Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 11.3, Problem 39P

(a)

To determine

The factor of safety associated with the yield strength σz=+16ksi.

(a)

Expert Solution
Check Mark

Answer to Problem 39P

The factor of safety associated with the yield strength σz=+16ksi is F.S.=2.33_.

Explanation of Solution

Given information:

The stress component along x direction is σx=8ksi.

The stress component along z direction is σz=0.

The shear stress component is τxz=14ksi.

The yield stress is σY=65ksi.

The principal stress is σz=+16ksi.

Calculation:

Sketch the state of stress in a machine component as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 39P , additional homework tip  1

Refer to Figure 1.

σx=8ksiσz=0τxz=14ksi

Apply the procedure to construct the Mohr’s circle as shown below.

  • Find the centre of the circle C located σavg=σx+σz2 from the origin.
  • Plot the reference points A having coordinates A(σx,τA).
  • Connect the point A with C and from the shaded triangle find the radius R of the circle.
  • Sketch the circle once R has been determined.

Construct the Mohr’s circle as shown below.

Calculate the centre of the circle (σavg) using average normal stress as shown below.

σavg=σx+σz2

Substitute 8ksi for σx and 0 for σz.

σavg=8+02=4ksi

The centre of the circle is C=4ksi.

Coordinates of the reference point z.

z=(σz,τxz)

Substitute 0 for σz and 14ksi for τxz.

z=(0, 14 ksi)

Coordinates of the reference point x.

x=(σx,τxz)

Substitute 8ksi for σx and 14ksi for τxz.

x=(8ksi,14ksi)

Calculate the radius (R) of the circle as shown below.

R=(σxσavg)2+(τxz)2

Substitute 8ksi for σx, 4ksi for σavg, and 14ksi for τxz.

R=(84)2+(14)2=212=14.56ksi

Sketch the Mohr’s circle as shown in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 11.3, Problem 39P , additional homework tip  2

Refer to Figure 2.

Calculate the principal stresses (σmaxand σmin) as shown below.

σmax,min=σavg±R

Substitute 4ksi for σavg and 14.56ksi for R.

σmax,min=4±14.56σa=18.56ksiσb=10.56ksi

Also the principal stress σc=σy.

Apply maximum distortion energy criterion as shown below.

(σaσb)2+(σbσc)2+(σcσa)2=2(σYF.S.)2 (1)

Substitute 18.56ksi for σa, 10.56ksi for σb, +16ksi for σc, 2.2 for F.S., and 65ksi for σY in Equation (1).

(18.56(10.56))2+((10.56)16)2+(1618.56)2=2×(65F.S.)2847.9744+705.4336+6.5536=8,450(F.S.)2(F.S.)2=5.4168F.S.=2.33

Hence, the factor of safety associated with the yield strength σz=+16ksi is F.S.=2.33_.

(b)

To determine

The factor of safety associated with the yield strength σz=16ksi.

(b)

Expert Solution
Check Mark

Answer to Problem 39P

The factor of safety associated with the yield strength σz=16ksi is F.S.=2.02_.

Explanation of Solution

Given information:

The stress component along x direction is σx=8ksi.

The stress component along z direction is σz=0.

The shear stress component is τxz=14ksi.

The yield stress is σY=65ksi.

The principal stress is σz=16ksi.

Calculation:

Refer to part (a).

The principal stresses σa=18.56ksi and σb=10.56MPa.

Apply maximum distortion energy criterion as shown below.

Substitute 18.56ksi for σa, 10.56ksi for σb, 16ksi for σc, 2.2 for F.S., and 65ksi for σY in Equation (1).

(18.56(10.56))2+((10.56)(16))2+((16)18.56)2=2×(65F.S.)2847.9744+29.5936+1,194.3936=8,450(F.S.)2(F.S.)2=4.0783F.S.=2.02

Therefore, the factor of safety associated with the yield strength σz=16ksi is F.S.=2.02_.

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Chapter 11 Solutions

Mechanics of Materials, 7th Edition

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