What entrepreneurial activities do you know?are you capable of doing entrepreneurial activities

A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of

Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0

3 years ago

What do you do when there’s a child in front of you?

disa [49]

If there is a random child, you should ask where they should be and where their parents went. if it your child, they prolly want food or sum.

6 0

3 years ago

What’s is 6 billion plus 6 quadrillions

goblinko [34]

Answer:

6.000006e+15 Or 6.000006000000000000000

Explanation:

I think pls brainliest me

3 0

3 years ago

Read 2 more answers

Why should a toolpath be verified on the screen of a CAM system prior to creating the program code?

NISA [10]

Answer:

The tool's trajectory in a CAM program refers to the places where the tool will be during the work. It is important to review it before generating the program for the following reasons

1. analyze the machining strategy and identify which one is better for each piece.

2.Avoid the collision of the tool holder with the work piece.

3.Avoid the shock of the tool with the piece.

4. Prevent the collision of the tool with elements that are not displayed on the CAM such as clamping flanges or screws.

5 0

3 years ago

An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flo

svetoff [14.1K]

Answer:

Air speed in the wind-tunnel $v_{2}$ = 27.5 m/s

Explanation:

Given data :

Manometer reading ; p1 - p2 = 45 mm of water

Pressure at section ( I ) p1 = 100 kPa ( abs )

temperature ( T1 ) = 25°C

Pw ( density of water ) = 999 kg/m3

g = 9.81 m/s^2

next we apply Bernoulli equation at section 1 and section 2

p1 - p2 = $\frac{PairV^{2} _{2} }{2}$ ---------- ( 1 )

considering ideal gas equation

Pair ( density of air ) = $\frac{P}{RT}$ ------------------- ( 2 )

R ( constant ) = 287 NM/kg.k

T = 25 + 273.15 = 298.15 k

P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2

substitute values into equation ( 2 )

= 100 * 10^3 / (287 * 298.15)

= 1.17 kg/m^3

Also note ; p1 - p2 = PwgΔh ------- ( 3 )

finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )

$\frac{PairV^{2} _{2} }{2}$ = PwgΔh

$V^{2} _{2}$ = $\frac{2*999* 9.81* 0.045}{1.17}$ = 753.86

$v_{2}$ = 27.5 m/s

5 0

3 years ago