# Brittle Fracture Self Study Question

• oxon88
In summary, to determine the maximum load that can be applied without causing brittle fracture, the fracture toughness of the material (120 MPa m-0.5) and the surface flaws on the tie bar (maximum depth of 1.5 mm and shape factor of 1.4) were taken into consideration. Using a factor of safety of 3, the maximum stress that can be applied was calculated to be 626.76 MPa. This corresponds to a force of 1200 Newtons, assuming a cross sectional area of 1963.5 mm2.
oxon88

## Homework Statement

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A tensile load F is to be applied to a tie bar. The tie bar is 50 mm diameter. The surface of the bar has defects resulting from the manufacturing process, these have a maximum depth of 1.5 mm and have a shape factor of 1.4.

If the fracture toughness of the material used is 120 MPa m–0.5, determine the maximum load which can be applied if brittle fracture is not to occur. Use a factor of safety of 3.

## Homework Equations

Fracture toughness, Kic = σf√Mc = 120 MPa m-0.5

shape factor, Q = 1.4

surface flaw shape, M = 1.21π / Q = 2.7153. Attempt at an Answer

not really sure where to start here. can anyone advise?

KIc=120 MPa m-0.5
Q = 1.4
M = 1.21π/1.4
120 MPa m-0.5 = σf √ ((1.21*π*1.5*10-3) / 1.4)

120 MPa m-0.5 = σf * 0.06382σf = 120 MPa m-0.5 / 0.06382

σf = 1880.29 MPasafety factor = 3

σf = 1880.29 / 3 = 626.76 MPacross sectional area = π*r2 = 3.14 * 252 = 1963.5 mm2

Force = Stress * Area

1963.5 * 626.76 = 1230643.26 pascals = 1.23Mpa1.23Mpa = 1200 Newtonswould this be correct?

can anyone help with this please?

anyone?

oxon88 said:
KIc=120 MPa m-0.5
Q = 1.4
M = 1.21π/1.4
120 MPa m-0.5 = σf √ ((1.21*π*1.5*10-3) / 1.4)

120 MPa m-0.5 = σf * 0.06382σf = 120 MPa m-0.5 / 0.06382

σf = 1880.29 MPasafety factor = 3

σf = 1880.29 / 3 = 626.76 MPacross sectional area = π*r2 = 3.14 * 252 = 1963.5 mm2

Force = Stress * Area

1963.5 * 626.76 = 1230643.26 pascals = 1.23Mpa1.23Mpa = 1200 Newtonswould this be correct?
how did you get 1200 Newtons from 1.23 Mpa?

## 1. What is brittle fracture?

Brittle fracture is a type of material failure where a solid material breaks or fractures into two or more pieces with little or no deformation. This usually occurs without warning and the material does not exhibit any plastic deformation before breaking.

## 2. What are the causes of brittle fracture?

The main causes of brittle fracture include high stress concentration, low ductility of the material, and environmental conditions such as temperature and humidity. Other factors such as manufacturing defects or improper design can also contribute to brittle fracture.

## 3. How is brittle fracture different from ductile fracture?

Brittle fracture differs from ductile fracture in that it occurs without any significant deformation, while ductile fracture involves plastic deformation before the material breaks. Additionally, brittle fracture typically occurs in materials with low ductility, while ductile fracture occurs in materials with high ductility.

## 4. What are some methods for preventing brittle fracture?

To prevent brittle fracture, it is important to choose materials with high ductility and toughness, and to avoid high stress concentrations. Proper design and manufacturing processes also play a crucial role in preventing brittle fracture. Additionally, environmental conditions such as temperature and humidity should be carefully considered.

## 5. How can I study brittle fracture?

There are several ways to study brittle fracture, including conducting experiments, analyzing case studies, and using simulation software. It is also important to understand the properties and behavior of different materials and how they are affected by stress and environmental conditions. Consulting with experts in the field and attending conferences and workshops can also be valuable for studying brittle fracture.

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