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Mechanics of Materials · Chapter 1 of 10 · Beginner

Concept of Stress and Strain

A force tells you nothing about whether a part survives until you spread it over an area. Stress is force per area, strain is the stretch it causes, and the two together decide if the part holds.

Readiness check

This chapter builds on statics and the elastic behaviour of materials. Tick only what you can do closed-notes.

Find the internal force on a cut section from a free body.
Compute the area of a circle and a rectangle.
Recall Hooke's law σ = Eε.
Read yield strength from a stress-strain curve.
Work in MPa (N/mm²) and kN.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview internal forces in Statics.

3 or more weak itemsRevisit elastic properties in Materials Chapter 5 first.

The core idea

Stress is the internal force spread over the area that carries it; strain is the deformation per unit length it produces. Compare the stress with the material's strength to judge safety.

σ = P/Aε = δ/L, σ = EεFS = σ_failure/σ_allow

The same force is harmless through a thick bar and dangerous through a thin wire, because stress, not force, is what the material feels. Normal stress acts perpendicular to a cut, shear stress along it, and bearing stress where parts press together. The deformation is measured as strain, related to stress by Hooke's law in the elastic range. Designing means keeping the working stress safely below the failure stress by a factor of safety.

The skill works when: you find the internal force, divide by the area that actually carries it, and compare with an allowable stress.

The skill breaks down when: the wrong area is used (shear vs normal), or force is judged without converting to stress.

The concept. Cut the loaded bar and the internal force P must be balanced by stress spread over the cut area A. Normal stress σ = P/A is that intensity; the smaller the area, the higher the stress.

The skills, taught in order

Mechanics of materials starts with stress, strain, and the strength to compare them against. Five skills lay the groundwork the whole course reuses.

1.1 Normal stress

Normal stress is the axial force divided by the cross-section it acts on: σ = P/A, positive in tension, negative in compression. It is assumed uniform for a centrally loaded member away from the ends. Units are pascals, almost always MPa (N/mm²) in practice.

1.2 Shear and bearing stress

Shear stress acts parallel to a cut, τ = V/A, as in a pin or bolt. A pin in double shear carries the load over two cross-sections, halving the stress. Bearing stress is the contact pressure where a pin presses on a hole, σ_b = P/(t·d), based on the projected area.

Stress	Formula	Acts
Normal (axial)	σ = P/A	perpendicular to the cut
Shear	τ = V/A	parallel to the cut
Bearing	σ_b = P/(t·d)	contact between pin and hole

1.3 Strain

Normal strain is the deformation per unit length, ε = δ/L, dimensionless (often given as μ-strain). Shear strain γ is the small change in a right angle, in radians. Strain is what the material actually experiences; stress is our way of relating it to force.

1.4 Hooke's law and the stress-strain diagram

In the elastic range, stress is proportional to strain: σ = Eε, with E the modulus of elasticity (stiffness). The stress-strain diagram shows the proportional limit, yield strength, ultimate strength, and fracture, the landmarks that define a material's usable range. Shear has its own law, τ = Gγ.

1.5 Factor of safety and allowable stress

Real parts are kept well below failure. The factor of safety is FS = σ_failure/σ_allow, so the allowable stress is σ_allow = σ_yield/FS and the required area is A ≥ P/σ_allow. Choosing FS balances uncertainty, consequence of failure, and cost.

Quantity	Relation
Factor of safety	FS = σ_failure/σ_allow
Allowable stress	σ_allow = σ_yield/FS
Design area	A ≥ P/σ_allow

Engineering connection: every member in the course, axial, torsional, bending, is sized by computing its stress and comparing it with an allowable; this chapter is the template.

Worked example 1: stresses in a pin connection

A 20 mm diameter rod carries a 40 kN tensile load and is attached by a 16 mm pin in double shear. Find the normal stress in the rod and the shear stress in the pin.

Figure 1. The rod carries normal stress over its cross-section; the pin carries shear over two cross-sections (double shear), so its area is doubled.

ProblemFind the rod normal stress and the pin shear stress in Figure 1.
Given / findP = 40 kN, rod d = 20 mm, pin d = 16 mm in double shear. Find σ_rod and τ_pin.
AssumptionsUniform normal stress in the rod; the pin shears on two planes equally.
ModelNormal stress is load over the rod area; shear stress is load over twice the pin area.
Equationsσ = P/A_rod τ = P/(2A_pin)
SolveA_rod = π(10)² = 314 mm², so σ = 40 000/314 = 127 MPa. A_pin = π(8)² = 201 mm², so τ = 40 000/(2 × 201) = 99.5 MPa.
CheckBoth stresses are in the right range for steel under load. Had the pin been in single shear, τ would have doubled to 199 MPa, which is why clevis joints use double shear.
ConclusionThe same 40 kN produces different stresses in different elements because each acts over a different area. Identifying the correct area, and whether the action is normal or shear, is the whole skill.

Result. Rod normal stress 127 MPa; pin shear stress 99.5 MPa (double shear).

Worked example 2: sizing for a factor of safety

A steel tie-rod must carry 50 kN. The steel yields at 250 MPa and a factor of safety of 2.5 is required. Find the allowable stress and the minimum rod diameter.

Figure 2. Dividing the yield strength by the factor of safety gives the allowable stress; the rod is then sized so its working stress sits at or below that line.

ProblemFind the allowable stress and minimum diameter for the rod in Figure 2.
Given / findP = 50 kN, σ_yield = 250 MPa, FS = 2.5. Find σ_allow and d_min.
AssumptionsAxial tension, uniform stress, failure defined by yield.
ModelAllowable stress from yield over the factor of safety, then the area and diameter that keep the working stress at the allowable.
Equationsσ_allow = σ_yield/FS A = P/σ_allow d = √(4A/π)
Solveσ_allow = 250/2.5 = 100 MPa. A = 50 000/100 = 500 mm². d = √(4 × 500/π) = 25.2 mm, so use 26 mm.
CheckAt d = 26 mm the actual stress is 50 000/(π × 13²) = 94 MPa, below the 100 MPa allowable, so the factor of safety is slightly better than 2.5, the safe side. Rounding diameter always down would have under-sized it.
ConclusionDesign inverts analysis: instead of finding the stress, you set it to the allowable and solve for size. The factor of safety is the cushion against everything you did not model.

Result. σ_allow = 100 MPa; minimum diameter 25.2 mm, rounded up to 26 mm.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Judging force, not stress	Thin and thick members treated alike	"What area carries this force?"	Divide by the area; stress, not force, governs failure.
Wrong area for shear	Pin shear uses the rod area	"Is this normal or shear, and over what plane?"	Shear uses the pin cross-section (twice it in double shear).
Forgetting double shear	Pin stress doubled	"How many planes carry the load?"	Double shear halves the stress; use 2A.
Rounding diameter down	Working stress exceeds allowable	"Did I round the size up?"	Always round a required size up to stay below the allowable.

Practice ladder

Level 1 · Direct skill

A 12 mm square bar carries 18 kN in tension. Find the normal stress.

Show answer

A = 12 × 12 = 144 mm²; σ = 18 000/144 = 125 MPa. Force over the actual cross-section.

Level 2 · Mixed concept

A 10 mm bolt in single shear carries 15 kN. What shear stress, and how does it change in double shear?

Show answer

Single shear: τ = 15 000/(π × 5²) = 191 MPa. Double shear halves it to 95.5 MPa, because two planes share the load.

Level 3 · Independent problem

A 25 mm rod (E = 200 GPa) carries 60 kN. Find the stress and the strain.

Show answer

A = π(12.5)² = 491 mm²; σ = 60 000/491 = 122 MPa. ε = σ/E = 122/200 000 = 6.1×10⁻⁴ (610 μ-strain). Hooke's law links the two in the elastic range.

Level 4 · Transfer to real engineering

Find a real loaded part (a bolt, a hanger rod, a bracket). Estimate the load and area, compute the stress, and judge it against a typical allowable for the material.

What good work looks like

The internal force found, the correct area used (normal, shear, or bearing), a stress computed, and a comparison with an allowable stress and factor of safety.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used the right area, normal versus shear versus bearing."

"Give me five loaded parts; I will identify the stress type and the area for each."

"Compute the stress." Choosing the area and stress type yourself is the skill.

"What size rod?" Setting the allowable and solving for area is the point.

Portfolio task

Analyse one real connection: compute the normal, shear, and bearing stresses, then size one element to a chosen factor of safety.

Must include: the internal force, the correct areas, three stresses, and an allowable-stress sizing.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Define normal, shear, and bearing stress.

σ = P/A (perpendicular), τ = V/A (parallel), σ_b = P/(t·d) (contact, projected area).

2. What is strain, and how does it relate to stress?

ε = δ/L (deformation per length); σ = Eε in the elastic range.

3. How does double shear change the stress?

It halves it; the load is carried over two cross-sections.

4. Write the factor of safety and allowable stress.

FS = σ_failure/σ_allow; σ_allow = σ_yield/FS.

5. Why size by stress rather than force?

The material fails at a stress; the same force is safe or unsafe depending on the area.

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

+1 dayRe-derive the pin stresses and the sizing from a blank page.

+3 daysOne analysis and one design problem.

+7 daysCarry stress and strain into axial loading, Chapter 2.

+30 daysReuse the allowable-stress method for every later member.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapters 1 and 2 (Stress, Strain)
Cross-reference	Hibbeler, Ch. 1 and 3 · Gere and Goodno, Ch. 1
Core topics	1.1 Normal stress · 1.2 Shear and bearing · 1.3 Strain · 1.4 Hooke's law · 1.5 Factor of safety
Engineering connection	The template for sizing every member in the course.
Read next	Chapter 2: Axial Loading.