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3 years ago

14

Part A Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. True or fa

lse?

2 answers:

sattari [20]3 years ago

6 0

Answer:

False

Explanation:

Engineering stress and strain are calculated utilizing the original gauge lengths and original area.

Galina-37 [17]3 years ago

5 0

Answer: True

Explanation:

Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.

While Engineering strain is the amount that a material deforms per unit length in a tensile test. It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.

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Who was the American founder and leader of the Shakers in the 1770’s who advocated equality, individual responsibility, and peac

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Answer: Ann Lee (1736-1784)

Explanation:

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3 years ago

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Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str

just olya [345]

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is $\frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}$

The shear strength= $0.58\sigma_y$ and in this case $\sigma_y=36 000 psi$

Shear strength= $\frac {Load}{Shear area}$ hence making load the subject then

Load=Shear area X Shear strength

Load= $\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi$

3 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago