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Romashka [77]
3 years ago
15

Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.

Engineering
1 answer:
lakkis [162]3 years ago
5 0

This question is incomplete, the complete question is;

A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N.

Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.

answer in mm please

Answer:

the minimum required wire radius is 5.3166 mm

Explanation:

Given that;

Load F = 11100N

N = 3

∝y = 1500 MPa

∝workmg = ∝y / N = 1500 / 3 = 500 MPa

now stress of Wire:

∝w = F/A

500 × 10⁶ = 11100 / A

A = 22.2 × 10⁻⁶ m²

so

(π/4)d² = A

(π/4)d² = 22.2 × 10⁻⁶

d² = 2.8265 × 10⁻⁵

d = 5.3165 7 × 10⁻³ m³

now we convert to mm(millimeters)

d = 5.3166 mm

Therefore the minimum required wire radius is 5.3166 mm

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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
The B-pillar may also be called the:
slega [8]

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

looked it up

6 0
3 years ago
Challenge:
goldenfox [79]

Answer:

what?

Explanation:

8 0
3 years ago
An airliner is flying at 34,000 ft cruise altitude on a standard day. Calculate the pressure difference between the cabin and th
nadya68 [22]

Answer:

\Delta P=61,952.8\ lb/ft^2

Explanation:

Given

Airline flying at 34,000 ft.

Cabin pressurized to an altitude 8,000 ft.

We know that at standard condition ,density of air

\rho =0.074\ lb/ft^3

We know that pressure difference    

ΔP=ρ g ΔZ

Here ΔZ=34,000-8,000  ft

        ΔZ=26,000 ft

g= 32.2\ ft/s^2

ΔP=0.074 x 32.2 x 26,000

\Delta P=61,952.8\ lb/ft^2

So pressure difference will be \Delta P=61,952.8\ lb/ft^2.

7 0
3 years ago
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