Materials Science and Manufacturing Engineering Miscellaneous
- Collapsible tubes are made by
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Collapsible tubes are made by extrusion process.
Correct Option: C
Collapsible tubes are made by extrusion process.
- The true stress (in MPa) versus true strain relationship for a metal is given by σ = 1020 σ0.4 The cross-sectional area at the start of a test (when the stress and strain values are equal to zero) is 100 mm². The cross-sectional area at the time of necking (in mm²) is _____(correct to two decimal places).
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σ = 1020 ∈0.4.
at UTS ⇒ ∈ = n = 0.4∈ = 0.4 = ln Ai Af 0.4 = ln 100 Af
Af = 67.032 mm²Correct Option: A
σ = 1020 ∈0.4.
at UTS ⇒ ∈ = n = 0.4∈ = 0.4 = ln Ai Af 0.4 = ln 100 Af
Af = 67.032 mm²
- A steel wire is drawn from an initial diameter (di) of 10 mm to a final diameter (df) of 7.5 mm. The half cone angle (α) of the die is 5° and the coefficient of friction (μ) between the die and the wire is 0.1. The average of the initial and final yield stress [(σy )avg] is 350 MPa. The equation for drawing stress σf. (in MPa) is given as:
The drawing stress (in MPa) required to carry out this operation is (correct to two decimal places).
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Di = 10 mm, Df = 7.5 mm
α = 5°, µ = 0.1
(σy)avg = 350 MPa
σf = 316.25 MPaCorrect Option: A
Di = 10 mm, Df = 7.5 mm
α = 5°, µ = 0.1
(σy)avg = 350 MPa
σf = 316.25 MPa
- The true stress (σ), true strain (ε) diagram of a strain hardening material is shown in figure. First, there is loading up to point A, i.e. up to stress of 500 MPa and strain of 0.5. then from point A, there is unloading up to point B, i.e. to stress of 100 MPa, Given that the Young's modulus E = 200 GPa, the natural strain at point B (εB) _____(correct to two decimal places).
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NA
Correct Option: A
NA
- The maximum reduction in cross-sectional area per pass (R) of a cold wire drawing process is R = 1 – e–(n + 1) where n represents the strain hardening coefficient. For the case of a perfectly plastic material, R is
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σd = σ0 l n .
For Maximum Reduction
σd = σ0.l = ln Ai Af
R = 0.63Correct Option: D
σd = σ0 l n .
For Maximum Reduction
σd = σ0.l = ln Ai Af
R = 0.63