EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220102804487
Author: BEER
Publisher: YUZU
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Chapter 7.5, Problem 97P

(a)

To determine

The normal stress σ0 at which rupture of the component should be expected using Mohr’s criterion.

(a)

Expert Solution
Check Mark

Answer to Problem 97P

The normal stress σ0=8ksi_.

Explanation of Solution

Given information:

The state of plane stress components are σx=σ0σy=12σ0, and τxy=0.

The uniaxial tension for the aluminum alloy is σUT=8ksi.

The uniaxial compression for the aluminum alloy is σUC=20ksi.

Calculation:

Calculate the average normal stress (σavg) as shown below.

σavg=σx+σy2

Substitute σ0 for σx and 12σ0 for σy.

σavg=σ0+12σ02=32σ02=34σ0

Calculate the value of τxy as shown below.

R=(σxσy2)2+τxy2

Substitute σ0 for σx, 12σ0 for σy, and 0 for τxy.

R=(σ012σ02)2+0=(14σ0)2=14σ0

Calculate principal stresses (σmaxandσmin) as shown below.

σmax,min=σavg±R

Substitute 34σ0 for σavg and 14σ0 for R.

σa,b=34σ0±14σ0σa=σ0σb=12σ0

Hence, stress point lies in the first quadrant.

Sketch the Mohr’s criterion for the machine component as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 7.5, Problem 97P

Refer to Figure 1.

The principal stresses for (a) in the first quadrant of boundary.

Calculate the value of σ0 s shown below.

σa=σ0=σUT

Substitute 8ksi for σUT.

σ0=8ksi

Therefore, the normal stress σ0=8ksi_.

(b)

To determine

The normal stress σ0 at which rupture of the component should be expected using Mohr’s criterion.

(b)

Expert Solution
Check Mark

Answer to Problem 97P

The normal stress σ0=6.67ksi_.

Explanation of Solution

Given information:

The state of plane stress components are σx=σ0σy=12σ0, and τxy=0.

The uniaxial tension for the aluminum alloy is σUT=8ksi.

The uniaxial compression for the aluminum alloy is σUC=20ksi.

Calculation:

Calculate the average normal stress (σavg) as shown below.

σavg=σx+σy2

Substitute σ0 for σx and 12σ0 for σy.

σavg=σ0+(12σ0)2=12σ02=14σ0

Calculate the value of τxy as shown below.

R=(σxσy2)2+τxy2

Substitute σ0 for σx, 12σ0 for σy, and 0 for τxy.

R=(σ0(12σ0)2)2+0=(34σ0)2=34σ0

Calculate principal stresses (σmaxandσmin) as shown below.

σmax,min=σavg±R

Substitute 14σ0 for σavg and 34σ0 for R.

σa,b=14σ0±34σ0σa=σ0σb=12σ0

Hence, stress point lies in the forth quadrant.

Refer to Figure 1.

The principal stresses for (b) in the forth quadrant of boundary.

Calculate the value of σ0 s shown below.

σaσUTσbσUC=1

Substitute σ0 for σa, 12σ0 for σb, 8ksi for σUT, and 20ksi for σUC.

σ08(12σ0)20=1σ0(18+140)=1640σ0=1σ0=6.67ksi

Therefore, the normal stress σ0=6.67ksi_.

(c)

To determine

The normal stress σ0 at which rupture of the component should be expected using Mohr’s criterion.

(c)

Expert Solution
Check Mark

Answer to Problem 97P

The normal stress σ0=8.89ksi_.

Explanation of Solution

Given information:

The state of plane stress components are σx=σ0σy=12σ0, and τxy=0.

The uniaxial tension for the aluminum alloy is σUT=8ksi.

The uniaxial compression for the aluminum alloy is σUC=20ksi.

Calculation:

Calculate the average normal stress (σavg) as shown below.

σavg=σx+σy2

Substitute σ0 for σx and 12σ0 for σy.

σavg=σ0+12σ02=12σ02=14σ0

Calculate the value of τxy as shown below.

R=(σxσy2)2+τxy2

Substitute σ0 for σx, 12σ0 for σy, and 0 for τxy.

R=(σ0(12σ0)2)2+0=(34σ0)2=34σ0

Calculate principal stresses (σmaxandσmin) as shown below.

σmax,min=σavg±R

Substitute 14σ0 for σavg and 34σ0 for R.

σa,b=14σ0±34σ0σa=12σ0σb=σ0

Hence, stress point lies in the forth quadrant.

Refer to Figure 1.

The principal stresses for (b) in the forth quadrant of boundary.

Calculate the value of σ0 s shown below.

σaσUTσbσUC=1

Substitute 12σ0 for σa, σ0 for σb, 8ksi for σUT, and 20ksi for σUC.

12σ08(σ0)20=1σ0(116+120)=1980σ0=1σ0=8.89ksi

Therefore, the normal stress σ0=8.89ksi_.

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Chapter 7 Solutions

EBK MECHANICS OF MATERIALS

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