Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 8, Problem 8.3P

(a)

To determine

The new arm radius.

(a)

Expert Solution
Check Mark

Answer to Problem 8.3P

  0.00638m would be the new radius for the arm.

Explanation of Solution

Given:

The radius of [resent control arm is 1cm .

The length of the present control arm is 30cm .

The maximum control force exerted on the arm is 7.85×104N .

The least ductility provided to sufficient fracture is 15%

Formula used:

The expression for the weight of the material is given by

  W1=ρ(πr2l) …… (I)

Here, W1 is the weight of the material, r is the radius, ρ is the density of low alloy steel and l is the length.

The expression for the new area is given as,

  An=F(I)σY …… (II)

Here, An is the new area, F is force, I is the percentage of metal yield stress and σY is the yield stress

The expression for arm radius is given as

  r2=Anπ …… (III)

Here, An is a new area.

Calculation:

The weight of material is calculated as,

Substitute 7.85×103kg/m3 for ρ , 30cm for l and 1cm for A in equation (I).

  W1=7.85×103kg/m3(π×( 1cm×0.01m/ cm 2 )×( 30cm×0.01m/ cm ))=(7.85× 103kg/ m 3)×(9.424× 10 3 m 2/cm)=0.7398kg( 10 3 × 10 3 m 3 m 3 )

The new area is calculated as,

Substitute 50% for I , 7.85×104N for F and 1225MPa for σY in equation (II).

  An=7 .85×104N( 50%)1225MPa=7 .85×104N( 50%)1225MPa×( 10 6 Pa 1MPa )=1.28×104m2

The arm radius is calculated as,

Substitute 1.28×104m2 for An in equation (III).

  r2=1.28× 10 4m2π=4.0734×104m2r=4..0734× 10 4m2=0.00638m

Conclusion:

Therefore, the new arm radius is 0.00638m .

(b)

To determine

The weight reduction that is provided by the new steel.

(b)

Expert Solution
Check Mark

Answer to Problem 8.3P

Areduction of 0.4386kg in the weight occurs by using the new steel.

Explanation of Solution

Formula used:

The expression for new weight of the material is given by,

  W2=ρ(Anl) …… (IV)

Here, An is new area, ρ is the density of low alloy steel and l is the length.

Calculation:

The weight of the material is calculated as,

Substitute 7.85×103kg/m3 for ρ , 30cm for l and 1.28×104m2 for An in equation(IV).

  W2=7.85×103kg/m3(1.28× 10 4m2×30cm×0.01m/cm)=7.85×103kg/m3×3.84×105m3=0.3014kg

The weight saving is calculated as,

  W=W1W2 …… (V)

Substitute 0.7398kg for W1 and 0.3014kg for in equation (V)

  W=0.7398kg0.3014kg=0.4386kg

Conclusion:

Therefore, the weight reduction that is provided by this new steel is 0.4386kg .

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Chapter 8 Solutions

Materials Science And Engineering Properties

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