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yan [13]
3 years ago
13

A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

stress ratio. (c) Compute the magnitude of the stress range. minimum stress levels.
Engineering
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

Explanation:

Given data in question

mean stress  = 50 MPa

amplitude stress  = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first  maximum stress  and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

\sigma _{m} =  \sigma _{maximum} + \sigma _{minimum}  / 2

50 =  \sigma _{maximum} + \sigma _{minimum}  / 2    ...........1

and

225 =  \sigma _{maximum} + \sigma _{minimum}  / 2      ...........2

from eqution 1 and 2 we get maximum and minimum stress

\sigma _{maximum} = 275 MPa        ............3

and \sigma _{minimum} = -175 MPa     ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio =  \sigma _{minimum} / \sigma _{maximum}

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = \sigma _{maximum} - \sigma _{minimum}  

magnitude = 275 - (-175) = 450 MPa

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<u>Option-(A)</u>

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Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

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<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

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<u>Area for shear head:</u>

                              A(head)=π×d×t

                                           =π×0.04×0.075

                                           =0.003×πm^2

<u>Area for plate:</u>

                               A(plate)=π×d×t  

                                            =π×0.08×0.03

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now we have to find shear stress for both head and plate

<u>For head:</u>

                                   T_allow=V/A(head)

                                    48 Mpa=P/0.003×π                 ..(V=P)

                                             P =48 Mpa×0.003×π

                                                =452.16 KN

<u>For plate:</u>

                                   T_allow=V/A(plate)

                                    48 Mpa=P/0.0024×π                 ..(V=P)

                                             P =48 Mpa×0.0024×π

                                                =361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

                                                P=361.91 KN

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