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Lina20 [59]
2 years ago
10

different lever designs can be engineered to alter the brake pedal effort required of the driver by using different levels of ?

Engineering
1 answer:
enot [183]2 years ago
8 0

Different lever designs can be engineered and developed to alter the brake pedal effort required of the driver by using different levels of <u>mechanical advantage</u>.

<h3>What is mechanical advantage?</h3>

Mechanical advantage can be defined as a ratio of the output force of a lever to the force acting on it (input force or effort), assuming no losses due to wear, flexibility, tear or friction.

This ultimately implies that, different lever designs can be suitably engineered and developed to alter the brake pedal effort (input force) that is required of the driver, especially by using different levels of <u>mechanical advantage</u>.

Read more on mechanical advantage here: brainly.com/question/18345299

#SPJ1

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3 years ago
A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa
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Answer:

stress  = 50MPa

Explanation:

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stress due to displacement in structural steel can be determined by using following relation

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7 0
4 years ago
If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
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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

maximum value of force P=??

<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

we know that,

                               V=P

<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

                                            =0.0024×πm^2

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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