2.2 Materials Materials

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# 2.2 Materials Materials - PowerPoint PPT Presentation

2.2 Materials Materials. Breithaupt pages 162 to 171. AQA AS Specification. Elasticity and springs. Elastic limit and Yield point. Elasticity and springs. Elastic limit and Yield point. Hooke’s Law.

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### 2.2 MaterialsMaterials

Breithaupt pages 162 to 171

Elasticity and springs

Elastic limit and Yield point

Elasticity and springs

Elastic limit and Yield point

Experiment to investigate Hooke’s Law

Obtain data to plot graphs on the same axes for

the following spring combinations:

Hooke’s law

The force (F ) needed to stretch a spring is directly proportional to the extension (ΔL ) of a spring from its natural length.

F α ΔL

F = k ΔL or F = -K ΔL

kis called the spring constant

The spring constant is the force required to produce an extension of one metre.

unit = Nm-1

Elastic limit

Up to a certain extension if the force is removed the spring will return to its original length. The spring is said to be behaving elastically.

If this critical extension is exceeded, known as the elastic limit, the spring will be permanently stretched.

Plastic behaviour then occurs and Hooke’s law is no longer obeyed by the spring.

Question

(b)F = k ΔL

→ ΔL = F / k

= 18N / 200 Nm-1

ΔL = 0.09 m

= 9 cm

And so the spring’s length

= 24 cm

A spring of natural length 15cm is extended by 3cm by a force of 6N. Calculate (a) the spring constant and (b) the length of the spring if a force of 18N is applied.

(a)F = k ΔL

→ k = F / ΔL

= 6N / 0.03m

spring constant, k

= 200 Nm-1

Tensile stress (σ)

A stretching force is also called a tensile force.

Tensile stress = tensile force

cross-section area

σ = F / A

unit – Pa (pascal) or Nm-2

Note: 1 Pa = 1 Nm-2

Breaking stress

This is the stress required to cause a material to break.

Tensile strain (ε)

Tensile strain = extension

original length

ε = ΔL / L

unit – none (it’s a ratio like pi)

Question

A wire of natural length 2.5 m and diameter 0.5 mm is extended by 5 cm by a force of 40 N. Calculate:

(a) the tensile strain ε = ΔL / L

(b) the tensile stress σ = F / A

(c) the force required to break the wire if its breaking stress is 1.5 x 109 Pa. σ = F / A

(a)ε = ΔL / L

= 0.05m / 2.5m

tensile strain, ε = 0.02

Question

(c)σ = F / A

→ F = σ A

= 1.5 x 109 Pa x 1.96 x 10-7 m2

Breaking Force, F = 294 N

(b)σ = F / A

A = Area

= π r2

= π x (0.25 x 10-3 )2 m2

= 1.96 x 10-7 m2

σ = 40N / 1.96 x 10-7 m2

stress, σ = 2.04 x 108Pa

Key points

1. Springs in series are ‘slacker’

Key points

1. Springs in series are ‘slacker’

Key points

1. Springs in series are ‘slacker’

Key points

2. Springs in parallel are ‘stiffer’

Key points

2. Springs in parallel are ‘stiffer’

Key points

2. Springs in parallel are ‘stiffer’

Key points

2. Springs in parallel are ‘stiffer’

• Define Hooke’s law. Quote the equation for Hooke’s law.
• What is meant by (a) the spring constant and (b) the elastic limit.
• A spring of natural length 40 cm is extended to 50 cm by a force of 2N. Calculate (a) the spring constant in Nm-1 (b) the expected length of the spring if it were to be extended by a force of 5N.
• Show that the overall spring constant, k for (a) springs in series is given by k = k1 + k2; (b) springs in parallel is given by 1 / k = 1 / k1 + 1 / k2 where k1 and k2 are the spring constants of the individual springs.
• Try Summary Questions 1, 2 & 3 on page 166