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exis [7]
3 years ago
11

Eeevegesgsesggesgrgrgsrgrgsr

Engineering
2 answers:
ikadub [295]3 years ago
8 0

xprggdsghdsfmjffnkljgsnncc

klemol [59]3 years ago
6 0

Answer:

nehshsisjsbsjsowokwwnbsjdjdsndn

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Choose the three questions that an engineer should ask himself or herself when identifying the need of a problem.
boyakko [2]

Answer:

What is the goal? What will a successful solution look like? What is it that the client expects me to accomplish?

Explanation:

5 0
3 years ago
Read 2 more answers
A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

                                                 = 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

8 0
3 years ago
1) What output force (Fout) is produced if the lever arm length (rout) is 100 mm?
Ierofanga [76]
The length of the arm is the main part of natur
6 0
3 years ago
Read 2 more answers
A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o
victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

crack length 2c = 0.5 mm at centre of specimen

\sigma _{applied} =  1000 N/cross sectional area

stress intensity factor  =  k  will be

\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

                   = 6.154\times 10^{6} Pa

we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

Kc = \sigma _{frac}(\sqrt{\pi C})

1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}

\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

LAOD = 6669.86 N

3 0
3 years ago
Các đặc điểm chính của đường dây dài siêu cao áp .
rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0
3 years ago
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