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sasho [114]
3 years ago
15

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

tic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.5 mm (0.22 in.) in diameter and 267 mm (10.52 in.) long is stressed in tension and found to elongate 7.0 mm (0.28 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Engineering
2 answers:
Ugo [173]3 years ago
7 0

Answer:

Computation of the load is not possible because E(test) >E(yield)

Explanation:

We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if

E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:

Calculation of E(test is as follows)

E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm

=0.0262

Computation of E(yield) is given below:

E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027

Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).

frosja888 [35]3 years ago
5 0

Answer:

It's not possible to compute the magnitude of the load

Explanation:

Given data:

Yield strength = 275Mpa

Tensile strength = 380Mpa

Elastic modulus = 103Gpa = 103 * 1000MPa

Diameter = 5.5 mm

Length = 267 mm

Elongation = 7.0 mm

Calculating the strain test value using the formula'

<h3>ε(test) = ΔL/Lo</h3>

Substituting, we have

ε = 7.0/267

  =  0.0262

Calculating the yield strain using the formula

<h3>ε(yield) = бy/E </h3>

Substituting, we have

ε(yield) = 275/ 103 * 1000

             = 275/103000

            = 0.00267

From the calculation above, the strain test is greater than the yield. Therefore, it's not possible to compute the magnitude of the load.

<h3>ε(test) ⊃  ε(yield)</h3><h3></h3>
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‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
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E = 2940 J

Explanation:

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A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower
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Answer:

γ_{xy} =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ_{xy} = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ_{xy}

By comapring both equations, we get

P/A = Gγ_{xy}   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

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6 0
3 years ago
8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of
san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

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To find modulus of elasticity apply'

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Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

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Material  K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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