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Kobotan [32]
2 years ago
13

Please help on two I will give brainiest​

Engineering
2 answers:
Readme [11.4K]2 years ago
7 0

Answer:

Prototype I believe

Explanation:

Delicious77 [7]2 years ago
4 0
Prototype correct me if i am wrong
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For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.
rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN,  Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

n_{p} = \frac{S_{p}A_{t}  }{CP_{max}+ F_{i}  }  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = \frac{380*20.1}{0.277*7500*5728.5} = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

n_{L} =  \frac{S_{p}A_{t}-F_{i}   }{C(P_{max} )}  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : n_{L} = 0.9191n_{L}  = 0.9191

<u>C)  Determine the factor of safety based on joint separation</u>

n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence n_{0} = 1.056

<u>D)  Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

4 0
2 years ago
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
2 years ago
What is one major life lesson you learned from the movie; ¨Spare Parts¨
allochka39001 [22]

Answer:

darts,” a smart, creative and highly-enjoyable drama about a team of intelligent, hard-working and ambitious high school students who enter a prestigious robotics competition, and their dedicated science teacher who mentors, educates, pushes and inspires them, is a rousing, uplifting, spirited–and excellent–film and a great start to the new film

6 0
3 years ago
Read 2 more answers
What is a chipping hammer used for? <br><br> State three things.
aleksley [76]

Answer:

i hope this helps.

Explanation:

they are used for breaking concrete, can be positioned to break vertical and overhead surfaces, allows precisely chip away only specific areas.

7 0
3 years ago
Read 2 more answers
Seperate real and imaginary parts tan(2x+i3y)
ahrayia [7]

Answer:

First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

3 0
3 years ago
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