ME 323: Mechanics of Materials Homework Set 8 solutions

Fall 2019 Due: Wednesday, October 23

Problem 8.1 (10 points)

A steel (𝐸 = 30,000𝑘𝑠𝑖) square beam with a side length𝑎 = 2" is subject to loading as shown in Fig. 7.4. Determine the deflection curve of the beam using superposition. Use the table of beam deflections from the R. R. Craig textbook distributed via email. Compare this with the result found using second order integration in Homework 7.

Fig. 8.1

Solution:

2’

6’

2’

Po = 100 lb/ft

x D

C B

A

(1)

(2)

(3)

=

_

+

𝑙0 = 2𝑓𝑡

4𝑙0 = 8ft

5𝑙0 = 10ft

The following parameters are given

𝑃0 = 100lb/ft

𝐸 = 3 × 10;ksi

𝐼 =112(2)@(2)in; = 1.33in; = 6.4 ∗ 10EFft;

Using the tables and solving for deflections for each of the sections we get the following:

(1)

The deflection for section AB is as follows 𝑣H(𝑥) =

JKLM

N;OP(6𝑙0N − 4𝑙0𝑥 + 𝑥N) , 0 ≤ 𝑥 ≤ 𝑙0

The deflection for the section CD is as follows

𝑣H(𝑥) =(JK)TUV

N;OP(4𝑥 − 𝑙0),𝑙0 ≤ 𝑥 ≤ 5𝑙0

(2)

The deflection for section AC is as follows

𝑣N(𝑥) = −JKLM

N;OP(6(4𝑙0)N − 4(4𝑙0)𝑥 + 𝑥N)

= − JKLM

N;OP(96𝑙0N − 16𝑙0𝑥 + 𝑥N) 0 ≤ 𝑥 ≤ 4𝑙0

The deflection for section CD is as follows

𝑣N(𝑥) = −JK(;TU)V

N;OP(4𝑥 − 4𝑙0), 4𝑙0 ≤ 𝑥 ≤ 5𝑙0

(3)

The deflection for the entire beam AD is given by

𝑣@(𝑥) =XYLM

ZOP(15𝑙0 − 𝑥), 0 ≤ 𝑥 ≤ 5𝑙0

Thus, by principle of superposition we can say that the deflections will be the sum of the deflections from the sections (1), (2) and (3) for the respective ranges of lengths:

JKLM

N;OP(−90𝑙0N + 12𝑙0𝑥) +

XYLM

ZOP(15𝑙0 − 𝑥),0 ≤ 𝑥 ≤ 𝑙0

𝜈(𝑥) = JKN;OP

(4𝑙0@𝑥 − 𝑙0; − 96𝑙0N𝑥N + 16𝑙0𝑥@ − 𝑥;) + XYLM

ZOP(15𝑙0 − 𝑥), 𝑙0 ≤ 𝑥 ≤ 4𝑙0

JKTKV

N;OP(−252𝑥 + 255𝑙0) +

XYLM

ZOP(15𝑙0 − 𝑥), 4𝑙0 ≤ 𝑥 ≤ 5𝑙0

Using the boundary condition that, the deflection at the roller D is 0,𝜈(5𝑙0) = 0 we solve for D]

𝑃0𝑙0@

24𝐸𝐼(−252(5𝑙0) + 255𝑙0) +

25𝐷_𝑙0N(15𝑙0 − 5𝑙0)6𝐸𝐼

= 0

𝑃0𝑙`@

24(−1005𝑙0) +

250𝐷_𝑙0@

6= 0

−1005𝑃 𝑙0; + 1000𝐷_𝑙0@ = 0

𝑫𝒚 = 𝟏.𝟎𝟎𝟓𝑷𝒐𝒍𝒐 = 𝟐𝟎𝟏𝒍𝒃

The final deflection can be described by the following equations:

JKLM

N;OP(−90𝑙0N + 12𝑙0𝑥) +

XYLM

ZOP(15𝑙0 − 𝑥),0 ≤ 𝑥 ≤ 𝑙0

𝜈(𝑥) = JKN;OP

(4𝑙0@𝑥 − 𝑙0; − 96𝑙0N𝑥N + 16𝑙0𝑥@ − 𝑥;) + XYLM

ZOP(15𝑙0 − 𝑥),𝑙0 ≤ 𝑥 ≤ 4𝑙0

JKTKV

N;OP(−252𝑥 + 255𝑙0) +

XYLM

ZOP(15𝑙0 − 𝑥), 4𝑙0 ≤ 𝑥 ≤ 5𝑙0

4.17𝑥N(24𝑥 − 360) + 33.5𝑥N(30 − 𝑥), 0 ≤ 𝑥 ≤ 2ft

EI𝜈(𝑥) = 4.17(−16 + 32𝑥 − 384𝑥N + 32𝑥@ − 𝑥;) + 33.5𝑥N(30 − 𝑥), 2ft ≤ 𝑥 ≤8ft

33.34(−252𝑥 + 510) + 33.5𝑥N(30 − 𝑥)  , 8ft ≤ 𝑥 ≤ 10ft

−495.12𝑥N + 66.52𝑥@,0 ≤ 𝑥 ≤ 2ft

EI𝜈(𝑥) = −4.17𝑥; + 99.84𝑥@ − 595.13𝑥N + 133.34𝑥 − 66.67,2ft ≤ 𝑥 ≤ 8ft

−8401.68𝑥 + 17003.4 + 1005𝑥N − 33.5𝑥@   ,8ft ≤ 𝑥 ≤ 10ft

EI𝜈H(𝑥) = −495. 12𝑥N + 66.52𝑥@ , 0 ≤ 𝑥 ≤ 2ft

EI𝜈N(𝑥) = −4.17𝑥; + 99.84𝑥@ − 595.13𝑥N + 133.34𝑥 − 66.67 ,2ft ≤ 𝑥 ≤ 8ft

EI𝜈@(𝑥) = −8401.68𝑥 + 17003.4 + 1005𝑥N − 33.5𝑥@ ,8ft ≤ 𝑥 ≤ 10ft

From HW problem 7.4 we have the following deflection equations:

EI𝜈H(𝑥) =𝑀o𝑥N

2+𝑉o𝑥@

6 ,0 ≤ 𝑥 ≤ 2ft

EI𝑣N(𝑥) =qrLM

N+ srLV

Z− NF

Z(𝑥 − 2); + 𝐶@𝑥 + 𝐶; ,2ft ≤ 𝑥 ≤ 8ft

EI𝑣@(𝑥) =qrLM

N+ srLV

Z− 100𝑥@ + 1500𝑥N + 𝐶F𝑥 + 𝐶Z ,8ft ≤ 𝑥 ≤ 10ft

Where,

𝑀o = −990lb. ft

𝑉o = 399lb

𝐶@ = 0

𝐶; = 0

𝐶F = −8400

𝐶Z = 17000

𝑉X = 201lb

The reaction force at D, 𝑉Xis exactly equal to the reaction force calculated from the superposition method, 𝑫𝒚 = 𝑽𝑫 = 𝟐𝟎𝟏𝐥𝐛

The final deflection of the beam from HW 7.4 is given by:

EI𝜈H(𝑥) = −495𝑥N + 66.5𝑥@ , 0 ≤ 𝑥 ≤ 2ft

EI𝜈N(𝑥) = −4.17𝑥; + 99.84𝑥@ − 595𝑥N + 133.34𝑥 − 66.67,2ft ≤ 𝑥 ≤ 8ft

EI𝜈@(𝑥) = −8400𝑥 + 17000 + 1005𝑥N − 33.5𝑥@ , 8ft ≤ 𝑥 ≤ 10ft

We can see from the numerical values that the deflection of the three sections of the beams calculated using method of superposition match very closely with the deflections calculated from the method of integration.

Further, if we compare and plot we get the following results.

Problem 8.2 (10 points)

The solid circular rod shown below has an axial force P0 acting at H and an axial force F0 acting at C. All components of the rod are made of steel (E = 200 GPa). Using P0 = 2 kN, F0 = 5 kN, d = 10 mm, and L = 200 mm, determine:

1) The strain energy stored in the rod. 2) The magnitude and direction of the horizontal displacement at H using Castigliano’s

theorem. 3) The magnitude and direction of the horizontal displacement at C using Castigliano’s

theorem.

Fig. 8.2

Solution:

Given:

𝐸 = 200,000MPa

𝑃0 = 2000N, 𝐹0 = 5000N

𝑑 = 10 = 1 × 10ENm

𝐿 = 200mm = 0.2m

𝐴H =𝜋𝑑N

4= 7.85 × 10EFmN

𝐴N = 𝜋𝑑N = 3.14 × 10E;mN

𝐴@ =9𝜋𝑑N

4= 7.065 × 10E;mN

Equilibrium in the shafts gives us following equations:

𝐹H = 𝑃0

𝐹H = 𝐹N = 𝑃0

𝐹@ = 𝑃0 − 𝐹0

To find the strain energy we use the following formulation:

𝑈 =12�

𝐹N

𝐸𝐴𝑑𝑥

�

`

Strain energy in shaft 1:

𝑈H = �𝑃0N

2𝐸𝐴H𝑑𝑥

;�

N�=𝑃0N2𝐿2𝐸𝐴H

= 50.9N.mm

Strain energy in shaft 2:

𝑈N = �𝑃0N

2𝐸𝐴N𝑑𝑥

N�

�=

𝑃0N𝐿2𝐸𝐴N

= 6.4N.mm

Strain energy in shaft 3:

𝑈@ = �(𝑃0 − 𝐹0)N

2𝐸𝐴@𝑑𝑥

�

`=(𝑃0 − 𝐹 )N𝐿2𝐸𝐴@

= 6.4N.mm

Total strain energy can then be calculated to be,

𝑈�0��T = 𝑈H + 𝑈N + 𝑈@ = 63.7N.mm = 0.0637J

To find the displacement at point H we take the partial derivative with respect to the applied force ( 𝑃0 ) at that point and get:

𝑢� =𝛿𝑈�0��T𝛿𝑃0

𝑢� =𝛿𝛿𝑃0

�𝑃0N2𝐿2𝐸𝐴H

+𝑃0N𝐿2𝐸𝐴N

+(𝑃0 − 𝐹 )N𝐿2𝐸𝐴@

�

𝑢� =2𝑃0𝐿𝐸𝐴H

+𝑃0𝐿𝐸𝐴N

+(𝑃0 − 𝐹0)𝐿𝐸𝐴@

𝑢� = 0.053𝑚𝑚 to the left

To find the displacement at point C we take the partial derivative with respect to the applied force at that point ( 𝐹0) and get:

𝑢� =���K�����K

𝑢� =𝛿𝛿𝐹0

�(𝑃0 − 𝐹0)N𝐿2𝐸𝐴@

�

𝑢� = −(𝑃0 − 𝐹0)𝐿𝐸𝐴@

𝑢� = 0.00424𝑚𝑚 to the right

Problem 8.3 (10 points)

The beam shown below has elastic modulus E, second area moment I, and cross-sectional area A. It is loaded by a point force P at H. Neglect the shear strain energy due to bending. Using Castigliano’s theorem, determine:

1) The magnitude and direction of the vertical displacement at D. 2) The magnitude and direction of the in-plane rotation angle at D. 3) The magnitude and direction of the horizontal displacement at H.

Express all answers in terms of P, E, I, A, and a.

Fig. 8.3

Solution:

In order to find the displacement at D, we apply a dummy moment 𝑀�and a dummy force 𝑃� at point D

∑𝑀� = 𝐶_𝑎 − 𝑃�(2𝑎) − 𝑃𝑎 + 𝑀�

𝐶_ = 2𝑃� + 𝑃 −1𝑎𝑀�

Making a cut through section DH,

∑𝑀��� = 0

𝑃𝑦 +𝑀H = 0

𝑴𝟏(𝒚) = 𝑷𝒚

Making a cut through section CD,

𝑭𝟐 = −𝑷

∑𝑀��� = 0

𝑴𝟐(𝒙𝟐) = −𝑷𝒅𝒙𝟐 +𝑴𝒅 − 𝑷𝒂

Making a cut through section BC,

∑𝑀��� = 0

−𝑀@(𝑥@) + 𝐶_𝑥@ − 𝑃�(𝑥@ + 𝑎) + 𝑀� − 𝑃𝑎 = 0

𝑀@(𝑥@) = ¡2𝑃� −1𝑎𝑀J¢ + 𝑃£𝑥@ − 𝑃�𝑥@ − 𝑃�𝑎 + 𝑀� − 𝑃𝑎

𝑴𝟑(𝒙𝟑) = 𝑷𝒅(𝒙𝟑 − 𝒂) +𝑴𝒅 ¥𝟏 −𝒙𝟑𝒂¦ + 𝑷(𝒙𝟑 − 𝒂)

𝑭𝟑 = −𝑷

𝑈 = �𝑀HN

2𝐸𝐼𝑑𝑦

�

`+ �

𝑀NN

2𝐸𝐼𝑑𝑥N

�

`+�

𝐹NN

2𝐸𝐴𝑑𝑥N

�

`+ �

𝑀@N

2𝐸𝐼𝑑𝑥@

�

`+�

𝐹@N

2𝐸𝐴𝑑𝑥@

�

`

𝛿𝑀H

𝛿𝑃�= 0,

𝛿𝑀N

𝛿𝑃�= 0,

𝛿𝐹N𝛿𝑃�

= 0,𝛿𝑀@

𝛿𝑃�= 𝑥@ − 𝑎,

𝛿𝐹@𝛿𝑃�

= 0

𝛿𝑀H

𝛿𝑀�= 0,

𝛿𝑀N

𝛿𝑀�= 1,

𝛿𝐹N𝛿𝑀�

= 0,𝛿𝑀@

𝛿𝑀�= 1 −

𝑥@𝑎,

𝛿𝐹@𝛿𝑀�

= 0

𝛿𝑀H

𝛿𝑃= 𝑦,

𝛿𝑀N

𝛿𝑃= −𝑎,

𝛿𝐹N𝛿𝑃

= −1,𝛿𝑀@

𝛿𝑃= 𝑥@ − 𝑎,

𝛿𝐹@𝛿𝑀�

= −1

𝛿𝑈𝛿𝑃�

= 𝑢X = 0 +�𝑀N

𝐸𝐼(−𝑥N)𝑑𝑥N

�

`+ 0 +�

𝑀@

𝐸𝐼(𝑥@ − 𝑎)𝑑𝑥@

�

`+ 0

𝑢X =1𝐸𝐼� 𝑃𝑎𝑥N𝑑𝑥N�

`+1𝐸𝐼� 𝑃(𝑥@ − 𝑎)N𝑑𝑥@�

`

𝑢X =𝑃𝑎@

2𝐸𝐼+𝑃𝐸𝐼� 𝑥@N − 2𝑥@𝑎 + 𝑎N𝑑𝑥@�

`

𝑢X =𝑃𝑎@

2𝐸𝐼+𝑃𝐸𝐼¡13𝑎@ − 𝑎@ + 𝑎@£

𝑢X =5𝑃𝑎@

6𝐸𝐼(𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)

𝛿𝑈𝛿𝑀�

= 𝜃X = 0 +�𝑀N

𝐸𝐼(1)𝑑𝑥N

�

`+ 0 + �

𝑀@

𝐸𝐼¥1 −

𝑥@𝑎¦ 𝑑𝑥@

�

`+ 0

𝜃X =1𝐸𝐼� −𝑃𝑎𝑑𝑥N�

`+1𝐸𝐼� 𝑃(𝑥@ − 𝑎) ¥1 −

𝑥@𝑎¦ 𝑑𝑥@

�

`

𝜃X =−𝑃𝑎N

𝐸𝐼+𝑃𝐸𝐼� 𝑥@ −

1𝑎𝑥@N − 𝑎 + 𝑥@𝑑𝑥@

�

`

𝜃X =−𝑃𝑎N

𝐸𝐼+𝑃𝐸𝐼¡𝑎N −

13𝑎𝑎@ − 𝑎N£

𝜃X = −4𝑃𝑎N

3𝐸𝐼(𝐶𝑊, 𝑠𝑖𝑛𝑐𝑒𝑀�𝑖𝑠𝐶𝐶𝑊𝑎𝑛𝑑𝜃X < 0)

𝛿𝑈𝛿𝑃

= 𝑢� =1𝐸𝐼� 𝑃𝑦N𝑑𝑦�

`+1𝐸𝐼� 𝑃𝑎N𝑑𝑥N�

`+1𝐸𝐴

� 𝑃𝑑𝑥N�

`+1𝐸𝐼� 𝑃(𝑥@ − 𝑎)N𝑑𝑥@�

`+1𝐸𝐴

� 𝑃𝑑𝑥@�

`

𝑢� =𝑃𝑎@

3𝐸𝐼+𝑃𝑎@

𝐸𝐼+𝑃𝑎𝐸𝐴

+𝑃𝐸𝐼� 𝑥@N − 2𝑥@𝑎 + 𝑎N𝑑𝑥@�

`+𝑃𝑎𝐸𝐴

𝑢� =𝑃𝑎@

3𝐸𝐼+𝑃𝑎@

𝐸𝐼+𝑃𝑎𝐸𝐴

+𝑃𝑎@

3𝐸𝐼+𝑃𝑎𝐸𝐴

𝑢� =5𝑃𝑎@

3𝐸𝐼+2𝑃𝑎𝐸𝐴

(𝑡𝑜𝑡ℎ𝑒𝑙𝑒𝑓𝑡)

Problem 8.4 (10 points)

Beam BCD is loaded with a distributed load w0 between B and C and a point force P at D. The beam has elastic modulus E and has a square cross section with side length b = L / 10. Neglect the shear strain energy due to bending.

Using Castigliano’s theorem, determine the reactions at the wall at B and the roller at C. Express your answer in terms of w0, P, L, and E.

Fig. 8.4

Solution:

Equilibrium:

∑𝑀o = 𝑀o −12𝑤0𝐿(𝐿) + 𝐵_𝐿 −

32𝑃𝐿

𝑀o −12𝑤0𝐿N + 𝐵_𝐿 −

32𝑃𝐿 = 0

∑𝐹_ = 𝐴_ + 𝐵_ − 𝑃 − 𝑤0𝐿 = 0

Making a cut at section BC we get:

∑𝑀��� = 0

−𝑀H − 𝑃𝑥H = 0

𝑀H = −𝑃𝑥H,𝛿𝑀H

𝛿�Y= 0

Making a cut at section AB we get:

∑𝑀��� = 0

−𝑀N −HN𝑤0𝑥NN + 𝐵_𝑥N − 𝑃 ¥𝑥N +

�N¦ = 0

𝑀N(𝑥) =HN𝑤0𝑥NN + 𝐵_𝑥N − 𝑃 ¥𝑥N +

�N¦,

𝛿𝑀N

𝛿𝐵_= 𝑥N

Total strain energy = 𝑈 = ∫ q³M

NOP

´M` 𝑑𝑥H + ∫

qMM

NOP�` 𝑑𝑥N

𝛿𝑈𝛿𝐵_

= 0 + �𝑀N

𝐸𝐼𝛿𝑀N

𝛿𝐵_𝑑𝑥N

�

`= 0

1𝐸𝐼� (−

12𝑤0𝑥N@ + 𝐵_𝑥NN − 𝑃𝑥NN −

12𝑃𝐿𝑥N)𝑑𝑥N

�

`= 0

−18𝑤0𝐿 +

13𝐵_ −

13𝑃 −

14𝑃 = 0

13𝐵_ =

18𝑤0𝐿 +

712𝑃

𝐵_ =38𝑤0𝐿 +

74𝑃

𝐴_ = 𝑃 +𝑤0𝐿 − 𝐵_ = 𝑃 + 𝑤0𝐿 −38𝑤0𝐿 −

74𝑃

𝐴_ = −34𝑃 +

58𝑤0𝐿

𝑀o =12𝑤0𝐿N − 𝐵_𝐿 +

32𝑃𝐿 =

12𝑤0𝐿N −

38𝑤0𝐿N −

74𝑃𝐿 +

32𝑃𝐿

𝑀o =18𝑤0𝐿N −

14𝑃𝐿