EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220102804487
Author: BEER
Publisher: YUZU
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Chapter 7.9, Problem 157P

The following state of strain has been determined on the surface of a cast-iron machine part:

x = -720 μ

y = -400 μ

γ x y = + 660 μ

Knowing that E = 69 GPa and G = 28 GPa, determine the principal planes and principal stresses (a) by determining the corresponding state of plane stress [use Eq. (2.36), Eq. (2.43), and the first two equations of Prob. 2.73] and then using Mohr's circle for stress, (b) by using Mohr's circle for strain to determine the orientation and magnitude of the principal strains and then determining the corresponding stresses.

(a)

Expert Solution
Check Mark
To determine

Find the direction and magnitude of the three principal strains.

Answer to Problem 157P

The maximum principal stress is is σa=29.8MPa_.

The intermediate principal stress is σb=70.9MPa_.

The orientation of the major principal stress is θa=32.1°_.

The orientation of the intermediate principal stress is θb=57.9°_.

Explanation of Solution

Given information:

Construct the Mohr’s circle for stress.

The modulus of elasticity of the material is E=69GPa.

The modulus of rigidity of the material is G=28GPa.

The normal strain in x-axis is εx=720μ×1061μ=7.20×104.

The normal strain in y-axis is εy=400μ×1061μ=4×104.

The shearing strain in xy-plane is γxy=+660μ×1061μ=+6.6×104.

Calculation:

Find the Poisson’s ratio (ν) using the relation.

G=E2(1+ν)

Here, the modulus of elasticity is E and the modulus of rigidity is G.

Substitute 69 GPa for E and 28 GPa for G.

28=692(1+ν)ν=0.2321

Find the normal stress in x-axis (σx) using the relation.

σx=E1ν2(εx+νεy)

Substitute 69 GPa for E, 0.2321 for ν, 7.20×104 for εx, and 4×104 for εy.

σx=69GPa×103MPa1GPa10.23212(7.20×104+0.2321(4×104))=59.28MPa

Find the normal stress in y-axis (σy) using the relation.

σy=E1ν2(εy+νεx)

Substitute 69 GPa for E, 0.2321 for ν, 7.20×104 for εx, and 4×104 for εy.

σy=69GPa×103MPa1GPa10.23212(4×104+0.2321(7.20×104))=41.36MPa

The principal stress in z-axis is σz=0.

Find the shearing stress in xy-plane (τxy) using the relation.

τxy=Gγxy

Substitute 28 GPa for G and +6.6×104 for γxy.

τxy=28GPa×103MPa1GPa×6.6×104=18.48MPa

Construct the Mohr circle as follows;

  • Plot the point X by –59.28 MPa to the left of the τ axis and 18.48 MPa below the σ axis.
  • Plot the point Y by –41.36 MPa to the left of the τ axis and 18.48 MPa above the σ axis.
  • Connect the points X and Y and the point C is known as abscissa.

Show the plotted Mohr’s circle diagram as in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 7.9, Problem 157P , additional homework tip  1

Find the average normal stress (σave) using the relation.

σave=σx+σy2

Substitute –59.28 MPa for σx and ­41.36 MPa for σy.

σave=59.28+(41.36)2=50.32MPa

Find the radius of the Mohr’s circle (R) using the relation.

R=(σxσy2)2+τxy2

Substitute –59.28 MPa for σx, ­41.36 MPa for σy, and 18.48 MPa for τxy.

R=(59.28(41.36)2)2+(18.48)2=20.54MPa

Find the maximum principal stress (σa) using the relation.

σa=σave+R

Substitute –50.32 MPa for σave and 20.54 MPa for R.

σa=50.32+20.54=29.8MPa

Find the minimum principal stress (σb) using the relation.

σb=σaveR

Substitute –50.32 MPa for σave and 20.54 MPa for R.

σb=50.3220.54=70.9MPa

Find the orientation (θa) of the maximum principal axis using the relation.

tan2θa=2τxyσxσy

Substitute –59.28 MPa for σx, ­41.36 MPa for σy, and 18.48 MPa for τxy.

tan2θa=2(18.48)59.28(41.36)θa=32.1°

Find the orientation of the intermediate principal axis (θb) using the relation.

θb=θa+90°

Substitute 32.1° for θa.

θb=32.1°+90°=57.9°

Therefore,

The maximum principal stress is is σa=29.8MPa_.

The intermediate principal stress is σb=70.9MPa_.

The orientation of the major principal stress is θa=32.1°_.

The orientation of the intermediate principal stress is θb=57.9°_.

(b)

Expert Solution
Check Mark
To determine

Find the direction and magnitude of the three principal strains.

Answer to Problem 157P

The maximum principal strain is εa=423μ_.

The intermediate principal strain is εb=951μ_.

The minimum principal strain is εc=222μ_.

The orientation of the major principal strain is θa=22.5°_.

The orientation of the intermediate principal strain is θb=67.5°_.

Explanation of Solution

Given information:

Construct the Mohr’s circle for strain.

The modulus of elasticity of the material is E=69GPa.

The modulus of rigidity of the material is G=28GPa.

The normal strain in x-axis is εx=720μ.

The normal strain in y-axis is εy=400μ.

The shearing strain in xy-plane is γxy=+660μ.

Calculation:

Find the Poisson’s ratio (ν) using the relation.

G=E2(1+ν)

Substitute 69 GPa for E and 28 GPa for G.

28=692(1+ν)ν=0.2321

Find the value of γxy2;

γxy2=6602=330μ

Construct the Mohr circle as follows;

  • Plot the point X by 720μ to the left of the γ2 axis and 330μ below the ε axis.
  • Plot the point Y by 400μ to the left of the γ2 axis and 330μ above the ε axis.
  • Connect the points X and Y and the point C is known as abscissa.

Show the plotted Mohr’s circle diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 7.9, Problem 157P , additional homework tip  2

Find the average normal strain (εave) using the relation.

εave=εx+εy2

Substitute 720μ for εx and 400μ for εy.

εave=7204002=560μ

Find the orientation (θa) of the maximum principal axis using the relation.

tan2θa=γxyεxεy

Substitute 660μ for γxy, 720μ for εx, and 400μ for εy.

tan2θa=660720(400)θa=32.1°

Find the orientation of the intermediate principal axis (θb) using the relation.

θb=θa+90°

Substitute 32.1° for θa.

θb=32.1°+90°=57.9°

Find the radius of the Mohr circle (R) using the equation.

R=(εxεy2)2+(γxy2)2

Substitute 660μ for γxy, 720μ for εx, and 400μ for εy.

R=(720(400)2)2+(6602)2=366.74μ

Find the maximum principal strain (εa) using the relation.

εa=εave+R

Substitute 560μ for εave and 366.74μ for R.

εa=560+366.74=193.26μ

Find the intermediate principal strain (εb) using the relation.

εb=εaveR

Substitute 560μ for εave and 366.74μ for R.

εb=560366.74=926.74μ

Find the maximum principal stress (σa) using the equation.

σa=E1ν2(εa+νεb)

Substitute 69 GPa for E, 0.2321 for ν, 193.26μ for εa, and 926.74μ for εb.

σa=69GPa×103MPa1GPa10.23212(193.26×106+0.2321(926.74×106))=59.28MPa

Find the intermediate principal stress (σb) using the equation.

σb=E1ν2(εb+νεa)

Substitute 69 GPa for E, 0.2321 for ν, 193.26μ for εa, and 926.74μ for εb.

σb=69GPa×103MPa1GPa10.23212(926.74×106+0.2321(193.26×106))=70.9MPa

Therefore,

The maximum principal stress is is σa=29.8MPa_.

The intermediate principal stress is σb=70.9MPa_.

The orientation of the major principal stress is θa=32.1°_.

The orientation of the intermediate principal stress is θb=57.9°_.

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Chapter 7 Solutions

EBK MECHANICS OF MATERIALS

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