OWL // Purdue Writing Lab · 2020. 12. 2. · Created Date: 20140309103846Z
62/87,21 D - Purdue University · 2020. 11. 4. · Microsoft Word - HW8_solution.docx Author: kylem...
Transcript of 62/87,21 D - Purdue University · 2020. 11. 4. · Microsoft Word - HW8_solution.docx Author: kylem...
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SOLUTION
a)
FBD analysis over the whole beam,
2 pt. for all FBDs
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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.
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c)
𝑅 =𝑣 ,( )
𝑣 ,( )=
4.8828
4.9203= 0.9924
1 pt.
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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
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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
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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.
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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.
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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