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Mechanical Properties OF Metals Problem

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fliud mechanics (2222222)

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Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the

CHAPTER 6

MECHANICAL PROPERTIES OF METALS

PROBLEM SOLUTIONS

Concepts of Stress and Strain

6 Using mechanics of materials principles (i., equations of mechanical equilibrium applied to a free-body diagram), derive Equations 6 and 6.

Solution

This problem asks that we derive Equations 6 and 6, using mechanics of materials principles. In Figure (a) below is shown a block element of material of cross-sectional area A that is

subjected to a tensile force P. Also represented is a plane that is oriented at an angle θ referenced to
the plane perpendicular to the tensile axis; the area of this plane is A' = A/cos θ. In addition, and the

forces normal and parallel to this plane are labeled as P' and V', respectively. Furthermore, on the left- hand side of this block element are shown force components that are tangential and perpendicular to

the inclined plane. In Figure (b) are shown the orientations of the applied stress σ, the normal stress to
this plane σ', as well as the shear stress τ' taken parallel to this inclined plane. In addition, two

coordinate axis systems in represented in Figure (c): the primed x and y axes are referenced to the inclined plane, whereas the unprimed x axis is taken parallel to the applied stress.

Normal and shear stresses are defined by Equations 6 and 6, respectively. However, we now chose to express these stresses in terms (i., general terms) of normal and shear forces (P and V) as

P

σ = A

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τ = V A

For static equilibrium in the x' direction the following condition must be met:

∑ Fx' = 0

which means that

P' − P cos θ= 0

Or that

P' = P cos θ

Now it is possible to write an expression for the stress σ' in terms of P' and A' using the above expression and the relationship between A and A' [Figure (a)]:

' '

P
σ ′= A

= cos = cos 2 cos

P P
A A
θ θ
θ
However, it is the case that P/A = σ; and, after making this substitution into the above expression, we

have Equation 6–that is

σ ′=σ cos 2 θ

Now, for static equilibrium in the y' direction, it is necessary that

∑Fy' = 0

= −V' + Psin θ

Or

V' = P sinθ

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6 (a) Equations 6 and 6 are expressions for normal ( σ′) and shear ( τ′) stresses,
respectively, as a function of the applied tensile stress ( σ) and the inclination angle of the
plane on which these stresses are taken ( θ of Figure 6). Make a plot on which is presented
the orientation parameters of these expressions (i., cos 2 θ and sin θ cos θ) versus θ.

(b) From this plot, at what angle of inclination is the normal stress a maximum? (c) Also, at what inclination angle is the shear stress a maximum?

Solution

(a) Below are plotted curves of cos 2 θ (for σ′) and sin θ cos θ (for τ') versus θ.

(b) The maximum normal stress occurs at an inclination angle of 0°. (c) The maximum shear stress occurs at an inclination angle of 45°.

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6 A cylindrical specimen of a titanium alloy having an elastic modulus of 108 GPa and an original diameter of 3 mm will experience only elastic deformation when a tensile load of 2000 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0 mm.

Solution We are asked to compute the maximum length of a cylindrical titanium alloy specimen

(before deformation) that is deformed elastically in tension. For a cylindrical specimen

2 0 0 2

A = π ⎛ d⎞
⎜⎝ ⎟⎠

where d 0 is the original diameter. Combining Equations 6, 6, and 6 and solving for l 0 leads to

0 2 02 0 0

2
4

l E d l l l l E l E d F F F E A

π π
∈ σ
∆ ⎛ ⎞
∆ ∆ ∆ ⎜⎝ ⎟⎠ ∆
= = = = =

0 10 3 m 108 10 N / m 9 2 ( ) 3 10 3 m 2 (4)(2000 N) 0 m 257 mm

= ( × − ) ( × ) π ( × − )

= =

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6 Consider a cylindrical specimen of a steel alloy (Figure 6) 15 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied.

Solution

This problem asks that we calculate the elongation ∆l of a specimen of steel the stress-strain behavior of which is shown in Figure 6. First it becomes necessary to compute the stress when a load of 20,000 N is applied using Equation 6 as

0 0 2 3 2

20000 N 113 MPa 15 10 2

F F
σ A π d
π
= = ⎛ ⎞ = ⎛ × −⎞ =
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠

Referring to Figure 6, at this stress level we are in the elastic region on the stress-strain curve, which corresponds to a strain of 0. Now, utilization of Equation 6 to compute the value of ∆l

∆ l = ∈l 0 = (0)(75 mm) =0 mm

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6 Figure 6 shows, for a gray cast iron, the tensile engineering stress–strain curve in the elastic region. Determine (a) the tangent modulus at 10 MPa, and (b) the secant modulus taken to 6 MPa.

Solution

(a) This portion of the problem asks that the tangent modulus be determined for the gray cast iron, the

stress-strain behavior of which is shown in Figure 6. In the figure below is shown a tangent drawn

on the curve at a stress of 10 MPa.

The slope of this line (i., ∆ σ/∆ ∈), the tangent modulus, is computed as follows:

15 MPa 5 MPa = 1410 MPa = 1 GPa 0 0.

σ
∈
∆ = −
∆ −

(b) The secant modulus taken from the origin is calculated by taking the slope of a secant drawn from the origin through the stress-strain curve at 6 MPa. This secant is drawn on the curve shown below:

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6 In Section 2 it was noted that the net bonding energy EN between two isolated positive and negative ions is a function of interionic distance r as follows:

EN = − A r + rB n (6)

where A, B, and n are constants for the particular ion pair. Equation 6 is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of the interionic force–separation curve at the equilibrium interionic separation; that is,

E ∝ ⎛ ⎝⎜ dFdr⎞ ⎠⎟ ro

Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the two-ion system) using the following procedure:

  1. Establish a relationship for the force F as a function of r, realizing that

F = dE drN 2. Now take the derivative dF/dr. 3. Develop an expression for r 0 , the equilibrium separation. Since r 0 corresponds to the value of r at the minimum of the EN-versus-r curve (Figure 2), take the derivative dEN/dr, set it equal to zero, and solve for r, which corresponds to r 0. 4. Finally, substitute this expression for r 0 into the relationship obtained by taking dF/dr.

Solution This problem asks that we derive an expression for the dependence of the modulus of

elasticity, E, on the parameters A, B, and n in Equation 6. It is first necessary to take dEN/dr in

order to obtain an expression for the force F; this is accomplished as follows:

= = +

N n

d A d B F dE r r dr dr dr

⎛⎜ − ⎞⎟ ⎛⎜ ⎞⎟
⎝ ⎠ ⎝ ⎠
= A

r 2

− nB r (n +1)

The second step is to set this dEN/dr expression equal to zero and then solve for r (= r 0 ). The algebra for this procedure is carried out in Problem 2, with the result that

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1/(1 ) 0 =

A n r nB

⎛ ⎞ −
⎜⎝ ⎟⎠

Next it becomes necessary to take the derivative of the force (dF/dr), which is accomplished as follows:

2 ( n1) d A d nB dF r r dr dr dr

⎛⎜⎝ ⎞⎟⎠ ⎛⎜⎝ − + ⎞⎟⎠
= +

3 ( 2)

2 ( )( 1)

n

A n n B r r +

= − + +

Now, substitution of the above expression for r 0 into this equation yields

3 /(1 ) ( 2) /(1 ) 0

2 ( )( 1)

r n n n

dF A n n B dr A A nB nB

− + −

⎛ ⎞ = − + +
⎜⎝ ⎟⎠ ⎛ ⎞ ⎛ ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠

which is the expression to which the modulus of elasticity is proportional.

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6 A cylindrical rod 100 mm long and having a diameter of 10 mm is to be deformed using a tensile load of 27,500 N. It must not experience either plastic deformation or a diameter reduction of more than 7 × 10 − 3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).

Material

Modulus of Elasticity (GPa)

Yield Strength (MPa) Poisson’s Ratio Aluminum alloy 70 200 0. Brass alloy 101 300 0. Steel alloy 207 400 0. Titanium alloy 107 650 0.

Solution

This problem asks that we assess the four alloys relative to the two criteria presented. The

first criterion is that the material not experience plastic deformation when the tensile load of 27,500 N

is applied; this means that the stress corresponding to this load not exceed the yield strength of the

material. Upon computing the stress

2 2 6 2 0 0 3

27,500 N 350 10 N / m 350 MPa 10 10 m 2

F F
σ A π d
π
= = ⎛ ⎞ = ⎛ × − ⎞ = × =
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa. Relative to the second criterion (i., that ∆d be less than 7 × 10 − 3 mm), it is necessary to calculate the change in diameter ∆d for these three alloys. From Equation 6.

x 0 z 0

d d E d d E

ν ∈
∈ σ σ
∆
= − = − = − ∆

Now, solving for ∆d from this expression,

d d 0 E

∆ = − ν σ

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For the steel alloy

= (0)(350 MPa)(10 mm) 3 = 5 10 3 mm d 207 10 MPa ∆ − − × − ×

Therefore, the steel is a candidate. For the Ti alloy

= (0)(350 MPa)(10 mm) 3 = 11 10 3 mm d 107 10 MPa ∆ − − × − ×

Hence, the titanium alloy is not a candidate.

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6 A cylindrical specimen of aluminum having a diameter of 12 mm and a gauge length of 50 mm is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a) through (f).

Load Length N mm 0 50. 7,330 50. 15,100 50. 23,100 50. 30,400 51. 34,400 51. 38,400 51. 41,300 51. 44,800 52. 46,200 53. 47,300 54. 47,500 55. 46,100 56. 44,800 57. 42,600 58. 36,400 59. Fracture

(a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0. (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience.

Solution

This problem calls for us to make a stress–strain plot for aluminum, given its tensile load– length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stress–strain curve, while for the second, the curve extends to just beyond the elastic region of deformation.

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(b) The elastic modulus is the slope in the linear elastic region (Equation 6) as

200 MPa 0 MPa 62 103 MPa 62 GPa E 0 0

σ
∈
= ∆ = − = × =
∆ −

(c) For the yield strength, the 0 strain offset line is drawn dashed. It intersects the stress–strain curve at approximately 285 MPa. (d) The tensile strength is approximately 370 MPa, corresponding to the maximum stress on the complete stress-strain plot.

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True Stress and Strain

6 Show that Equations 6 and 6 are valid when there is no volume change during deformation.

Solution

To show that Equation 6 is valid, we must first rearrange Equation 6 as

i 0 i

A A l = l

Substituting this expression into Equation 6 yields

0 0 0 T i i i

F F l l

σ A A l σ l
= = ⎛ ⎞ = ⎛ ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠

But, from Equation 6.

0

∈ = lli − 1

Or

0

lli = ∈+ 1

Thus,

0

σ T = σ ⎛⎜ lli⎞⎟= σ ( ∈+ 1)
⎝ ⎠

For Equation 6

∈ T = ln (1 + ∈)

is valid since, from Equation 6.

T = ln 0 i l

∈ l
⎛ ⎞
⎜⎝ ⎟⎠

and

0

lli = ∈ + 1

from above.

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6 Demonstrate that Equation 6, the expression defining true strain, may also be represented by

T ln 0 i

A
∈ A
= ⎛ ⎞
⎜ ⎟
⎝ ⎠

when specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why?

Solution This problem asks us to demonstrate that true strain may also be represented by

T ln 0 i

A
∈ A
= ⎛ ⎞
⎜ ⎟
⎝ ⎠

Rearrangement of Equation 6 leads to

li l 0 =

A 0

Ai

Thus, Equation 6 takes the form

T 0 0

ln i ln i

l A

∈ l A
= ⎛ ⎞ = ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

The expression T ln 0 i

A
∈ A
= ⎛ ⎞
⎜ ⎟
⎝ ⎠

is more valid during necking because Ai is taken as the area of

the neck.

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Mechanical Properties OF Metals Problem

Course: fliud mechanics (2222222)

16 Documents
Students shared 16 documents in this course

University: جامعة كفر الشيخ

Was this document helpful?
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
CHAPTER 6
MECHANICAL PROPERTIES OF METALS
PROBLEM SOLUTIONS
Concepts of Stress and Strain
6.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to
a free-body diagram), derive Equations 6.4a and 6.4b.
Solution
This problem asks that we derive Equations 6.4a and 6.4b, using mechanics of materials
principles. In Figure (a) below is shown a block element of material of cross-sectional area A that is
subjected to a tensile force P. Also represented is a plane that is oriented at an angle
θ
referenced to
the plane perpendicular to the tensile axis; the area of this plane is A' = A/cos
θ
. In addition, and the
forces normal and parallel to this plane are labeled as P' and V', respectively. Furthermore, on the left-
hand side of this block element are shown force components that are tangential and perpendicular to
the inclined plane. In Figure (b) are shown the orientations of the applied stress
σ
, the normal stress to
this plane
σ
', as well as the shear stress
τ
' taken parallel to this inclined plane. In addition, two
coordinate axis systems in represented in Figure (c): the primed x and y axes are referenced to the
inclined plane, whereas the unprimed x axis is taken parallel to the applied stress.
Normal and shear stresses are defined by Equations 6.1 and 6.3, respectively. However, we
now chose to express these stresses in terms (i.e., general terms) of normal and shear forces (P and V)
as
P
A
σ
=

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