fall 2019 problem 3.1 (10 points): a rigid bar is

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ME 323: Mechanics of Materials Homework Set 3

Fall 2019

Problem 3.1 (10 points): A rigid bar is supported by three elastic Aluminum alloy (Young’s

Modulus E) posts with identical cylindrical cross-section of area A. The center post is slightly

shorter than the two outer posts by an amount 𝛿, as shown in Fig. 3.1. A load P acts downward

directly over the center of the middle post.

A = 200 mm2; L = 2 m; E = 72 GPa.

a) What is the width 𝛿 of the gap closed by a load P = 24 kN?

b) If the gap was 𝛿 = 1 mm, what is the new length of each bar when a load P = 40 kN is

applied?

Fig. 3.1

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Solution

Part a)

Free Body Diagram (FBD)

Looking at the forces on the rigid bar in Fig. 1A

Fig. 1A

Equilibrium equations

Force balance along the y-axis is:

Σ𝐹𝑦 = 0 => −𝑃 + 2𝐹 = 0

=> 𝐹 =1

2𝑃

[1.1]

The gap 𝛿 must be such that the force applied 𝑃 = 24 𝑘𝑁 would just close it without applying

any force on the post (2).

Therefore, only the forces applied by posts (1) act upon the rigid bar as shown in the FBD in fig.

𝑒1 =𝐹𝐿1

𝐸𝐴=

𝑃𝐿1

2𝐸𝐴=

24 ∗ 103 ∗ 2

2 ∗ 72 ∗ 109 ∗ 200 ∗ 10−6

=> 𝑒1 = 1.67 𝑚𝑚

[1.2]

Part b)

There are many ways one can try to solve this part but the method which is highly encouraged to

be followed is using compatibility equations as depicted in Part b.2) below

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Part b.1)

First, it should be assessed if the rigid bar would deform post (2) when a load 𝑃 = 40 𝑘𝑁 is

applied while a gap 𝛿 = 1 𝑚𝑚 is present between the rigid bar and end of the post (2).

So, using the same FBD in Fig. 1A, the change in length can be calculated using the Eqn. 1.2:

𝑒1 =𝐹𝐿1

𝐸𝐴=

𝑃𝐿1

2𝐸𝐴=

40 ∗ 103 ∗ 2

2 ∗ 72 ∗ 109 ∗ 200 ∗ 10−6

=> 𝑒1 = 2.78 𝑚𝑚

[1.3]

The change in length is definitely higher than 1 mm, hence the rigid post not only makes contact

with the post (2) but also contracts it. Hence, the new FBD can be drawn after the rigid bar

makes contact with the post (2) as in Fig. 1B:

Fig. 1B

Equilibrium of the forces can be written as:

Σ𝐹𝑥 = 0 ⟹ 2𝐹1 + 𝐹2 = 𝑃 [1.4]

Since, the posts are all made of same materials 𝐹1 = 𝐹2 = 𝐹 =1

3𝑃

The change in length of the post (2) can be calculated as follows:

𝑒2 =𝐹𝐿2

𝐸𝐴=

𝑃𝐿2

3𝐸𝐴=

40 ∗ 103 ∗ (2 − 0.001)

3 ∗ 72 ∗ 109 ∗ 200 ∗ 10−6

=> 𝑒2 = 1.85 𝑚𝑚

[1.5]

Hence, the final lengths of all the three posts is calculated as:

𝐿𝑓𝑖𝑛𝑎𝑙 = 𝐿𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝛿 − 𝑒2 = 2 − 0.001 − 0.00185

=> 𝐿𝑓𝑖𝑛𝑎𝑙 = 1997.15 𝑚𝑚

[1.6]

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Part b.2) – The best way to solve this part




Fig. 1B

Equilibrium of the forces can be written as:

Σ𝐹𝑥 = 0 ⟹ 2𝐹1 + 𝐹2 = 𝑃

⟹ 𝐹2 = 𝑃 − 2𝐹1

[1.4]

The compatibility conditions can be obtained from the figure below:

Fig. 1C

𝑒2 + 𝛿 = 𝑒1 [1.3]

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−𝐹2𝐿2

𝐴𝐸+ 𝛿 = −

𝐹1𝐿1

𝐴𝐸

⟹(𝑃 − 2𝐹1)𝐿2

𝐴𝐸− 𝛿 =

𝐹1𝐿1

𝐴𝐸

⟹ 𝐹1 =𝑃𝐿2 − 𝛿𝐴𝐸

𝐿1 + 2𝐿2=

40 ∗ 103 ∗ 1.999 − 0.001 ∗ 200 ∗ 10−6 ∗ 72 ∗ 10−9

2 + 2 ∗ 1.999

⟹ 𝐹1 = 10.93 𝑘𝑁

[1.4]

𝑒1 =𝐹1𝐿1

𝐸𝐴=

10.93 ∗ 103 ∗ 2

200 ∗ 10−6 ∗ 72 ∗ 109= 1.52 𝑚𝑚

[1.5]

𝐿𝑓𝑖𝑛𝑎𝑙 = 𝐿1 − 𝑒1 = 2 − 0.00152 = 1998.47 𝑚𝑚 [1.6]

Part b.3)




Fig. 1B

The force required to close the gap 𝛿 as seen from the FBD in Fig. 1B can be calculated as:

𝐹1 =𝛿𝐸𝐴

𝐿1= 0.001 ∗ 72 ∗ 109 ∗ 200 ∗

10−6

2= 7.2 𝑘𝑁

[1.3]

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The force acting on the post (2) can be calculated as follows:

𝐹2 =𝑃 − 2𝐹1

3=

40 − 2 ∗ 7.2

3= 8.53 𝑘𝑁

[1.4]

𝑒2 =𝐹2𝐿2

𝐸𝐴=

8.53 ∗ 103 ∗ (2 − 0.001)

72 ∗ 109 ∗ 200 ∗ 10−6= 1.18 𝑚𝑚

[1.5]

𝐿𝑓𝑖𝑛𝑎𝑙 = 𝐿 − 𝛿 − 𝑒2 = 2000 − 1 − 1.18 = 1.9978 𝑚𝑚 [1.6]

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Problem 3.2 (10 points): A three-segment rod that is initially stress-free is attached to rigid

supports at ends A, B, and D. Elastic segments (1) and (2) are joined by the left hand side of rigid

connector C, whereas elastic segment (3) is joined by the right hand side. Segment (1) has solid

circular cross-section with diameter d. Segments (2) and (3) have circular tube cross-sections.

Segment (2) has an inner diameter 2d and an outer diameter 3d. Segment (3) has an inner diameter

d and an outer diameter 2d. Segment (1) has a length of 2L, whereas segments (2) and (3) have

lengths of L. Each segment has a Young’s modulus of E. An external load 2P is applied on the

connector C.

a) Determine the axial stresses induced in segments (1), (2), and (3).

b) Determine the displacement of the connector C.

Express your results in terms of P, d, L, E and π.

Fig. 3.2

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Solution


From the FBD of the connector C

Fig. 2A

Equilibrium equations

Force balance is shown below along x-axis:

Σ𝐹𝑥 = 0 => 𝐹3 − 𝐹1 − 𝐹2 − 2𝑃 = 0

[2.1]

There are 3 unknowns and only 1 equation, hence it’s an indeterminate problem.

Force-elongation equations for each of the three rods:

𝑒1 =𝐹1𝐿1

𝐸𝐴=

2𝐹1𝐿

𝐸 (𝜋𝑑2

4)

=8𝐹1𝐿

𝐸𝜋𝑑2

𝑒2 =𝐹2𝐿2

𝐸𝐴=

𝐹2𝐿

𝐸 (5𝜋𝑑2

4)

=4𝐹2𝐿

5𝐸𝜋𝑑2

𝑒3 =𝐹3𝐿3

𝐸𝐴=

𝐹3𝐿

𝐸 (3𝜋𝑑2

4)

=4𝐹3𝐿

3𝐸𝜋𝑑2

[2.2]

Compatibility Equations:

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𝑢𝐴 + 𝑒1 = 𝑢𝑐

𝑢𝐵 + 𝑒2 = 𝑢𝑐

𝑢𝑐 + 𝑒3 = 𝑢𝐷

[2.3]

where, 𝑢𝐴 = 𝑢𝐵 = 𝑢𝐷 = 0 as these walls are immovable. Therefore:

𝑒1 = 𝑒2

𝑒1 = −𝑒3

𝑒1 = 𝑒2 = −𝑒3

[2.4]

Using the Eqns in 2.4, Eqns 2.2 can be re-written as follows:

8𝐹1𝐿

𝐸𝜋𝑑2=

4𝐹2𝐿

5𝐸𝜋𝑑2 ⟹ 𝐹2 = 10𝐹1

8𝐹1𝐿

𝐸𝜋𝑑2=

−4𝐹3𝐿

3𝐸𝜋𝑑2 ⟹ 𝐹3 = −6𝐹1

[2.5]

Sub. 𝐹2 and 𝐹3 in Eqn. 2.1 to get:

𝐹1 = −2

17𝑃

𝐹2 = −20

17𝑃

𝐹3 = +12

17𝑃

[2.6]

Part b)

The displacement of the connector C would be equal to the elongation of one of the bars

(according to Eqn. 2.4). Therefore,

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𝑢𝑐 = 𝑒1

𝑢𝑐 =8𝐹1𝐿

𝐸𝜋𝑑2= −

16𝑃𝐿

17𝐸𝜋𝑑2

[2.7]

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Problem 3.3 (10 points): Three elastic members form the pin-jointed truss in Fig. 3.3. The joint

at A is constrained by a slider block to move only in the horizontal direction (i.e., vA = 0). All

members have the same cross-sectional area A = 525 mm2 and a modulus of elasticity E = 150

GPa. The pin-joints B, C and D are colinear.

a) Determine the member forces F1, F2 and F3 which result in moving the joint A to the right

by 13 mm (i.e., uA = 13 mm).

b) Determine the horizontal load P.

c) Determine the vertical reaction between the slider block at A and the track.

Fig. 3.3

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Solution


The FBD at point A is drawn below:

Fig. 3A

Force balance equations:

𝛴𝐹𝑥 = 0 => −3

5𝐹1 − 𝐹2 −

1

2𝐹3 + 𝑃 = 0

[3.1]

𝛴𝐹𝑦 = 0 => 𝐴𝑦 +4

5𝐹1 −

√3

2𝐹3 = 0

[3.2]

There are 5 unknowns and only 2 equations, hence this is an indeterminate problem.

Part a)

Force-elongation equations:

𝐹1 =𝐴𝐸

𝐿1𝑒1

𝐹2 =𝐴𝐸

𝐿2𝑒2

𝐹3 =𝐴𝐸

𝐿3𝑒3

[3.3]

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Elongations of the elastic members (1), (2) and (3) can be represented along with the motion of

pin joint A as:

Fig. 3B

From the Fig. 3B, there is no vertical displacement and hence the elongations can be represented

only in the horizontal displacement of the joint A as:

𝑒1 =3

5𝑢𝐴 ⟹ 𝐹1 =

3𝐴𝐸

5𝐿1𝑢𝐴

𝑒2 = 𝑢𝐴 ⟹ 𝐹2 =𝐴𝐸

𝐿2𝑢𝐴

𝑒3 =1

2𝑢𝐴 ⟹ 𝐹3 =

𝐴𝐸

2𝐿3𝑢𝐴

[3.4]

[OR] – It can be obtained using:

𝑒1 = 𝑢𝐴𝑐𝑜𝑠𝜃1 + 𝑣𝐴𝑠𝑖𝑛𝜃1; 𝜃1 = 306.87𝑜

𝑒2 = 𝑢𝐴𝑐𝑜𝑠𝜃2 + 𝑣𝐴𝑠𝑖𝑛𝜃2; 𝜃2 = 0𝑜

𝑒3 = 𝑢𝐴𝑐𝑜𝑠𝜃3 + 𝑣𝐴𝑠𝑖𝑛𝜃3; 𝜃3 = 60𝑜

[3.4]

𝐿1 = 5 𝑚; 𝐿2 = 3 𝑚; 𝐿3 = 6 𝑚; 𝐴 = 525 𝑚𝑚2; 𝐸 = 150 𝐺𝑃𝑎

Substituting the above values in Eqn. 3.4, gives:

𝐹1 = 122.85 𝑘𝑁 [3.5]

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𝐹2 = 341.25 𝑘𝑁

𝐹3 = 85.31 𝑘𝑁

Part b)

Substitute the above force values in Eqn. 3.1 to calculate the value of 𝑃

𝑃 =3

5𝐹1 + 𝐹2 +

1

2𝐹3 = 457.615 𝑘𝑁

[3.3]

Part c)

Substitute the above force values in Eqn. 3.2 to calculate 𝐴𝑦

𝐴𝑦 =√3

2𝐹3 −

4

5𝐹1 = −24.4 𝑘𝑁

[3.4]

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Problem 3.4 (10 points): The 50 ft long steel rails on a train track are laid with a small gap between

them to allow for thermal expansion.

(a) Determine the required gap so that the rails just touch one another when the temperature is

increased from T1 = -20 °F to T2 = 120 °F.

Hint: since all rails expand equally, each rail needs to effectively expand by 𝛿.

(b) Using the gap determined in (a), what would be the axial force in the rails if the temperature

were to rise to T3 = 140°F? The cross-sectional area of each rail is 5.10 in2.

Hint: since all rails expand equally, study a rail of length 50 ft at T1 between two rigid walls

located at 50 ft + 𝛿 from each other.

The coefficient of thermal expansion 𝛼 = 6.60×10-6 /°F, and Young’s modulus E = 29×103 ksi.

Fig. 3.4

Solution

Part a)

All the rails equally have a gap of 𝛿 and when the temperature increases, all of them expand

equally by 𝛿

2 on both the sides, so a total of 𝛿 increase can be considered for a single rail.

When is gap is just closed, there wouldn’t be any force acting on the rail. Therefore,

𝛿 = 𝑒 = 𝛼Δ𝑇𝐿 = 6.6 ∗ 10−6 ∗ (120 − (−20)) ∗ 50 = 0.554 𝑖𝑛 = 0.0462 𝑓𝑡

[4.1]

Part b)

After the gap is filled, any temperature increase would result in experience of stress in the rail as

there isn’t any more free space to expand into. Therefore, the expansion can be expressed as:

𝑒 =𝑃𝐿

𝐸𝐴+ 𝛼Δ𝑇𝐿

𝑒 = 0 here and hence the above equation can be re-written as:

𝑃 = −𝛼Δ𝑇𝐸𝐴 = 6.6 ∗ 10−6 ∗ (140 − 120) ∗ 29 ∗ 106 ∗ 5.1 = 19.522 𝑘𝑖𝑝𝑠

[4.2]

fall 2019 problem 3.1 (10 points): a rigid bar is

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