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3 years ago

How many millimeters are there in a centimeter?

Engineering

1 answer:

artcher [175]3 years ago

3 0

Explanation:

There are 10 millimeters in a centimeter.

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For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in.

makkiz [27]

Answer:

nP ≈ 4.9

nL = 1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = $\frac{Kb}{Kb + Km}$ = $\frac{3}{3+2}$ = 0.2

A) yielding factor of safety

nP = $\frac{sPAt}{Cp + Fi}$ = $\frac{120* 0.1419}{0.2*14.17 + 12.771}$

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

$nL = \frac{SpAt - Fi}{CP}$ = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

3 0

3 years ago

Many farms and ranches use electric fences to keep animals from getting into or out of specific pastures. When switched on, an e

Nikolay [14]

Answer:

Aluminum

Explanation:

The best material to use when creating an electric fence would be Aluminum. Aluminum wiring is incredibly durable and can be easily obtained. Since aluminum is a non-magnetic metal its conducting capabilities far exceed other metallic options in the market and is also why companies choose aluminum for their high tension cable wiring. Aside from being more expensive than other feasible options its durability and conducting capabilities make it easily the best option.

7 0

3 years ago

Read 2 more answers

Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point

mezya [45]

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, $\sigma _c$ = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, $\epsilon _c$ = 0.2% = 0.002

The inelatic strain $\epsilon_c^{in}$ is given as follows;

$\epsilon_c^{in}$ = $\epsilon _c$ - $\sigma _c$ /∈

Therefore, we have;

$\epsilon_c^{in}$ = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, $\epsilon_c^{in}$ = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

7 0

3 years ago

Given the following code, what indexes must be passed to the substring method to produce the new String with the value "SCORE"?

Ierofanga [76]

Answer:

For expr1 = index 5, length 5

For expr2 = index 0, length 4 and index 21, length 5

string quote = "Four score and seven years ago";

string expr1 = quote.Substring(5, 5).ToUpper(); // "SCORE"

string expr2 = quote.Substring(0, 4) + quote.Substring(21, 5).ToLower(); // "fouryears"

Console.WriteLine(expr1);

Console.WriteLine(expr2);

Explanation:

Then code is written in c# and it produces SCORE and f

ouryears

Substring takes 2 arguments, the start of the specific character and the length

6 0

3 years ago

Use Routh's stability criterion to determine how many roots with positive real parts the following equations have:

Pavlova-9 [17]

Answer:

a) no roots not in LHP

b) 2 roots not in LHP

c) 2 roots not in the LHP

d) 2 roots not in the LHP

e) 2 roots not in LHP

Explanation:

$a) s^4 + 8s^3 + 32s^2 + 80s + 100 = 0\\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:32\:\:\:\:\:\:100\\s^3:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\\s^2:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:100\\s^1:\:\:\:80-\frac{800}{22} =43.6\\s^0:\:\:\:100$

No roots not in the LHP

$b) s^5 + 10s^4 + 30s^3 + 80s^2+344s + 480 =0 \\\\s^5:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:30\:\:\:\:\:\:344\\s^4:\:\:\:10\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\:\:\:\:\:\:480\\s^3:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:296\\s^2:\:\:\:-545\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:480\\s^1:\:\:\:490\\s^0:\:\:\:480$

2 roots not in the LHP

$c) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^3:\:\:\:2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2\\s^2:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^1:\:\:\:-4\\s^0:\:\:\:8$

There are roots in the RHP (not all coefficients are greater than 0).

2 roots not in the LHP

$d) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^3:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:20\\s^2:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:78\\s^1:\:\:\:-58\\s^0:\:\:\:78$

There are two sign changes in the first column of the Routh array.

2 roots not in the LHP

$e) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^3:\:\:\:4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: new \:\:row \\s^2:\:\:\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^1:\:\:\:12-\frac{100}{3}=-21.3 \\s^0:\:\:\:25$

2 roots not in LHP

check:

$a (s) = 0$ ⇒

$s^2 = -3 \limits^+_- 4j = 5e^{j(\pi \limits^+_- 0.92)}\\\\s = \sqrt5 e^{j( \frac{\pi}{2} \limits^+_- 0.46)+n\pi j},\:\:\:\:\: n= 0, 1\\$

3 0

3 years ago