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Chart C4 Flexural Strength

The document outlines formulas and criteria for evaluating flexural strength and compactness of structural elements. It provides conditions for determining whether a section is compact, non-compact, or slender, along with corresponding moment capacities. Additionally, it includes calculations for torsional constants and load factors for different structural shapes.

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Claire Aves
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0% found this document useful (0 votes)
26 views1 page

Chart C4 Flexural Strength

The document outlines formulas and criteria for evaluating flexural strength and compactness of structural elements. It provides conditions for determining whether a section is compact, non-compact, or slender, along with corresponding moment capacities. Additionally, it includes calculations for torsional constants and load factors for different structural shapes.

Uploaded by

Claire Aves
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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𝐿𝑏 = 𝑢𝑛𝑏𝑟𝑎𝑐𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝐿𝑝 = 1,76𝑟𝑦 (𝐸/𝐹𝑦) FLEXURAL STRENGTH

𝐸 𝐽𝑐 𝐽𝑐
2
0.7𝐹𝑦
2 Check compactness
𝐿𝑟 = 1.95𝑟𝑡𝑠 + + 6.76 𝜆 < 𝜆𝑟 𝑜𝑟 𝜆 ≤ 𝜆𝑝 : COMPACT
0.7𝐹𝑦 𝑆𝑥 ℎ𝑜 𝑆𝑥 ℎ𝑜 𝐸
𝜆𝑝 < 𝜆 ≤ 𝜆𝑟 𝑜𝑟 𝜆 > 𝜆𝑟 : NONCOMPACT or SLENDER

COMPACT: LTB NON-COMPACT: LTB & FLB; if noncompact due to flange: Mn is smaller of LTB & FLB
𝐿𝑏 ≤ 𝐿𝑝 𝑁𝑂 𝐿𝑇𝐵 : 𝑀𝑛 = 𝑀𝑝 = 𝐹𝑦 𝑍 LTB: same as in COMPACT (Lb ≤ Lp, Lp < Lb ≤ Lr, Lb > Lr)
𝐿 𝑏 − 𝐿𝑝 FLB (FLANGE LOCAL BUCKLING):
𝐿𝑝 < 𝐿𝑏 ≤ 𝐿𝑟 𝑖𝑛𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝐿𝑇𝐵 : 𝑀𝑛 = 𝐶𝑏 𝑀𝑝 − 𝑀𝑝 − 0.7𝐹𝑦 𝑆𝑥 ≤ 𝑀𝑝
𝐿𝑟 − 𝐿𝑝 𝜆 ≤ 𝜆𝑝𝑓 : NO FLB 𝜆 > 𝜆𝑟 (𝑆𝐿𝐸𝑁𝐷𝐸𝑅 𝐹𝐿𝐴𝑁𝐺𝐸𝑆): 𝑀𝑛 = 0.9𝐸𝑘𝑐 𝑆𝑥 /𝜆2
𝐿𝑏 > 𝐿𝑟 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝐿𝑇𝐵 : 𝑀𝑛 = 𝐹𝑐𝑟 𝑆𝑥 ≤ 𝑀𝑝 𝜆 − 𝜆𝑝𝑓
𝜆𝑝𝑓 < 𝜆 ≤ 𝜆𝑟𝑓 FLANGE IS NON − COMPACT : 𝑀𝑛 = 𝑀𝑝 − 𝑀𝑝 − 0.7𝐹𝑦 𝑆𝑥
𝜆𝑟 − 𝜆𝑝𝑓

LRFD ASD
𝑀𝑢 = 𝜙𝑏 𝑀𝑛 ; 𝜙𝑏 = 0.90 𝑀𝑎 = 𝑀𝑛 /Ω𝑏 ; Ω𝑏 = 1.67
Torsional Constant, J c Cb: Smply Supported Cb: Smply Supported
For double-symetric I-shapes: c =1 Concentrated Distirbuted
2𝑏𝑓 𝑡𝑓3 +𝑑′ 𝑡𝑤3 For a channel: c = (ho/2)√(Iy/Cw) - None (loads at Midspan): Cb = 1.32 - None: Cb = 1.14
wide flange: 𝐽 = ; 𝑑 ′ = 𝑑 − 𝑡𝑓 Load Point: Cb = 1.67 - Midpoint: Cb = 1.30
3 For double-symetric w/ rectangluar flange: rts2 = Iyho/2Sx
2𝑏′ 𝑡𝑓 3+𝑑′ 𝑡𝑤3 - None (loads at Third pts): Cb = 1.14 - Third pts: Cb = 1.01
channel: 𝐽 = ; 𝑑 ′ = 𝑑 − 𝑡𝑓 ; 𝑏′ = 𝑏 − 𝑡𝑤 /2 𝑏𝑓 𝐼𝑦 ℎ𝑜 2
3 𝑟𝑡𝑠 = ; 𝐶𝑤 = Load pts: Cb = 1.00 - Quarter pts: Cb = 1.06
𝑏𝑓 𝑡𝑓 3+𝑑′ 𝑡𝑤3 4 - None (Load at Quarter pts): Cb = 1.14
tee: 𝐽 = ; 𝑑 ′ = 𝑑 − 𝑡𝑓 /2 1 ℎ𝑡𝑤 - Fifth pts: Cb = 1.00
3 12 1 + Load pts: Cb = 1.11
6 𝑏𝑓 𝑡𝑓
(𝑏′ +𝑑′ )𝑡 3
angle section: 𝐽 = ; 𝑏 ′ = 𝑑 − 0.5𝑡; 𝑑 ′ = 𝑑 − 0.5𝑡
3
12.5 𝑀𝑚𝑎𝑥
𝐶𝑏 = 𝑅 ≤ 3.0
2.5 𝑀𝑚𝑎𝑥 + 3𝑀𝐴 + 4𝑀𝐵 + 3𝑀𝐶 𝑚

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