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A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a bronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.

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klefenz

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003  m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

lhmarianateixeira

In this exercise we have to use the knowledge of force to calculate that:

[tex]153.6 kN[/tex]

Then from the formula of the elastic constant we get that:

[tex]k = E * A/l\\k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m[/tex]

With that, now using Hooke's law we find that:

[tex]F = x * k\\F = 0.0003 * 729.6*10^6 = 218.88 kN[/tex]

The compressive load will generate a stress of

[tex]\sigma = F / A\\F = \sigma * A\\F = 80*10^6 * 0.048 * 0.04 = 153.6 kN[/tex]

See more about force at brainly.com/question/26115859