A state of plane stress (σ3=0) occurs at a critical point of a mild steel machine component (see Figure 1 below). As a result of several tension tests, it has been found that the tensile yield strength is = 250MPa for the grade of steel used. (a) Determine the factor of safety (FS) with respect to yield using (i) the maximum shear stress criterion and (ii) the maximum distortion energy (von Mises) criterion. (b) Graph the Tresca Hexagon and the von Mises oval for this material and note the location of the given state of plane stress as a point on the graph

Q: The minimum strain at failure in tension steel having yield stress modulus E = 200 GPa, as per Limit…

A:

Q: 2. The Goodman diagram relates oa and om for fatigue failure after a specific number of cycles N₁,…

A: Given,   Fatigue strength under zero mean stress , σa(0) = 0.45 σTS   Tensile strength, σTS = 375…

Q: The failure stress of a homogeneous steel bar having a diameter of 10 mm is 250 MPa. Using a factor…

A:

Q: The strain gauge measurement of a steel plate indicate the principle strains are 0.003 and 0.001.…

A: Principle strains- ε1=0.003ε2=0.001E=200 GPaμ=0.33

Q: fracture. Determine the following for the two metals (show all calculations and units): a.…

A: Stress vs strain curve is given. BY doing matriculation we can find the unknown.

Q: The stresses on the surface of a cast iron machine component are sx = 85 MPa, =-230 MPa, and fy =…

A:

Q: (b) (i) A tensile test specimen made from 0.4% C steel has a circular cross section of diameter d mm…

A: Shape of the specimen used is circle of diameter, dGauge length of the specimen = 25 mmLoad applied…

Q: A component made of Aluminium 6061-T651 has an edge crack with length equal to 15 mm, as shown The…

A:

Q: 2. The Goodman diagram relates oa and om for fatigue failure after a specific num of cycles Nf,…

A: Fatigue strength under zero mean stress ,   Tensile strength,  Number of cycles to failure, N = 107…

Q: A steel specimen is tested in tension. The specimen is 25 mm wide by 12.5 mm thick in the test…

A: Area=25×12.5mm2=312.5mm2

Material Properties

Construction in civil engineering demands high strength, durability, and longevity. These targets can be achieved by better design and analysis, along with the correct choice of materials. Different materials have different properties and hence, they respond and performs differently under the various condition of external loadings. It is therefore a critical task of an engineer to select the right material based on the property that the material is supposed to function in real-time. When speaking about the property, there are different categories of material properties like chemical property, physical property, mechanical property, electrical property, magnetic property, thermal property, etc. These material properties form the basis of material selection in various applications.