SOLUTION

a)

FBD analysis over the whole beam,

2 pt. for all FBDs

Σ𝑀 = 𝐹𝐿

2− 𝑃𝐿 = 0 ⇒ 𝐹 = 2𝑃

0 < 𝑥 < 0.5𝐿:

Σ𝐹 = 𝑉 − 𝑃 + 𝐹 = 0 ⇒ 𝑉 = −𝑃

Σ𝑀 = −𝑀 − 𝑃(𝐿 − 𝑥) + 𝐹𝐿

2− 𝑥 = 0 ⇒ 𝑀 = −𝑃𝑥

0.5𝐿 < 𝑥 < 𝐿:

Σ𝐹 = 𝑉 − 𝑃 = 0 ⇒ 𝑉 = 𝑃

Σ𝑀 = −𝑀 − 𝑃(𝐿 − 𝑥) = 0 ⇒ 𝑀 = 𝑃(𝑥 − 𝐿)

Flexural energy due to bending,

𝑈 =𝑀1𝑥 𝑑𝑥

2𝐸𝐼

.

+𝑀2𝑥 𝑑𝑥

2𝐸𝐼.=

1

2

𝑃 𝑥 𝑑𝑥

𝐸𝐼

.

+1

2

𝑃 (𝑥 − 𝐿) 𝑑𝑥

𝐸𝐼.=

𝑃 𝐿

24𝐸𝐼

Using Castigliano’s Theorem,

𝑣 =𝜕𝑈

𝜕𝑃=

𝑃𝐿

12𝐸𝐼= 4.8828 × 10 𝑚

b)

Shear energy due to bending,

𝑈 =𝑓

𝑠𝑉

1𝑥𝑑𝑥

2𝐺𝐴

.

+𝑓

𝑠𝑉

2𝑥𝑑𝑥

2𝐺𝐴.=

𝑓𝑠𝑃 𝑑𝑥

2𝐺𝐴=

𝑓𝑠𝑃 𝐿

2𝐺𝐴

Combining energy terms,

𝑈 = 𝑈 + 𝑈 =𝑃 𝐿

24𝐸𝐼+

𝑓𝑠𝑃 𝐿

2𝐺𝐴

Using Castigliano’s Theorem,

𝑣 =𝜕𝑈

𝜕𝑃=

𝑃𝐿

12𝐸𝐼+

𝑓𝑠𝑃𝐿

𝐺𝐴= 4.9203 × 10 𝑚

1 pt.

1 pt.

1 pt.

1 pt.

1 pt.

1 pt.

1 pt.

c)

𝑅 =𝑣 ,( )

𝑣 ,( )=

4.8828

4.9203= 0.9924

1 pt.

Solution:

Equilibrium: Beam is statically indeterminate.

∑𝑀 = 𝑀 −1

2𝑤 𝐿

𝐿

4+

5

4𝐵 𝐿 −

3

2𝑃𝐿

𝑀 −1

8𝑤 𝐿 +

5

4𝐵 𝐿 −

3

2𝑃𝐿 = 0

∑𝐹 = 𝐴 + 𝐵 − 𝑃 −1

2𝑤 𝐿 = 0

2 pt.

For all FBDs

1 pt.

𝑥

Making a cut at section 0 < 𝑥 < we get:

∑𝑀 = 0

−𝑀 − 𝑃𝑥 = 0

𝑀 (𝑥) = −𝑃𝑥,𝛿𝑀

𝛿= 0

Making a cut at section < 𝑥 < 𝐿 we get:

∑𝑀 = 0

−𝑀 + 𝐵 𝑥 −𝐿

4− 𝑃𝑥 = 0

𝑀 (𝑥) = 𝐵 𝑥 −𝐿

4− 𝑃𝑥,

𝛿𝑀

𝛿= 𝑥 −

𝐿

4

1 pt.

1 pt.

1 pt. for 3 correct cuts

𝑥

𝑥

Making a cut at section 𝐿 < 𝑥 < 𝐿 we get:

∑𝑀 = 0

−𝑀 −1

2𝑤 (𝑥 − 𝐿) + 𝐵 𝑥 −

𝐿

4− 𝑃𝑥 = 0

𝑀 (𝑥) = −1

2𝑤 (𝑥 − 𝐿) + 𝐵 𝑥 −

𝐿

4− 𝑃𝑥,

𝛿𝑀

𝛿𝐵= 𝑥 −

𝐿

4

Total strain energy = 𝑈 = ∫ 𝑑𝑥 + ∫ 𝑑𝑥 + ∫ 𝑑𝑥

𝛿𝑈

𝛿𝐵= 0 +

𝑀

𝐸𝐼

𝛿𝑀

𝛿𝐵𝑑𝑥 +

𝑀

𝐸𝐼

𝛿𝑀

𝛿𝐵𝑑𝑥 = 0

1

𝐸𝐼𝐵 𝑥 −

𝐿

4− 𝑃𝑥 𝑥 −

𝐿

4𝑑𝑥 +

1

𝐸𝐼−

1

2𝑤 (𝑥 − 𝐿) + 𝐵 𝑥 −

𝐿

4− 𝑃𝑥 𝑥 −

𝐿

4𝑑𝑥 = 0

9𝐿

1282𝐵 − 3𝑃 +

𝐿

384196𝐵 − 9𝐿𝑤 − 244𝑃 = 0

128

192𝐵 =

3

128𝑤 𝐿 +

325

384𝑃

𝐵 =9

256𝑤 𝐿 +

325

256𝑃

Optional solutions for 𝑨𝒚 and 𝑴𝒂

1 pt.

1 pt.

1 pt.

1 pt.

𝑥

𝐴 = 𝑃 + 𝑤 𝐿 − 𝐵 = 𝑃 + 𝑤 𝐿 −9

256𝑤 𝐿 −

325

256𝑃

𝐴 =247

256𝑤 𝐿 −

69

256𝑃

𝑀 =1

8𝑤 𝐿 −

5

4𝐵 𝐿 +

3

2𝑃𝐿 =

1

8𝑤 𝐿 −

45

1024𝑤 𝐿 −

1625

1024𝑃𝐿 +

3

2𝑃𝐿

𝑀 =83

1024𝑤 𝐿 −

89

1024𝑃𝐿

SOLUTION

FBD analysis on Section AB,

Σ𝐹 = 𝑉 + 𝑃 = 0 ⇒ 𝑉 = −𝑃

Σ𝑀 = 𝑀 (𝑦) + 𝑃(𝐿 − 𝑦) = 0 ⇒ 𝑀 (𝑦) = 𝑃(𝑦 − 𝐿)

Σ𝑀 = 𝑀 − 𝑃𝐿 + 𝑀 = 0 ⇒ 𝑀 = 𝑃𝐿 − 𝑀

FBD analysis on Section BC,

Σ𝐹 = −𝑉(𝑥) + 𝑃 = 0 ⇒ 𝑉(𝑥) = 𝑃

Σ𝑀 = 𝑀 − 𝑀(𝑥) − 𝑃(𝐿 − 𝑥) = 0 ⇒ 𝑀(𝑥) = 𝑀 + 𝑃(𝑥 − 𝐿)

Energy on Section AB,

𝑈 =𝑀𝑜𝑥 𝑑𝑦

2𝐸𝐼+

𝑀𝑜𝑦 𝑑𝑦

2𝐺𝐼=

1

2

𝑃 (𝑦 − 𝐿)

𝐸𝐼𝑑𝑦 +

1

2

(𝑃𝐿 − 𝑀𝑑) 𝐿

𝐺𝐼𝑑𝑦

Energy on Section BC,

𝑈 =𝑀(𝑥) 𝑑𝑥

2𝐸𝐼=

1

2

(𝑀𝑑 + 𝑃(𝑥 − 𝐿))

𝐸𝐼𝑑𝑥

𝑈 = 𝑈 + 𝑈 =1

2

𝑃 (𝑦 − 𝐿)

𝐸𝐼𝑑𝑦 +

1

2

(𝑃𝐿 − 𝑀𝑑)

𝐺𝐼𝑑𝑦 +

1

2

(𝑀𝑑 + 𝑃(𝑥 − 𝐿))

𝐸𝐼𝑑𝑥

Using Castigliano’s Theorem,

𝐿 𝐿

2 pt. for all FBDs

1 pt.

1 pt.

1 pt.

1 pt.

a)

𝑣 =𝜕𝑈

𝜕𝑃|

=1

𝐸𝐼𝑃(𝑦 − 𝐿) 𝑑𝑦 +

1

𝐺𝐼(𝑃𝐿 − 𝑀 )𝐿 𝑑𝑦 +

1

𝐸𝐼(𝑀𝑑 + 𝑃(𝑥 − 𝐿))(𝑥 − 𝐿) 𝑑𝑥

=1

𝐸𝐼𝑃(𝑦 − 𝐿) 𝑑𝑦 +

1

𝐺𝐼𝑃𝐿2 𝑑𝑦 +

1

𝐸𝐼𝑃(𝑥 − 𝐿)2 𝑑𝑥

=𝑃𝐿

3𝐸𝐼+

𝑃𝐿

𝐺𝐼𝑝+

𝑃𝐿

3𝐸𝐼

=2𝑃𝐿3

3𝐸𝐼+

𝑃𝐿

𝐺𝐼𝑝

𝑆𝑖𝑛𝑐𝑒 𝐼 =𝜋𝑑

64, 𝐼 =

𝜋𝑑

32

⇒ 𝑣 =32𝑃𝐿3

𝜋𝑑4(

4

3𝐸+

1

𝐺)

b)

𝜃 =𝜕𝑈

𝜕𝑀| = 0 +

−1

𝐺𝐼(𝑃𝐿 − 𝑀 )𝑑𝑦 +

1

𝐸𝐼(𝑀 + 𝑃(𝑥 − 𝐿))𝑑𝑥

=−1

𝐺𝐼𝑃𝐿𝑑𝑦 +

1

𝐸𝐼𝑃(𝑥 − 𝐿) 𝑑𝑥

= −𝑃𝐿

𝐺𝐼𝑝−

𝑃𝐿

2𝐸𝐼

=−32𝑃𝐿2

𝜋𝑑4(

1

𝐺+

1

𝐸)

1 pt.

1 pt.

1 pt.

1 pt.

Justification:

{0 < 𝑥 < 𝐿}, {𝐿 < 𝑥 < 1.5𝐿}, {1.5𝐿 < 𝑥 < 2𝐿}, 𝑎𝑛𝑑 {2𝐿 < 𝑥 < 3𝐿} each have unique equations for bending moment. Integration is performed section by section so that the sum of integrals spans the region of interest.

5 pt.

Partial credit case-by-case