Flexural and Torsional Rigidity - Glass Beams

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2 LAYERS 3 LAYERS 4 LAYERS = 1, 2 = 1, 2, 3 = 1, 2, 3, 4 FLEXURAL STIFFNESS = 1++ 2 1+ 2 = 2 = 2 2 = βˆ‘ =βˆ‘ 3 12 2 = = = 2 2 = 1 2 2 =( 1 2 2 + 2 2 2 2 ) = 2 = 2 2 = βˆ™ 1 2 + 2 2 2 3 1 + 2 2 TORSIONAL STIFFNESS = (βˆ‘ 3 3 + ) = (1 βˆ’ 2 β„Ž 2 ) = (1 βˆ’ 1 1 β„Ž 1 2 βˆ’ 2 2 β„Ž 2 2 ) = 4 =√ βˆ™ 2 =√ βˆ™ 1 1 1 =√ βˆ™ 3 1 + 2 βˆ’ 2 1 2 2 =√ βˆ™ 3 1 + 2 + 2 1 2 = √9 1 2 βˆ’ 2 1 2 + 2 2 = 3 1 2 2 + 4 1 2 ( 1 2 βˆ’ 1 2 )+ 2 2 2 2 1 2 + 2 2 2 1 =1+ 2 =1βˆ’

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Transcript of Flexural and Torsional Rigidity - Glass Beams

Page 1: Flexural and Torsional Rigidity - Glass Beams

2 LAYERS 3 LAYERS 4 LAYERS

𝑖 = 1, 2 𝑖 = 1, 2, 3 𝑖 = 1, 2, 3, 4

FLEXURAL STIFFNESS

𝐸𝐼𝑓 =1 + 𝛼 + πœ‹2𝛼𝛽

1 + πœ‹2𝛽𝐸𝐼𝑠

𝛼 =𝐷𝑓

𝐷2 𝛽 =

𝐷2

𝐷𝑄𝐿2 𝐷𝑓 = βˆ‘ 𝐸𝐼𝑖

𝑖

= βˆ‘ 𝐸𝑏𝑑𝑖

3

12𝑖

𝐷2 = 𝐸𝐼𝑠 𝐷𝑄 = 𝐴𝐺𝑖𝑛𝑑

𝐼𝑠 =𝑏𝑑𝑑2

2 𝐼𝑠 =

𝑏𝑑1𝑑2

2 𝐼𝑠 = 𝑏 (

𝑑1𝑑2

2+

𝑑2π‘Ž22

2)

𝐴 =𝑏𝑑2

𝑑𝑖𝑛𝑑 𝐴 =

𝑏𝑑2

2 𝑑𝑖𝑛𝑑 𝐴 =

𝑏𝑑

π‘‘π‘–π‘›π‘‘βˆ™

𝑑1𝑑2 + 𝑑2π‘Ž22

3 𝑑1𝑑 + 𝑑2π‘Ž2

TORSIONAL STIFFNESS

𝐺𝐽𝑑 = 𝐺 (βˆ‘π‘π‘‘π‘–

3

3+ π½π‘π‘œπ‘šπ‘

𝑖

)

π½π‘π‘œπ‘šπ‘ = 𝐽𝑠 (1 βˆ’2

πœ†π‘π‘‘π‘Žπ‘›β„Ž

πœ†π‘

2) π½π‘π‘œπ‘šπ‘ = 𝐽𝑠 (1 βˆ’

𝛾1

πœ†1π‘π‘‘π‘Žπ‘›β„Ž

πœ†1𝑏

2βˆ’

𝛾2

πœ†2π‘π‘‘π‘Žπ‘›β„Ž

πœ†2𝑏

2)

𝐽𝑠 = 4 𝐼𝑠

πœ† = βˆšπΊπ‘–π‘›π‘‘

πΊβˆ™

2

𝑑𝑖𝑛𝑑𝑑 πœ† = √

𝐺𝑖𝑛𝑑

πΊβˆ™

1

𝑑𝑖𝑛𝑑𝑑1 πœ†1 = √

𝐺𝑖𝑛𝑑

πΊβˆ™

3 𝑑1 + 𝑑2 βˆ’ 𝛿

2 𝑑𝑖𝑛𝑑𝑑1𝑑2

πœ†2 = βˆšπΊπ‘–π‘›π‘‘

πΊβˆ™

3 𝑑1 + 𝑑2 + 𝛿

2 𝑑𝑖𝑛𝑑𝑑1𝑑2

𝛿 = √9 𝑑12 βˆ’ 2 𝑑1𝑑2 + 𝑑2

2

νœ€ =3 𝑑1

2𝑑2 + 4 𝑑1𝑑2(π‘Ž1π‘Ž2 βˆ’ π‘Ž12) + 𝑑2

2π‘Ž22

𝑑1𝑑2 + 𝑑2π‘Ž22

𝛾1 = 1 +νœ€

𝛿 𝛾2 = 1 βˆ’

νœ€

𝛿

Page 2: Flexural and Torsional Rigidity - Glass Beams

5 LAYERS

𝑖 = 1, 2, 3, 4, 5

FLEXURAL STIFFNESS

𝐸𝐼𝑓 =1 + 𝛼 + πœ‹2𝛼𝛽

1 + πœ‹2𝛽𝐸𝐼𝑠

𝛼 =𝐷𝑓

𝐷2 𝛽 =

𝐷2

𝐷𝑄𝐿2 𝐷𝑓 = βˆ‘ 𝐸𝐼𝑖

𝑖

= βˆ‘ 𝐸𝑏𝑑𝑖

3

12𝑖

𝐷2 = 𝐸𝐼𝑠 𝐷𝑄 = 𝐴𝐺𝑖𝑛𝑑

𝐼𝑠 = 𝑏 (𝑑1𝑑2

2+ 2 𝑑2π‘Ž2

2)

𝐴 =𝑏𝑑

4 π‘‘π‘–π‘›π‘‘βˆ™

𝑑1𝑑2 + 4 𝑑2π‘Ž22

𝑑1𝑑 + 𝑑2π‘Ž2

TORSIONAL STIFFNESS

𝐺𝐽𝑑 = 𝐺 (βˆ‘π‘π‘‘π‘–

3

3+ π½π‘π‘œπ‘šπ‘

𝑖

)

π½π‘π‘œπ‘šπ‘ = 𝐽𝑠 (1 βˆ’π›Ύ1

πœ†1π‘π‘‘π‘Žπ‘›β„Ž

πœ†1𝑏

2βˆ’

𝛾2

πœ†2π‘π‘‘π‘Žπ‘›β„Ž

πœ†2𝑏

2)

𝐽𝑠 = 4 𝐼𝑠

πœ†1 = βˆšπΊπ‘–π‘›π‘‘

πΊβˆ™

2 𝑑1 + 𝑑2 βˆ’ 𝛿

2 𝑑𝑖𝑛𝑑𝑑1𝑑2

πœ†2 = βˆšπΊπ‘–π‘›π‘‘

πΊβˆ™

2 𝑑1 + 𝑑2 + 𝛿

2 𝑑𝑖𝑛𝑑𝑑1𝑑2

𝛿 = √4 𝑑12 + 𝑑2

2

νœ€ =2 𝑑1

2𝑑2 + 4 𝑑1𝑑2(π‘Ž22 + 2 π‘Ž1π‘Ž2 βˆ’ π‘Ž1

2) + 4 𝑑22π‘Ž2

2

𝑑1𝑑2 + 4 𝑑2π‘Ž22

𝛾1 = 1 +νœ€

𝛿 𝛾2 = 1 βˆ’

νœ€

𝛿