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A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm

Answers

Answer 1

Answer:

Answer:

tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa

Explanation:

Given data in question

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

to find out

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

solution

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield = yield load / area

tensile stress at yield = 80 ×10³ / $\pi$ /4 × D²

tensile stress at yield = 80 ×10³ / $\pi$ /4 × 0.020²

tensile stress at yield = 254 MPa

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³ / $\pi$ /4 × 0.020²

ultimate stress = 477 MPa

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³ / $\pi$ /4 × d²

average stress = 70 ×10³ / $\pi$ /4 × 0.010²

average stress = 892 MPa

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How much computer memory (in bytes) in minimum would be required to store 10 seconds of a sensor signal sampled by a 12-bit A/D converter operating at a sampling rate of 5 kHz?

Answers

Answer:

73.24 K byte

Explanation:

Assuming that

N = total number of samples

N = 10 * 5kHz

N = 50*10^3

Also, the total number of bits, T

T = 12 * N

T = 12 * 50*10^3

T = 600 * 10^3

And then, finally, the total number of byte,

B = 600*10^(3/8)

B = 75*10^3 byte

75*10^3 byte = 75*10^3/1024 kilo byte

And on converting to decimal, we will have

= 73.24 K byte

Therefore, the memory required = 73.24 K byte

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

Answers

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

$E(t)=E_o[1-a(T)/(T_m)]$

where,

E(t) is the modulus of elasticity at any temperature 'T'

$E_o$ is the modulus of elasticity at absolute zero.

$T_(m)$ is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

Will mark brainliest if correctWhen a tractor is driving on a road, it must have a SMV sign prominently displayed.

True
False

Answers

Answer: true

Explanation:

Convert the angles of a triangle to radians.Part A31∘43′53′′, 90∘32′11′′, 57∘43′56′′Express your answers, separated by commas, to six significant figures.nothingrad, rad, radRequest AnswerPart B94∘22′19′′, 40∘54′53′′, 44∘42′48′′Express your answers, separated by commas, to six significant figures.

Answers

Answer:

Explanation:

To convert to radians

A31∘43′53′′, 90∘32′11′′, 57∘43′56′′

using DMS approach ; 1degree = 60minutes = 3600 seconds

1° = 60' = 3600"

And degree to radian = multiply by π/180

A) 31∘43′53′′ = 31degree + 43minutes + 53 seconds

= 31 degree + 43minutes + 53/60

= 31 degree + 43.88minutes

= 31 degree + 43.88/60 = 31.73 degree x π/180 = 0.5534radians

FOR 90∘32′11′′ = 90 degree + 32minutes + 11seconds

= 90degree + 32minutes + 11/60

= 90 degree + 32.183minutes

= 90 degree + 32.183/60 = 90.54degree x π/180

= 1.580radians

FOR 57∘43′56′′ = 57degree + 43minutes+ 56seconds

= 57degree + 43minutes + 56/60

= 57 degree + 43.93minutes

= 57degree + 43.93/60 = 57.73degree X π/180

= 1.00radians

PART B

FOR 94∘22′19′′ = 94degree + 22minutes + 19seconds

= 94degree + 22minutes + 19/60

= 94degree + 22.32minutes

= 94degree + 22.32/60

= 94.37degree X π/180 = 1.65radians

FOR 40∘54′53′′ = 40degree + 54minutes + 53seconds

= 40 degree + 54minutes + 53/60

= 40 degree + 54.88minutes = 40 degree + 54.88/60

= 40.91degree X π/180 = 0.714radians

FOR 44∘42′48′′ = 44degree + 42minutes + 48seconds

= 44degree + 42.8minutes

= 44.71degree X π/180 = 0.780radians

Answer:

0.176270π rad, 0.502980π rad, 0.320735π rad

0.524289π rad, 0.227304π rad, 0.248407π rad

Explanation:

We know that,

1° = 60' 180° = π

1 ' = 1°/60 1° = π/180

a. 31°43'53''

Step 1

53'' = 53 * 1/60

= 53'/60

Step 2

43'53''

= 43'+53'/60

= (2580+43)/60

= 2623'/60

-------- Convert to degrees

= 2623/60 * 1/60

= 2623/3600

Step 3

31°43'53''

= 31+ 2623/3600

= (111600 + 2623)/3600

= 114223°/3600

Now, we convert to radians

= 114223/3600 * π/180°

= 0.176270π rad

90°32'11''

Step 1.

11' = 11 * 1/60

= 11/60

Step 2

32'11'

= 32 + 11/60

= 1931/60

-------- Convert to degrees

= 1931/60 * 1/60

= 1931/3600

Step 3

90°31'11''

= 90 + 1931/3600

= 325931°/3600

Now we convert to radians

= 325931°/3600 * π/180°

= 0.502980π rad

57°43'56''

Step 1

56' = 56 * 1/60

= 56/60

= 14/15

Step 2

43'56''

= 43 + 14/15

= 659/15

Now we convert to degrees

= 659/15 * 1/60

= 659°/900

Step 3

57°43'56''

= 57 + 659/900

= 51959/900

Now we convert to radians

= 51959°/900 * π/180°

= 0.320735π rad

94∘22′19′′

Step 1

19'' = 19/60

Step 2

22'19''

= 22 + 19/60

= 1339/60

Now we convert to degrees

= 1339/60 * 1/60

= 1339°/3600

Step 3

94°22'19"

= 94 + 1339/3600

= 339739°/3600

Now we convert to radians

= 339739°/3600 * π/180

= 0.524289π rad

40∘54′53′′

Step 1

53" = 53/60

Step 2

54'53"

= 54'+ 53/60

= 3293/60

Now we convert to degrees

= 3293/60 * 1/60

= 3293/3600

Step 3

40°54'53"

= 40 + 3293/3600

= 147293/3600

Now we convert to radians

= 147293/3600 * π/180

= 0.227304π rad

44∘42′48′

Step 1

48' = 48/69

= 4/5

Step 2

42'48"

= 42 + 4/5

=214/5

Nowz we convert to degrees

= 214/5 * 1/60

= 107/150

Step 3

44°42'48"

= 44 + 107/150

= 6707/150

Now we convert to radians

= 6707/150 * π/180

= 0.248407π rad

An interest buydown program offers to reduce interest rates by 4% from the base rate. Suppose the base rate for a loan of $8000 is 8% for 10 years. What is the monthly payment before and after the buydown? In this case, use monthly compounding, that is, the term is 120 payment periods, and the interest per month is 0.667% before and 0.333% after the buydown.

Answers

Answer: The monthly payment before the buydown is $71.3

The monthly payment after the buydown is $68.9

Explanation: The payment is compounding so we use compound interest;

A= P[1+(r/n)^nt]

Where;

A= Compounded amount

P = principal

r= interest rate per payment

n= number of payment per period

t= number of period.

NOTE: from our questions, the period is yearly and the payment is monthly. Therefore;

number of payment per period (n) is 12

number of payment period (t) is 10

P=$8000, r= 0.667% or 0.333%

FIND MONTHLY PAYMENT BEFORE BUYDOWN:

Step 1: find the Compounded amount to pay.

A= $8000[1+(0.00667÷12)^(12×10)]=

$8551.64 this is the total amount he has to pay for a period of 10years

Step 2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months

$8551.64÷120= $71.3 monthly

FIND MONTHLY PAYMENT AFTER BUYDOWN:

Step 1: find the compounded amount to pay.

A= 8000[1+(0.00333÷12)^(12×10)=

$8270.85 this is the total amount he has to pay for a period of 10years

Step2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months;

$8270.85÷120= $68.9 monthly

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace