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3 years ago

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A stainless-steel specimen from the same material characterized up above, was formed into a rectangular cross-section of dimensi

ons (3/4 in. x 1/8 in). The specimen is subject to a tensile force of 72,000 N.
(a) Determine the elastic and plastic strain values when the specimen is subject to the 72,000 N tensile load.
(b) If the original length is 24 in., what will be its final length after the 72,000 N tensile load is released

1 answer:

AlexFokin [52]3 years ago

7 0

Our dates are in different Units, we will do a unique system. I will use International System Units.

F=72000N

$\frac{3}{8}in= 19,05mm$

$\frac{1}{8}in=3,175mm$

We can now calculate the stress with the force applied, that is

$\sigma = \frac{F}{A} = \frac{72000}{(3,175)(19,05)*10^{-6}}$

$\sigma = 1190.40Mpa$

A) The elastic straint is given by,

$\epsilon_e = \frac{\sigma}{E}$

Where E is the Elastic coefficient. This coefficient for Steel is 201Gpa or $207*10^3Mpa$

$\epsilon_e=\frac{1190.40}{207*10^3}$

$\epsilon_e =5.7507*10^{-3}$

We know for the tables of Strain Steel that the Total Strain is $\epsilon_f=0.01$

Then the Plastic Strain is given by,

$\epsilon_p=\epsilon_f-\epsilon_e$

$\epsilon_p= 0.01-5.7507*10^{-3}$

$\epsilon_p = 4.2492*10^{-3}$

B) Now with the original length of 24in, i.e 609,6mm we have that

$l_i = l_0 (1+\epsilon_p)$

$l_i = (609.6mm)(1+4.294*10^{-3})$

$l_i = 612.2176mm$

<em>Therefore the final length after the 72000N will be 612.2176mm</em>

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2 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

Ratling [72]

Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

where;

$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

FOS = 2.183

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2$

$\dfrac{350}{2 FOS} =88.51$

$\dfrac{350}{ FOS} =2 \times 88.51$

${ FOS} =\dfrac{350}{2 \times 88.51}$

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

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3 years ago

What are two factors that determine the thermal energy of a substance

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Mass and chemical composition

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3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

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Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

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An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

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Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

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$Y=\frac {K}{\sigma_c \sqrt {a\pi}}$

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, $\sigma_c$ is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682$

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making K the subject

$K=\sigma_c Y \sqrt {a\pi}$ and substituting 260 MPa for $\sigma_c$ while a is taken as 0.003m and Y is already known

$K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa$

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3 years ago