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3 years ago

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For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

GPa (15.0 ×106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in2 ) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?

1 answer:

Alona [7]3 years ago

3 0

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

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