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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Question
Chapter 4.6, Problem 92P

(a)

To determine

Find the residual stress at y=2in.

(a)

Expert Solution
Check Mark

Answer to Problem 92P

The residual stress is σres=13.36ksi_.

Explanation of Solution

Given information:

The yield stress for the beam is σY=42ksi.

The Young’s modulus of steel is E=29×106psi.

Calculation:

Show the cross-section of the beam as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 4.6, Problem 92P , additional homework tip  1

Refer to Figure 1.

Calculate the area of the cross section (A) as shown below.

A=bd (1)

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the portion (1) (A1) as shown below.

Substitute 3in. for b and 1in. for d in Equation (1).

A1=3×1=3in.2

Calculate the area of the portion (2) (A2) as shown below.

Substitute 1in. for b and 1in. for d in Equation (1).

A2=1×1=1in.2

Calculate the moment of inertia (I) as shown below.

I=bd312+Ah2 (2)

Calculate the moment of inertia of portion (1) (I1) as shown below.

Substitute 3in. for b, 1in. for d, 3in.2 for A and 1.5in. for h in Equation (2).

I1=3×1312+3×1.52=7in.4

Hence, I3=7in.4

Calculate the moment of inertia of portion (2) (I2) as shown below.

Substitute 1in. for b, 2in. for d, 2in.2 for A and 0 for h in Equation (2).

I2=1×2312+0=0.667in.4

Calculate the total moment of inertia (I) as shown below.

I=I1+I2+I3

Substitute 7in.4 for I1, 0.667in.4 for I2, and 7in.4 for I3.

I=7+0.667+7=14.667in.4

Calculate the centroid (c) as shown below.

c=h2

Substitute 4in. for h.

c=42=2in.

Sketch the stress acting on the cross-section of the beam as shown in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 4.6, Problem 92P , additional homework tip  2

Refer Figure 2.

Calculate the area of the portion (2) (A2) as shown below.

A2=12bh

Substitute 1in. for b and 1in. for d.

A2=12×1×1=0.5in.2

Calculate the reaction applied to portion (1) (R1) as shown below.

R1=σYA1

Substitute 42ksi for σY and 3in.2 for A1.

R1=42×3=126kips

Calculate the reaction applied to portion (2) (R2) as shown below.

R2=σYA2

Substitute 42ksi for σY and 0.5in.2 for A2.

R2=42×0.5=21kips

Calculate the moment (M) as shown below.

M=2(R1y1+R2y2)

Substitute 126kips for R1, (1+12)in. for y1, 21kips for R2, and (23×1)in. for y2.

M=2(126×(1+12)+21×(23×1))=406kipin.

Calculate the stress (σ) as shown below.

σ=McI

Substitute 406kipin. for M, 2in. for c, and 14.667in.4 for I.

σ=406×214.667=55.362ksi

Calculate the stress (σ) as shown below.

σ=MyI

Substitute 406kipin. for M, 1in. for y, and 14.667in.4 for I.

σ=406×114.667=27.681ksi

Calculate the residual stress at y=c as shown below.

σres=σσY

Substitute 55.362ksi for σ and 42ksi for σY.

σres=55.36242=13.362ksi

Calculate the residual stress at y=yY as shown below.

σres=σσY

Substitute 27.681ksi for σ and 42ksi for σY.

σres=27.68142=14.319ksi

Sketch the stress distribution as shown in Figure 3.

Mechanics of Materials, 7th Edition, Chapter 4.6, Problem 92P , additional homework tip  3

Hence, the residual stress is σres=13.36ksi_.

(b)

To determine

Find the point where the residual stress is zero.

(b)

Expert Solution
Check Mark

Answer to Problem 92P

The point where the residual stress is zero is y0=1.517in., 0, 1.517in_.

Explanation of Solution

Given information:

The yield stress for the beam is σY=42ksi.

The Young’s modulus for steel is E=29×106psi.

Calculation:

Consider that the residual stress is σres=0.

Calculate the yield stress (σ) as shown below.

σ=My0I

Calculate the point where the residual stress is zero as shown below.

σres=σσY

Substitute My0I for σ and 0 for σres.

0=My0IσYσY=My0Iy0=σYIM

Substitute 42ksi for σY, 406kipin. for M, and 14.667in.4 for I.

y0=42×14.667406=1.517in.

Therefore, the point where the residual stress is zero is y0=1.517in., 0, 1.517in_.

(c)

To determine

Find the radius of curvature corresponding to the permanent deformation of the bar.

(c)

Expert Solution
Check Mark

Answer to Problem 92P

The radius of curvature is ρ=168.8ft_.

Explanation of Solution

Given information:

The yield stress for the beam is σY=42ksi.

The Young’s modulus of steel is E=29×106psi.

Calculation:

Refer to part (a).

The residual stress σres=14.32ksi.

Calculate the radius of curvature (ρ) as shown below.

σ=My0I

Calculate the point where the residual stress is zero as shown below.

ρ=Eyσ

Substitute 29×106psi for E, 14.32ksi for σ and 1in. for y.

ρ=29×106psi×1ksi1,000psi×1in.14.32ksi=2,025.14in.×1ft12in.=168.8ft

Therefore, the radius of curvature is ρ=168.8ft_.

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Chapter 4 Solutions

Mechanics of Materials, 7th Edition

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