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2 years ago

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

a) Determine the yielding factor of safety

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

b) Determine the overload factor of safety

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 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 : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

C) Determine the factor of safety based on joint separation

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

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

input values into equation above

Hence $n_{0}$ = 1.056

D) Determine the fatigue factor of safety using the Goodman criterion.

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? 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

Please help on two I will give brainiest​

Please help on two I will give brainiest