the axial (8.0), so one face flips into tension, the warning sign that the load is outside the middle third (e = 40 > h/6 = 25). A wider section or smaller eccentricity would keep it all in compression.
ConclusionEccentric loading superposes axial and bending stress and can create unexpected tension, critical for concrete, masonry, and bolted joints that cannot take tension.

Result. −20.8 MPa (compression) on the near face, +4.8 MPa (tension) on the far face.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Adding normal and shear directly	Wrong combined stress	"Are these like components?"	Superpose normal with normal, shear with shear, then transform.
Hoop equals longitudinal	Pressure vessel under-designed	"Which stress is twice the other?"	For a cylinder, hoop = pr/t = twice the longitudinal pr/2t.
Ignoring the moment from eccentricity	Tension face missed	"Is the load on the centroid?"	An off-center load adds M = Pe; superpose its bending.
Wrong critical point	Stress checked where loads cancel	"Where do the contributions add?"	Pick the point where axial, bending, and torsion reinforce.

Practice ladder

Level 1 · Direct skill

A spherical tank of 2 m diameter, 10 mm wall, holds 1.8 MPa. Find the wall stress.

Show answer

σ = pr/2t = 1.8 × 1000/(2 × 10) = 90 MPa, equal in all directions. Spheres are the most efficient pressure vessel for this reason.

Level 2 · Mixed concept

For the Worked Example 2 post, what eccentricity keeps the whole section in compression?

Show answer

Tension is avoided while Mc/I ≤ P/A, i.e. e ≤ h/6 = 150/6 = 25 mm (the middle-third rule). At e = 25 mm the far face just reaches zero stress; beyond it, tension appears.

Level 3 · Independent problem

A 50 mm diameter shaft carries a bending moment giving 70 MPa and a torque giving 35 MPa shear at the surface. Find the maximum principal stress and τ_max.

Show answer

σ_avg = 35, R = √(35² + 35²) = 49.5. σ₁ = 35 + 49.5 = 84.5 MPa, σ₂ = −14.5 MPa, τ_max = 49.5 MPa. Superpose then transform, the standard shaft check.

Level 4 · Transfer to real engineering

Find a real combined-loading part (a pressurised pipe with wind load, a crane hook, a bike crank). Identify the loads, superpose the stresses at the critical point, and transform if needed.

What good work looks like

Each load's stress computed, like components summed at the critical point, and a transformation to principal stress or τ_max where normal and shear coexist.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I added like stress components and picked the point where they reinforce."

"Give me five parts; I will list the loads and the critical point for each."

"Combine these stresses." Choosing the critical point and superposing is the skill.

"Is the vessel safe?" Identifying hoop as the governing stress is the point.

Portfolio task

Analyse one combined-loading part: superpose the axial, bending, torsion, or pressure stresses at the critical point, then transform to the governing stress.

Must include: each load's stress, the superposed σ_x, σ_y, τ_xy, and a principal-stress or τ_max result.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. State the superposition principle for stress.

In the elastic range, total stress is the sum of the stresses from each load acting alone (like components added).

2. Give the hoop and longitudinal stresses for a cylinder.

σ_h = pr/t and σ_l = pr/2t; the hoop is twice.

3. How is an eccentric load handled?

As a central load plus M = Pe; σ = P/A ± Mc/I.

4. What is the middle-third rule?

Keep eccentricity within h/6 so no tension develops on the section.

5. What is the general combined-loading procedure?

Find each stress at the critical point, superpose like components, then transform to principal stresses and τ_max.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the vessel and eccentric-load stresses from a blank page.

+3 daysOne pressure-vessel and one shaft (bending plus torsion) problem.

+7 daysMove to beam deflection, Chapter 9.

+30 daysApply combined loading to a real design with a factor of safety.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 8 (Principal Stresses under a Given Loading) and thin-walled vessels
Cross-reference	Hibbeler, Ch. 8 · Gere and Goodno, Ch. 8
Core topics	8.1 Superposition · 8.2 Pressure vessels · 8.3 Eccentric loading · 8.4 Bending plus torsion · 8.5 General procedure
Engineering connection	Pressure vessels, eccentric columns and C-frames, and power shafts.
Read next	Chapter 9: Deflection of Beams.