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NeX [460]
1 year ago
8

For some alloy, the yield stress is 345-MPa (50,000-psi) and the elastic modulus (E) is 103-GPa (15x106 psi). What is the maximu

m load that may be applied to a specimen with a cross-sectional area of 130-mm2 (0.2-in2) without plastic deformation
Engineering
1 answer:
OverLord2011 [107]1 year ago
4 0

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N

<h3>How to find Elastic Modulus?</h3>

We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.

Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;

We are given the parameters;

Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa

Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa

Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²

Formula for stress without Plastic deformation is;

σ = F_max/Area

where;

σ is stress

F_max is maximum force

Area is Area

Thus making maximum force the subject of the formula gives;

F_max = σ * A

Plugging in the relevant values for stress and area gives us;

F_max = 345 * 10⁶ * 103 * 10⁻⁶

F_max = 35535 N

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N

Read more about Elastic Modulus at; brainly.com/question/6864866

#SPJ1

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The EMV - Ending Market Value is given as:
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The EMV is given as:

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Hence the EMV = 2,000,000 x ( 1 + 20%)

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brainly.com/question/1350233

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Hidden lines

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2 years ago
A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla
creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed =  0.5 m

cohesion = 25 kPa

internal friction angle =  5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

and we know formula for N_{\gamma } is

N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

so here Ф is very less  N_{\gamma } should be nearest to zero

and its value can be 0.5

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3 years ago
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

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Colt1911 [192]

Answer:

Explanation:

The given equation is :

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