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3 years ago

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A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 84998 N , and experienc

es an axial elongation of 0.18 mm . Assuming that the deformation is entirely elastic, calculate the elastic modulus of this steel in GPa.Answer Format: X (no decimal places)

2 answers:

german3 years ago

4 0

Answer:

Elastic Modulus = 227 GPa

Explanation:

Given,

Load = 84998 N

Length of bar = 300 mm = 0.3 m

Elongation = 0.18 mm = 0.00018 m

Cross sectional Area of the bar = (25mm × 25mm) = 0.025 × 0.025 = 0.000625 m²

From Hooke's law, the stress experienced by a material is proportional to the strain experienced by the same body, as long as the elastic limit isn't exceeded.

Stress ∝ strain

The coefficient of proportionality is the elastic modulus, E.

Stress = E × (Strain)

Stress = (Load)/(Cross sectional Area)

Stress = (84998 ÷ 0.000625) = 135,996,800 N/m²

Strain = (Change in length)/(Original length)

Strain = (ΔL/L) = 0.00018 ÷ 0.3 = 0.0006

E = (Stress/Strain)

E = 135,996,800 ÷ 0.0006 = 226,661,333,333.3 Pa = (2.267 × 10¹¹) Pa

1 GPa = 10⁹ Pa

(2.267 × 10¹¹) = 2.267 × 10² × 10⁹ = 226.7 GPa = 227 GPa to the nearest GPa. (No decimal place)

Hope this Helps!!!

Paraphin [41]3 years ago

4 0

Answer:

227 Gpa

Explanation:

∆L = PL/AE

E = PL/A∆L

E is Elastic Modulus

L is length

A is Area

L = 300 mm = 300* 10^-3

A = (25 * 10 ^-3)^2

P = 84998N

∆L = 0.18mm = 0.18*10^-3

E = 84998*300*10^-3/((25*10^-3)^2*0.18*10^-3

E = 226661333333.3Pa

= 226.7 * 10^9Pa

10^9Pa = 1 GPA

E = 226.7 Gpa

E = 227 no decimal

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Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

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$n$ - Safety factor, dimensionless.

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$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

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$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

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$V = \frac{P}{5}$

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The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

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The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

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The diameter is now cleared and computed:

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$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

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