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Strength of Materials: Stress, Strain and Shape

Stress = force ÷ area (MPa). Strain = extra length ÷ original length (no unit). Up to the elastic limit a part springs back; beyond the yield point it stays stretched (plastic); at the ultimate stress it breaks. E = stress ÷ strain. Safety factor = ultimate stress ÷ working stress. Sharp corners raise stress, so designers round them.

🎬 Step-by-step story

  1. A steel rod hangs from a beam. No load, so no stress and no stretch.
  2. We hang 40 kN on a thin rod of 4 cm². Stress = force ÷ area = 100 MPa. The rod stretches a little (drawn bigger so you can see). Strain = extra length ÷ original length.
  3. Now we make the rod thicker, 10 cm². The same load is shared by more area. Stress falls to 40 MPa and the stretch is smaller.
  4. We add more load on the thin rod. At 250 MPa it passes the yield point. The rod turns orange: it will stay longer even after we remove the load.
  5. Look at two shafts with a step. At the sharp corner the stress piles up (red dot) and a crack can start. The rounded corner spreads the stress (green dot).
  6. Free play: change load and thickness. Watch stress, strain, the colour and the safety factor. Can you make the rod break?

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why does a thicker rod stretch less under the same load?

The load is shared by more area, so stress is smaller, and strain follows stress. Step 2 shows stress dropping from 100 to 40 MPa.

Why does strain have no unit?

It is extra length divided by original length, both in the same unit, so the unit cancels.

Why does the rod turn orange and stay longer?

Beyond the yield point (250 MPa here) the stretch is permanent. A blue (elastic) rod would go back to its old length.

Why do cracks start at sharp corners?

The force lines crowd together at a sharp corner, so the stress there is much bigger than elsewhere. A rounded corner lets the lines spread out.

Why do we need a safety factor?

Real loads can be bigger than planned and the material can have hidden faults. The gap before breaking protects people. Slide the load in free play and watch the factor fall.

Can I break the rod in the 3D?

Yes. Raise the load or make the rod thinner until stress goes above 400 MPa. The rod turns red and the weight falls.

Stress and strain in machine parts

Stress is the force on each unit of area inside a part: stress = F ÷ A. Units: pascal (Pa) or N/mm² (1 N/mm² = 1 MPa). A thin rod has more stress than a thick rod under the same load.

Strain is how much the part stretches compared with its length: strain = extension ÷ original length. It has no unit because it is a ratio.

Example: a 2 m wire that stretches by 1 mm has strain 0.001 ÷ 2 = 0.0005.

The stress–strain relation

For small loads, stress and strain go up together in a straight line. This is Hooke's law. The ratio is a number for each material: E = stress ÷ strain (Young's modulus). Steel has E about 200 GPa, so steel stretches very little.

As the load grows, a part passes through these points:

Machine parts must always work in the elastic region.

Safety factor

Designers never load a part up to its breaking value. They keep a gap called the factor of safety: factor of safety = ultimate stress ÷ working stress. A factor of 4 means the part can carry four times the normal stress before it breaks. We use bigger factors where a failure can hurt people, like lifts and cranes, and because loads and materials are never perfect.

Shapes of machine parts

Shape matters as much as material. Stress piles up at sharp inside corners, small holes and deep scratches. This is called stress concentration, and cracks often start there. Good designs:

Try it: paper clip test

Take a paper clip and bend it a tiny bit. It springs back (elastic). Bend it far and it stays bent (plastic). Bend it back and forth and it snaps. Now cut a notch in the edge of a paper strip and pull; it tears at the notch. This is stress concentration, as in step 4 of the 3D.

Key formulas and definitions

Worked examples

1. A rod of area 5 cm² carries 20 kN. Find the stress.

Area = 500 mm². Force = 20 000 N. Stress = 20 000 ÷ 500 = 40 MPa.

2. A 2 m wire stretches 1 mm. Find the strain.

Strain = 1 mm ÷ 2000 mm = 0.0005.

3. Stress is 100 MPa and strain is 0.0005. Find E.

E = 100 ÷ 0.0005 = 200 000 MPa = 200 GPa (steel).

4. A steel rod 3 m long has 100 MPa stress. E = 200 GPa. Find the extension.

Strain = 100 ÷ 200 000 = 0.0005. Extension = 0.0005 × 3000 mm = 1.5 mm.

5. A steel breaks at 400 MPa. The working stress is 100 MPa. Find the safety factor.

Factor = 400 ÷ 100 = 4.

6. A rod of 6 cm² may carry at most 80 MPa. Find the safe load.

Area = 600 mm². Load = 80 × 600 = 48 000 N = 48 kN.

Common mistakes

Practice quiz

1. Stress is:
2. Strain has:
3. After the yield point a part:
4. Factor of safety =
5. A rounded fillet helps because it:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is the difference between stress and strain?

Stress is the force per unit area inside the part. Strain is how much the part stretches compared with its original length.

What is the yield point?

It is the stress after which a material does not fully return to its shape when the load is removed.

Why do shafts have rounded steps?

A sharp corner makes stress pile up there and cracks start. A smooth round corner spreads the stress out.

Where this is taught

Japan高校(専門学科)1〜3年Machine Design

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