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 7.2, Problem 37P

Solve Prob. 7.15, using Mohr's circle.

7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.

Chapter 7.2, Problem 37P, Solve Prob. 7.15, using Mohr's circle. 7.13 through 7.16 For the given state of stress, determine

Fig P7 15

(a)

Expert Solution
Check Mark
To determine

The normal and shearing stresses after the element has been rotated through 25° clockwise using Mohr’s circle.

Answer to Problem 37P

The normal stresses are σx=9.022ksi_ and σy=13.022ksi_.

The shear stress is τxy=3.80ksi_.

Explanation of Solution

Given information:

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

The stress component along y direction as σy=12ksi.

The shear stress component as τxy=6ksi.

The orientation of the principal plane as θ=25°.

Calculation:

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

  • Find the centre of the circle C located σavg=σx+σy2 from the origin.
  • Plot the reference points A having coordinates A(σx,τA).
  • Connect the point A with C and from the shaded triangle and 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 strain as shown below.

σavg=σx+σy2

Substitute 8ksi for σx and 12ksi for σy.

σavg=8122=2ksi

The centre of the circle is C=2ksi.

Coordinates of the reference point X.

X=(σx,τxy)

Substitute 8ksi for σx and 6ksi for τxy.

X=(8ksi,(6ksi))=(8ksi, 6 ksi)

Coordinates of the reference point Y.

Y=(σy,τxy)

Substitute 12ksi for σy and 6ksi for τxy.

Y=(12ksi,6ksi)

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

R=(σxσavg)2+(τxy)2

Substitute 8ksi for σx, 2ksi for σavg, and 6ksi for τxy,

R=(8(2))2+(6)2=136=11.66ksi

Sketch the Mohr’s circle as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 7.2, Problem 37P , additional homework tip  1

Refer to Figure 1.

Calculate the principle plane (θp) as shown below.

tan2θp=FXFCtan2θp=6102θp=tan1(0.6)2θp=30.96°

Calculate the angle φ as shown below.

φ=2θ2θp

Substitute 30.96° for 2θp and 25° for θ

φ=2×25°30.96°=19.04°

Calculate the normal stress along x direction (σx) as shown below.

σx=σavg+Rcosφ (1)

Substitute 2ksi for σavg, 11.66ksi for R, and 19.04° for θ in Equation (1).

σx=2+11.66cos19.04°=2+11.022=9.022ksi

Hence, the normal stress σx=9.022ksi_.

Calculate the normal stress along y direction (σy) as shown below.

σy=σavg+Rcosφ (2)

Substitute 2ksi for σavg, 11.66ksi for R, and 19.04° for θ in Equation (2).

σy=211.66cos19.04°=211.022=13.022ksi

Hence, the normal stress σy=13.022ksi_.

Calculate the shear stress (τxy) as shown below.

τxy=Rsinφ

Substitute 11.66ksi for R and 19.04° for θ in Equation (3).

τxy=11.66sin19.04°=3.80ksi

Therefore, the shear stress τxy=3.80ksi_.

(b)

Expert Solution
Check Mark
To determine

The normal and shearing stresses after the element has been rotated through 10° counter clockwise using Mohr’s circle.

Answer to Problem 37P

The normal stresses are σx=5.34ksi_ and σy=9.34ksi_.

The shear stress is τxy=9.06ksi_.

Explanation of Solution

Given information:

The stress component along x direction σx=8ksi.

The stress component along y direction σy=12ksi.

The shear stress component τxy=6ksi.

The orientation of the principal plane θ=10°.

Calculation:

Refer to part (a).

Coordinates of the reference point X=(8ksi,6ksi)

Coordinates of the reference point Y=(12ksi,6ksi)

The principal plane 2θp=30.96°.

The average normal stress σavg=2ksi.

The radius of the Mohr’s circle R=11.66ksi.

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

Mechanics of Materials, 7th Edition, Chapter 7.2, Problem 37P , additional homework tip  2

Refer to Figure 2.

Calculate the angle φ as shown below.

φ=2θ+2θp

Substitute 30.96° for 2θp and 10° for θ

φ=2×10°+30.96°=50.96°

Calculate the normal stress along x direction (σx) as shown below.

Substitute 2ksi for σavg, 11.66ksi for R, and 50.96° for θ in Equation (1).

σx=2+11.66cos50.96°=2+7.34=5.34ksi

Hence, the normal stress σx=5.34ksi_.

Calculate the normal stress along y direction (σy) as shown below.

σy=σavg+Rcosφ (2)

Substitute 2ksi for σavg, 11.66ksi for R, and 50.96° for θ in Equation (2).

σy=211.66cos50.96°=27.34=9.34ksi

Hence, the normal stress σy=9.34ksi_.

Calculate the shear stress (τxy) as shown below.

τxy=Rsinφ

Substitute 11.66ksi for R and 50.96° for θ.

τxy=11.66sin50.96°=9.06ksi

Therefore, the shear stress τxy=9.06ksi_.

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Chapter 7 Solutions

Mechanics of Materials, 7th Edition

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Understanding Stress Transformation and Mohr's Circle; Author: The Efficient Engineer;https://www.youtube.com/watch?v=_DH3546mSCM;License: Standard youtube license