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‫‪www.icivil.ir‬‬ ‫پرتال جامع داوشجًیان ي مهىدسیه عمران‬ ‫ارائه کتابها ي جسيات رایگان مهىدسی عمران‬ ‫بهتریه ي برتریه مقاالت ريز عمران‬ ‫اوجمه های تخصصی مهىدسی عمران‬ ‫فريشگاه تخصصی مهىدسی عمران‬ Urmia University, M. Sheidaii 1 ‫ﺳﺎﺯﻩ ﻫﺎﻱ ﻓﻮﻻﺩﻱ‬ ‫‪Steel Structures‬‬ ‫ﻣﺤﻤﺪﺭﺿﺎ ﺷﻴﺪﺍﺋﻲ ﺍﺳﺘﺎﺩﻳﺎﺭ ﮔﺮﻭﻩ ﻣﻬﻨﺪﺳﻲ ﻋﻤﺮﺍﻥ ﺩﺍﻧﺸﮕﺎﻩ ﺍﺭﻭﻣﻴﻪ‬ ‫ﻫﺪﻑ‪ :‬ﺁﺷﻨﺎﻳﻲ ﺑﺎ ﻣﺸﺨﺼﺎﺕ ﻣﺼﺎﻟﺢ‪ ،‬ﺭﻭﺷﻬﺎﻱ ﻃﺮﺍﺣﻲ ﻭ ﻣﻘﺮﺭﺍﺕ‬ ‫ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﺍﻱ ﻣﺮﺑﻮﻁ ﺑﻪ ﻓﻮﻻﺩ‬ ‫‪To familiarize the student with the material properties, design‬‬ ‫‪procedures, and code requirements for steel.‬‬ ‫‪2‬‬ ‫ﻣﻨﺎﺑﻊ‬ References :‫ﻛﺘﺐ ﺩﺭﺳﻲ‬ Steel Design, fourth edition by William T. Segui, , 2007 Specification for Structural Steel Buildings, AISC 2005. ‫ ﺩﻓﺘﺮ ﺗﺪﻭﻳﻦ ﻭﺗﺮﻭﻳﺞ‬،‫ ﻣﺒﺤﺚ ﺩﻫﻢ ﻃﺮﺡ ﻭﺍﺟﺮﺍی ﺳﺎﺧﺘﻤﺎﻧﻬﺎی ﻓﻮﻻﺩی‬:‫ ﻣﻘﺮﺭﺍﺕ ﻣﻠﯽ ﺳﺎﺧﺘﻤﺎﻥ ﺍﻳﺮﺍﻥ‬ 1387 ،‫ ﻭﺯﺍﺭﺕ ﻣﺴﮑﻦ ﻭﺷﻬﺮﺳﺎﺯی‬،‫ﻣﻘﺮﺭﺍﺕ ﻣﻠﯽ ﺳﺎﺧﺘﻤﺎﻥ‬ :‫ﻣﺮﺍﺟﻊ ﭘﻴﺸﻨﻬﺎﺩﻱ‬ Structural Steel Design, fourth edition by Jack C. McCormac, Prentice Hall, 2008. Steel Structures: Design and Behavior. 5th ed. by Salmon, G. Charles, Johnson, E. John and Malhas A. Faris, Prentice Hall, 2008. Design of Steel Structures. 3rd ed. By Gaylord, E.H., Gaylord, C.N. and Stallmeyer J.E., McGraw-Hill, 1992. ‫ ﺩﻛﺘﺮ ﻓﺮﻳﺪﻭﻥ ﺍﻳﺮﺍﻧﻲ‬، (LRFD) ‫ ﻃﺮﺍﺣﻲ ﺳﺎﺯﻩ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺑﻪ ﺭﻭﺵ ﺿﺮﺍﻳﺐ ﺑﺎﺭ ﻭ ﻣﻘﺎﻭﻣﺖ‬ Urmia University, M. Sheidaii 3 ‫ﺳﺮﻓﺼﻞ ﻫﺎﻱ ﻣﻄﺎﻟﺐ ﺩﺭﺳﻲ‬ Topics covered Introduction to Structural Steel Design Design Methods Design of Tension Members Design of Compression Members Design of Built-up Columns Design of Base plates Design of Beams Design of Beam-Columns Design of Bolted Connections Design of Welded Connections Design of Plate Girders Design of Castellated and Composite Beams Design of Braced Frames Urmia University, M. Sheidaii ‫ﻣﻘﺪﻣﻪ ﺍﻱ ﺑﺮ ﻃﺮﺍﺣﻲ ﺳﺎﺯﻩ ﻫﺎﻱ ﻓﻮﻻﺩﻱ‬ ‫ﺭﻭﺷﻬﺎﻱ ﻃﺮﺍﺣﻲ‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ‬ ‫ﻃﺮﺍﺣﻲ ﺻﻔﺤﺎﺕ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎ‬ ‫ﺳﺘﻮﻧﻬﺎ‬-‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫ﻃﺮﺍﺣﻲ ﺍﺗﺼﺎﻻﺕ ﭘﻴﭽﻲ‬ ‫ﻃﺮﺍﺣﻲ ﺍﺗﺼﺎﻻﺕ ﺟﻮﺷﻲ‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮﻭﺭﻗﻬﺎ‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎﻱ ﻻﻧﻪ ﺯﻧﺒﻮﺭﻱ ﻭ ﻣﺨﺘﻠﻂ‬ ‫ﻃﺮﺍﺣﻲ ﻗﺎﺏ ﻫﺎﻱ ﻣﻬﺎﺭﺑﻨﺪﻱ ﺷﺪﻩ‬ 4 Introduction to Structural Steel Design ‫ ﻣﻘﺪﻣﻪ ﺍﻱ ﺑﺮ ﻃﺮﺍﺣﻲ ﺳﺎﺯﻩ ﻫﺎﻱ ﻓﻮﻻﺩﻱ‬: ‫ﻓﺼﻞ ﺍﻭﻝ‬ Urmia University, M. Sheidaii 5 ‫ﻓﻮﻻﺩ ﺳﺎﺧﺘﻤﺎﻧﻲ‬ ‫‪Structural Steel‬‬ ‫ ﻓﻮﻻﺩ ﺁﻟﻴﺎژﻱ ﺍﺳﺖ ﻣﺮﻛﺐ ﺍﺯ ﺁﻫﻦ ‪ ،‬ﻛﺮﺑﻦ )‪ 0.15‬ﺗﺎ ‪ ( % 2‬ﻭ ﻣﻘﺪﺍﺭ‬ ‫ﺟﺰﺋﻲ ﺍﺯ ﻋﻨﺎﺻﺮ ﺩﻳﮕﺮ )ﻣﻨﮕﻨﺰ ‪ ،‬ﻧﻴﻜﻞ ﻭ ‪( ...‬‬ ‫ ﻛﺮﺑﻦ ﺑﺎﻋﺚ ﺍﻓﺰﺍﻳﺶ ﻣﻘﺎﻭﻣﺖ ﻓﻮﻻﺩ ﺷﺪﻩ ﺍﻣﺎ ﺩﺭ ﻋﻴﻦ ﺣﺎﻝ ﺷﻜﻞ ﭘﺬﻳﺮﻱ‬ ‫ﺭﺍ ﻛﺎﻫﺶ ﻣﻲ ﺩﻫﺪ‪.‬‬ ‫‪6‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺁﺯﻣﺎﻳﺶ ﻛﺸﺶ‬ The Tension test Urmia University, M. Sheidaii 7 ‫ﻣﺸﺨﺼﻪ ﻫﺎﻱ ﻓﻮﻻﺩ‬ Steel Properties : ‫ ﻣﻬﻤﺘﺮﻳﻦ ﻣﺸﺨﺼﻪ ﻫﺎﻱ ﻓﻮﻻﺩ ﺍﺯ ﻧﻜﺘﻪ ﻧﻈﺮ ﻃﺮﺍﺣﻲ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ‬ (yield stress (Fy) ) ‫ ﺗﻨﺶ ﺗﺴﻠﻴﻢ‬ ‫ ﺗﻨﺶ ﻧﻬﺎﻳﻲ‬ (modulus of elasticity (E) ) ‫ ﺿﺮﻳﺐ ﺍﺭﺗﺠﺎﻋﻲ‬ (percent elongation (H) ) ‫ ﺩﺭﺻﺪ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﻧﺴﺒﻲ‬ (coefficient of thermal expansion (D) ) ‫ ﺿﺮﻳﺐ ﺍﻧﺒﺴﺎﻁ ﺣﺮﺍﺭﺗﻲ‬ (ultimate stress (Fu) ) Urmia University, M. Sheidaii 8 ‫ﻣﺸﺨﺼﻪ ﻫﺎﻱ ﻣﻄﻠﻮﺏ ﻓﻮﻻﺩ ﺍﺯ ﻧﻈﺮ ﺭﻓﺘﺎﺭ ﻣﺼﺎﻟﺢ‬ Attractive Material Properties of Steels (a linear elastic range) ‫ ﻭﺟﻮﺩ ﻳﻚ ﻣﺤﺪﻭﺩﻩ ﺭﻓﺘﺎﺭ ﺍﺭﺗﺠﺎﻋﻲ ﺧﻄﻲ‬ (A well-defined yield (except Heat treated)) ‫ ﻭﺟﻮﺩ ﻧﻘﻄﻪ ﺗﺴﻠﻴﻢ ﻣﺸﺨﺺ‬ (‫)ﻣﮕﺮ ﺩﺭ ﻓﻮﻻﺩﻫﺎﻱ ﺑﺎﺯﭘﺨﺖ ﺷﺪﻩ‬ (strain hardening) (significant ductility ) Urmia University, M. Sheidaii ‫ ﺳﺨﺖ ﺷﺪﮔﻲ ﻛﺮﻧﺸﻲ‬ ‫ ﺷﻜﻞ ﭘﺬﻳﺮﻱ ﻗﺎﺑﻞ ﺗﻮﺟﻪ‬ 9 ‫ﻣﺰﺍﻳﺎﻱ ﻓﻮﻻﺩ ﺑﻪ ﻋﻨﻮﺍﻥ ﻳﻚ ﻣﺼﺎﻟﺢ ﺳﺎﺯﻩ ﺍﻱ‬ Advantages of Steel as a Structural Material ( High Strength to Weight ) ‫ﻣﻘﺎﻭﻣﺖ ﺑﺎﻻ ﺑﻪ ﻭﺯﻥ‬ ( Uniformity & Permanence) (‫ﻳﻜﻨﻮﺍﺧﺘﻲ ﻭ ﺩﻭﺍﻡ )ﺧﻮﺍﺹ ﺛﺎﺑﺖ ﺑﺎ ﺯﻣﺎﻥ‬ ( Elasticity ) ‫ﺧﺎﺻﻴﺖ ﺍﺭﺗﺠﺎﻋﻲ‬ ( Highly Ductile ) ‫ﺷﻜﻞ ﭘﺬﻳﺮﻱ ﺑﺎﻻ‬ (Toughness ) (‫ﭘﺮﻃﺎﻗﺘﻲ)ﺷﻜﻞ ﭘﺬﻳﺮﻱ ﻭﻣﻘﺎﻭﻣﺖ ﺑﺎﻻ ﺑﻄﻮﺭ ﻫﻤﺰﻣﺎﻥ‬ ( Easily Constructed and Modified Structures ) ‫ﺳﻬﻮﻟﺖ ﺳﺎﺧﺖ ﻭ ﺗﻮﺳﻌﻪ‬ ( Easily recycled ) ‫ﺑﺎﺯﻳﺎﺑﻲ ﺁﺳﺎﻥ‬ ‫ ﭘﻴﭻ ﻭ ﭘﺮچ‬،‫ ﺍﻣﻜﺎﻥ ﺍﺗﺼﺎﻝ ﺗﻮﺳﻂ ﺍﺑﺰﺍﺭ ﺳﺎﺩﻩ ﺍﺗﺼﺎﻝ ﻧﻈﻴﺮ ﺟﻮﺵ‬ ( Simple Connection Devices such as Welds, Rivets and Bolts ) ( Possibility of Prefabrication ) ‫ ﺍﻣﻜﺎﻥ ﭘﻴﺶ ﺳﺎﺧﺘﻪ ﻛﺮﺩﻥ‬ ( Speed of Erection ) ‫ ﺳﺮﻋﺖ ﺩﺭ ﻧﺼﺐ‬ Urmia University, M. Sheidaii 10 ‫ﻣﻌﺎﻳﺐ ﻓﻮﻻﺩ ﺑﻪ ﻋﻨﻮﺍﻥ ﻳﻚ ﻣﺼﺎﻟﺢ ﺳﺎﺯﻩ ﺍﻱ‬ Disadvantages of Steel as a Structural Material ( Maintenance Costs ) ( Requires Fireproofing ) (Often Results in Slender members ) (→ Susceptible to Buckling ) ( Fatigue ) (Brittle Fracture ) Urmia University, M. Sheidaii ‫ ﻫﺰﻳﻨﻪ ﻧﮕﻬﺪﺍﺭﻱ‬ ‫ ﻧﻴﺎﺯ ﺑﻪ ﻣﻘﺎﻭﻡ ﺳﺎﺯﻱ ﺩﺭ ﺑﺮﺍﺑﺮ ﺁﺗﺶ‬ ‫ ﺍﻏﻠﺐ ﻣﻨﺠﺮ ﺑﻪ ﺍﻋﻀﺎﻳﻲ ﻻﻏﺮ ﻣﻲ ﺷﻮﻧﺪ‬ ‫← ﺧﻄﺮ ﻛﻤﺎﻧﺶ‬ ‫ ﺧﺴﺘﮕﻲ‬ ‫ ﺧﻄﺮ ﺷﻜﺴﺖ ﺗﺮﺩ‬ 11 ‫ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﻓﻮﻻﺩ‬ ‫‪Steel Applications‬‬ ‫ ﻓﻮﻻﺩ ﻳﻜﻲ ﺍﺯ ﻣﺼﺎﻟﺢ ﺑﺴﻴﺎﺭ ﭘﺮ ﻛﺎﺭﺑﺮﺩ ﺑﺮﺍﻱ ﺳﺎﺯﻩ ﻣﻲ ﺑﺎﺷﺪ ‪:‬‬ ‫‪12‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 13 Urmia University, M. Sheidaii 14 Urmia University, M. Sheidaii 15 Urmia University, M. Sheidaii 16 Urmia University, M. Sheidaii 17 Urmia University, M. Sheidaii 18 Urmia University, M. Sheidaii 19 Urmia University, M. Sheidaii 20 ‫ﺭﻭﻧﺪ ﻃﺮﺍﺣﻲ ﺳﺎﺯﻩ ﻓﻮﻻﺩﻱ‬ Steel Structure Design Process Urmia University, M. Sheidaii 21 ‫ﺭﻭﻧﺪ ﻃﺮﺍﺣﻲ‬ ‫‪The Design Process‬‬ ‫ ﺑﻪ ﺻﻮﺭﺕ ﻳﻚ ﭼﺮﺧﻪ ﺍﺳﺖ ) ‪(a cycle‬‬ ‫ ﺟﻮﺍﺑﻬﺎﻱ ﺻﺤﻴﺢ ﺑﺴﻴﺎﺭﻱ ﻣﻤﻜﻦ ﺍﺳﺖ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ‬ ‫ ﻛﻠﻴﻪ ﻣﻼﺣﻈﺎﺕ ﻃﺮﺍﺣﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﻋﻤﻠﻲ ﻧﺒﺎﺷﻨﺪ ) ‪( not all technical‬‬ ‫‪22‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻬﻨﺪﺳﻲ ﺳﺎﺯﻩ ﭼﻴﺴﺖ ؟‬ ‫?‪What is Structural Engineering‬‬ ‫ﻣﻬﻨﺪﺳﻲ ﺳﺎﺯﻩ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ ‪:‬‬ ‫ﻫﻨﺮ ﻗﺎﻟﺐ ﺩﻫﻲ ﻣﺼﺎﻟﺤﻲ ﻛﻪ ﺷﻨﺎﺧﺖ ﻛﺎﻣﻠﻲ ﺍﺯ ﺁﻧﻬﺎ ﻧﺪﺍﺭﻳﻢ‬ ‫ﺑﻪ ﺷﻜﻞ ﻫﺎﻳﻲ ﻛﻪ ﻧﻤﻲ ﺗﻮﺍﻧﻴﻢ ﺑﻪ ﺩﻗﺖ ﺁﻧﻬﺎ ﺭﺍ ﺗﺤﻠﻴﻞ ﻛﻨﻴﻢ‪،‬‬ ‫ﺑﻪ ﻧﺤﻮﻱ ﻛﻪ ﻣﻘﺎﺑﻠﻪ ﻛﻨﻨﺪ ﺑﺎ ﻧﻴﺮﻭ ﻫﺎﻳﻲ ﻛﻪ ﻗﺎﺩﺭ ﺑﻪ ﺍﺭﺯﻳﺎﺑﻲ ﻣﻘﺪﺍﺭ ﻭﺍﻗﻌﻲ ﺁﻧﻬﺎ ﻧﻴﺴﺘﻴﻢ‪،‬‬ ‫ﺑﻪ ﻃﺮﻳﻘﻲ ﻛﻪ ﻫﻴﭻ ﻛﺲ ﺑﻪ ﺍﻧﺒﻮﻩ ﻋﺪﻡ ﺁﮔﺎﻫﻲ ﻣﺎ ﻇﻨﻴﻦ ﻧﺸﻮﺩ ‪.‬‬ ‫ﺍﻗﺘﺼﺎﺩ‬ ‫‪ECONOMY‬‬ ‫‪23‬‬ ‫ﺍﻳﻤﻨﻲ‬ ‫ﭘﺎﻳﺪﺍﺭﻱ‬ ‫‪SAFETY‬‬ ‫‪STABIILIITY‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺴﺌﻮﻟﻴﺖ ﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ ‫‪Design Responsibilities‬‬ ‫ ﺍﻳﻤﻨﻲ ) ‪( safety‬‬ ‫ ﻭﺟﻮﺩ ﻣﻘﺎﻭﻣﺖ ﻛﺎﻓﻲ ﺩﺭ ﻫﻨﮕﺎﻡ ﺍﺟﺮﺍ ﻭ ﭘﺲ ﺍﺯ ﺁﻥ‬‫ ﻣﻘﺎﻭﻣﺖ ﺩﺭ ﺑﺮﺍﺑﺮ ﺧﺮﺍﺑﻲ ﭘﻴﺸﺮﻭﻧﺪﻩ ) ‪( Progressive collapse‬‬‫ ﺍﻳﻤﻦ ﺍﺭﺯﻳﺎﺑﻲ ﺷﺪﻥ ﺍﺯ ﻃﺮﻑ ﻛﺎﺭﺑﺮﺍﻥ‬‫ ﺍﺟﺮﺍﻳﻲ ﺑﻮﺩﻥ ) ‪( Practical‬‬ ‫‪ -‬ﺑﻪ ﻫﻤﺎﻥ ﺻﻮﺭﺕ ﻛﻪ ﻃﺮﺍﺣﻲ ﺷﺪﻩ ‪ ،‬ﻗﺎﺑﻞ ﺳﺎﺧﺖ ﺑﺎﺷﺪ‬ ‫ ﺭﻭﺍﺩﺍﺭﻳﻬﺎﻱ ﻣﻨﻄﻘﻲ ﺑﺎﻳﺪ ﻣﺠﺎﺯ ﺑﺎﺷﺪ ) ‪( Reasonable tolerances‬‬‫ ﻧﺼﺐ ﻭ ﺑﺮ ﭘﺎ ﻧﻤﻮﺩﻥ ﺳﺎﺯﻩ ﺑﻪ ﺻﻮﺭﺕ ﻏﻴﺮﻣﻨﻄﻘﻲ ﭘﻴﭽﻴﺪﻩ ﻳﺎ ﺧﻄﺮﻧﺎﻙ ﻧﺒﺎﺷﺪ‪.‬‬‫‪24‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺴﺌﻮﻟﻴﺖ ﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ ‫‪Design Responsibilities continued‬‬ ‫ ﺍﻗﺘﺼﺎﺩ ) ‪( Economy‬‬ ‫‪ -‬ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺼﺎﻟﺢ ﻭ ﻣﻘﺎﻃﻊ ﺩﺭ ﺩﺳﺘﺮﺱ ‪ -‬ﻣﺸﻮﺭﺕ ﺑﺎ ﻛﺎﺭﭘﺮﺩﺍﺯ!‬ ‫ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﻫﺰﻳﻨﻪ ﻛﻞ ﺳﺎﺯﻩ‪ :‬ﻣﺼﺎﻟﺢ ﻭ ﺳﺎﺧﺖ ‪ -‬ﺗﺼﻮﺭ ﻧﺸﻮﺩ ﻛﻪ ﺳﺒﻚ ﺗﺮﻳﻦ ﺍﺭﺯﺍﻥ ﺗﺮﻳﻦ‬‫ﺍﺳﺖ‬ ‫ ﺍﺳﺘﻌﻤﺎﻝ ﺣﺪﺍﻗﻞ ﺍﺟﺰﺍ ﻣﻮﺭﺩ ﻧﻴﺎﺯ ‪ -‬ﺑﻜﺎﺭﮔﻴﺮﻱ ﻫﺮ ﺟﺰ ﻧﻴﺎﺯﻣﻨﺪ ﺻﺮﻑ ﻫﺰﻳﻨﻪ ﺍﺳﺖ‪.‬‬‫ ﺳﺎﺩﻩ ﺗﺮﻳﻦ ﺩﺗﺎﻳﻞ ﻫﺎ ﻛﻪ ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﻛﺎﺭ ﻫﺴﺘﻨﺪ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ ﻭ ﺍﺯ ﻣﺰﺍﻳﺎﻱ‬‫ﺗﻜﺮﺍﺭ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ ﺷﻮﺩ‪.‬‬ ‫‪25‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻓﻮﻻﺩ ﭼﻴﺴﺖ؟‬ What is STEEL? (C) ‫( ﻭ ﻛﺮﺑﻦ‬Fe) ‫ ﻓﻮﻻﺩ ﺁﻟﻴﺎﮋﻱ ﺍﺳﺖ ﺍﺯ ﺁﻫﻦ‬ ‫ ﻳﻚ ﻳﺎ ﭼﻨﺪ ﻋﻨﺼﺮ ﺁﻟﻴﺎﮋﻱ ﺩﻳﮕﺮ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﻪ‬، ‫ﺑﺮﺍﻱ ﺩﺳﺘﻴﺎﺑﻲ ﺑﻪ ﻣﺸﺨﺼﺎﺕ ﻣﻜﺎﻧﻴﻜﻲ ﻣﻮﺭﺩ ﻧﻈﺮ‬ .‫ﻓﻮﻻﺩ ﺍﺿﺎﻓﻪ ﺷﻮﺩ‬ : ‫ﻣﻬﻤﺘﺮﻳﻦ ﺍﻳﻦ ﻋﻨﺎﺻﺮ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ‬ manganese (Mn), silicon (Si), aluminum (Al), nickel (Ni), chromium (Cr), molybdenum (Mo), copper (Cu), vanadium (V), niobium (Nb), and titanium (Ti) boron (B). Urmia University, M. Sheidaii 26 ‫ﻓﻮﻻﺩ ﭼﻴﺴﺖ؟‬ ‫?‪What is STEEL‬‬ ‫ ﻛﺮﺑﻦ‬ ‫ ﺍﻓﺰﺍﻳﺶ ‪ :‬ﻣﻘﺎﻭﻣﺖ ‪ ،‬ﺗﻨﺶ ﺗﺴﻠﻴﻢ ‪ ،‬ﺗﺮﺩﺷﻜﻨﻲ‬ ‫ ﻛﺎﻫﺶ ‪ :‬ﺷﻜﻞ ﭘﺬﻳﺮﻱ ‪ ،‬ﻣﻘﺎﻭﻣﺖ ﺩﺭ ﺑﺮﺍﺑﺮ ﺿﺮﺑﻪ ﻭ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺩﻳﻨﺎﻣﻴﻜﻲ ‪ ،‬ﻗﺎﺑﻠﻴﺖ ﺟﻮﺷﻜﺎﺭﻱ‬ ‫ ﻣﻨﮕﻨﺰ‬ ‫ ﺍﻓﺰﺍﻳﺶ ‪ :‬ﺷﻜﻞ ﭘﺬﻳﺮﻱ ‪ ،‬ﻣﻘﺎﻭﻣﺖ ﺩﺭ ﺑﺮﺍﺑﺮ ﺳﺎﻳﺶ ﻭ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺿﺮﺑﻪ ﺍﻱ‬ ‫ ﻣﻮﻟﻴﺒﺪﻥ‬ ‫ ﺍﻓﺰﺍﻳﺶ ‪ :‬ﺳﺨﺘﻲ ‪ ،‬ﻣﻘﺎﻭﻣﺖ ﺣﺮﺍﺭﺗﻲ‬ ‫ ﻛﺎﻫﺶ ‪ :‬ﺷﻜﻞ ﭘﺬﻳﺮﻱ‬ ‫ ﻛﺮﻭﻡ ‪ ،‬ﺳﻴﻠﻴﻜﻮﻥ ﻭ ﻧﻴﻜﻞ )ﺑﺠﺎﻱ ﺍﻓﺰﻭﺩﻥ ﻛﺮﺑﻦ( ‪:‬‬ ‫ ﻣﺰﺍﻳﺎ ‪ :‬ﻣﻘﺎﻭﻣﺖ ﺑﺴﻴﺎﺭ ﺑﻴﺸﺘﺮ ‪ ،‬ﺷﻜﻞ ﭘﺬﻳﺮﻱ ‪ ،‬ﻗﺎﺑﻠﻴﺖ ﺟﻮﺷﻜﺎﺭﻱ‬ ‫ ﻣﻌﺎﻳﺐ ‪ :‬ﮔﺮﺍﻥ ﻗﻴﻤﺖ ‪ ،‬ﻋﺪﻡ ﺳﻬﻮﻟﺖ ﺳﺎﺧﺖ‬ ‫ ﻣﺲ‬ ‫ ﺍﻓﺰﺍﻳﺶ ‪ :‬ﻣﻘﺎﻭﻣﺖ ﺑﻪ ﺧﻮﺭﺩﮔﻲ )ﻓﻮﻻﺩﻫﺎﻱ ﺿﺪ ﻓﺮﺳﺎﻳﺶ ‪( A242 ، A558‬‬ ‫‪27‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ‬ ‫‪STEEL Manufacturing‬‬ ‫ ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ ﺑﻪ ﺩﻭ ﺭﻭﺵ ﺻﻮﺭﺕ ﻣﻲ ﮔﻴﺮﺩ ‪:‬‬ ‫¾‬ ‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻛﻮﺭﻩ ﺫﻭﺏ ﺁﻫﻦ )‪(Blast Furnace‬‬ ‫– ﺭﻭﺵ ﻛﻬﻦ )ﺍﻣﺮﻭﺯﻩ ﻏﻴﺮ ﺍﻗﺘﺼﺎﺩﻱ(‬ ‫¾‬ ‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻛﻮﺭﻩ ﻗﻮﺱ ﺍﻟﻜﺘﺮﻳﻜﻲ )‪(Electric Arc Furnaces‬‬ ‫– )ﺭﻭﺷﻲ ﺟﺪﻳﺪ(‬ ‫‪28‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ‬ STEEL Manufacturing 29 ‫ﺭﻭﺵ ﻛﻮﺭﻩ ﺫﻭﺏ ﺁﻫﻦ – ﻣﺠﺘﻤﻊ ﻛﺎﺭﺧﺎﻧﻪ ﻫﺎﻱ ﻓﻮﻻﺩ‬ ‫‪Blast Furnace Method – Integrated Steel Plants‬‬ ‫ﮔﺎﻡ ﻫﺎﻱ ﺍﺳﺎﺳﻲ ﺍﻳﻦ ﺭﻭﺵ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪:‬‬ ‫‪.1‬‬ ‫‪.2‬‬ ‫‪.3‬‬ ‫‪.4‬‬ ‫‪.5‬‬ ‫‪.6‬‬ ‫‪.7‬‬ ‫‪.8‬‬ ‫‪.9‬‬ ‫‪30‬‬ ‫ﻣﺼﺎﻟﺢ ﺧﺎﻡ )‪ (raw material‬ﺫﺧﺎﻳﺮ ﻃﺒﻴﻌﻲ ﺳﻨﮓ ﺁﻫﻦ )‪ (iron ore‬ﻫﺴﺘﻨﺪ ‪ :‬ﻣﺎﮔﻨﺘﻴﺖ )‪ (Fe3O4‬ﻭ ﻳﺎ ﻫﻤﺎﺗﻴﺖ‬ ‫ﺩﺭ ﻳﻚ ﻛﻮﺭﻩ ﺫﻭﺏ ﺁﻫﻦ ‪ ،‬ﻃﻲ ﭘﺮﻭﺳﻪ ﺫﻭﺏ ﻛﺮﺩﻥ ‪ ،‬ﺁﻫﻦ ﺍﺯ ﺳﻨﮓ ﺁﻫﻦ ﺧﺎﺭﺝ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﻋﻤﻞ ﺫﻭﺏ ﺑﺎﻋﺚ ﺍﻳﺠﺎﺩ ﺗﻔﺎﻟﻪ ﻫﺎﻳﻲ )‪ (slags‬ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺑﺮ ﺭﻭﻱ ﺳﻄﺢ ﺁﻫﻦ ﻣﺬﺍﺏ ﺷﻨﺎﻭﺭﻧﺪ‪.‬‬ ‫ﺑﻪ ﺁﻫﻦ ﻣﺬﺍﺑﻲ ﻛﻪ ﺗﻔﺎﻟﻪ ﻫﺎ ﺭﻭﻱ ﺁﻥ ﺷﻨﺎﻭﺭﻧﺪ ‪ ،‬ﺁﻫﻦ ﺧﺎﻡ ﻣﻲ ﮔﻮﻳﻨﺪ‪.‬‬ ‫ﺁﻫﻦ ﺧﺎﻡ ﺑﻪ ﺷﻜﻞ ﺑﻠﻮﻙ ﻫﺎﻱ ﻓﻮﻻﺩ ﺑﺎﺭﻳﻚ ﺷﺪﻩ ﺩﺭ ﻣﻲ ﺁﻳﺪ‪ .‬ﻛﺮﺑﻦ ﺁﻫﻦ ﺧﺎﻡ ﺧﻴﻠﻲ ﺑﻴﺶ ﺍﺯ ﻣﻘﺪﺍﺭﻱ ﺍﺳﺖ ﻛﻪ ﺑﺘﻮﺍﻥ‬ ‫ﺍﺯ ﺁﻥ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺼﺎﻟﺢ ﺳﺎﺧﺘﻤﺎﻧﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪ .‬ﺑﺪﻳﻦ ﻣﻨﻈﻮﺭ ﻻﺯﻡ ﺍﺳﺖ ﺁﻫﻦ ﺧﺎﻡ ﭘﺎﻻﻳﺶ ﺷﻮﺩ‪.‬‬ ‫ﺍﺯ ﻳﻚ ﻛﻮﺭﻩ ﺍﻛﺴﻴﮋﻥ ﺑﺮﺍﻱ ﭘﺎﻻﻳﺶ ﻓﻮﻻﺩ ﻣﺬﺍﺏ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﺁﻫﻦ ﻣﺬﺍﺏ ﺩﺭ ﺣﺎﻻﺕ ﻣﺎﻳﻊ ﺗﺤﺖ ﭘﺎﻻﻳﺶ ﺑﻴﺸﺘﺮ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﻭ ﺑﺪﺍﻥ ﺁﻟﻴﺎژﻫﺎﻱ ﺁﻫﻦ )‪ (Ferro alloys‬ﺍﻓﺰﻭﺩﻩ‬ ‫ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﺳﭙﺲ ﻋﻤﻠﻴﺎﺕ ﺭﻳﺨﺘﻪ ﮔﺮﻱ ﺑﺮ ﺭﻭﻱ ﻓﻮﻻﺩ ﻣﺬﺍﺏ ﺍﻧﺠﺎﻡ ﮔﺮﻓﺘﻪ ﻭ ﺷﻤﺶ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺩﺭ ﺍﺷﻜﺎﻝ ﻣﺨﺘﻠﻒ ﺗﻮﻟﻴﺪ‬ ‫ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﺳﭙﺲ ﺷﻤﺶ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ ﺗﺎ ﻓﻠﺰﻱ ﭼﻜﺶ ﺧﻮﺍﺭ ﻭ ﻧﺮﻡ ﻭ ﺍﻧﻌﻄﺎﻑ ﭘﺬﻳﺮ ﺍﻳﺠﺎﺩ ﺷﺪﻩ ﻭ ﺩﺭ ﺍﺩﺍﻣﻪ‬ ‫ﻋﻤﻠﻴﺎﺕ ﻧﻮﺭﺩ )ﺑﺮﺍﻱ ﺗﻮﻟﻴﺪ ﻭﺭﻕ ﻫﺎ ﻭ ﻧﻴﻤﺮﺥ ﻫﺎ( ﻳﺎ ﻋﻤﻠﻴﺎﺕ ﻛﺸﺶ )ﺑﺮﺍﻱ ﺗﻮﻟﻴﺪ ﺳﻴﻢ ﻫﺎ( ﺍﻧﺠﺎﻡ ﻣﻲ ﮔﻴﺮﺩ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺵ ﻛﻮﺭﻩ ﻗﻮﺱ ﺍﻟﻜﺘﺮﻳﻜﻲ – ﻛﺎﺭﺧﺎﻧﻪ ﻫﺎﻱ ﻓﻮﻻﺩ ﻛﻮﭼﻚ‬ ‫‪Electric Arc Furnace Method – Mini Steel Plants‬‬ ‫ﻣﺮﺍﺣﻞ ﺍﺳﺎﺳﻲ ﺍﻳﻦ ﺭﻭﺵ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪:‬‬ ‫‪ .1‬ﻣﺼﺎﻟﺢ ﺧﺎﻡ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪ :‬ﺁﻫﻦ ﻗﺮﺍﺿﻪ‪ ،‬ﺁﻫﻦ ﺟﺎﻣﺪ ﻓﻠﺰﻱ‪ ،‬ﻭ ﺁﻟﻴﺎژﻫﺎﻱ ﺁﻫﻦ‪.‬‬ ‫‪ .2‬ﺫﻭﺏ ﻣﺼﺎﻟﺢ ﺑﻮﺳﻴﻠﻪ ﺍﻟﻜﺘﺮﻭﺩﻫﺎﻱ ﻛﺮﺑﻨﻲ ﺗﺤﺖ ﺟﺮﻳﺎﻥ ﺑﺴﻴﺎﺭ ﺷﺪﻳﺪ ﻭ ﻳﻚ ﻗﻮﺱ ﺍﻟﻜﺘﺮﻳﻜﻲ ﺑﻴﻦ ﺍﻟﻜﺘﺮﻭﺩﻫﺎ‬ ‫ﺍﻧﺠﺎﻡ ﻣﻲ ﮔﻴﺮﺩ‪.‬‬ ‫‪ .3‬ﻓﻮﻻﺩ ﻣﺬﺍﺏ ﺩﺭ ﻛﻮﺭﻩ ﺗﻮﻟﻴﺪ ﺷﺪﻩ ﻭ ﻧﻈﻴﺮ ﺣﺎﻟﺖ ﻗﺒﻞ ‪ ،‬ﻻﺯﻡ ﺍﺳﺖ ﻋﻤﻠﻴﺎﺕ ﭘﺎﻻﻳﺶ ﺑﺮ ﺭﻭﻱ ﺁﻥ ﺻﻮﺭﺕ‬ ‫ﮔﻴﺮﺩ‪.‬‬ ‫‪ .4‬ﺩﺭ ﺍﺩﺍﻣﻪ ﺭﻭﻧﺪ ﺗﻮﻟﻴﺪ ‪ ،‬ﻣﺮﺍﺣﻞ ‪ 8 ، 7‬ﻭ ‪ 9‬ﺭﻭﺵ ﻛﻮﺭﻩ ﺫﻭﺏ ﺁﻫﻦ )ﺭﻭﺵ ﻗﺒﻞ( ﺩﻧﺒﺎﻝ ﻣﻲ ﮔﺮﺩﺩ‪.‬‬ ‫ﺭﻭﺵ ﻛﻮﺭﻩ ﻗﻮﺱ ﺍﻟﻜﺘﺮﻳﻜﻲ ﺍﻗﺘﺼﺎﺩﻱ ﺗﺮﻳﻦ ﺭﻭﺵ ﺑﺮﺍﻱ ﺗﻮﻟﻴﺪ ﻣﺼﺎﻟﺢ ﻓﻮﻻﺩ ﺳﺎﺯﻩ ﺍﻱ ﺍﺳﺖ‪.‬‬ ‫ﻣﺠﻤﻮﻋﻪ ﻫﺎﻱ ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ ﻛﻪ ﺷﺎﻣﻞ ﻛﻮﺭﻩ ﻗﻮﺱ ﺍﻟﻜﺘﺮﻳﻜﻲ ﻫﺴﺘﻨﺪ ﺑﺴﻴﺎﺭ ﻛﻮﭼﻜﺘﺮ ﺍﺯ ﻣﺠﻤﻮﻋﻪ ﻫﺎﻳﻲ ﻫﺴﺘﻨﺪ ﻛﻪ‬ ‫ﺷﺎﻣﻞ ﻛﻮﺭﻩ ﺫﻭﺏ ﺁﻫﻦ ﻣﻲ ﺑﺎﺷﻨﺪ ‪ ،‬ﺍﺯ ﺍﻳﻨﺮﻭ ﺳﺖ ﻛﻪ ﺑﻪ ﭼﻨﻴﻦ ﻣﺠﻤﻮﻋﻪ ﻫﺎﻳﻲ ﻛﺎﺭﺧﺎﻧﻪ ﻓﻮﻻﺩ ﻛﻮﭼﻚ ﮔﻔﺘﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪31‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ‬ ‫‪STEEL Manufacturing‬‬ ‫ ﺩﻱ ﺍﻛﺴﻴﺪﺍﺳﻴﻮﻥ )‪: (Deoxidizing‬‬ ‫ﭘﺲ ﺍﺯ ﺗﻜﻤﻴﻞ ﺷﺪﻥ ﻋﻤﻠﻴﺎﺕ ﭘﺎﻻﻳﺶ ‪ ،‬ﺩﻱ ﺍﻛﺴﻴﺪ ﻛﻨﻨﺪﻩ ﻫﺎ ﺩﺭ ﻃﻲ ﻋﻤﻠﻴﺎﺕ ﺭﻭﺍﻥ‬ ‫ﺳﺎﺯﻱ )‪ (pouring‬ﻭ ﺟﺎﻣﺪﺳﺎﺯﻱ )‪ (solidification‬ﺑﻪ ﻓﻮﻻﺩ ﺍﻓﺰﻭﺩﻩ ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫ﺣﺬﻑ ﺍﻛﺴﻴﮋﻥ ﺍﺯ ﻣﻮﺍﺩ ﻣﺬﺍﺏ ﺭﺍ ﺍﻣﺤﺎ )‪ (killing‬ﻣﻲ ﻧﺎﻣﻨﺪ ﺯﻳﺮﺍ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻥ‬ ‫ﺩﻱﺍﻛﺴﻴﺪﻛﻨﻨﺪﻩ ﻫﺎ ﻣﻮﺟﺐ ﺗﻮﻗﻒ ﻳﺎ ﻣﺤﻮ ﺗﺸﻜﻴﻞ ﺣﺒﺎﺏ ﻫﺎﻱ ﮔﺎﺯ ﻣﻲ ﺷﻮﺩ ‪:‬‬ ‫ﻓﻮﻻﺩ ﺑﺎ ﺍﻣﺤﺎﻱ ﻛﺎﻣﻞ )‪ : (Fully-Killed Steel‬ﺑﺎﻻﺗﺮﻳﻦ ﺳﻄﺢ ﺩﻱ ﺍﻛﺴﻴﺪﺍﺳﻴﻮﻥ‬ ‫ﻓﻮﻻﺩ ﺑﺎ ﺍﻣﺤﺎﻱ ﻣﺘﻮﺳﻂ )‪ : (Semi-Killed Steel‬ﺳﻄﺢ ﻣﺘﻮﺳﻂ ﺩﻱ ﺍﻛﺴﻴﺪﺍﺳﻴﻮﻥ‬ ‫ﻓﻮﻻﺩ ﺑﺎ ﺍﻣﺤﺎﻱ ﺟﺰﺋﻲ )‪ : (Rimmed Steel‬ﺣﺬﻑ ﺍﻛﺴﻴﮋﻥ ﺟﺰﺋﻲ‬ ‫‪32‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻳﺨﺘﻪ ﮔﺮﻱ ﻓﻮﻻﺩ ﻭ ﻧﻮﺭﺩ ﮔﺮﻡ‬ Steel Casting & Hot Rolling Urmia University, M. Sheidaii 33 ‫ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ‬ ‫‪STEEL Manufacturing‬‬ ‫ ﺭﻳﺨﺘﻪ ﮔﺮﻱ ﻗﺎﻟﺐ )‪(Ingot Casting‬‬ ‫ﻓﻮﻻﺩ ﻣﺬﺍﺏ ﺑﻌﺪ ﺍﺯ ﻋﻤﻠﻴﺎﺕ ﺩﻱ ﺍﻛﺴﻴﺪﺍﺳﻴﻮﻥ ﺑﻪ ﺷﻜﻞ ﻗﺎﻟﺒﻬﺎﻱ ﺑﺎﺭﻳﻜﻲ ﺩﺭ ﻣﻲ ﺁﻳﺪ )ﺑﻪ‬ ‫ﻃﻮﻝ ‪ 7‬ﻓﻮﺕ ﻭ ﻣﻘﻄﻊ ﻣﺮﺑﻌﻲ ﺑﻪ ﻃﻮﻝ ﺿﻠﻊ ‪ 2‬ﻓﻮﺕ ﻭ ﻭﺯﻥ ﺗﻘﺮﻳﺒﻲ ‪ 7‬ﺗﻦ(‬ ‫ﺳﭙﺲ ﻗﺎﻟﺒﻬﺎ ﺑﻪ ﻛﻮﺭﻩ ﺍﻱ ﻛﻪ ‪ soaking pit‬ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ‪ ،‬ﺣﻤﻞ ﻣﻲ ﺷﻮﻧﺪ ﻭ ﺗﺎ‬ ‫ﻫﻨﮕﺎﻣﻲ ﻛﻪ ﺩﺭﺟﻪ ﺣﺮﺍﺭﺕ ﻳﻜﺴﺎﻧﻲ ﺩﺭ ﺗﻤﺎﻡ ﺣﺠﻢ ﺷﺎﻥ ﺍﻳﺠﺎﺩ ﺷﻮﺩ ﺩﺭ ﻫﻤﺎﻥ ﺟﺎ ﺑﺎﻗﻲ‬ ‫ﻣﻲ ﻣﺎﻧﻨﺪ ‪ ،‬ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﻗﺎﻟﺐ ﺁﻣﺎﺩﻩ ﺑﺮﺍﻱ ﺍﻧﺠﺎﻡ ﻋﻤﻠﻴﺎﺕ ﻧﻮﺭﺩ ﻣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫‪34‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ‬ ‫‪STEEL Manufacturing‬‬ ‫ ﻧﻮﺭﺩ‬ ‫)‪(Rolling‬‬ ‫ﺩﺭ ﻋﻤﻠﻴﺎﺕ ﻧﻮﺭﺩ ‪ ،‬ﻗﺎﻟﺐ ﺑﺎ ﻓﺸﺎﺭ ﺍﺯ ﺑﻴﻦ ﻏﻠﺘﻚ ﻫﺎﻱ ﭼﺮﺧﺎﻥ ﻋﺒﻮﺭ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ‪ .‬ﻗﺎﻟﺐ ﻫﺎ ﺑﻴﺶ ﺍﺯ ﻳﻜﺒﺎﺭ ﺍﺯ ﺑﻴﻦ‬ ‫ﻏﻠﺘﻚ ﻫﺎ ﻋﺒﻮﺭ ﺩﺍﺩﻩ ﺷﺪﻩ ﻭ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺁﻧﻬﺎ ﻛﺎﻫﺶ ﻳﺎﻓﺘﻪ ﻭ ﻃﻮﻝ ﺷﺎﻥ ﺍﻓﺰﺍﻳﺶ ﻣﻲ ﻳﺎﺑﺪ‪) .‬ﺗﻮﺟﻪ ﺷﻮﺩ ﻛﻪ ﺣﺠﻢ‬ ‫ﻗﺎﻟﺐ ﺛﺎﺑﺖ ﺍﺳﺖ‪(.‬‬ ‫ﭘﺲ ﺍﺯ ﻋﻤﻠﻴﺎﺕ ﻣﻘﺪﻣﺎﺗﻲ ﻏﻠﺘﻚ ﺯﺩﻥ ‪ ،‬ﻓﻮﻻﺩ ﺑﺎ ﻛﻴﻔﻴﺖ ﭘﺎﻳﻴﻦ ﻣﻌﻤﻮﻻ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻗﺎﻟﺐ ﺟﻤﻊ ﻣﻲ ﺷﻮﺩ‪ .‬ﺍﺯ‬ ‫ﺍﻳﻨﺮﻭﺳﺖ ﻛﻪ ﺍﻧﺘﻬﺎﻫﺎﻱ ﻗﺎﻟﺐ ﺑﺮﻳﺪﻩ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﻋﻤﻠﻴﺎﺕ ﻧﻮﺭﺩ ﺍﻭﻟﻴﻪ ﺭﺍ ‪ Blooming‬ﻭ ‪ Slabbing‬ﮔﻮﻳﻨﺪ‪ .‬ﺍﻳﻦ ﻋﻤﻠﻴﺎﺕ ﺗﻮﺳﻂ ﻏﻠﺘﻚ ﻫﺎﻱ ‪ Blooming‬ﻭ‬ ‫‪ Slabbing‬ﺻﻮﺭﺕ ﻣﻲ ﮔﻴﺮﺩ‪.‬‬ ‫‪ Slabs ، Billers ، Blooms‬ﻣﺤﺼﻮﻻﺗﻲ ﻧﻴﻤﻪ ﭘﺮﺩﺍﺧﺖ ﺷﺪﻩ ﻫﺴﺘﻨﺪ‪ .‬ﺩﺭ ﺍﺻﻞ ﺍﻳﻨﻬﺎ ﻣﺤﺼﻮﻻﺕ ﭘﺎﻳﻪ ﻫﺴﺘﻨﺪ‬ ‫ﻛﻪ ﺑﻪ ﻏﻠﺘﻜﻬﺎﻱ ﭘﺮﺩﺍﺧﺘﻜﺎﺭﻱ ﺍﺭﺳﺎﻝ ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫‪35‬‬ ‫ﭼﻮﻥ ﻏﻠﺘﻚ ﻛﺎﺭﻱ ﺑﺮ ﺭﻭﻱ ﻣﺤﺼﻮﻻﺗﻲ ﻛﻪ ﻫﻨﻮﺯ ﺩﺍﻍ ﻫﺴﺘﻨﺪ ﺍﻧﺠﺎﻡ ﻣﻲ ﺷﻮﺩ ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﺗﻮﻟﻴﺪ ﺷﺪﻩ ﺭﺍ ﺍﻏﻠﺐ‬ ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻧﻮﺭﺩ ﮔﺮﻡ )‪ (hot-rolled shapes‬ﻣﻲ ﻧﺎﻣﻨﺪ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ DIN ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﮔﺮﻡ ﻧﻮﺭﺩ ﺷﺪﻩ ﺍﺭﻭﭘﺎﻳﻲ‬ Hot-Rolled European Standard Cross-Sectional Shapes (DIN) T IPB IPBl INP IPE C L IPBv Pipe Urmia University, M. Sheidaii Tube Bars Plate 36 ASTM ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﮔﺮﻡ ﻧﻮﺭﺩ ﺷﺪﻩ ﺁﻣﺮﻳﻜﺎﻳﻲ‬ Hot-Rolled American Standard Cross-Sectional Shapes (ASTM) Urmia University, M. Sheidaii 37 ASTM ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﮔﺮﻡ ﻧﻮﺭﺩ ﺷﺪﻩ ﺁﻣﺮﻳﻜﺎﻳﻲ‬ Hot-Rolled American Standard Cross-Sectional Shapes (ASTM) Urmia University, M. Sheidaii 38 ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺳﺮﺩﻛﺎﺭ ﺷﺪﻩ‬ ‫‪Cold-Formed Steel Shapes‬‬ ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺳﺮﺩ ﻛﺎﺭ ﺷﺪﻩ ﺑﻪ ﺷﻜﻠﻬﺎﻱ ﻣﺨﺘﻠﻒ ﺩﺭ ﺩﻣﺎﻱ ﻣﻌﻤﻮﻟﻲ )ﻣﺤﻴﻂ( ﺑﺎ ﻋﻤﻠﻴﺎﺕ‬ ‫ﺧﻢﻛﺎﺭﻱ ﺍﻳﺠﺎﺩ ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫‪39‬‬ ‫ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺳﺮﺩﻛﺎﺭ ﺷﺪﻩ‬ ‫‪Cold-Formed Steel Shapes‬‬ ‫ ﺑﺎ ﺧﻢ ﻛﺮﺩﻥ ﻭﺭﻗﻪ ﻫﺎﻱ ﻧﺎﺯﻙ ﻓﻮﻻﺩﻱ ﻛﺮﺑﻨﻲ ﻳﺎ ﻛﻢ ﺁﻟﻴﺎژ ﺳﺎﺧﺘﻪ ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﺳﺒﻚ ﺑﺎﻡ ﻫﺎ ‪ ،‬ﻛﻒ ﻫﺎ ﻭ ﺩﻳﻮﺍﺭﻫﺎ ﻣﻤﻜﻦ ﺍﺳﺖ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ‪.‬‬ ‫ ﺿﺨﺎﻣﺖﺷﺎﻥ ﺍﺯ ‪ ./25 mm‬ﺗﺎ ﺑﻴﺶ ﺍﺯ ‪ 6/5 mm‬ﻣﺘﻐﻴﺮ ﺍﺳﺖ‪.‬‬ ‫ ﺍﮔﺮﭼﻪ ﻋﻤﻠﻴﺎﺕ ﺳﺮﺩﻛﺎﺭﻱ ﺗﺎ ﺣﺪﻱ ﺷﻜﻞ ﭘﺬﻳﺮﻱ ﺭﺍ ﻛﺎﻫﺶ ﻣﻲ ﺩﻫﺪ ﻭﻟﻲ ﺩﺭ ﻋﻴﻦ ﺣﺎﻝ‬ ‫ﻣﻘﺎﻭﻣﺖ ﺭﺍ ﻗﺪﺭﻱ ﺍﻓﺰﺍﻳﺶ ﻣﻲ ﺩﻫﺪ‪.‬‬ ‫‪40‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺒﻘﻪ ﺑﻨﺪﻱ ﺍﻋﻀﺎ ﺑﺮﺍﺳﺎﺱ ﺑﺎﺭﮔﺬﺍﺭﻱ‬ MEMBER CLASSIFICATION WITH LOADING Urmia University, M. Sheidaii 41 ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺘﺪﺍﻭﻝ‬ Typical Tension Members Urmia University, M. Sheidaii 42 ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺘﺪﺍﻭﻝ‬ Typical Compression Members Urmia University, M. Sheidaii 43 ‫ﺍﻋﻀﺎﻱ ﺗﻴﺮﻱ ﻣﺘﺪﺍﻭﻝ‬ Typical Beam Members Urmia University, M. Sheidaii 44 ‫ﺍﻧﻮﺍﻉ ﻓﻮﻻﺩ ﺳﺎﺯﻩ ﺍﻱ‬ Types of Structural Steels ‫ ﻓﻮﻻﺩ ﻛﺮﺑﻨﻲ ﻧﺮﻡ‬ (Mild carbon steel) ‫ ﻓﻮﻻﺩ ﭘﺮ ﻣﻘﺎﻭﻣﺖ ﻛﻢ ﺁﻟﻴﺎژ‬ (High Strength Low Alloy (HS)) (‫ ﻓﻮﻻﺩ ﺁﻟﻴﺎژﻱ ﺁﺑﺪﻳﺪﻩ ﺑﺎﺯﭘﺨﺖ ﺷﺪﻩ )ﻓﻮﻻﺩ ﺧﺸﻜﻪ‬ (Quenched and tempered Alloy Steel) Urmia University, M. Sheidaii 45 ‫ ﻓﻮﻻﺩ ﻣﻌﻤﻮﻟﻲ‬،‫ ﻓﻮﻻﺩ ﺳﺎﺧﺘﻤﺎﻧﻲ‬،‫ﻓﻮﻻﺩ ﻧﺮﻣﻪ‬ Mild carbon steel 0.15 to 0.29 Carbon Percent 2200 ≤ Fy ≤ 2600 kgf/cm2 3400 ≤ Fu ≤ 3800 kgf/cm2 ‫( ﺍﺭﺯﺍﻥ ﻗﻴﻤﺖ‬Low Cost) Works Easily( ‫)ﺑﺨﺎﻃﺮ ﺷﻜﻞ ﭘﺬﻳﺮﻱ ﺯﻳﺎﺩ‬ A36 ASTM American Standard ST37 DIN Germany Standard Urmia University, M. Sheidaii 46 ‫ﻓﻮﻻﺩ ﭘﺮ ﻣﻘﺎﻭﻣﺖ ﻛﻢ ﺁﻟﻴﺎژ‬ ‫)‪High Strength Low Alloy (HS‬‬ ‫ ﺑﺎ ﺍﻓﺰﻭﺩﻥ ﻣﻘﺎﺩﻳﺮ ﺟﺰﺋﻲ ﻋﻨﺎﺻﺮ ﺁﻟﻴﺎژﻱ ﻧﻈﻴﺮ ﻛﺮﻡ ‪ ،‬ﻣﺲ ‪ ،‬ﻣﻮﻟﻴﺒﺪﻥ ﻭ ﻧﻴﻜﻞ ﺑﻪ ﻓﻮﻻﺩﻫﺎﻱ ﻛﺮﺑﻨﻲ‬ ‫ﺗﻮﻟﻴﺪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ‪2750 ≤ Fy ≤ 4800 kgf/cm2‬‬ ‫ ‪4600 ≤ Fu ≤ 5400 kgf/cm2‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺑﺎﻻ‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺑﺎﻻ ﺩﺭ ﺑﺮﺍﺑﺮ ﺧﻮﺭﺩﮔﻲ – ﻓﻮﻻﺩﻫﺎﻱ ﺿﺪ ﻓﺮﺳﺎﻳﺶ )ﺿﺪ ﺧﻮﺭﺩﮔﻲ(‬ ‫)‪(Increased Corrosion Resistance –Weathering Steels‬‬ ‫ ﺑﻌﻀﻲ ﺍﺯ ﺍﻳﻦ ﻓﻮﻻﺩﻫﺎ ﺟﻮﺷﻜﺎﺭﻱ ﻛﺮﺩﻧﺸﺎﻥ ﺑﺴﻴﺎﺭ ﻣﺸﻜﻞ ﺍﺳﺖ‪.‬‬ ‫ ‪A242, A441, A572,A588,A992‬‬ ‫‪ASTM American Standard‬‬ ‫ ‪ST52‬‬ ‫‪DIN Germany Standard‬‬ ‫‪47‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻓﻮﻻﺩ ﺁﻟﻴﺎژﻱ ﺁﺑﺪﻳﺪﻩ ﺑﺎﺯ ﭘﺨﺖ ﺷﺪﻩ‬ ‫‪Quenched and tempered Alloy Steel‬‬ ‫‬ ‫‬ ‫ﺑﺎ ﺍﻧﺠﺎﻡ ﻋﻤﻠﻴﺎﺕ ﺣﺮﺍﺭﺗﻲ ﺯﻳﺮ ﺑﺮ ﺭﻭﻱ ﻓﻮﻻﺩﻫﺎﻱ ﻛﻢ ﺁﻟﻴﺎژ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﻨﺪ ‪:‬‬ ‫‪.I‬‬ ‫ﺁﺑﺪﻳﺪﻩ ﻛﺮﺩﻥ )ﺗﺒﺮﻳﺪ( ‪ :‬ﻓﻮﻻﺩ ﺭﺍ ﺍﺯ ﺣﺮﺍﺭﺕ ﺣﺪﻭﺩ ‪ 900 Ԩ‬ﺳﺮﻳﻌﺎ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺁﺏ ﻳﺎ ﺭﻭﻏﻦ ﺑﻪ ﺣﺪﻭﺩ ‪ 200Ԩ‬ﻣﻲﺭﺳﺎﻧﻨﺪ‪.‬‬ ‫‪.II‬‬ ‫ﺑﺎﺯﭘﺨﺖ ‪ :‬ﺣﺮﺍﺭﺕ ﻓﻮﻻﺩ ﺭﺍ ﻣﺠﺪﺩﺍ ﺗﺎ ﺣﺪﻭﺩ ‪ 600Ԩ‬ﺑﺎﻻ ﺑﺮﺩﻩ ﻭﻟﻲ ﻣﻲ ﮔﺬﺍﺭﻧﺪ ﺗﺎ ﺑﺘﺪﺭﻳﺞ ﺧﻨﻚ ﺷﻮﺩ‪.‬‬ ‫ﺍﻳﻦ ﻧﻮﻉ ﻓﻮﻻﺩﻫﺎ ﺣﺪ ﺗﺴﻠﻴﻢ ﻣﺸﺨﺼﻲ ﻧﺪﺍﺭﻧﺪ ﻭ ﻟﺬﺍ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﻣﻌﻤﻮﻻ ﺑﺎ ﺭﺳﻢ ﺧﻄﻲ ﺍﺯ ﻛﺮﻧﺶ ‪ 0/002‬ﺑﻪ ﻣﻮﺍﺯﺍﺕ‬ ‫ﺗﻐﻴﻴﺮﺍﺕ ﺧﻄﻲ ﺑﺨﺶ ﺍﺭﺗﺠﺎﻋﻲ ﻓﻮﻻﺩ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ‪.‬‬ ‫‬ ‫‪Fu ≥ 8000 kgf/cm2‬‬ ‫‬ ‫ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺑﺴﻴﺎﺭ ﺑﺎﻻﺗﺮ‬ ‫‬ ‫ﻗﻴﻤﺖ ﺑﺎﻻﺗﺮ ﺑﻌﻠﺖ ﻋﻤﻠﻴﺎﺕ ﺣﺮﺍﺭﺗﻲ‬ ‫‬ ‫ﺷﻜﻞ ﭘﺬﻳﺮﻱ ﻛﻤﺘﺮ‬ ‫‬ ‫ﻟﺰﻭﻡ ﭘﻴﺶ ﮔﺮﻡ ﻛﺮﺩﻥ ﻗﺒﻞ ﺍﺯ ﺟﻮﺷﻜﺎﺭﻱ‬ ‫‬ ‫‪48‬‬ ‫‪A514‬‬ ‫;‬ ‫‪Fy ≥ 6000 kgf/cm2‬‬ ‫‪ASTM American Standard‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 49 Urmia University, M. Sheidaii 50 ‫ﺁﺯﻣﺎﻳﺶ ﻛﺸﺶ‬ ‫‪The Tension Test‬‬ ‫ﻣﻨﺤﻨﻲ ﺗﻨﺶ‪-‬‬ ‫ﻛﺮﻧﺶ ﺣﻘﻴﻘﻲ‬ ‫ﻣﻨﺤﻨﻲ ﺗﻨﺶ‪-‬‬ ‫ﻛﺮﻧﺶ ﻣﻬﻨﺪﺳﻲ‬ ‫‪51‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺸﺨﺼﺎﺕ ﻓﻮﻻﺩﻫﺎﻱ ﺳﺎﺯﻩ ﺍﻱ‬ Properties of Structural Steels E = 2.03u106 kgf/cm2 (=29000 ksi) G = 0.77u106 kgf/cm2 Q = 0.3 J = 7850 kgf/m3 D = 12 u10-6 /ºC Urmia University, M. Sheidaii 52 (‫ﻓﻮﻻﺩﻫﺎﻱ ﻭﺳﺎﻳﻞ ﺍﺗﺼﺎﻻﺕ ) ﭘﻴﭻ ﻭ ﭘﺮچ‬ Fastener Steels A) Carbon Steel Bolts (A-307): These are common non-structural fasteners with minimum tensile strength (Fu) of 60 ksi. B) High Strength Bolts (A-325): These are structural fasteners (bolts) with low carbon, their ultimate tensile strength could reach 105 ksi. C) Quenched and Tempered Bolts (A-449): These are similar to A-307 in strength but can be produced to large diameters exceeding 1.5 inch, D) Heat Treated Structural Steel Bolts (A-490): These are in carbon content (upto 0.5%) and has other alloys. They are quenched and re-heated (tempered) to 900oF. The minimum yield strength (Fy) for these bolts ranges from 115 ksi upto 130 ksi. 53 Specifications, Loads and Design Methods ‫ ﺑﺎﺭﻫﺎ ﻭ ﺭﻭﺷﻬﺎﻱ ﻃﺮﺍﺣﻲ‬،‫ ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞﻫﺎ‬: ‫ﻓﺼﻞ ﺩﻭﻡ‬ Urmia University, M. Sheidaii 1 Specifications and Building Codes ‫ﺩﺳﺘﻮﺭ ﺍﻟﻌﻤﻞﻫﺎ ﻭ ﺁﻳﻴﻦ ﻧﺎﻣﻪﻫﺎﻱ ﺳﺎﺧﺘﻤﺎﻧﻲ‬ Urmia University, M. Sheidaii 2 ‫ﺩﺳﺘﻮﺭ ﺍﻟﻌﻤﻞ ﻫﺎ‬ ‫‪Specifications‬‬ ‫ ﺗﻮﺳﻂ ﺳﺎﺯﻣﺎﻥ ﻫﺎﻳﻲ ﻧﻈﻴﺮ ‪ ASCE ، ACI ، AISC‬ﺑﺴﻂ ﻭ ﺗﻮﺳﻌﻪ‬ ‫ﻣﻲ ﻳﺎﺑﻨﺪ‪.‬‬ ‫ ﺭﺍﻫﻨﻤﺎﻳﻲﻫﺎﻳﻲ ﺭﺍ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﺳﺎﺯﻩ ﺍﻱ ﻭ ﺍﺗﺼﺎﻻﺕ ﺁﻧﻬﺎ ﺍﺭﺍﺋﻪ‬ ‫ﻣﻲ ﻧﻤﺎﻳﻨﺪ‪.‬‬ ‫ ﺁﻧﻬﺎ ﺑﻪ ﺧﻮﺩﻱ ﺧﻮﺩ ﺩﺍﺭﺍﻱ ﺍﻟﺰﺍﻡ ﻗﺎﻧﻮﻧﻲ ﻧﺒﻮﺩﻩ ‪ ،‬ﺍﻣﺎ ﺑﻪ ﺭﺍﺣﺘﻲ ﻣﻤﻜﻦ‬ ‫ﺍﺳﺖ ﺑﻪ ﻋﻨﻮﺍﻥ ﺑﺨﺸﻲ ﺍﺯ ﻳﻚ ﺁﺋﻴﻦ ﻧﺎﻣﻪ ﺳﺎﺧﺘﻤﺎﻧﻲ ﻣﺮﺟﻊ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ‪.‬‬ ‫ ﻫﻤﻪ ﻣﻮﺍﺭﺩ ﻣﻤﻜﻦ ﺭﺍ ﺩﺭ ﺑﺮ ﻧﻤﻲﮔﻴﺮﻧﺪ ) ﺍﻣﻜﺎﻥ ﺍﻳﻨﻜﻪ ﺑﺘﻮﺍﻧﻨﺪ ﻫﻤﻪ‬ ‫ﺣﺎﻻﺕ ﻣﻤﻜﻦ ﻃﺮﺍﺣﻲ ﺭﺍ ﭘﻮﺷﺶ ﺩﻫﻨﺪ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ (‪.‬‬ ‫‪3‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﺎﺯﻣﺎﻥ ﻫﺎ‬ Organizations AISC = American Institute of Steel Construction ASCE = American Society of Civil Engineers AASHTO = American Association of State Highway and Transportation Officials ACI = American Concrete Institute ‫ ﺑﺎﺭﻫﺎﻱ ﻭﺍﺭﺩ ﺑﺮ ﺳﺎﺧﺘﻤﺎﻥ‬:‫ ﻣﺒﺤﺚ ﺷﺸﻢ‬،‫ ﻣﻘﺮﺭﺍﺕ ﻣﻠﻲ ﺳﺎﺧﺘﻤﺎﻥ ﺍﻳﺮﺍﻥ‬ ‫ ﻃﺮﺡ ﻭﺍﺟﺮﺍﻱ ﺳﺎﺧﺘﻤﺎﻧﻬﺎﻱ ﻓﻮﻻﺩﻱ‬:‫ ﻣﺒﺤﺚ ﺩﻫﻢ‬،‫ ﻣﻘﺮﺭﺍﺕ ﻣﻠﻲ ﺳﺎﺧﺘﻤﺎﻥ ﺍﻳﺮﺍﻥ‬ Urmia University, M. Sheidaii 4 ‫ﺁﺋﻴﻦ ﻧﺎﻣﻪ ﻫﺎﻱ ﺳﺎﺧﺘﻤﺎﻧﻲ‬ ‫‪Building Codes‬‬ ‫ ﺳﻨﺪﻱ ﻗﺎﻧﻮﻧﻲ ﺩﺭ ﺑﺮﺩﺍﺭﻧﺪﻩ ﻣﻠﺰﻭﻣﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﻣﻮﺍﺭﺩﻱ ﺍﺯ ﻗﺒﻴﻞ ﺍﻳﻤﻨﻲ‬ ‫ﺳﺎﺯﻩ‪ ،‬ﺍﻳﻤﻨﻲ ﺁﺗﺶ ﺳﻮﺯﻱ‪ ،‬ﻟﻮﻟﻪﻛﺸﻲ ﻭ ﺗﻬﻮﻳﻪ‬ ‫ ﺍﻟﺰﺍﻡ ﻗﺎﻧﻮﻧﻲ ﺩﺍﺷﺘﻪ ﻭ ﻧﻈﺎﺭﺕ ﺑﺮ ﺁﻥ ﺗﻮﺳﻂ ﺳﺎﺯﻣﺎﻧﻬﺎﻱ ﺷﻬﺮﻱ‪ ،‬ﻛﺸﻮﺭﻱ ﻳﺎ‬ ‫ﺩﻳﮕﺮ ﺳﺎﺯﻣﺎﻧﻬﺎﻱ ﺩﻭﻟﺘﻲ ﺍﻧﺠﺎﻡ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺭﻭﻧﺪ ﻃﺮﺍﺣﻲ ﺩﺭ ﺁﻥ ﺁﻭﺭﺩﻩ ﻧﺸﺪﻩ ﺍﺳﺖ‪ ،‬ﺍﻣﺎ ﺩﺭ ﺁﻥ ﺑﻪ ﺫﻛﺮ ﺍﻟﺰﺍﻣﺎﺕ ﻃﺮﺍﺣﻲ‬ ‫ﭘﺮﺩﺍﺧﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪5‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺁﺋﻴﻦ ﻧﺎﻣﻪ ﻫﺎﻱ ﺳﺎﺧﺘﻤﺎﻧﻲ‬ Building Codes UBC = Uniform Building Code IBC = International Building Code The ASCE7-05, Minimum Design Loads for Building and Other Structures, is another accepted document the United States of America. (519 ‫ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﺣﺪﺍﻗﻞ ﺑﺎﺭ ﻭﺍﺭﺩﻩ ﺑﺮ ﺳﺎﺧﺘﻤﺎﻧﻬﺎ ﻭ ﺍﺑﻨﻴﻪ ﻓﻨﻲ)ﺍﺳﺘﺎﻧﺪﺍﺭﺩ‬ (2800-84 ‫ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻃﺮﺍﺣﻲ ﺳﺎﺧﺘﻤﺎﻧﻬﺎ ﺩﺭ ﺑﺮﺍﺑﺮ ﺯﻟﺰﻟﻪ )ﺍﺳﺘﺎﻧﺪﺍﺭﺩ‬ Urmia University, M. Sheidaii 6 Design Loads ‫ﺑﺎﺭﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ Urmia University, M. Sheidaii 7 ‫ﺑﺎﺭﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ ‫‪Design Loads‬‬ ‫ ﻋﻤﻮﻣﺎً ﺗﻮﺳﻂ ﺁﺋﻴﻦﻧﺎﻣﻪﻫﺎﻳﻲ ﻫﻤﭽﻮﻥ ‪ ، UBC ، AISC‬ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ‪519‬‬ ‫ﻣﺸﺨﺺ ﻣﻲﺷﻮﻧﺪ‪.‬‬ ‫ ﻣﻘﺎﺩﻳﺮ ﺁﻳﻴﻦﻧﺎﻣﻪ ﺣﺪﺍﻗﻞ ﻣﻘﺎﺩﻳﺮ ﺑﺎﺭﻫﺎ ﺑﻮﺩﻩ ﻭ ﻛﻠﻴﻪ ﻭﺿﻌﻴﺖﻫﺎ ﻭ ﻧﻴﺰ ﺳﺎﺯﻩﻫﺎ‬ ‫ﺭﺍ ﺩﺭﺑﺮﻧﻤﻲﮔﻴﺮﻧﺪ‪.‬‬ ‫ ﻣﺴﺌﻮﻟﻴﺖﺷﺎﻥ ﺑﺮ ﻋﻬﺪﻩ ﻃﺮﺍﺡ ﻣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫ ﺩﺭ ﺍﻳﺮﺍﻥ ﻣﺒﺤﺚ ﺷﺸﻢ ﻣﻘﺮﺭﺍﺕ ﻣﻠﻲ ﺳﺎﺧﺘﻤﺎﻥ‪ ،‬ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﺩ‪.‬‬ ‫‪8‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻧﻮﺍﻉ ﺑﺎﺭﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ Types of Design Loads ( Dead Loads ) ‫ ﺑﺎﺭﻫﺎﻱ ﻣﺮﺩﻩ‬ ( Live Loads ) ‫ ﺑﺎﺭﻫﺎﻱ ﺯﻧﺪﻩ‬ Snow/Rain ) ‫ ﺑﺎﺭ ﺑﺮﻑ ﻭ ﺑﺎﺭﺍﻥ‬ ( Load ( Wind Load ) ‫ ﺑﺎﺭ ﺑﺎﺩ‬ ( Seismic Load ) ‫ ﺑﺎﺭ ﺯﻟﺰﻟﻪ‬ ( Earth Pressure )‫ ﻓﺸﺎﺭ ﺧﺎﻙ‬ ( Other ) ‫ ﺳﺎﻳﺮ‬ Urmia University, M. Sheidaii 9 ‫ﺑﺎﺭﻫﺎﻱ ﻣﺮﺩﻩ‬ ‫‪Dead Loads‬‬ ‫ ﺑﺎﺭﻫﺎﻳﻲ ﻫﺴﺘﻨﺪ ﻛﻪ ﻧﺎﺷﻲ ﺍﺯ ﻭﺯﻥ ﺳﺎﺯﻩ ﻣﻲ ﺑﺎﺷﻨﺪ ) ﺷﺎﻣﻞ ﻭﺯﻥ ﺍﺳﻜﻠﺖ ‪ ،‬ﺩﻳﻮﺍﺭﻫﺎ ‪ ،‬ﺳﺮﻭﻳﺲﻫﺎﻱ‬ ‫ﻣﻜﺎﻧﻴﻜﻲ ﻭ ﺍﻟﻜﺘﺮﻳﻜﻲ ﺩﺍﺋﻤﻲ ﻭ ‪.( ....‬‬ ‫ ﺷﺎﻣﻞ ﺍﺟﺰﺍ ﺑﺎﺭﺑﺮ ﻭ ﻏﻴﺮ ﺑﺎﺭﺑﺮ ﺳﺎﺯﻩ ﻫﺴﺘﻨﺪ‪.‬‬ ‫ ﺑﺎﺭﻫﺎﻱ ﺳﺎﻛﻦ ﺑﺎ ﻣﻘﺪﺍﺭ ﺳﺎﻛﻦ ﻫﺴﺘﻨﺪ‪.‬‬ ‫ ﻏﺎﻟﺒﺎً ﺑﺎ ﻗﻄﻌﻴﺖ ﻣﻌﻘﻮﻝ ﻭ ﺑﺎﻻﻳﻲ ﻗﺎﺑﻞ ﺗﻌﻴﻴﻦ ﻫﺴﺘﻨﺪ‪.‬‬ ‫ ﻭﺯﻥ ﺧﻮﺩ ﺳﺎﺯﻩ ﻗﺒﻞ ﺍﺯ ﻃﺮﺡ ﺳﺎﺯﻩ ﻣﻌﻠﻮﻡ ﻧﻴﺴﺖ‪.‬‬ ‫‪10‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺭﻫﺎﻱ ﺯﻧﺪﻩ‬ ‫‪Live Loads‬‬ ‫ ﺑﺎﺭﻫﺎﻱ ﻣﺘﺤﺮﻙ ﻳﺎ ﺑﺎﺭﻫﺎﻳﻲ ﻛﻪ ﻣﻘﺪﺍﺭﺷﺎﻥ ﺗﻐﻴﻴﺮ ﻣﻲ ﻛﻨﺪ‪.‬‬ ‫ ﺑﺎﺭﻫﺎﻳﻲ ﻛﻪ ﻧﺎﺷﻲ ﺍﺯ ﺳﻜﻮﻧﺖ ﻳﺎ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ ﺳﺎﺧﺘﻤﺎﻥ ‪ ،‬ﺳﺎﺧﺖ ‪،‬ﻳﺎ ﺗﻐﻴﻴﺮ ﺳﺎﺯﻩ ﻫﺴﺘﻨﺪ ‪ .‬ﺍﻳﻦ ﺑﺎﺭﻫﺎ‬ ‫ﻣﻤﻜﻦ ﺍﺳﺖ ﻧﺎﺷﻲ ﺍﺯ ﻣﻮﺍﺭﺩ ﺫﻳﻞ ﺑﺎﺷﻨﺪ ‪:‬‬ ‫ ﺳﺎﻛﻨﻴﻦ ﺳﺎﺧﺘﻤﺎﻥ‬ ‫ ﺍﺛﺎﺛﻴﻪ‬ ‫ ﻣﺼﺎﻟﺢ ﺍﻧﺒﺎﺭ ﺷﺪﻩ‬ ‫ ﺗﺠﻬﻴﺰﺍﺕ ﻣﺘﺤﺮﻙ‬ ‫ ﺑﺎﺭ ﺗﺮﺍﻓﻴﻚ ﻭ ﺑﺎﺭ ﭘﻴﺎﺩﻩ ﺭﻭ ﭘﻠﻬﺎ‬ ‫ ﺍﻧﻔﺠﺎﺭ‬ ‫ ﺿﺮﺑﻪ‬ ‫ ﻭﺳﺎﻳﻞ ﻧﻘﻠﻴﻪ‬ ‫ ﺍﮔﺮ ﺑﺎﺭ ﺑﻪ ﺻﻮﺭﺕ ﻧﺎﮔﻬﺎﻧﻲ ﻭﺍﺭﺩ ﺷﻮﺩ ‪ ،‬ﺍﺛﺮﺍﺕ ﺿﺮﺑﻪ ﺑﺎﻳﺪ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻩ ﺷﻮﺩ‪.‬‬ ‫ ﺍﮔﺮ ﺑﺎﺭ ﺩﺭ ﺻﻮﺭﺕ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ ﺍﺯ ﺳﺎﺯﻩ ﺑﺎﺭ ﻣﺰﺑﻮﺭ ﺑﻪ ﻛﺮﺍﺕ ﻭﺍﺭﺩ ﺷﺪﻩ ﻭ ﺑﺮﺩﺍﺷﺘﻪ ﺷﻮﺩ ‪ ،‬ﺗﻨﺶ ﺧﺴﺘﮕﻲ‬ ‫) ‪ ( Fatigue‬ﺑﺎﻳﺪ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻩ ﺷﻮﺩ ‪.‬‬ ‫‪11‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺭ ﺑﺮﻑ‬ ‫‪Snow Load‬‬ ‫ ﺑﻪ ﻣﻮﻗﻌﻴﺖ ﺟﻐﺮﺍﻓﻴﺎﻳﻲ ﺳﺎﺧﺘﻤﺎﻥ ﻭ ﺷﻜﻞ ﺑﺎﻡ ﺳﺎﺧﺘﻤﺎﻥ ﺑﺴﺘﮕﻲ ﺩﺍﺭﺩ‬ ‫‪Pr = Cs Ps‬‬ ‫ﺑﺎﺭ ﺑﺮﻑ ﺭﻭﻱ ﺑﺎﻡ = ‪Pr‬‬ ‫ﺑﺎﺭ ﺑﺮﻑ ﻣﺒﻨﺎ = ‪Ps‬‬ ‫ﺿﺮﻳﺐ ﺍﺛﺮ ﺷﻴﺐ = ‪Cs‬‬ ‫‪12‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺴﻴﺮ ﺑﺎﺭ ﺛﻘﻠﻲ‬ Gravity Load Path Urmia University, M. Sheidaii 13 ‫ ﺗﻴﺮﭼﻪ‬/ ‫ﺳﻄﺢ ﺑﺎﺭ ﮔﻴﺮ ﺗﻴﺮ‬ Beam/Joist Tributary Area Urmia University, M. Sheidaii 14 ‫ﺳﻄﺢ ﺑﺎﺭﮔﻴﺮ ﺷﺎﻫﺘﻴﺮ‬ Girder Tributary Area Urmia University, M. Sheidaii 15 ‫ﺳﻄﺢ ﺑﺎﺭﮔﻴﺮ ﺳﺘﻮﻥ‬ Column Tributary Area Urmia University, M. Sheidaii 16 ‫ﺑﺎﺭ ﺑﺎﺩ‬ ‫‪Wind Load‬‬ ‫ ﻓﺸﺎﺭ ﻳﺎ ﻣﻜﺶ ﺑﻪ ﺳﻄﻮﺡ ﺧﺎﺭﺟﻲ ﺳﺎﺧﺘﻤﺎﻥ ﺍﻋﻤﺎﻝ ﻣﻲ ﻧﻤﺎﻳﺪ ‪.‬‬ ‫‪P = Ce C q q‬‬ ‫‪q = 0.005V2‬‬ ‫ﻓﺸﺎﺭ ﻳﺎ ﻣﻜﺶ ﻧﺎﺷﻲ ﺍﺯ ﺑﺎﺩ = ‪P‬‬ ‫ﺿﺮﻳﺐ ﺍﺛﺮ ﺗﻐﻴﻴﺮ ﺳﺮﻋﺖ = ‪Ce‬‬ ‫ﺿﺮﻳﺐ ﺷﻜﻞ = ‪Cq‬‬ ‫ﻓﺸﺎﺭ ﻣﺒﻨﺎﻱ ﺑﺎﺩ = ‪q‬‬ ‫ﺳﺮﻋﺖ ﻣﺒﻨﺎﻱ ﺑﺎﺩ = ‪V‬‬ ‫‪17‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺭ ﺯﻟﺰﻟﻪ‬ Seismic Load ‫ ﺍﺛﺮﺍﺕ ﺣﺮﻛﺖ ﺯﻣﻴﻦ ﺑﺮ ﺳﺎﺧﺘﻤﺎﻥ ﺑﺎ ﺳﻴﺴﺘﻤﻲ ﺍﺯ ﻧﻴﺮﻭﻫﺎﻱ ﺍﻓﻘﻲ ﺷﺒﻴﻪ‬ . ‫ﺳﺎﺯﻱ ﻣﻲ ﺷﻮﺩ‬ V=CW Ft Fn Fn-1 C = ABI / R Fi Fi= (V-Ft) Wihi/ΣWjhj Wi hi F1 V Urmia University, M. Sheidaii 18 ‫ﭼﺮﺍ ﺍﺯ ﻣﻬﺎﺭﺑﻨﺪﻱ ﺟﺎﻧﺒﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ؟‬ Why Use Lateral Bracing? Urmia University, M. Sheidaii 19 ‫ﺟﺮﻳﺎﻥ ﺑﺎﺭ ﺟﺎﻧﺒﻲ ﺩﺭ ﻳﻚ ﺳﺎﺧﺘﻤﺎﻥ‬ Lateral Load Flow on a building Urmia University, M. Sheidaii 20 Design Methods ‫ﺭﻭﺵ ﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ Urmia University, M. Sheidaii 21 ‫ﺣﺎﺷﻴﻪ ﺍﻳﻤﻨﻲ‬ ‫‪Margin of Safety‬‬ ‫]ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ[ – ]ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ[ = ]ﺣﺎﺷﻴﻪ ﺍﻳﻤﻨﻲ[‬ ‫ﻋﺪﻡ ﻗﻄﻌﻴﺖ ﻫﺎﻱ ﺗﺎﺛﻴﺮ ﮔﺬﺍﺭ‪:‬‬ ‫‪ .1‬ﻣﻘﺎﻭﻣﺖ ﻣﺼﺎﻟﺢ‪ :‬ﺗﻐﻴﻴﺮﺍﺕ ﺍﻭﻟﻴﻪ ‪ ،‬ﺧﺰﺵ ‪ ،‬ﺧﻮﺭﺩﮔﻲ ﻭ ﺧﺴﺘﮕﻲ‬ ‫‪ .2‬ﺭﻭﺵ ﺗﺤﻠﻴﻞ‬ ‫‪ .3‬ﻓﺎﺟﻌﻪ ) ﺯﻟﺰﻟﻪ ‪ ،‬ﻃﻮﻓﺎﻥ (‬ ‫‪ .4‬ﺗﻨﺶ ﻫﺎﻱ ﺳﺎﺧﺖ ﻭ ﺍﺟﺮﺍ‬ ‫‪ .5‬ﺗﻐﻴﻴﺮ ﺩﺭ ﺑﺎﺭ ﺯﻧﺪﻩ ﻭ ﺑﺎﺭ ﺗﺮﺍﻓﻴﻚ ﺑﻮﺍﺳﻄﻪ ﭘﻴﺸﺮﻓﺖ ﻫﺎﻱ ﺗﻜﻨﻮﻟﻮژﻳﻚ‬ ‫‪ .6‬ﺗﺨﻤﻴﻦ ﺑﺎﺭ ﺯﻧﺪﻩ‬ ‫ﺗﻨﺶ ﭘﺴﻤﺎﻧﺪ ‪ ،‬ﺗﻤﺮﻛﺰ ﺗﻨﺶ ﻭ ﺗﻐﻴﻴﺮ ﺩﺭ ﺍﺑﻌﺎﺩ‬ ‫‪ .7‬ﺳﺎﻳﺮ ﻣﻮﺍﺭﺩ‬ ‫‪22‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻓﻠﺴﻔﻪ ﻫﺎﻱ ﻃﺮﺍﺣﻲ‬ Design Philosophies ‫ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ‬ .i Allowable Strength Design Method (ASD) ‫ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ﺿﺮﺍﻳﺐ ﺑﺎﺭ ﻭ ﻣﻘﺎﻭﻣﺖ‬ .ii Load and Resistance Factor Design (LRFD) Urmia University, M. Sheidaii 23 ‫ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ‬ ‫)‪Allowable Stress Design (ASD‬‬ ‫ ‪ ASD‬ﺭﻭﺵ ﻣﻘﺪﻣﺎﺗﻲ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺳﺎﺯﻩ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺑﻮﺩﻩ ﺍﺳﺖ ‪ .‬ﺑﻪ ﺍﻳﻦ ﺭﻭﺵ ﻫﻤﭽﻨﻴﻦ‪ ،‬ﺭﻭﺵ ﻃﺮﺍﺣﻲ‬ ‫ﺍﺭﺗﺠﺎﻋﻲ ) ‪ ( elastic design‬ﻳﺎ ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ﺗﻨﺶ ﻫﺎﻱ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ ) ‪(working stress design‬‬ ‫ﻧﻴﺰ ﻣﻲ ﮔﻮﻳﻨﺪ ‪.‬‬ ‫ ﺍﻳﻦ ﺭﻭﺵ ﻋﻤﻮﻣﺎ ﺑﺮ ﺍﺳﺎﺱ ﺗﺤﻠﻴﻞ ﺍﺭﺗﺠﺎﻋﻲ ﺳﺎﺯﻩﻫﺎ ﻣﻲﺑﺎﺷﺪ– ﺍﻋﻀﺎ ﺑﺎﻳﺪ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﺭﻓﺘﺎﺭ ﺍﺭﺗﺠﺎﻋﻲ ﺑﺎﺷﻨﺪ‪.‬‬ ‫ ﻳﻚ ﻋﻀﻮ ﺑﻪ ﻧﺤﻮﻱ ﺍﻧﺘﺨﺎﺏ ﻣﻲﺷﻮﺩ ﻛﻪ ﻧﻴﺮﻭﻱ ﺣﺪﺍﻛﺜﺮ ﻧﺎﺷﻲ ﺍﺯ ﺑﺎﺭﻫﺎﻱ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ‬ ‫ﻃﺮﺍﺣﻲ ﺗﺠﺎﻭﺯ ﻧﻜﻨﺪ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ ‪ d‬ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ‬ ‫‪required strength d allowable strength‬‬ ‫ ﻛﻪ ﻣﻘﺎﻭﻣﺖ ﻣﻲﺗﻮﺍﻧﺪ ﻧﻴﺮﻭﻱ ﻣﻘﺎﻭﻡ ﻣﺤﻮﺭﻱ ﻳﺎ ﻧﻴﺮﻭﻱ ﻣﻘﺎﻭﻡ ﺧﻤﺸﻲ ﻳﺎ ﻧﻴﺮﻭﻱ ﻣﻘﺎﻭﻡ ﺑﺮﺷﻲ ﺑﺎﺷﺪ‪.‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ ﺑﺎ ﺗﻘﺴﻴﻢ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ )ﺗﺌﻮﺭﻳﻚ( ﺑﺮ ﻳﻚ ﺿﺮﻳﺐ ﺍﻳﻤﻨﻲ ﺑﻪ ﺩﺳﺖ ﺁﻳﺪ‪.‬‬ ‫‪allowable strength = nominal strength / factor of safety‬‬ ‫ﺿﺮﻳﺐ ﺍﻳﻤﻨﻲ ‪ /‬ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ = ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ‬ ‫ ﺍﮔﺮ ﺩﺭ ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﺑﺠﺎﻱ ﻧﻴﺮﻭﻫﺎ ﻳﺎ ﻟﻨﮕﺮﻫﺎ ‪ ،‬ﺍﺯ ﺗﻨﺸﻬﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ ،‬ﺭﻭﺵ ﻣﺰﺑﻮﺭ ﺭﻭﺵ ﺗﻨﺶ ﻣﺠﺎﺯ ﻧﺎﻣﻴﺪﻩ‬ ‫ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﺗﻨﺶ ﻣﺠﺎﺯ ‪ d‬ﺗﻨﺶ ﻭﺍﺭﺩﻩ ﺣﺪﺍﻛﺜﺮ‬ ‫ ‪ ASD 24‬ﻳﻜﻲ ﺍﺯ ﺭﻭﺵﻫﺎﻱ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪ AISC‬ﺍﺳﺖ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ﻣﻘﺎﻭﻣﺖ ﻣﺠﺎﺯ‬ Allowable Stress Design (ASD) : ‫ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‬ASD ‫ ﻣﻌﺎﺩﻟﻪ ﻛﻠﻲ ﺭﻭﺵ‬ Ra d Rn / : : ‫ﻛﻪ‬ Ra =∑ Qi = required strength = sum of service loads ( ‫) ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ‬ Rn = nominal strength of member ( ‫) ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ‬ Ω = safety factor ( ‫) ﺿﺮﻳﺐ ﺍﻳﻤﻨﻲ‬ = 1.67 for limit-states involving yielding ( ‫) ﺑﺮﺍﻱ ﺣﺎﻻﺕ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻛﺸﺸﻲ‬ = 2.00 for limit-states involving rupture ( ‫) ﺑﺮﺍﻱ ﺣﺎﻻﺕ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻛﺸﺸﻲ‬ = 1.5 / I Urmia University, M. Sheidaii 25 ‫ﺍﺧﺘﺼﺎﺭﺍﺕ ﺑﺎﺭ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ‬ Service Load Notation ¾ D = Dead Load ¾ L = Live Load ¾ Roof Loads Lr = Roof Live Load S = Snow R = Rain or Ice (Does not include ponding) ¾ E = Earthquake Load ¾ W = Wind Load ¾ H = Load due to lateral earth pressure, ground water pressure, or pressure of bulk materials Urmia University, M. Sheidaii 26 ‫ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭ ‪ASD‬‬ ‫‪Load Combinations for ASD‬‬ ‫ﻫﺸﺖ ﺗﺮﻛﻴﺐ ﺑﺎﺭ ﺯﻳﺮ ﺑﺮﺍﻱ ﺑﺪﺳﺖ ﺁﻭﺭﺩﻥ ﺷﺪﻳﺪﺗﺮﻳﻦ ﺣﺎﻻﺕ ﺑﺎﺭ ﮔﺬﺍﺭﻱ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲ ﮔﻴﺮﻧﺪ‬ ‫) ‪:(ASCE05 Eqs 2.4.1.1 to 2.4.1.8‬‬ ‫‪D‬‬ ‫‪D+H+L‬‬ ‫)‪D + H + (Lr or S or R‬‬ ‫)‪D + H + 0.75L + 0.75(Lr or S or R‬‬ ‫)‪D + H + (W or 0.7E‬‬ ‫)‪D + H + 0.75(W or 0.7E) + 0.75L + 0.75(Lr or S or R‬‬ ‫‪0.6D + W + H‬‬ ‫‪0.6D + 0.7E + H‬‬ ‫ﺗﻮﺟﻪ ‪ :‬ﺑﺎﺭﻫﺎﻱ ﺑﺎﺩ ﻭ ﺯﻟﺰﻟﻪ ﻣﻤﻜﻦ ﺍﺳﺖ ﻋﻼﻣﺖ ﻫﺎﻱ ﻣﺜﺒﺖ ﻳﺎ ﻣﻨﻔﻲ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ‪.‬‬ ‫‪27‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫)‪1‬‬ ‫)‪2‬‬ ‫)‪3‬‬ ‫)‪4‬‬ ‫)‪5‬‬ ‫)‪6‬‬ ‫)‪7‬‬ ‫)‪8‬‬ ‫ﺗﺮﻛﻴﺐ ﺑﺎﺭﻫﺎﻱ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺁﻳﻴﻦﻧﺎﻣﻪ ﺍﻳﺮﺍﻥ‬ ‫‪28‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﺮﻛﻴﺐ ﺑﺎﺭﻫﺎﻱ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺁﻳﻴﻦﻧﺎﻣﻪ ﺍﻳﺮﺍﻥ‪ -‬ﺍﺩﺍﻣﻪ‬ ‫‪29‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example 1 GIVEN: A flat roof is framed with 24’-0” long W18x40 beams spaced 8’-0” o.c. The service applied roof dead load is 25 PSF and the applied service roof live load = 20 PSF. The service wind load on the flat roof is -8 PSF (uplift). REQUIRED: 1) Determine the maximum ASD factored uniform load on the beam, w. 2) Determine the maximum ASD factored moment on the beam, Mmasx. Urmia University, M. Sheidaii 30 Urmia University, M. Sheidaii 31 ( LRFD ) ‫ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ﺿﺮﺍﻳﺐ ﺑﺎﺭ ﻭ ﻣﻘﺎﻭﻣﺖ‬ Load and Resistance Factor Design (LRFD) ‫ ﻳﻚ ﻋﻀﻮ ﭼﻨﺎﻥ ﺍﻧﺘﺨﺎﺏ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﻣﻘﺎﻭﻣﺖ ﻣﻮﺟﻮﺩ ﻳﺎ ﺿﺮﻳﺒﺪﺍﺭ ﺁﻥ ﺑﻴﺶ ﺍﺯ‬ ( ultimate loads ‫ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ ﺑﺮ ﺍﺳﺎﺱ ﺑﺎﺭﻫﺎﻱ ﺿﺮﻳﺒﺪﺍﺭ ﻭﺍﺭﺩﻩ ) ﺑﺎﺭﻫﺎﻱ ﻧﻬﺎﻳﻲ‬ .‫ﺑﺎﺷﺪ‬ :‫ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‬LRFD ‫ ﻣﻌﺎﺩﻟﻪ ﻛﻠﻲ ﺭﻭﺵ‬ Σ λi.Qi d I.Rn Required Strength (ultimate load) Urmia University, M. Sheidaii Design Strength (available capacity) Qi = service (working) load – the best estimation of the loads acting on the structure, λi = load factor for the service load i, Rn = nominal strength– the calculated strength of the structure or element, I = the capacity reduction factor for nominal strength. 32 Urmia University, M. Sheidaii 50 Tension Members ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬: ‫ﻓﺼﻞ ﺳﻮﻡ‬ Urmia University, M. Sheidaii 1 ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Tension Members‬‬ ‫ ﺍﻋﻀﺎﻱ ﺳﺎﺯﻩ ﺍﻱ ﻛﻪ ﺗﺤﺖ ﻧﻴﺮﻭﻱ ﻛﺸﺸﻲ ﻣﺤﻮﺭﻱ ﻫﺴﺘﻨﺪ )ﺍﻋﻀﺎﻱ‬ ‫ﺧﺮﭘﺎ ‪ ،‬ﻛﺎﺑﻠﻬﺎ ﺩﺭ ﭘﻠﻬﺎﻱ ﻣﻌﻠﻖ ‪ ،‬ﻣﻬﺎﺭﺑﻨﺪﻱ ﺳﺎﺧﺘﻤﺎﻧﻬﺎ ‪( ... ،‬‬ ‫‪2‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻧﻮﺍﻉ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Types of Tension Members‬‬ ‫ ﻫﺮ ﺷﻜﻞ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﺩ‪ ،‬ﭼﻮﻥ‬ ‫ﺗﻨﻬﺎ ﻋﺎﻣﻞ ﺗﺎﺛﻴﺮﮔﺬﺍﺭ ﺩﺭ ﻣﻘﺎﻭﻣﺖ ﻋﻀﻮ‪ ،‬ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﺍﺳﺖ‪.‬‬ ‫ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻧﻮﺭﺩﺷﺪﻩ ﻧﺒﺸﻲ ﺷﻜﻞ ﻭ ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺩﺍﻳﺮﻩﺍﻱ‬ ‫ﺑﺴﻴﺎﺭ ﻣﺘﺪﺍﻭﻝ ﺍﺳﺖ‪.‬‬ ‫‪3‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Holes in Tension Members‬‬ ‫ ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﻭ ﺍﺛﺮ ﻣﻬﻢ ﺩﺍﺭﻧﺪ ‪:‬‬ ‫‪ (1‬ﺗﻤﺮﻛﺰ ﺗﻨﺶ )‪(Stress Concentration‬‬ ‫‪ (2‬ﻛﺎﻫﺶ ﻣﻘﻄﻊ ﻋﺮﺿﻲ )‪(area reduction‬‬ ‫ ﺭﻭﺵ ﻣﺘﺪﺍﻭﻝ ﺳﻮﺭﺍﺧﻜﺎﺭﻱ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﻣﺘﻪ ﺯﺩﻥ ﻳﺎ ﭘﺎﻧﭻ ﻛﺮﺩﻥ ﺳﻮﺭﺍﺥ ﻫﺎﻱ ﺍﺳﺘﺎﻧﺪﺍﺭﺩﻱ ﺑﺎ‬ ‫ﻗﻄﺮﻱ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ‪1.5‬ﻣﻴﻠﻴﻤﺘﺮ) ‪ 1/16‬ﺍﻳﻨﭻ( ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻗﻄﺮ ﭘﻴﭻ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻥ ﻧﺎﻫﻤﻮﺍﺭﻱ ﻫﺎ ﻭ ﺯﺑﺮﻱ ﺩﻭﺭ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ‪ AISC ،‬ﻣﻠﺰﻡ ﻣﻲ ﻧﻤﺎﻳﺪ ‪ 1.5‬ﻣﻴﻠﻴﻤﺘﺮ‬ ‫)‪ 1/16‬ﺍﻳﻨﭻ( ﺩﻳﮕﺮ ﻋﻤﻼ ﺑﺮﺍﻱ ﻗﻄﺮ ﺳﻮﺭﺍﺥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﻮﺩ‪.‬‬ ‫ ﺑﻨﺎﺑﺮﺍﻳﻦ ﻗﻄﺮ ﻣﻮﺛﺮ ﺳﻮﺭﺍﺥ‪ 3 ،‬ﻣﻴﻠﻴﻤﺘﺮ )‪ 1/8‬ﺍﻳﻨﭻ( ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻗﻄﺮ ﭘﻴﭻ ﻣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫‪4‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ﻭ ﺳﻄﺢ ﻣﻘﻄﻊ‬ ‫ﺧﺎﻟﺺ )‪(Ag, An‬‬ ‫‪Gross and Net Areas‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ )‪ (Ag‬ﻳﻚ ﻋﻀﻮ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻛﻞ‬ ‫ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺁﻥ‪.‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ = ‪Ag = Gross Area‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ = ‪An = Net Area‬‬ ‫‪An = Ag - Aholes‬‬ ‫‪5‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ﻭ ﺧﺎﻟﺺ ﻭﺭﻕ ‪ 1.5×20 cm‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ‪.‬‬ ‫ﺍﺗﺼﺎﻝ ﻭﺭﻕ ﺩﺭ ﺍﻧﺘﻬﺎ ﺑﺎ ﺩﻭ ﺭﺩﻳﻒ ﭘﻴﭻ ‪ 1.9‬ﺳﺎﻧﺘﻴﻤﺘﺮﻱ ﺍﻧﺠﺎﻡ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪ Ag = 20 ×1.5 = 30 mm2‬‬ ‫‪ An = 30 - 2(1.9 + 0.3)(1.5) = 23.4 mm2‬‬ ‫‪6‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Design Strength of Tension Members‬‬ ‫ ﺩﻭ ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ ﻣﻬﻢ ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪ .‬ﺍﻳﻦ ﺣﺎﻟﺖ ﻫﺎﻱ ﺣﺪﻱ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪:‬‬ ‫‪ (1‬ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ)‪(yielding on the gross area‬‬ ‫‪ (2‬ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪(fracture on the net section‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ﻭ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﻫﺎﻱ ﺍﺿﺎﻓﻲ ﻧﺎﺷﻲ ﺍﺯ ﺁﻥ‪ ،‬ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫)‪ (Ag‬ﺑﺎﻳﺪ ﻛﻮﭼﻜﺘﺮ ﺍﺯ ﺗﻨﺶ ﺗﺴﻠﻴﻢ )‪ (Fy‬ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪ ،‬ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪ (An‬ﺑﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﺗﻨﺶ‬ ‫ﻧﻬﺎﻳﻲ)‪ (Fu‬ﺑﺎﺷﺪ‪.‬‬ ‫‪7‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ‬ ‫‪Nominal Strength‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺩﺭ ﺗﺴﻠﻴﻢ ‪،‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺩﺭ ﮔﺴﻴﺨﺘﮕﻲ ‪،‬‬ ‫‪P n = F y Ag‬‬ ‫‪Pn = Fu Ae‬‬ ‫‪ Ae‬ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺼﻲ ﺍﺳﺖ ﻛﻪ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﺩﺭ ﻣﻘﻄﻊ ﮔﺬﺭﻧﺪﻩ ﺍﺯ‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎ ﺩﺭ ﺑﺮﺍﺑﺮ ﻛﺸﺶ ﻣﻘﺎﻭﻣﺖ ﻣﻲ ﻛﻨﺪ‪.‬‬ ‫‪ Ae‬ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﺨﺎﻃﺮ ﻋﻮﺍﻣﻠﻲ ﻫﻤﭽﻮﻥ ﺗﻤﺮﻛﺰ ﺗﻨﺶ ﻭ ﺑﺮﺧﻲ ﻋﻮﺍﻣﻞ‬ ‫ﺩﻳﮕﺮ ‪ ،‬ﻗﺪﺭﻱ ﻛﻤﺘﺮ ﺍﺯ ‪ An‬ﺑﺎﺷﺪ‪.‬‬ ‫‪8‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫‪Yielding on Gross Area‬‬ ‫‪ = It Pn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ‪ LRFD‬ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﺩﺭ ﻛﺸﺶ‬ ‫‪ =Pn / Ω‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﺠﺎﺯ ‪ ASD‬ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﺩﺭ ﻛﺸﺶ‬ ‫ﻛﻪ ‪:‬‬ ‫)‪It = 0.90 (LRFD‬‬ ‫)‪Ω = 1.67 (ASD‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ = ‪Pn‬‬ ‫‪= FyAg‬‬ ‫‪9‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫‪Fracture on Net Section‬‬ ‫‪ = It Pn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ‪ LRFD‬ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﻛﺸﺶ‬ ‫‪ =Pn / Ω‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﺠﺎﺯ ‪ ASD‬ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﻛﺸﺶ‬ ‫ﻛﻪ ‪:‬‬ ‫)‪It = 0.75 (LRFD‬‬ ‫)‪Ω = 2.00(ASD‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ = ‪Pn‬‬ ‫‪= FuAe‬‬ ‫ﺿﺮﻳﺐ ﻛﺎﻫﺶ ﻣﻘﺎﻭﻣﺖ ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻛﻤﺘﺮ ﺍﺯ ﺗﺴﻠﻴﻢ ﺍﺳﺖ‪ .‬ﭼﻮﻥ ﺭﺳﻴﺪﻥ ﺑﻪ ﺣﺎﻟﺖ‬ ‫ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺑﺴﻴﺎﺭ ﺧﻄﺮﻧﺎﻛﺘﺮ ﺍﺳﺖ‪.‬‬ ‫‪10‬‬ Tension Members ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬: ‫ﻓﺼﻞ ﺳﻮﻡ‬ Urmia University, M. Sheidaii 1 ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Tension Members‬‬ ‫ ﺍﻋﻀﺎﻱ ﺳﺎﺯﻩ ﺍﻱ ﻛﻪ ﺗﺤﺖ ﻧﻴﺮﻭﻱ ﻛﺸﺸﻲ ﻣﺤﻮﺭﻱ ﻫﺴﺘﻨﺪ )ﺍﻋﻀﺎﻱ‬ ‫ﺧﺮﭘﺎ ‪ ،‬ﻛﺎﺑﻠﻬﺎ ﺩﺭ ﭘﻠﻬﺎﻱ ﻣﻌﻠﻖ ‪ ،‬ﻣﻬﺎﺭﺑﻨﺪﻱ ﺳﺎﺧﺘﻤﺎﻧﻬﺎ ‪( ... ،‬‬ ‫‪2‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻧﻮﺍﻉ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Types of Tension Members‬‬ ‫ ﻫﺮ ﺷﻜﻞ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﺩ‪ ،‬ﭼﻮﻥ‬ ‫ﺗﻨﻬﺎ ﻋﺎﻣﻞ ﺗﺎﺛﻴﺮﮔﺬﺍﺭ ﺩﺭ ﻣﻘﺎﻭﻣﺖ ﻋﻀﻮ‪ ،‬ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﺍﺳﺖ‪.‬‬ ‫ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻧﻮﺭﺩﺷﺪﻩ ﻧﺒﺸﻲ ﺷﻜﻞ ﻭ ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺩﺍﻳﺮﻩﺍﻱ‬ ‫ﺑﺴﻴﺎﺭ ﻣﺘﺪﺍﻭﻝ ﺍﺳﺖ‪.‬‬ ‫‪3‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Holes in Tension Members‬‬ ‫ ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﻭ ﺍﺛﺮ ﻣﻬﻢ ﺩﺍﺭﻧﺪ ‪:‬‬ ‫‪ (1‬ﺗﻤﺮﻛﺰ ﺗﻨﺶ )‪(Stress Concentration‬‬ ‫‪ (2‬ﻛﺎﻫﺶ ﻣﻘﻄﻊ ﻋﺮﺿﻲ )‪(area reduction‬‬ ‫ ﺭﻭﺵ ﻣﺘﺪﺍﻭﻝ ﺳﻮﺭﺍﺧﻜﺎﺭﻱ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﻣﺘﻪ ﺯﺩﻥ ﻳﺎ ﭘﺎﻧﭻ ﻛﺮﺩﻥ ﺳﻮﺭﺍﺥ ﻫﺎﻱ ﺍﺳﺘﺎﻧﺪﺍﺭﺩﻱ ﺑﺎ‬ ‫ﻗﻄﺮﻱ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ‪1.5‬ﻣﻴﻠﻴﻤﺘﺮ) ‪ 1/16‬ﺍﻳﻨﭻ( ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻗﻄﺮ ﭘﻴﭻ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻥ ﻧﺎﻫﻤﻮﺍﺭﻱ ﻫﺎ ﻭ ﺯﺑﺮﻱ ﺩﻭﺭ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ‪ AISC ،‬ﻣﻠﺰﻡ ﻣﻲ ﻧﻤﺎﻳﺪ ‪ 1.5‬ﻣﻴﻠﻴﻤﺘﺮ‬ ‫)‪ 1/16‬ﺍﻳﻨﭻ( ﺩﻳﮕﺮ ﻋﻤﻼ ﺑﺮﺍﻱ ﻗﻄﺮ ﺳﻮﺭﺍﺥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﻮﺩ‪.‬‬ ‫ ﺑﻨﺎﺑﺮﺍﻳﻦ ﻗﻄﺮ ﻣﻮﺛﺮ ﺳﻮﺭﺍﺥ‪ 3 ،‬ﻣﻴﻠﻴﻤﺘﺮ )‪ 1/8‬ﺍﻳﻨﭻ( ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻗﻄﺮ ﭘﻴﭻ ﻣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫‪4‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ﻭ ﺳﻄﺢ ﻣﻘﻄﻊ‬ ‫ﺧﺎﻟﺺ )‪(Ag, An‬‬ ‫‪Gross and Net Areas‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ )‪ (Ag‬ﻳﻚ ﻋﻀﻮ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻛﻞ‬ ‫ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺁﻥ‪.‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ = ‪Ag = Gross Area‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ = ‪An = Net Area‬‬ ‫‪An = Ag - Aholes‬‬ ‫‪5‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ﻭ ﺧﺎﻟﺺ ﻭﺭﻕ ‪ 1.5×20 cm‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ‪.‬‬ ‫ﺍﺗﺼﺎﻝ ﻭﺭﻕ ﺩﺭ ﺍﻧﺘﻬﺎ ﺑﺎ ﺩﻭ ﺭﺩﻳﻒ ﭘﻴﭻ ‪ 1.9‬ﺳﺎﻧﺘﻴﻤﺘﺮﻱ ﺍﻧﺠﺎﻡ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪ Ag = 20 ×1.5 = 30 mm2‬‬ ‫‪ An = 30 - 2(1.9 + 0.3)(1.5) = 23.4 mm2‬‬ ‫‪6‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Design Strength of Tension Members‬‬ ‫ ﺩﻭ ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ ﻣﻬﻢ ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪ .‬ﺍﻳﻦ ﺣﺎﻟﺖ ﻫﺎﻱ ﺣﺪﻱ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪:‬‬ ‫‪ (1‬ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ)‪(yielding on the gross area‬‬ ‫‪ (2‬ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪(fracture on the net section‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ﻭ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﻫﺎﻱ ﺍﺿﺎﻓﻲ ﻧﺎﺷﻲ ﺍﺯ ﺁﻥ‪ ،‬ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫)‪ (Ag‬ﺑﺎﻳﺪ ﻛﻮﭼﻜﺘﺮ ﺍﺯ ﺗﻨﺶ ﺗﺴﻠﻴﻢ )‪ (Fy‬ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪ ،‬ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪ (An‬ﺑﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﺗﻨﺶ‬ ‫ﻧﻬﺎﻳﻲ)‪ (Fu‬ﺑﺎﺷﺪ‪.‬‬ ‫‪7‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ‬ ‫‪Nominal Strength‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺩﺭ ﺗﺴﻠﻴﻢ ‪،‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺩﺭ ﮔﺴﻴﺨﺘﮕﻲ ‪،‬‬ ‫‪P n = F y Ag‬‬ ‫‪Pn = Fu Ae‬‬ ‫‪ Ae‬ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺼﻲ ﺍﺳﺖ ﻛﻪ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﺩﺭ ﻣﻘﻄﻊ ﮔﺬﺭﻧﺪﻩ ﺍﺯ‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎ ﺩﺭ ﺑﺮﺍﺑﺮ ﻛﺸﺶ ﻣﻘﺎﻭﻣﺖ ﻣﻲ ﻛﻨﺪ‪.‬‬ ‫‪ Ae‬ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﺨﺎﻃﺮ ﻋﻮﺍﻣﻠﻲ ﻫﻤﭽﻮﻥ ﺗﻤﺮﻛﺰ ﺗﻨﺶ ﻭ ﺑﺮﺧﻲ ﻋﻮﺍﻣﻞ‬ ‫ﺩﻳﮕﺮ ‪ ،‬ﻗﺪﺭﻱ ﻛﻤﺘﺮ ﺍﺯ ‪ An‬ﺑﺎﺷﺪ‪.‬‬ ‫‪8‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫‪Yielding on Gross Area‬‬ ‫‪ = It Pn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ‪ LRFD‬ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﺩﺭ ﻛﺸﺶ‬ ‫‪ =Pn / Ω‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﺠﺎﺯ ‪ ASD‬ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﺩﺭ ﻛﺸﺶ‬ ‫ﻛﻪ ‪:‬‬ ‫)‪It = 0.90 (LRFD‬‬ ‫)‪Ω = 1.67 (ASD‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ = ‪Pn‬‬ ‫‪= FyAg‬‬ ‫‪9‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫‪Fracture on Net Section‬‬ ‫‪ = It Pn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ‪ LRFD‬ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﻛﺸﺶ‬ ‫‪ =Pn / Ω‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﺠﺎﺯ ‪ ASD‬ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﻛﺸﺶ‬ ‫ﻛﻪ ‪:‬‬ ‫)‪It = 0.75 (LRFD‬‬ ‫)‪Ω = 2.00(ASD‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ = ‪Pn‬‬ ‫‪= FuAe‬‬ ‫ﺿﺮﻳﺐ ﻛﺎﻫﺶ ﻣﻘﺎﻭﻣﺖ ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻛﻤﺘﺮ ﺍﺯ ﺗﺴﻠﻴﻢ ﺍﺳﺖ‪ .‬ﭼﻮﻥ ﺭﺳﻴﺪﻥ ﺑﻪ ﺣﺎﻟﺖ‬ ‫ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺑﺴﻴﺎﺭ ﺧﻄﺮﻧﺎﻛﺘﺮ ﺍﺳﺖ‪.‬‬ ‫‪10‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ )‪(LRFD‬‬ ‫‪Design Strength‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ ﻛﻪ ﻫﺪﻑ ﺍﺯ ﺍﻋﻤﺎﻝ ﺁﻥ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺍﻓﺰﺍﻳﺶ‬ ‫ﻃﻮﻝ ﺑﻴﺶ ﺍﺯ ﺣﺪ ﻋﻀﻮ ﺍﺳﺖ ‪:‬‬ ‫‪ It = 0.90‬؛ ‪Pu ≤ It Fy Ag‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺳﻮﺭﺍﺧﻬﺎﻱ ﭘﻴﭻ ﻳﺎ‬ ‫ﭘﺮچ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ‪:‬‬ ‫‪ It = 0.75‬؛ ‪Pu ≤ It Fu Ae‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ﻛﻤﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺍﺯ ﺩﻭ ﻣﻘﺪﺍﺭ ﻓﻮﻕ ﺍﺳﺖ‪.‬‬ ‫‪11‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ )‪(Ae‬‬ ‫‪Effective Net Area‬‬ ‫ ﻫﻨﮕﺎﻣﻲ ﻛﻪ ﻫﻤﻪ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺍﺗﺼﺎﻝ ﻧﻴﺎﻓﺘﻪ ﺑﺎﺷﻨﺪ )ﻣﺜﻼ ‪ :‬ﻓﻘﻂ ﻳﻚ ﭘﺎﻱ‬ ‫ﻧﺒﺸﻲ ﺑﻪ ﻭﺭﻕ ﻟﭽﻜﻲ ﭘﻴﭽﻜﺎﺭﻱ ﺷﻮﺩ( ‪ ،‬ﭘﺪﻳﺪﻩ ﺗﺎﺧﻴﺮ ﺑﺮﺷﻲ)‪ (shear lag‬ﭘﻴﺶ‬ ‫ﻣﻲﺁﻳﺪ‪.‬‬ ‫ ﺍﺟﺰﺍﻱ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ ﺗﺤﺖ ﺑﺎﺭ ﺑﻴﺸﺘﺮﻱ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﻧﺪ ﻭ ﻗﺴﻤﺖ ﻫﺎﻱ ﺍﺗﺼﺎﻝ ﻧﻴﺎﻓﺘﻪ‬ ‫ﺗﺤﺖ ﺗﻨﺶ ﻛﺎﻣﻞ ﻗﺮﺍﺭ ﻧﻤﻲﮔﻴﺮﻧﺪ)‪.(not fully stressed‬‬ ‫ ﺑﺮﺍﻱ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻥ ﺍﻳﻦ ﭘﺪﻳﺪﻩ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﻳﻚ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﻛﺎﻫﺶﻳﺎﻓﺘﻪ ‪ ،‬ﻳﺎ ﻣﻮﺛﺮ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫‪12‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ )‪(Ae‬‬ ‫‪Effective Net Area‬‬ ‫‪Ae = U An‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ ‪Ae = effective net area‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪An = net area (see AISC p. 16.1-14‬‬ ‫ﺿﺮﻳﺐ ﺗﺎﺧﻴﺮ ﺑﺮﺵ ”‪U = reduction factor considering “shear lag‬‬ ‫)ﻳﺎ ﺟﺪﻭﻝ ‪ 1-3-2-10‬ﺹ ‪ 163‬ﻣﺒﺤﺚ ﺩﻫﻢ( ﻳﺎ ‪= See AISC Table D3.1 p. 16.1-29‬‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻛﺸﺸﻲ ﻣﺴﺘﻘﻴﻤﺎ ﺑﻪ ﻫﺮ ﺟﺰء ﺗﻮﺳﻂ ﭘﻴﭻ ﻫﺎ ﻳﺎ ﺟﻮﺵﻫﺎ ﺍﻧﺘﻘﺎﻝ ﻳﺎﺑﺪ ‪= 1.0‬‬ ‫‪x‬‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻛﺸﺸﻲ ﺑﻪ ﺗﻌﺪﺍﺩﻱ ﺍﺯ ﺍﺟﺰﺍ )ﻭ ﻧﻪ ﻫﻤﻪ ﺁﻧﻬﺎ( ﺗﻮﺳﻂ ﭘﻴﭽﻬﺎ ﻳﺎ ﺟﻮﺵﻫﺎ ﺍﻧﺘﻘﺎﻝ ﻳﺎﺑﺪ ‪= 1‬‬ ‫"‬ ‫ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺍﺗﺼﺎﻝ ‪x = connection eccentricity‬‬ ‫ﻃﻮﻝ ﺍﺗﺼﺎﻝ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ = "‬ ‫‪13‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 14 Urmia University, M. Sheidaii 15 Urmia University, M. Sheidaii 16 Urmia University, M. Sheidaii 17 ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ‬ ‫‪ x‬ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺍﺗﺼﺎﻝ ﺍﺳﺖ )ﻓﺎﺻﻠﻪ ﻣﺮﻛﺰ ﺳﻄﺢ ﻗﻄﻌﻪ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ ﺗﺎ ﺻﻔﺤﻪ ﺍﺗﺼﺎﻝ(‬ ‫ﺍﮔﺮ ﻋﻀﻮﻱ ﺩﺍﺭﺍﻱ ﺩﻭ ﺻﻔﺤﻪ ﺍﺗﺼﺎﻝ‪ ،‬ﻛﻪ ﺑﻄﻮﺭ ﻣﺘﻘﺎﺭﻥ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪﺍﻧﺪ‪ ،‬ﺑﺎﺷﺪ ‪ x‬ﺍﺯ ﻣﺮﻛﺰ ﻧﺰﺩﻳﻜﺘﺮﻳﻦ‬ ‫ﻧﻴﻤﻪ ﻣﻘﻄﻊ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‬ ‫‬ ‫‪18‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ‬ ‫ﻃﻮﻝ ﺍﺗﺼﺎﻝ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ = "‬ ‫‪19‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪ Ae‬ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﭘﻴﭽﻲ ﻭ ﭘﺮﭼﻲ‬ ‫‪Ae for Bolted and Riveted Connections‬‬ ‫‪Ae = U An‬‬ ‫ ﺍﮔﺮ ﻫﻤﻪ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﺘﺼﻞ ﺷﺪﻩ ﺑﺎﺷﻨﺪ ‪U=1 :‬‬ ‫ ﺩﺭ ﻏﻴﺮ ﺍﻳﻦ ﺻﻮﺭﺕ ﺍﺯ ﻣﻘﺎﺩﻳﺮ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺿﺮﻳﺐ ﻛﺎﻫﺸﻲ ‪ U‬ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ ‪:‬‬ ‫)‪(AISC05 Table D3.1 case 2 or cases 7,8‬‬ ‫ ﺑﺮﺍﻱ ﺻﻔﺤﺎﺕ ﻭﺻﻠﻪ ﭘﻴﭽﻲ ﺩﺭ ﺍﺗﺼﺎﻻﺕ ‪:‬‬ ‫‪(AISC05 J4.1 p16.1-112): U=1, Ae=And0.85Ag‬‬ ‫‪20‬‬ ‫‪Urmia University, M. Sheidaii‬‬ U ‫ﻣﻘﺎﺩﻳﺮ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺑﺮﺍﻱ‬ Recommended Values for U Urmia University, M. Sheidaii 21 ‫‪ Ae‬ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﺟﻮﺷﻲ‬ ‫‪Ae for welded connections‬‬ ‫‪transverse‬‬ ‫‪ Ae = U Ag‬‬ ‫‪longitudinal‬‬ ‫‪ (1‬ﺑﺮﺍﻱ ﻫﺮ ﻧﻴﻤﺮﺥ ‪ I‬ﻳﺎ ‪ T‬ﺷﻜﻞ ﻛﻪ ﻓﻘﻂ ﺑﺎ ﺟﻮﺵ ﺟﺎﻧﺒﻲ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ‬ ‫ﺑﺎﺷﺪ ‪:‬‬ ‫‪U=1‬‬ ‫ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﺍﺟﺰﺍﻳﻲ ﻛﻪ ﻣﺴﺘﻘﻴﻤﺎ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ ﺍﻧﺪ = ‪Ae‬‬ ‫‪22‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪ Ae‬ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﺟﻮﺷﻲ‪ -‬ﺍﺩﺍﻣﻪ‬ ‫‪Ae for welded connections‬‬ ‫‪ (2‬ﺑﺮﺍﻱ ﻭﺭﻗﻬﺎ ﻳﺎ ﻣﻴﻠﻪ ﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﺍﻧﺘﻬﺎﻳﺸﺎﻥ ﺑﺎ ﺟﻮﺷﻬﺎﻱ ﻃﻮﻟﻲ ﺍﺗﺼﺎﻝ‬ ‫ﻳﺎﻓﺘﻪ ﺍﻧﺪ ‪:‬‬ ‫‪L t 2w‬‬ ‫‪2w ! L t 1.5w‬‬ ‫‪1.5w ! L t w‬‬ ‫‪U=1‬‬ ‫‪U=0.87‬‬ ‫‪U=0.75‬‬ ‫¾‬ ‫¾‬ ‫¾‬ ‫‪ t w‬ﻃﻮﻝ ﺯﻭﺝ ﺟﻮﺷﻬﺎ = ‪L‬‬ ‫ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺟﻮﺷﻬﺎ = ‪w‬‬ ‫‪23‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example 1: Determine the tensile design strength of a IPB 20 with two lines of ¾ inch(1.9cm) diameter in each flange using ST37 steel with Fy=2333 kgf/cm2 and Fu=3700 kgf/cm2.There are assumed at least 3 bolts in each line 4 in on center. PL 210×300×10 Pu/2 IPB 20 Solution. Pu Pu/2 Using an IPB 20 (Ag=78.1cm2, d=20cm, bf=20cm, tf=15mm) (a) Gross section yield Pu ≤ Øt Fy Ag = (0.9) (2333) (78.1) = 164 ton (b) Net section fracture An = 78.1 - (4) (1.9 + 0.3) (1.5) = 64.9 cm2 bf > 2d /3 → U= 0.9 Ae = U An = 0.9 (64.9) = 58.41 cm2 Pu ≤ Øt Fu Ae = (0.75) (3700) (58.41) = 162 ton ← USE 24 Example 2: The tension member of Example 1 is assumed to be connected at its ends with two 300×10mm plates, as shown in Figure. Determine the design tensile force which the plates PL 210×300×10 can transfer. Pu/2 IPB 20 Pu Pu/2 Solution. (a) Gross section yield Pu ≤ Øt Fy Ag = (0.9) (2333) (2×1×30) = 125982 kgf ← USE (b) Net section fracture An of 2 plates= [(1×30) - (2.2×2×1)](2) = 51.2 cm2 An≤ 0.85 Ag= 0.85 (2×1×30) = 51.0 cm2 → An = 51.0 cm2 Pu ≤ Øt Fu An = (0.75) (3700) (51.0) = 141525 kgf Pu ≤ 125982 kgf ≈ 126 ton 25 26 Urmia University, M. Sheidaii 27 ‫ﺳﻮﺭﺍﺧﻬﺎﻱ ﻧﺎﻣﻨﻈﻢ‬ ‫‪Staggered Holes‬‬ ‫ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺑﻴﺸﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺭﺍ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ ﺍﮔﺮ ﭘﻴﭽﻬﺎ ﺩﺭ ﻳﻚ ﺧﻂ ﺑﺎﺷﻨﺪ‪.‬‬ ‫ ﺍﮔﺮ ﺑﻴﺶ ﺍﺯ ﻳﻚ ﺧﻂ ﭘﻴﭻ ﺿﺮﻭﺭﺕ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ )ﻃﻮﻝ ﺍﺗﺼﺎﻝ ﻣﺤﺪﻭﺩ( ‪ ،‬ﺍﻳﺠﺎﺩ‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎﻱ ﻧﺎﻣﻨﻈﻢ )ﻗﻄﺮﻱ ﻳﺎ ﺯﻳﮕﺰﺍگ( ﺑﺎﻋﺚ ﺑﻪ ﺣﺪﺍﻗﻞ ﺭﺳﻴﺪﻥ ﻛﺎﻫﺶ ﺩﺭ ﺳﻄﺢ‬ ‫ﻣﻘﻄﻊ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﻣﺴﻴﺮﻫﺎﻱ ﺧﺮﺍﺑﻲ ﻣﻤﻜﻦ ‪ABC or ABDE :‬‬ ‫ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺣﺪﺍﻗﻞ ﺍﻧﺘﺨﺎﺏ ﺷﻮﺩ‪.‬‬ ‫‪28‬‬ ‫ﺭﻭﺵ ﺗﺠﺮﺑﻲ ﻣﺤﺎﺳﺒﻪ ﻣﻘﻄﻊ ﺧﺎﻟﺺ‬ ‫‪net width = gross width – 6 d + 6 s2 / 4g‬‬ ‫ﻋﺮﺽ ﻛﻞ‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫ﻋﺮﺽ ﺧﺎﻟﺺ‬ ‫‪ d‬ﻗﻄﺮ ﺳﻮﺭﺍﺥ ) " ‪ dh + 1/16‬ﻳﺎ ‪(db + 1/8 " ,‬‬ ‫‪ s2/4g‬ﺑﻪ ﺍﺯﺍﻱ ﻫﺮ ﮔﺎﻡ ﻋﺮﺿﻲ ﺩﺭ ﺯﻧﺠﻴﺮﻩ ﻣﻮﺭﺩ ﻧﻈﺮ ‪ ،‬ﺍﺿﺎﻓﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪) s‬ﮔﺎﻡ ﻃﻮﻟﻲ( ‪ :‬ﻓﺎﺻﻠﻪ ﻃﻮﻟﻲ ﻣﺮﻛﺰ ﺑﻪ ﻣﺮﻛﺰ ﻫﺮ ﺩﻭ ﺳﻮﺭﺍﺥ ﻣﺠﺎﻭﺭ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ‬ ‫‪) g‬ﮔﺎﻡ ﻋﺮﺿﻲ( ‪ :‬ﻓﺎﺻﻠﻪ ﻣﺮﻛﺰ ﺑﻪ ﻣﺮﻛﺰ ﺳﻮﺭﺍﺧﻬﺎ ﻋﻤﻮﺩ ﺑﺮ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ‬ ‫ﺿﺨﺎﻣﺖ ﻭﺭﻕ × ﻋﺮﺽ ﺧﺎﻟﺺ = )‪ (An‬ﻣﻘﻄﻊ ﺧﺎﻟﺺ‬ ‫‪ (Ae) = U An‬ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﻣﻮﺛﺮ‬ ‫)‪ = Øt Ae Fu (Øt = 0.75‬ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ‬ ‫‪29‬‬ Example : Compute the smallest net area for the plate shown below: Plate thickness = 12 mm Bolt diameter = 19 mm Solution. Urmia University, M. Sheidaii 30 Example : Compute the Minimum pitch (smin) in the other words determine the pitch will give a net area DEFG equal to the one along ABC. Solution. cm Urmia University, M. Sheidaii cm 31 Example : Compute the smallest net area for the angle shown below: Bolt diameter = 19 mm Solution. The unfolding is done at the middle surface to obtain an equivalent plate (with gross width equal to the sum of the leg lengths minus the angle thickness). mm mm cm2 Urmia University, M. Sheidaii 32 Example : Determine the net area along route ABCDEF for the C380×100×54.5 (Ag =69.39 cm2). Holes are for 19 mm Bolts. Solution. The unfolding is done at the middle surface to obtain an equivalent plate. /2 – tf /2 /2 – 16 /2 136.8 136.8 Urmia University, M. Sheidaii 33 ‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪BLOCK SHEAR‬‬ ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻫﻤﭽﻨﻴﻦ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﻪ ﻋﻠﺖ ﭘﺎﺭﮔﻲ )‪ (tear-out‬ﻣﺼﺎﻟﺢ ﺩﺭ ﻧﻮﺍﺣﻲ‬ ‫ﻣﺠﺎﻭﺭ ﺍﺗﺼﺎﻝ ﺩﭼﺎﺭ ﺧﺮﺍﺑﻲ ﺷﻮﻧﺪ‪ .‬ﺑﻪ ﺍﻳﻦ ﭘﺪﻳﺪﻩ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻣﻲ ﮔﻮﻳﻨﺪ‪.‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻋﻀﻮ ﻛﺸﺸﻲ‬ ‫ﺗﻚ ﻧﺒﺸﻲ‬ ‫‪34‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪BLOCK SHEAR‬‬ ‫ ﻣﺜﺎﻝﻫﺎﻱ ﻣﺘﺪﺍﻭﻝ ﺩﺭ ﺷﻜﻞﻫﺎﻱ ﻣﻘﺎﺑﻞ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫ ﺧﺮﺍﺑﻲ ﺩﺭ ﻃﻮﻝ ﻣﺴﻴﺮﻫﺎﻱ ﺯﻳﺮ ﺍﺗﻔﺎﻕ ﻣﻲﺍﻓﺘﺪ‪:‬‬ ‫‪ .i‬ﻛﺸﺶ ﺩﺭ ﺻﻔﺤﻪ ﺍﻱ ﻋﻤﻮﺩ ﺑﺮ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭ‬ ‫‪ .ii‬ﺑﺮﺵ ﺩﺭ ﺻﻔﺤﻪ ﺍﻱ ﻣﻮﺍﺯﻱ ﺑﺎ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭ‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻣﺠﻤﻮﻉ ‪:‬‬ ‫‪ .i‬ﻣﻘﺎﻭﻣﺖ ﺑﺮﺷﻲ ﺩﺭ ﻳﻚ ﻣﺴﻴﺮ ﺧﺮﺍﺑﻲ‬ ‫‪ .ii‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﺭﻭﻱ ﻗﻄﻌﻪﺍﻱ ﻋﻤﻮﺩ ﺑﺮ‬ ‫ﻣﺴﻴﺮ ﺧﺮﺍﺑﻲ ﻓﻮﻕ ﺍﻟﺬﻛﺮ‬ ‫‪35‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬-‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‬ BLOCK SHEAR Urmia University, M. Sheidaii 36 ‫ﻣﻘﺎﻭﻣﺖ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﻣﻘﺎﻭﻣﺖ ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪Block Shear Rupture Strength‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ )‪ (Rn‬ﺩﺭ ﻃﻮﻝ ﺳﻄﺢ ﻳﺎ ﺳﻄﻮﺡ ﺑﺮﺷﻲ ﻭ ﻛﺸﺸﻲ‬ ‫ﻋﻤﻮﺩ ﺑﺮ ﺁﻥ ﺍﺯ ﺭﺍﺑﻄﻪ ﺫﻳﻞ ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﺩ )‪ AISC p. 16.1-112‬ﻭ ﻳﺎ ﺻﻔﺤﻪ ‪ 318‬ﻣﺒﺤﺚ ﺩﻫﻢ(‬ ‫‪Rn = 0.6FuAnv + UbsFuAnt ≤ 0.6FyAgv + UbsFuAnt‬‬ ‫ﻭ ﺑﺮ ﺍﻳﻦ ﺍﺳﺎﺱ ‪:‬‬ ‫ﻛﻪ ‪:‬‬ ‫‪37‬‬ ‫‪LRFD available block shear rupture strength = Ø Rn‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻣﻮﺟﻮﺩ ‪LRFD‬‬ ‫‪ASD allowable block shear rupture strength = Rn / Ω‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻣﺠﺎﺯ ‪ASD‬‬ ‫)‪Ø = 0.75 (LRFD‬‬ ‫)‪Ω = 2.00 (ASD‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺗﺤﺖ ﺑﺮﺵ ‪Anv = net area subjected to shear‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺗﺤﺖ ﻛﺸﺶ ‪Ant = net area subjected to tension‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻠﻲ ﺗﺤﺖ ﺑﺮﺵ ‪Agv = gross area subjected to shear‬‬ ‫ﺿﺮﻳﺐ ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﻛﺸﺸﻲ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ‪Ubs = 1.0‬‬ ‫ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﻏﻴﺮﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﻛﺸﺸﻲ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ‪= 0.5‬‬ ‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪BLOCK SHEAR‬‬ ‫ﺿﺮﻳﺐ ﻛﺎﻫﺶ ‪ Ubs‬ﺩﺭ ﻣﻌﺎﺩﻟﻪ‪ ،‬ﺑﺮﺍﻱ ﺗﻘﺮﻳﺐ ﺳﺎﺯﻱ ﺗﻮﺯﻳﻊ ﻏﻴﺮ ﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﺩﺭ ﺳﻄﺢ ﻛﺸﺸﻲ ﺁﻭﺭﺩﻩ‬ ‫ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪38‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example : A steel angle L6x6x½ using A36 steel is subjected to tensile load. Determine the block shear rupture strength. Urmia University, M. Sheidaii 39 Solution: Anv = net area subjected to shear (in2) = (Matl. Thickness)[Lv – (# bolts)(Bolt dia. + 1/8”)] = ½”[10” – (2.5 holes)(¾” + 1/8”)] = 3.91 in2 Ant = net area subjected to tension (in2) = (Matl. Thickness)[Lt – (# bolts)(Bolt dia. + 1/8”)] = ½”[2½” – (0.5 holes)(¾” + 1/8”)] = 1.03 in2 Agv = gross area subjected to shear (in2) = (Matl. Thickness)(Lv) = (½”)(10”) = 5.0 in2 Rn = Available block shear = 0.6FuAnv + UbsFuAnt = 0.6(58 KSI)(3.91 in2) + (1.00)(58 KSI)(1.03 in2) = 195.8 Kips Check if Rn ≤ 0.6FyAgv + UbsFuAnt ≤ 0.6(36 KSI)(5 in2) + (1.00)(58 KSI)(1.03 in2) ≤ 167.7 Kips Pbs = Øt Rn = 0.75 (167.7) = 125.8 Kips 40 ‫ﺳﺨﻦ ﺁﺧﺮ‬ ‫‪Bottom line:‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪41‬‬ ‫ﻫﺮ ﻛﺪﺍﻡ ﺍﺯ ﺳﻪ ﺣﺎﻟﺖ ﺣﺪﻱ )ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪ ،‬ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ‬ ‫ﺧﺎﻟﺺ ‪ ،‬ﻳﺎ ﺧﺮﺍﺑﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ( ﻣﻲ ﺗﻮﺍﻧﺪ ﺣﺎﻛﻢ ﺑﺮ ﻃﺮﺍﺣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺑﺮﺍﻱ ﻫﺮ ﺳﻪ ﺣﺎﻟﺖ ﺣﺪﻱ ﺑﺎﻳﺪ ﻣﺤﺎﺳﺒﻪ ﺷﻮﺩ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ) ‪ ØPn‬ﻳﺎ ‪ (Pn /Ω‬ﻛﻤﺘﺮﻳﻦ ﺳﻪ ﻣﻘﺪﺍﺭ ﻣﺤﺎﺳﺒﻪ‬ ‫ﺷﺪﻩ ﻓﻮﻕ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ﺑﺎﻳﺪ ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ ﻋﻀﻮ ﻛﺸﺸﻲ‬ ‫ﺑﺎﺷﺪ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 42 Urmia University, M. Sheidaii 43 Urmia University, M. Sheidaii 44 Urmia University, M. Sheidaii 45 ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Design of Tension Members‬‬ ‫ﻃﺮﺡ ﻋﻀﻮ ﻛﺸﺸﻲ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﭘﻴﺪﺍ ﻛﺮﺩﻥ ﺳﺒﻚﺗﺮﻳﻦ ﻣﻘﻄﻊ ﻓﻮﻻﺩﻱ )ﻧﺒﺸﻲ‪،I ،‬‬ ‫ﻧﺎﻭﺩﺍﻧﻲ ﻭ ‪ ( ...‬ﺑﺎ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ‪.‬‬ ‫ﺑﻪ ﺭﻭﺵ ‪Ø Pn ≥ Pu LRFD‬‬ ‫ﺑﻪ ﺭﻭﺵ ‪Pn / Ω ≥ Pa ASD‬‬ ‫ ‪ Pn‬ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻃﺮﺍﺣﻲ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪،‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﻭ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﺍﺳﺖ‪.‬‬ ‫ ‪ Pu‬ﻣﻘﺎﻭﻣﺖ ﻧﻬﺎﻳﻲ ﻻﺯﻡ )ﺑﺎﺭ ﻧﻬﺎﻳﻲ( ﻛﻪ ﺍﺯ ﺗﺤﻠﻴﻞ ﺳﺎﺯﻩ ﺗﺤﺖ ﺗﺮﻛﻴﺒﺎﺕ‬ ‫ﺑﺎﺭﮔﺬﺍﺭﻱ ﺿﺮﻳﺒﺪﺍﺭ )ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭ ‪ (LRFD‬ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ‪.‬‬ ‫ ‪ Pa‬ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ )ﺑﺎﺭ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ( ﻛﻪ ﺍﺯ ﺗﺤﻠﻴﻞ ﺳﺎﺯﻩ ﺗﺤﺖ ﺗﺮﻛﻴﺒﺎﺕ‬ ‫ﺑﺎﺭﮔﺬﺍﺭﻱ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ )ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭ ‪ (ASD‬ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ‪.‬‬ ‫‪46‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﻪ ﺭﻭﺵ ‪LRFD‬‬ ‫‪Design of Tension Members - LRFD method‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪ØPn = 0.9 Ag Fy ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.9 Ag Fy ≥ Pu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ‪Ag ≥ Pu / 0.9 Fy‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ‪ØPn = 0.75 Ae Fu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.75 Ae Fu ≥ Pu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪ Ae ≥ Pu / 0.75 Fu ،‬ﻳﺎ‬ ‫‪An ≥ Pu / 0.75 Fu U‬‬ ‫ ﻛﻨﺘﺮﻝ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ‪،‬‬ ‫) ‪ØRn = 0.75( 0.6FuAnv + UbsFuAnt)≤ 0.75(0.6FyAgv + UbsFuAnt‬‬ ‫¾‬ ‫‪47‬‬ ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ‪ØRn ≥ Pu ،‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﻪ ﺭﻭﺵ ‪ASD‬‬ ‫‪Design of Tension Members - ASD method‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪Pn/Ω = AgFy /1.67 = 0.6AgFy ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.6Ag Fy ≥ Pa ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ‪Ag ≥ Pa / 0.6 Fy‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ‪Pn/Ω = AeFu /2.00 = 0.5AeFu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.5Ae Fu ≥ Pa ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪Ae ≥ Pa / 0.5Fu ،‬‬ ‫ ﻛﻨﺘﺮﻝ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ‪،‬‬ ‫) ‪Rn/Ω = Rn/2.00 = 0.5( 0.6FuAnv + UbsFuAnt)≤ 0.5(0.6FyAgv + UbsFuAnt‬‬ ‫¾‬ ‫‪48‬‬ ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ‪Rn/Ω ≥ Pa ،‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‪ -‬ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﻻﻏﺮﻱ‬ ‫‪Design of Tension Members - Slenderness Limitations‬‬ ‫ ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪ AISC‬ﺣﺪﻱ ﺑﺮﺍﻱ ﻻﻏﺮﻱ ﺣﺪﺍﻛﺜﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‪.‬‬ ‫ ﺍﻣﺎ ﺍﮔﺮ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﺩﺭ ﻳﻚ ﻋﻀﻮ ﻛﺸﺸﻲ ﻻﻏﺮ ﺑﺮﺩﺍﺷﺘﻪ ﺷﺪﻩ ﻭ ﺑﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻛﻮﭼﻜﻲ ﺍﻋﻤﺎﻝ‬ ‫ﺷﻮﺩ ‪ ،‬ﺍﺭﺗﻌﺎﺷﺎﺕ ﻭ ﺧﻴﺰﻫﺎﻱ ﻧﺎﻣﻄﻠﻮﺑﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﭘﺪﻳﺪ ﺁﻳﺪ‪ .‬ﻟﺬﺍ ‪ AISC‬ﭘﻴﺸﻨﻬﺎﺩ ﻣﻲ ﻧﻤﺎﻳﺪ ‪:‬‬ ‫)‪L/r t 300 ( not for cables or rods‬‬ ‫ﻛﻪ ‪ r‬ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﺣﺪﺍﻗﻞ ﻭ ‪ L‬ﻃﻮﻝ ﻋﻀﻮ ﺍﺳﺖ‪.‬‬ ‫ ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻓﻮﻻﺩ ﺍﻳﺮﺍﻥ )ﺻﻔﺤﻪ ‪ 160‬ﻣﺒﺤﺚ ﺩﻫﻢ( ‪ :‬ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺣﺪﺍﻛﺜﺮ‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻧﺒﺎﻳﺪ ﺍﺯ ‪ 300‬ﺗﺠﺎﻭﺯ ﻛﻨﺪ‪.‬‬ ‫ ﺩﺭ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎﻱ ﻛﺸﺸﻲ ﻛﻪ ﺩﺍﺭﺍﻱ ﭘﻴﺶ ﺗﻨﻴﺪﮔﻲ ﺍﻭﻟﻴﻪ ﺑﻪ ﻣﻘﺪﺍﺭ ﻛﺎﻓﻲ ﺑﺎﺷﻨﺪ ‪ ،‬ﺭﻋﺎﻳﺖ‬ ‫ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﻻﻏﺮﻱ ﻻﺯﻡ ﻧﻴﺴﺖ‪ ،‬ﻟﻴﻜﻦ ﻧﺴﺒﺖ ﻃﻮﻝ ﺑﻪ ﻗﻄﺮ ﺍﻳﻦ ﺍﻋﻀﺎ ﻧﺒﺎﻳﺪ ﺍﺯ ‪ 300‬ﺗﺠﺎﻭﺯ‬ ‫ﻛﻨﺪ‪.‬‬ ‫‪49‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺭﺯﻭﻩ ﺷﺪﻩ ﻭ ﻛﺎﺑﻠﻬﺎ‬ ‫‪Threaded Rods and Cables‬‬ ‫‬ ‫‬ ‫‬ ‫ﺭﺷﺘﻪ‬ ‫‪50‬‬ ‫ﻛﺎﺑﻞ ﺳﻴﻤﻲ‬ ‫‬ ‫ﺍﮔﺮ ﻻﻏﺮﻱ ﻋﻀﻮ ﻣﻮﺭﺩ ﺗﻮﺟﻪ ﻗﺮﺍﺭ ﻧﮕﻴﺮﺩ ‪ ،‬ﺍﻏﻠﺐ ﺍﺯ‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎ ﻳﺎ ﻛﺎﺑﻞﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲﺷﻮﺩ‪) .‬ﺁﻭﻳﺰﻫﺎ ‪،‬‬ ‫ﭘﻠﻬﺎﻱ ﻣﻌﻠﻖ(‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎ ﻋﻨﺎﺻﺮﻱ ﺑﺎ ﻣﻘﻄﻊ ﺗﻮﭘﺮ ﺑﻮﺩﻩ ﻭ ﻛﺎﺑﻞﻫﺎ‬ ‫ﺍﺯ ﺭﺷﺘﻪ ﻫﺎﻱ )ﺳﻴﻢ ﺑﺎﻓﺘﻪ( ﺟﺪﺍﮔﺎﻧﻪ ﭘﻴﭽﻴﺪﻩ ﺑﻪ‬ ‫ﻫﻢ ﺳﺎﺧﺘﻪ ﻣﻲﺷﻮﻧﺪ‪.‬‬ ‫ﻫﺮ ﺭﺷﺘﻪ ﺍﺯ ﺳﻴﻢﻫﺎﻱ ﺟﺪﺍﮔﺎﻧﻪﺍﻱ ﻛﻪ ﺩﻭﺭ ﻳﻚ‬ ‫ﻫﺴﺘﻪ ﻣﺮﻛﺰﻱ ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﺭﭘﻴﭽﻲ ﭘﻴﭽﻴﺪﻩﺍﻧﺪ‬ ‫ﺗﺸﻜﻴﻞ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﻳﻚ ﻛﺎﺑﻞ ﺳﻴﻤﻲ ﺧﻮﺩ ﺍﺯ ﭼﻨﺪﻳﻦ ﺭﺷﺘﻪ ﻛﻪ ﺩﻭﺭ‬ ‫ﻳﻚ ﻫﺴﺘﻪ ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﺭﭘﻴﭽﻲ ﭘﻴﭽﺎﻧﺪﻩ ﺷﺪﻩﺍﻧﺪ‬ ‫ﺗﺸﻜﻴﻞ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪Threaded Rod‬‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺭﺯﻭﻩ ﺷﺪﻩ ﻭ ﻛﺎﺑﻠﻬﺎ‬ ‫‪Threaded Rods and Cables‬‬ ‫ﻣﻌﻤﻮﻻ ﺍﻳﻦ ﺍﻋﻀﺎ ﺑﻪ ﺳﺎﺯﻩ ﺑﻪ ﻃﺮﻕ ﻣﺨﺘﻠﻔﻲ ﻫﻤﭽﻮﻥ ﺟﻮﺷﻜﺎﺭﻱ ﻳﺎ ﺍﺗﺼﺎﻻﺕ ﭘﻨﺠﻪ‬ ‫ﻣﻔﺼﻠﻲ )‪ (bolted clevises‬ﻭ ﭘﻴﭻﻫﺎ‪ ،‬ﻭﺻﻞ ﻣﻲﺷﻮﻧﺪ‪.‬‬ ‫‪51‬‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺭﺯﻭﻩ ﺷﺪﻩ ﻭ ﻛﺎﺑﻠﻬﺎ‬ ‫‪Threaded Rods and Cables‬‬ ‫ ﺭﺯﻭﻩ ﻛﺮﺩﻥ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻲ ﺷﻮﺩ )ﻗﻄﻮﺭ ﻛﺮﺩﻥ ﺍﻧﺘﻬﺎﻱ‬ ‫ﻋﻀﻮ ﺍﺯ ﭼﻨﻴﻦ ﻛﺎﺭﻱ ﺟﻠﻮﮔﻴﺮﻱ ﻣﻲ ﻛﻨﺪ ‪ ،‬ﻭﻟﻲ ﭘﺮﻫﺰﻳﻨﻪ ﺍﺳﺖ(‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻛﺸﺸﻲ ﺑﺮﺍﻱ ﻳﻚ ﻣﻴﻠﮕﺮﺩ ﺭﺯﻭﻩ ﺷﺪﻩ ‪:‬‬ ‫)‪I Pn = 0.75 (0.75 AD Fu‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﺭﺯﻭﻩ ﻧﺸﺪﻩ )ﻛﻞ( ﻣﻴﻠﮕﺮﺩ = ‪AD‬‬ ‫ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪AD ≥ Pu / Ø 0.75 Fu ; Ø=0.75‬‬ ‫ ﺑﻪ ‪ Example 3.14 p.70 Segui‬ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ‪.‬‬ ‫‪52‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﺭ ﺳﻘﻒ ﺧﺮﭘﺎﻳﻲ‬ ‫‪Tension Members in Roof Truss‬‬ ‫ ﺧﺮﭘﺎﻫﺎ ﺩﺭ ﻣﻮﺍﺭﺩﻱ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﻧﺪ ﻛﻪ ﻗﻴﻤﺖ ﻭ ﻭﺯﻥ‬ ‫ﺗﻴﺮ ﮔﺮﺍﻥ ﻭ ﺑﺎﺯﺩﺍﺭﻧﺪﻩ ﺑﺎﺷﺪ‪) .‬ﺩﻫﺎﻧﻪ ﻫﺎﻱ ﻃﻮﻳﻞ(‬ ‫ ﻣﻲ ﺗﻮﺍﻥ ﺧﺮﭘﺎ ﺭﺍ ﺗﻴﺮ ﻋﻤﻴﻘﻲ ﺗﺼﻮﺭ ﻛﺮﺩ ﻛﻪ ﺑﺨﺶ ﻋﻤﺪﻩ ﺍﻱ ﺍﺯ ﺟﺎﻥ‬ ‫ﺁﻥ ﺑﺮﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﺭ ﺳﻘﻒﻫﺎﻱ ﺧﺮﭘﺎﻳﻲ ﺷﺎﻣﻞ ﺑﻌﻀﻲ ﺍﺯ ﺍﻋﻀﺎﻱ ﺧﺮﭘﺎ ﻭ‬ ‫ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪53‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ Sag Rods Urmia University, M. Sheidaii 54 ‫ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ ‫‪Sag Rods‬‬ ‫ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﺑﺮﺍﻱ ﺗﺎﻣﻴﻦ ﺗﻜﻴﻪﮔﺎﻩ ﺟﺎﻧﺒﻲ ﺑﺮﺍﻱ ﻻﭘﻪﻫﺎ ﺑﻜﺎﺭ ﻣﻲﺭﻭﻧﺪ )ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ‬ ‫ﺷﻜﻢ ﺩﺍﺩﻥ ﻻﭘﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﻮﺍﺯﻱ ﺑﺎ ﺷﻴﺐ ﺑﺎﻡ ﺗﺤﺖ ﺑﺎﺭﻫﺎﻱ ﻗﺎﺋﻢ ﻭﺍﺭﺩﻩ(‬ ‫‪55‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ ‫‪Design of Sag Rods‬‬ ‫ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﺑﺮﺍﻱ ﺗﺤﻤﻞ ﺑﺨﺸﻲ ﺍﺯ ﺑﺎﺭﻫﺎﻱ ﺑﺎﻡ ﻛﻪ ﺑﻪ ﻣﻮﺍﺯﺍﺕ ﺑﺎﻡ ﻋﻤﻞ ﻣﻲ ﻛﻨﻨﺪ ﻃﺮﺍﺣﻲ‬ ‫ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫ ﺑﻨﺎ ﺑﻪ ﻓﺮﺽ ﻫﺮ ﻗﻄﻌﻪﺍﻱ ﺍﺯ ﻣﻴﻞ ﻣﻬﺎﺭ ﺑﻴﻦ ﻻﭘﻪﻫﺎ ‪ ،‬ﺑﺎﺭ ﺗﻤﺎﻡ ﻗﺴﻤﺖﻫﺎﻱ ﺯﻳﺮﻳﻦ ﺭﺍ ﺗﺤﻤﻞ‬ ‫ﻣﻲﻛﻨﺪ ؛ ﻟﺬﺍ ﻣﻴﻞ ﻣﻬﺎﺭ ﻓﻮﻗﺎﻧﻲ ﺑﺎﻳﺪ ﺑﺮﺍﻱ ﺳﻄﺢ ﺑﺎﺭﮔﻴﺮ ﻣﻴﻠﻪ ﻛﻪ ﺍﺯ ﺭﺍﺱ ﺗﺎ ﭘﺎﻱ ﺧﺮﭘﺎ‬ ‫ﺍﺳﺖ ‪ ،‬ﻃﺮﺍﺣﻲ ﺷﻮﺩ‪.‬‬ ‫‪56‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ ‫‪Design of Sag Rods‬‬ ‫ ﻣﻴﻞ ﻣﻬﺎﺭ ﻭﺍﻗﻊ ﺩﺭ ﺭﺍﺱ ﺧﺮﭘﺎ ﺑﺎﻳﺪ ﺑﺎﺭ ﻫﻤﻪ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺩﻳﮕﺮ ﺩﺭ ﻫﺮ ﺩﻭ ﻃﺮﻑ ﺧﺮﭘﺎ ﺭﺍ‬ ‫ﺗﺤﻤﻞ ﻛﻨﺪ‪.‬‬ ‫‪57‬‬ ‫‪T2 = T1 / Cosα‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 58 ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﻟﻮﻻﻳﻲ‬ ‫‪PIN-CONNECTED MEMBERS‬‬ ‫ ﺑﺮﺍﻱ ﺍﻳﺠﺎﺩ ﻳﻚ ﺍﺗﺼﺎﻝ ﻋﺎﺭﻱ ﺍﺯ ﺧﻤﺶ ‪ ،‬ﺳﻮﺭﺍﺧﻲ ﺩﺭ ﻋﻀﻮ ﻭ ﻗﻄﻌﺎﺕ ﻣﺘﺼﻞ ﺑﻪ‬ ‫ﺁﻥ ﺍﻳﺠﺎﺩ ﻛﺮﺩﻩ ﻭ ﻳﻚ ﭘﻴﻦ ﺍﺯ ﺳﻮﺭﺍﺥ ﮔﺬﺭﺍﻧﺪﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ‪ ØPn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ‪ ، LRFD‬ﻭ ﻧﻴﺰ ‪ Pn /Ω‬ﻣﻘﺎﻭﻣﺖ‬ ‫ﻛﺸﺸﻲ ﻣﺠﺎﺯ ﺑﻪ ﺭﻭﺵ ‪ ASD‬ﺑﺎﻳﺪ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻻﺕ ﺣﺪﻱ ﺯﻳﺮ ﺗﻌﻴﻴﻦ ﺷﻮﺩ ‪:‬‬ ‫)‪ AISC05 D5.1 p.16.1-28‬ﻭ ﺻﻔﺤﻪ ‪ 167‬ﻣﺒﺤﺚ ﺩﻫﻢ(‬ ‫ﺍﻟﻒ( ﮔﺴﻴﺨﺘﮕﻲ ﻛﺸﺸﻲ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ ‪:‬‬ ‫‪Øt = 0.75 , Ωt= 2.00 , Pn = 2tbeffFu‬‬ ‫ﺏ( ﮔﺴﻴﺨﺘﮕﻲ ﺑﺮﺷﻲ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ‪:‬‬ ‫‪Øsf = 0.75, Ωsf= 2.00 , Pn = 0.6Fu Asf‬‬ ‫‪59‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﻟﻮﻻﻳﻲ‬ ‫‪PIN-CONNECTED MEMBERS‬‬ ‫پ( ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﺩﺭ ﺳﻄﺢ ﺗﺼﻮﻳﺮ ﺷﺪﻩ ﻗﻠﻢ ﻟﻮﻻ )ﭘﻴﻦ( ‪:‬‬ ‫‪Ø = 0.75 , Ω = 2.00 , Pn = 1.8Fy Apb‬‬ ‫ﺕ( ﺗﺴﻠﻴﻢ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ‪:‬‬ ‫‪Øt = 0.75, Ωt= 1.67 , Pn = Fy Ag‬‬ ‫ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺭﻭﺍﺑﻂ ‪:‬‬ ‫ﺿﺨﺎﻣﺖ ﻭﺭﻕ = ‪t‬‬ ‫‪beff = 2t + 0.63, in. (= 2t + 16, mm) d b‬‬ ‫ﻓﺎﺻﻠﻪ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ﺗﺎ ﻟﺒﻪ ﻋﻀﻮ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﻋﻤﻮﺩ ﺑﺮ ﻧﻴﺮﻭﻱ ﻭﺍﺭﺩﻩ = ‪b‬‬ ‫‪= 2t(a + d/2),‬‬ ‫ﻛﻮﺗﺎﻩ ﺗﺮﻳﻦ ﻓﺎﺻﻠﻪ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ﺗﺎ ﻟﺒﻪ ﻋﻀﻮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭ =‬ ‫ﻗﻄﺮ ﭘﻴﻦ =‬ ‫‪ = dt‬ﺳﻄﺢ ﺗﺼﻮﻳﺮ ﺷﺪﻩ ﭘﻴﻦ =‬ ‫‪60‬‬ ‫‪Asf‬‬ ‫‪a‬‬ ‫‪d‬‬ ‫‪Apb‬‬ ‫ﺍﻋﻀﺎﻱ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﻟﻮﻻﻳﻲ – ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﺍﺑﻌﺎﺩﻱ ﺗﺴﻤﻪ ﻫﺎﻱ ﻟﻮﻻ ﺷﺪﻩ‬ ‫‪PIN-CONNECTED MEMBERS - DIMENSIONAL REQUIREMENTS‬‬ ‫ ﻣﺤﺪﻭﺩﻳﺖ ﻫـﺎﻱ ﺍﺑﻌـﺎﺩﻱ ﺑـﺮﺍﻱ ﺍﻋﻀـﺎﻱ ﺑـﺎ‬ ‫ﺍﺗﺼــــــــﺎﻻﺕ ﻣﻔﺼــــــــﻠﻲ ﻃﺒــــــــﻖ‬ ‫‪ AISC05 D5.2 p.16.1-30‬ﻭ ﻧﻴﺰ ﻣﺒﺤـﺚ‬ ‫ﺩﻫﻢ ﺻﻔﺤﻪ ‪ 166‬ﺩﺭ ﺷﻜﻞ ﺯﻳـﺮ ﻧﺸـﺎﻥ ﺩﺍﺩﻩ‬ ‫ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫‪61‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﺴﻤﻪ ﺳﺮﭘﻬﻦ‬ ‫‪EYEBARS‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪62‬‬ ‫ﺗﺴﻤﻪ ﺳﺮﭘﻬﻦ ﻧﻮﻉ ﺧﺎﺻﻲ ﺍﺯ ﻋﻀﻮ ﺑﺎ ﺍﺗﺼﺎﻝ ﻣﻔﺼﻠﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﻗﻄﺮ ﺍﻧﺘﻬﺎﻱ ﺳﻮﺭﺍﺧﺪﺍﺭ ﺗﻮﺳﻌﻪ‬ ‫ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﻮﺟﻮﺩ ﺗﺴﻤﻪ ﻫﺎﻱ ﺳﺮﭘﻬﻦ ﺑﻪ ﻃﺮﻳﻘﻲ ﻣﺸﺎﺑﻪ ﺣﺎﻟﺖ ﻛﻠﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺗﻌﻴﻴﻦ‬ ‫ﻣﻲ ﺷﻮﺩ‪) .‬ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪،( AISC D2‬‬ ‫ﺑﺎ ﺍﻳﻦ ﺗﻔﺎﻭﺕ ﻛﻪ‪ Ag ،‬ﺑﺮﺍﺑﺮ ﺑﺎ ﺳﻄﺢ ﻣﻘﻄﻊ ﺑﺪﻧﻪ ﺗﺴﻤﻪ ﺩﺭﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﺍﺑﻌﺎﺩﻱ ﺑﺮﺍﻱ ﺗﺴﻤﻪ ﻫﺎﻱ ﺳﺮﭘﻬﻦ ﻃﺒﻖ ﺿﻮﺍﺑﻂ ‪ AISC05 D6.2 p.16.1-30‬ﻭ ﻧﻴﺰ‬ ‫ﻣﺒﺤﺚ ﺩﻫﻢ ﺻﻔﺤﻪ ‪ 171‬ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ‪-‬‬ ‫ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪AISC‬‬ ‫‪Built-Up Tension Members-According to AISC D.4 p16.1-28‬‬ ‫ ﻫﺮﭼﻨﺪ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ﭼﻨﺪﺍﻥ ﻣﺘﺪﺍﻭﻝ ﻧﻴﺴﺘﻨﺪ ﻟﻴﻜﻦ ﺿﻮﺍﺑﻂ ‪AISC‬‬ ‫ﺳﺎﺧﺖ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻗﻴﺪﻫﺎﻱ )ﺑﺴﺖ ﻫﺎ‪ ،‬ﺻﻔﺤﺎﺕ(‬ ‫ﻣﻮﺍﺯﻱ‪ ،‬ﻣﻮﺭﺏ ﻭ ﺻﻔﺤﺎﺕ ﭘﻮﺷﺸﻲ ﺳﻮﺭﺍﺧﺪﺍﺭ ﺭﺍ ﻣﺠﺎﺯ ﻣﻲﺩﺍﻧﺪ‪.‬‬ ‫ ﺩﺭ ﺭﺍﺑﻄﻪ ﺑﺎ ﻣﺤﺪﻭﺩﻳﺖﻫﺎﻱ ﻭﺿﻊ ﺷﺪﻩ ﺩﺭ ﺯﻣﻴﻨﻪ ﻓﺎﺻﻠﻪ ﻃﻮﻟﻲ ﺍﺑﺰﺍﺭ ﺍﺗﺼﺎﻝ‬ ‫)‪ (connectors‬ﺍﺟﺰﺍﻱ ﺑﺎ ﺍﺗﺼﺎﻝ ﻣﻤﺘﺪ ﻣﺘﺸﻜﻞ ﺍﺯ ﻳﻚ ﻭﺭﻕ ﻭ ﻳﻚ ﻧﻴﻤﺮﺥ‪ ،‬ﻳﺎ ﺩﻭ‬ ‫ﻭﺭﻕ‪ ،‬ﺑﻪ ﺿﻮﺍﺑﻂ ‪ AISC J3.5‬ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ‪.‬‬ ‫ ﺩﺭ ﻭﺟﻪ ﺑﺎﺯ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻭﺭﻕ ﻫﺎﻱ ﭘﻮﺷﺸﻲ ﺳﻮﺭﺍﺧﺪﺍﺭ ﻭ ﻳﺎ‬ ‫ﺑﺴﺖﻫﺎﻱ ﻣﻮﺍﺯﻱ ﺑﺪﻭﻥ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻣﺠﺎﺯ ﺍﺳﺖ‪.‬‬ ‫‪63‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ‪-‬‬ ‫ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪AISC‬‬ ‫‪Built-Up Tension Members-According to AISC D.4 p16.1-28‬‬ ‫ ﻃﻮﻝ ﺻﻔﺤﺎﺕ ﺑﺴﺖ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﺩﻭ ﺳﻮﻡ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺧﻄﻮﻁ ﺟﻮﺵ ﻳﺎ ﭘﻴﭻ ﻫﺎﻱ ﺍﺗﺼﺎﻝ ﺁﻧﻬﺎ ﺑﻪ ﺍﺟﺰﺍﻱ ﻋﻀﻮ ﻣﺮﻛﺐ ﺑﺎﺷﺪ‪.‬‬ ‫ ﺿﺨﺎﻣﺖ ﺍﻳﻦ ﺻﻔﺤﺎﺕ ﺑﺴﺖ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﻳﻚ ﭘﻨﺠﺎﻫﻢ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺧﻄﻮﻁ ﺟﻮﺵ ﻳﺎ ﭘﻴﭽﻬﺎﻱ ﺍﺗﺼﺎﻝ ﺁﻧﻬﺎ ﺑﻪ ﺍﺟﺰﺍﻱ ﻋﻀﻮ‬ ‫ﻣﺮﻛﺐ ﺑﺎﺷﺪ‪.‬‬ ‫ ﻓﺎﺻﻠﻪ ﻃﻮﻟﻲ ﺟﻮﺵ ﻫﺎ ﻳﺎ ﭘﻴﭽﻬﺎﻱ ﻣﺘﻨﺎﻭﺏ ﺩﺭ ﺻﻔﺤﺎﺕ ﺑﺴﺖ ﻧﺒﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ‪ 150‬ﻣﻴﻠﻴﻤﺘﺮ )‪ 6‬ﺍﻳﻨﭻ( ﺑﺎﺷﺪ‪.‬‬ ‫ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺍﺑﺰﺍﺭ ﺍﺗﺼﺎﻝ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺑﺎﻳﺪ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﺍﻱ ﺑﺎﺷﺪ ﻛﻪ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﻫﺮﻳﻚ ﺍﺯ ﺍﺟﺰﺍﻱ ﻛﺸﺸﻲ ﻣﺘﺼﻞ ﺷﺪﻩ ﺑﻴﻦ ﺍﻳﻦ‬ ‫ﺍﺗﺼﺎﻻﺕ ﺗﺮﺟﻴﻬﺎ ﺍﺯ ‪ 300‬ﺑﻴﺸﺘﺮ ﻧﺸﻮﺩ‪.‬‬ ‫ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻓﻮﻻﺩ ﺍﻳﺮﺍﻥ ﻧﻴﺰ ﺿﻮﺍﺑﻂ ﺗﻘﺮﻳﺒﺎ ﻣﺸﺎﺑﻬﻲ ﺭﺍ ﺩﺭ ﺻﻔﺤﻪ ‪ 166‬ﻣﺒﺤﺚ ﺩﻫﻢ ﻣﻄﺮﺡ ﻧﻤﻮﺩﻩ ﺍﺳﺖ‪.‬‬ ‫‪t 2h/3‬‬ ‫‪t h/50‬‬ ‫‪h‬‬ ‫‪150 mm‬‬ ‫‪64‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Compression Members ‫ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬: ‫ ﻗﺴﻤﺖ ﺍﻭﻝ‬-‫ﻓﺼﻞ ﭼﻬﺎﺭﻡ‬ Urmia University - M.Sheidaii 1 ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬ ‫‪Compression Members‬‬ ‫ ﺍﻋﻀﺎﻱ ﺳﺎﺯﻩ ﺍﻱ ﻛﻪ ﺗﺤﺖ ﻧﻴﺮﻭﻱ ﻓﺸﺎﺭﻱ ﻣﺤﻮﺭﻱ ﻫﺴﺘﻨﺪ)ﺳﺘﻮﻧﻬﺎ‪،‬‬ ‫ﺍﻋﻀﺎﻱ ﺧﺮﭘﺎ‪ ،‬ﺳﻴﺴﺘﻢ ﻫﺎﻱ ﻣﻬﺎﺭﺑﻨﺪﻱ ﻭ ‪.(...‬‬ ‫ ﺑﻪ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﺿﻌﻴﻒﺗﺮ ﺑﻌﻀﻲ ﻣﻮﺍﻗﻊ ‪ strut‬ﮔﻔﺘﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪2‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬ ‫‪Compression Members‬‬ ‫ ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﺳﺘﻮﻥ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﻣﺤﺎﺳﺒﻪ ﻛﺮﺩ ‪:‬‬ ‫‪ƒ = P/A‬‬ ‫ﻛﻪ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﺑﻪ ﺻﻮﺭﺕ ﻳﻜﻨﻮﺍﺧﺖ ﺩﺭ ﻛﻞ ﻣﻘﻄﻊ ﺗﻮﺯﻳﻊ ﺷﺪﻩ ﺑﺎﺷﺪ‪.‬‬ ‫ ﺍﻳﻦ ﻭﺿﻌﻴﺖ ﺍﻳﺪﻩ ﺁﻝ ﻫﻴﭽﮕﺎﻩ ﺩﺭ ﻋﻤﻞ ﭘﻴﺶ ﻧﺨﻮﺍﻫﺪ ﺁﻣﺪ‪ .‬ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﻋﻀﻮ ﺑﻪ ﺧﺎﻃﺮ ﻋﻮﺍﻣﻞ ﺯﻳﺮ‬ ‫ﻏﻴﺮ ﻳﻜﻨﻮﺍﺧﺖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ‪:‬‬ ‫ ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺗﺼﺎﺩﻓﻲ ﺑﺎﺭ ﮔﺬﺍﺭﻱ ﻧﺴﺒﺖ ﺑﻪ ﻣﺮﻛﺰ ﺳﻄﺢ ﻋﻀﻮ‪،‬‬ ‫ ﻣﺴﺘﻘﻴﻢ ﻧﺒﻮﺩﻥ )ﺧﻤﻴﺪﮔﻲ( ﻋﻀﻮ‪ ،‬ﻳﺎ‬ ‫ ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻋﻀﻮ ﻧﺎﺷﻲ ﺍﺯ ﻋﻤﻠﻴﺎﺕ ﺳﺎﺧﺖ‪.‬‬ ‫ ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺗﺼﺎﺩﻓﻲ ﻭ ﺧﻤﻴﺪﮔﻲ ﻋﻀﻮ ﻣﻲ ﺗﻮﺍﻧﺪ ﺑﺎﻋﺚ ﺍﻳﺠﺎﺩ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺩﺭ ﻋﻀﻮ ﺷﻮﺩ‪ .‬ﻫﺮ ﭼﻨﺪ ﻛﻪ‪،‬‬ ‫ﺍﻳﻨﻬﺎ ﻟﻨﮕﺮ ﺛﺎﻧﻮﻳﻪ ﺑﻮﺩﻩ ﻭ ﺍﻏﻠﺐ ﻣﻮﺭﺩ ﺻﺮﻑ ﻧﻈﺮ ﻗﺮﺍﺭ ﻣﻲ ﮔﻴﺮﻧﺪ‪.‬‬ ‫ ﺍﮔﺮ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺑﺮ ﻋﻀﻮ ﻭﺍﺭﺩ ﺷﻮﺩ ﻧﻤﻲ ﺗﻮﺍﻥ ﺍﺯ ﺍﺛﺮﺍﺕ ﺧﻤﺸﻲ ﺻﺮﻓﻨﻈﺮ ﻛﺮﺩ‪ ،‬ﺍﻋﻀﺎﻳﻲ ﻛﻪ ﺗﺤﺖ ﺍﺛﺮ‬ ‫ﻫﻤﺰﻣﺎﻥ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﻭ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻫﺴﺘﻨﺪ ﺗﻴﺮ‪-‬ﺳﺘﻮﻥ)‪ (beam-column‬ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫‪3‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ‬ ‫‪Residual Stresses‬‬ ‫ ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ ﻭ ﺗﻮﺯﻳﻊ ﺁﻧﻬﺎ‪ ،‬ﻋﻮﺍﻣﻞ ﺑﺴﻴﺎﺭ ﻣﻬﻢ ٴﺗﺎﺛﻴﺮﮔﺬﺍﺭ ﺩﺭ ﻣﻘﺎﻭﻣﺖ ﺳﺘﻮﻧﻬﺎﻱ‬ ‫ﻓﻮﻻﺩﻱ ﺗﺤﺖ ﺑﺎﺭ ﻣﺤﻮﺭﻱ ﻫﺴﺘﻨﺪ)ﺑﺨﺼﻮﺹ ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ ﻣﺘﺪﺍﻭﻝ ﺑﺎ ﺿﺮﻳﺐ ﻻﻏﺮﻱ‬ ‫ﻣﺘﻐﻴﻴﺮ ﺑﻴﻦ ‪ 40‬ﺗﺎ ‪(120‬‬ ‫ ﺍﺟﺘﻨﺎﺏ ﻧﺎﭘﺬﻳﺮ ‪ -‬ﻧﺎﺷﻲ ﺍﺯ ﺳﺮﺩ ﺷﺪﻥ ﻏﻴﺮ ﻳﻜﻨﻮﺍﺧﺖ ﻧﻴﻤﺮﺥﻫﺎ ﭘﺲ ﺍﺯ ﻧﻮﺭﺩ ﮔﺮﻡ‪ ،‬ﻳﺎ‬ ‫ﺟﻮﺷﻜﺎﺭﻱ ﻣﻘﺎﻃﻊ ﻣﺮﻛﺐ‪.‬‬ ‫ ﻣﻲ ﺗﻮﺍﻧﻨﺪ ﻣﻘﺪﺍﺭ ﺑﺴﻴﺎﺭ ﻗﺎﺑﻞ ﺗﻮﺟﻬﻲ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ)ﺗﺎ ‪( 0.5Fy‬‬ ‫‪4‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ‬ ‫‪Residual Stresses‬‬ ‫ ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ ﺩﺭ ﻣﻘﻄﻊ ﻳﻚ ﻧﻴﻤﺮﺥ ﺑﺎﻝ ﭘﻬﻦ ‪:‬‬ ‫ ﻧﺤﻮﻩ ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﺩﺭ ﺟﺎﻥ ﻋﻀﻮ ﺩﺭ ﻃﻲ ﺑﺎﺭﮔﺬﺍﺭﻱ ‪:‬‬ ‫ ﻣﻘﺎﻳﺴﻪ ﻣﻨﺤﻨﻲ ﺑﺎﺭ‪-‬ﺗﻐﻴﻴﺮﻣﻜﺎﻥ ﻣﻨﺘﺠﻪ ﺑﺎ ﺭﻓﺘﺎﺭ ﻧﻤﻮﻧﻪ ﺍﻳﺪﻩ ﺁﻝ‬ ‫‪5‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ‬ Residual Stresses Urmia University - M.Sheidaii 6 ‫ﺍﺛﺮﺍﺕ ﺗﻨﺶ ﻫﺎﻱ ﻓﺸﺎﺭﻱ ﭘﺴﻤﺎﻧﺪ‬ ‫‪Effects of Residual Compressive Stresses‬‬ ‫ ﺗﺴﻠﻴﻢ ﺯﻭﺩ ﻫﻨﮕﺎﻡ)‪ (early yield‬ﺩﺭ ﺑﺨﺶﻫﺎﻳﻲ ﺍﺯ ﻣﻘﻄﻊ ﻋﺮﺿﻲ‬ ‫ ﺍﻳﺠﺎﺩ ﺭﻓﺘﺎﺭ ﻏﻴﺮﺧﻄﻲ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪ ﻗﺒﻞ ﺍﺯ ﺗﺴﻠﻴﻢ ﻛﺎﻣﻞ ﻣﻘﻄﻊ ﻋﺮﺿﻲ‬ ‫ ﺭﻓﺘﺎﺭ ﺳﺘﻮﻥ ﻣﺸﺎﺑﻪ ﺳﺘﻮﻧﻲ ﺑﺎ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻛﺎﻫﺶ ﻳﺎﻓﺘﻪ ﺍﺳﺖ)ﻓﻘﻂ ﻗﺴﻤﺖ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫ﻣﻘﻄﻊ ﺳﺘﻮﻥ(‬ ‫ ﺍﻳﻦ ﺍﻣﺮ ﻣﻤﻜﻦ ﺍﺳﺖ ﺩﺭ ﺑﻌﻀﻲ ﻣﻮﺍﺭﺩ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ‪ 25‬ﺩﺭﺻﺪﻱ ﺩﺭ ﻣﻘﺎﻭﻣﺖ ﻋﻀﻮ‬ ‫ﺷﻮﺩ‪.‬‬ ‫‪7‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺳﺘﻮﻧﻬﺎ ﭼﻄﻮﺭ ﻣﻲﺗﻮﺍﻧﻨﺪ ﺧﺮﺍﺏ ﺷﻮﻧﺪ؟‬ How can columns fail? (Compression Yielding) ‫ ﺗﺴﻠﻴﻢ ﻓﺸﺎﺭﻱ‬ (Global Buckling)‫ ﻛﻤﺎﻧﺶ ﻛﻠﻲ‬ (Flexural or Lateral Buckling)‫ﺧﻤﺸﻲ‬ ‫• ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﻳﺎ‬ ( Flexural-Torsional Buckling)‫ﺧﻤﺸﻲ‬-‫• ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ‬ (Torsional Buckling)‫• ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ‬ (Local Buckling)‫ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‬ Urmia University - M.Sheidaii 8 ‫ﺍﻧﻮﺍﻉ ﻛﻤﺎﻧﺶ‬ Buckling Types 9 ‫ﻣﻘﺎﻃﻊ ﻣﺘﺪﺍﻭﻝ ﺑﺮﺍﻱ ﺳﺘﻮﻥ‬ Sections used for Column Urmia University - M.Sheidaii 10 ‫ﻃﺒﻘﻪ ﺑﻨﺪﻱ ﺭﻓﺘﺎﺭ ﺳﺘﻮﻥ‬ Column Behavior Classification (Short Columns - Compression failure)‫ ﺳﺘﻮﻧﻬﺎﻱ ﻛﻮﺗﺎﻩ – ﺧﺮﺍﺑﻲ ﻓﺸﺎﺭﻱ‬ (Slender (Long) Columns – Buckling failure)‫ﻛﻤﺎﻧﺸﻲ‬ ‫ﺧﺮﺍﺑﻲ‬-(‫ ﺳﺘﻮﻧﻬﺎﻱ ﻻﻏﺮ)ﺑﻠﻨﺪ‬ ‫ ﺧﺮﺍﺑﻲ ﺗﺮﻛﻴﺒﻲ ﻓﺸﺎﺭﻱ ﻭ ﭘﺎﻳﺪﺍﺭﻱ‬-‫ ﺳﺘﻮﻧﻬﺎﻱ ﻣﺘﻮﺳﻂ‬ (Intermediate Columns – combined compression and stability failure) Urmia University - M.Sheidaii 11 ‫ﺳﺘﻮﻧﻬﺎﻱ ﻛﻮﺗﺎﻩ‬ ‫‪Short Columns‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺗﺴﻠﻴﻢ ﻭ ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺗﻌﻴﻴﻦﻛﻨﻨﺪﻩ ﻇﺮﻓﻴﺖ ﺳﺘﻮﻥ‬ ‫ﺍﺳﺖ‪.‬‬ ‫ ﺩﺭ ﺳﺎﺯﻩ ﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺧﻴﻠﻲ ﻣﺘﺪﺍﻭﻝ ﻧﻴﺴﺘﻨﺪ‪.‬‬ ‫ ﻇﺮﻓﻴﺖ ﺍﺳﻤﻲ ﻋﻀﻮ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪:‬‬ ‫‪Pn = Fy Ag‬‬ ‫‪12‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺳﺘﻮﻧﻬﺎﻱ ﻻﻏﺮ )ﺑﻠﻨﺪ(‬ ‫‪Slender (Long) Columns‬‬ ‫ ﻏﺎﻟﺒﺎً ﻛﻤﺎﻧﺶ ﺧﻤﺸﻲ ﺗﻌﻴﻴﻦ ﻛﻨﻨﺪﻩ ﻇﺮﻓﻴﺖ ﻋﻀﻮ ﺍﺳﺖ‪.‬‬ ‫ ﻇﺮﻓﻴﺖ ﻋﻀﻮ‪ ،‬ﺑﺮ ﺍﺳﺎﺱ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ‪ ،‬ﻣﺴﺘﻘﻞ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﺗﺴﻠﻴﻢ‬ ‫ﻣﺼﺎﻟﺢ ﺍﺳﺖ‪.‬‬ ‫ ﻛﻤﺎﻧﺶ ﺑﻪ ﺻﻮﺭﺕ ﺍﺭﺗﺠﺎﻋﻲ ﺍﺳﺖ‪.‬‬ ‫ ﻇﺮﻓﻴﺖ ﻋﻀﻮ ﺑﺴﻴﺎﺭ ﺣﺴﺎﺱ ﺑﻪ ﺷﺮﺍﻳﻂ ﺍﻧﺘﻬﺎﻳﻲ ﻋﻀﻮ ﺍﺳﺖ ‪:‬‬ ‫‪13‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺳﺘﻮﻧﻬﺎﻱ ﻣﺘﻮﺳﻂ‬ ‫‪Intermediate Columns‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻭ ﭘﺎﻳﺪﺍﺭﻱ ﺗﻌﻴﻴﻦ ﻛﻨﻨﺪﻩ ﻇﺮﻓﻴﺖ ﻋﻀﻮ ﺍﺳﺖ‪.‬‬ ‫ ﻛﻤﺎﻧﺶ ﺑﻪ ﺻﻮﺭﺕ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ ﺍﺳﺖ‪.‬‬ ‫ ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ ﺑﺴﻴﺎﺭ ﺗﺎٴﺛﻴﺮﮔﺬﺍﺭ ﺩﺭ ﻇﺮﻓﻴﺖ ﻋﻀﻮ ﻫﺴﺘﻨﺪ‪.‬‬ ‫‪14‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ‬ ‫‪The Euler Formula‬‬ ‫ﻣﻌﺎﺩﻟﻪ ﺍﺳﺎﺱ ﺣﺎﻛﻢ ﺑﺮ ﻛﻤﺎﻧﺶ ﺧﻤﺸﻲ ﺳﺘﻮﻥ ﺩﻭ ﺳﺮ ﻣﻔﺼﻠﻲ ﺩﺭ ﺯﻳﺮ ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫ﺣﻞ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ )ﻫﻤﮕﻦ‪ ،‬ﻣﺮﺗﺒﻪ ﺩﻭ‪،‬ﺑﺎ ﺿﺮﺍﻳﺐ ﺛﺎﺑﺖ( ﻋﺒﺎﺗﺴﺖ ﺍﺯ ‪:‬‬ ‫‪15‬‬ ‫ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ‬ ‫‪The Euler Formula‬‬ ‫ﺿﺮﺍﻳﺐ ﺛﺎﺑﺖ ‪ A‬ﻭ ‪ B‬ﺑﺮ ﺍﺳﺎﺱ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺳﺘﻮﻥ ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﺩ ‪:‬‬ ‫ﺑﺮ ﺍﺳﺎﺱ ﺭﺍﺑﻄﻪ ﻓﻮﻕ‪ ،‬ﺑﺮﺍﻱ ﺑﺪﺳﺖ ﺣﻞ ﻛﻤﺎﻧﺸﻲ‪ ،‬ﺑﺎﺭ ﺑﺎﻳﺪ ﺑﻪ ﻳﻚ ﻣﻘﺪﺍﺭ ﺑﺤﺮﺍﻧﻲ ﺑﺮﺳﺪ ‪:‬‬ ‫ﺑﺮﺍﻱ ﺍﻫﺪﺍﻑ ﻃﺮﺍﺣﻲ‪ ،‬ﺑﻬﺘﺮ ﺍﺳﺖ ﻋﺒﺎﺭﺕ ﻓﻮﻕ ﺑﺮ ﺣﺴﺐ ﺗﻨﺶ ﺑﺤﺮﺍﻧﻲ ﻧﻮﺷﺘﻪ ﺷﻮﺩ)ﺑﻪ ﺟﺎﻱ‬ ‫ﺑﺎﺭ ﺑﺤﺮﺍﻧﻲ( ‪:‬‬ ‫‪16‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ‬ ‫‪The Euler Formula‬‬ ‫ﺑﺮﺍﻱ ﺷﺮﺍﻳﻂ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﺑﺴﻴﺎﺭ ﻛﻠﻲ ﺗﺮ‪ ،‬ﺭﻭﺵ ﺗﺤﻠﻴﻞ ﻳﻜﺴﺎﻥ ﺑﻮﺩﻩ ‪ -‬ﺗﻨﻬﺎ ﺗﻔﺎﻭﺕ ﺩﺭ‬ ‫ﻋﺒﺎﺭﺕ ﻻﻏﺮﻱ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪ .‬ﺍﻳﻦ ﺗﻔﺎﻭﺕ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﺗﻌﺮﻳﻒ ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﺆﺛﺮ ‪ ،K‬ﻛﻪ‬ ‫ﺩﺭ ﺭﺍﺑﻄﻪ ﻛﻤﺎﻧﺶ ﺍﺭﺗﺠﺎﻋﻲ ﺯﻳﺮ ﺁﻣﺪﻩ ﺍﺳﺖ‪ ،‬ﺍﻋﻤﺎﻝ ﻧﻤﻮﺩ ‪:‬‬ ‫ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ‪ ،‬ﺭﺍﺑﻄﻪ ﺍﻱ ﺍﺳﺎﺳﻲ ﺑﺮﺍﻱ ﻓﻬﻢ ﺭﻓﺘﺎﺭ ﻛﻤﺎﻧﺸﻲ ﺳﺘﻮﻥ ﺑﻮﺩﻩ ﻭ ﺑﻪ ﺻﻮﺭﺕ‬ ‫ﺗﺮﺳﻴﻤﻲ ﺯﻳﺮ ﻗﺎﺑﻞ ﻧﻤﺎﻳﺶ ﺍﺳﺖ‪:‬‬ ‫‪17‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Elastic Buckling‬‬ ‫ﺑﺎﺭ ﻛﻤﺎﻧﺶ ﺑﺤﺮﺍﻧﻲ ‪ ،Pcr‬ﺑﺮﺍﻱ ﻳﻚ ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﻻﻏﺮ ﺑﻠﻨﺪ ﺑﻪ ﺻﻮﺭﺕ ﺗﺌﻮﺭﻳﻚ ﺑﺎ‬ ‫ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﺩ‪:‬‬ ‫‪S2 EI‬‬ ‫‪(KL) 2‬‬ ‫‪Pcr‬‬ ‫‪Pcr‬‬ ‫‪A‬‬ ‫‪Fcr‬‬ ‫‪S2 E‬‬ ‫‪(KL / r ) 2‬‬ ‫ﻛﻪ ‪:‬‬ ‫‪ = r‬ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ‬ ‫‪ = KL/r‬ﺿﺮﻳﺐ ﻻﻏﺮﻱ‬ ‫‪ = K‬ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﺆﺛﺮ ﺑﺮ ﺍﺳﺎﺱ ﺷﺮﺍﻳﻂ‬ ‫ﻣﺮﺯﻱ ﺍﻧﺘﻬﺎﻱ‬ ‫‪18‬‬ ‫ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﺆﺛﺮ‪K ،‬‬ ‫‪Effective Length Factor, K‬‬ ‫ ﺷﺮﺍﻳﻂ ﺍﻧﺘﻬﺎﻳﻲ ﻭ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﺗﻌﻴﻴﻦ ﻛﻨﻨﺪﻩ ﻃﻮﻝ ﻣﺆﺛﺮ ﺍﺳﺖ‪:‬‬ ‫¾ ﻗﻴﻮﺩ ﺩﻭﺭﺍﻧﻲ‬ ‫¾‬ ‫ﺍﻧﺘﻘﺎﻝ ﻳﺎ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ‬ ‫ ﻣﻘﺎﺩﻳﺮ ‪ K‬ﺗﻘﺮﻳﺒﻲ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﻋﻤﻠﻲ ﺩﺭ ‪ AISC05 p. 16.1-240‬ﺁﻣﺪﻩ‬ ‫ﺍﺳﺖ)ﺍﺳﻼﻳﺪ ﺑﻌﺪﻱ(‪.‬‬ ‫ ﻣﻘﺎﺩﻳﺮ ﺗﺌﻮﺭﻳﻚ ﻭ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪ K‬ﺑﺎ ﻫﻢ ﻓﺮﻕ ﺩﺍﺭﻧﺪ‪ ،‬ﺯﻳﺮﺍ ﻭﺿﻌﻴﺖ ﻫﺎﻱ ﮔﻴﺮﺩﺍﺭﻱ ﻭ‬ ‫ﻟﻮﻻﻳﻲ ﻛﺎﻣﻞ ﻋﻤﻼً ﺍﻣﻜﺎﻥ ﭘﺬﻳﺮ ﻧﻴﺴﺖ‪.‬‬ ‫‪19‬‬ ‫‪Urmia University - M.Sheidaii‬‬ K ،‫ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﺆﺛﺮ‬ Effective Length Factor, K Urmia University - M.Sheidaii 20 ‫ﻣﺤﻮﺭ ﻛﻤﺎﻧﺶ‬ ‫‪Buckling Axis‬‬ ‫ ﺳﺘﻮﻧﻬﺎ ﻣﻌﻤﻮﻻً ﺑﻪ ﺻﻮﺭﺕ ﺟﺎﻧﺒﻲ ﻛﻤﺎﻧﻪ ﻣﻲ ﻛﻨﻨﺪ)ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻣﻲ ﻧﻤﺎﻳﺪ(‪.‬‬ ‫ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﺣﻮﻝ ﻣﺤﻮﺭﻱ ﻛﻪ ﺑﻴﺸﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ‪ L / r‬ﺭﺍ ﺩﺍﺭﺩ ﺍﺗﻔﺎﻕ ﻣﻲ ﺍﻓﺘﺪ‪.‬‬ ‫ ﺍﻳﻦ ﻣﺤﻮﺭ ﻣﻌﻤﻮﻻ ﻣﺤﻮﺭ ‪ y-y‬ﺍﺳﺖ‪ ،‬ﺍﻣﺎ ﺑﺎﻳﺴﺘﻲ ﻫﺮ ﺩﻭ ﻣﺤﻮﺭ ﻣﻮﺭﺩ ﺑﺮﺭﺳﻲ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ‪.‬‬ ‫‪21‬‬ ‫ﻣﺤﻮﺭ ﻛﻤﺎﻧﺶ‬ ‫‪Buckling Axis‬‬ ‫ ﺍﮔﺮ ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﺩﺭ ﻛﻤﺎﻧﺶ ﺣﻮﻝ ﺩﻭ ﻣﺤﻮﺭ ﺍﺻﻠﻲ ﺷﺮﺍﻳﻂ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﻣﺘﻔﺎﻭﺗﻲ‬ ‫ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ‪ ،‬ﻃﻮﻝ ﻣﺆﺛﺮ ﺑﺮﺍﻱ ﺩﻭ ﺭﺍﺳﺘﺎ ﻣﺘﻔﺎﻭﺕ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﻭ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺑﺰﺭﮔﺘﺮ‬ ‫ﺑﺎﻳﺪ ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﻇﺮﻓﻴﺖ ﻋﻀﻮ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﺩ‪.‬‬ ‫‪22‬‬ 23 ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻣﻬﺎﺭﺑﻨﺪﻱ‬ ‫‪Design of Bracing Members‬‬ ‫ ﺍﻋﻀﺎﻱ ﻣﻬﺎﺭﺑﻨﺪﻱ ﺑﺎﻳﺪ ﻗﺎﺩﺭ ﺑﻪ ٴﺗﺎﻣﻴﻦ ﻧﻴﺮﻭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻻﺯﻡ ﺑﺎﺷﻨﺪ‪ ،‬ﺑﺪﻭﻥ ﺍﻳﻨﻜﻪ‬ ‫ﺧﻮﺩﺷﺎﻥ ﻛﻤﺎﻧﻪ ﻛﻨﻨﺪ‪.‬‬ ‫ ﻧﻴﺮﻭﻫﺎﻳﻲ ﻛﻪ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻣﻬﺎﺭﺑﻨﺪﻱ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ ﻛﺎﻣﻼً ﻛﻮﭼﻚ‬ ‫ﺑﻮﺩﻩ ﻭ ﺍﻏﻠﺐ ﻣﻘﺪﺍﺭ ﻣﺤﺎﻓﻈﻪ ﻛﺎﺭﺍﻧﻪ ﺍﻱ ﺑﺮﺍﺑﺮ ﺑﺎ ‪ 0.02‬ﺑﺎﺭ ﻃﺮﺍﺣﻲ ﺳﺘﻮﻥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ‬ ‫ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﻃﺮﺡ ﻭ ﺍﻧﺘﺨﺎﺏ ﻣﻬﺎﺭﻫﺎ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﻧﻈﻴﺮ ﺳﺎﻳﺮ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﺍﻧﺠﺎﻡ ﺩﺍﺩ‪.‬‬ ‫ ﺍﻧﺘﻬﺎﻱ ﺩﻳﮕﺮ ﻋﻀﻮ ﻣﻬﺎﺭﺑﻨﺪ ﺑﺎﻳﺪ ﺑﻪ ﺍﻋﻀﺎﻱ ﺩﻳﮕﺮ ﺑﻪ ﻧﺤﻮﻱ ﺍﺗﺼﺎﻝ ﻳﺎﺑﺪ ﻛﻪ ﺍﺯ ﺟﺎﺑﺠﺎﻳﻲ‬ ‫ﺟﺎﻧﺒﻲ ﺳﺘﻮﻥ ﺟﻠﻮﮔﻴﺮﻱ ﻧﻤﺎﻳﺪ‪.‬‬ ‫‪24‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Inelastic Buckling‬‬ ‫ ﻣﻨﺤﻨﻲ ﺍﻭﻟﺮ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﻭﻗﺘﻲ ﻻﻏﺮﻱ‬ ‫ﻋﻀﻮ ﺑﻪ ﺻﻔﺮ ﻧﺰﺩﻳﻚ ﻣﻲ ﺷﻮﺩ ﻇﺮﻓﻴﺖ ﺁﻥ ﺑﻪ‬ ‫ﺑﻲ ﻧﻬﺎﻳﺖ ﻣﻲ ﺭﺳﺪ‪.‬‬ ‫ ﺍﻣﺎ ﺩﺭ ﻭﺍﻗﻊ ﺍﻣﻜﺎﻥ ﺗﺠﺎﻭﺯ ﻇﺮﻓﻴﺖ ﺳﺘﻮﻥ ﺍﺯ ﻇﺮﻓﻴﺖ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‪.‬‬ ‫ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ ﺗﺎ ﻫﻨﮕﺎﻣﻲ ﺻﺎﺩﻕ ﺍﺳﺖ ﻛﻪ ﻣﺼﺎﻟﺢ ﺩﺭ ﺗﻤﺎﻡ ﻣﻘﻄﻊ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫ﺑﺎﺷﻨﺪ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺿﺮﺍﻳﺐ ﻻﻏﺮﻱ ﻛﻮﭼﻜﺘﺮ‪ ،‬ﻣﺼﺎﻟﺢ ﻭﺍﺭﺩ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ ﺷﺪﻩ ﻭ ﻟﺬﺍ ﻣﻌﺎﺩﻟﻪ‬ ‫ﺍﻭﻟﺮ ﻗﺎﺑﻞ ﺍﺳﺘﻔﺎﺩﻩ ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫‪25‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Inelastic Buckling‬‬ ‫ﻋﻮﺍﻣﻞ ﻣﺘﻌﺪﺩ ﺩﻳﮕﺮﻱ ﻧﻴﺰ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ ﻭﺟﻮﺩ ﺩﺍﺭﻧﺪ ﻛﻪ ﺗﺤﻠﻴﻞ ﺳﺎﺩﻩ ﻛﻤﺎﻧﺶ ﺍﻭﻟﺮ ﺁﻧﻬﺎ‬ ‫ﺭﺍ ﺑﻪ ﺣﺴﺎﺏ ﻧﻴﺎﻭﺭﺩﻩ ﺍﺳﺖ ‪:‬‬ ‫‪26‬‬ ‫‬ ‫ﻣﺴﺘﻘﻴﻢ ﻧﺒﻮﺩﻥ ﻋﻀﻮ ٴﺗﺎﺛﻴﺮ ﻗﺎﺑﻞ ﺗﻮﺟﻬﻲ ﺑﺮ ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺸﻲ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫ﺩﺍﺭﺩ‪.‬‬ ‫‬ ‫ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ ﺩﺭ ﻋﻀﻮ ﻧﺎﺷﻲ ﺍﺯ ﻋﻤﻠﻴﺎﺕ ﺳﺎﺧﺖ‪ ،‬ﺑﺎﻋﺚ ﻣﻲ ﺷﻮﺩ ﺗﺴﻠﻴﻢﺷﺪﮔﻲ ﺩﺭ‬ ‫ﻣﻘﻄﻊ ﺑﺴﻴﺎﺭ ﺯﻭﺩﺗﺮ ﺍﺯ ﺍﻳﻨﻜﻪ ﺗﻨﺶ ﻳﻜﻨﻮﺍﺧﺖ ‪ f = P/A‬ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ‪ Fy‬ﺑﺮﺳﺪ ﭘﺪﻳﺪ‬ ‫ﻣﻲ ﺁﻳﺪ‪.‬‬ ‫‬ ‫ﺷﻜﻞ ﻣﻘﻄﻊ ﻋﺮﺿﻲ)‪ C، I‬ﻭ ‪ (...‬ﻧﻴﺰ ﺑﺮ ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺶ ﺗﺎٴﺛﻴﺮﮔﺬﺍﺭ ﺍﺳﺖ‪.‬‬ ‫‬ ‫ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ‪ ،‬ﻣﺼﺎﻟﺢ ﻓﻮﻻﺩﻱ ﻣﻲﺗﻮﺍﻧﻨﺪ ﺑﻪ ﺳﺨﺖﺷﺪﮔﻲ ﻛﺮﻧﺸﻲ ﺑﺮﺳﻨﺪ‪.‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Inelastic Buckling‬‬ ‫ ﻣﺪﻝ ﺍﻭﻟﻴﻪ ﺍﻱ ﻛﻪ ﺩﺭ ﺁﻥ ﺭﻓﺘﺎﺭ ﺳﺘﻮﻧﻬﺎﻱ ﻛﻮﺗﺎﻫﺘﺮ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﺑﺮ ﭘﺎﻳﻪ‬ ‫ﻣﻼﺣﻈﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﮔﺴﺘﺮﺵ ﺗﺴﻠﻴﻢ ﺷﺪﮔﻲ ﻧﺎﺷﻲ ﺍﺯ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ‬ ‫ﻋﻀﻮ ﺍﺳﺘﻮﺍﺭ ﮔﺮﺩﻳﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺭﻓﺘﺎﺭ ﺗﺴﻠﻴﻢ ﺷﺪﮔﻲ ﻣﻘﻄﻊ ﺷﺪﻳﺪﺍً ﻣﺘﺎٴﺛﺮ ﺍﺯ ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ ﺍﺳﺖ‪.‬‬ ‫ ﻫﻤﺎﻧﻄﻮﺭ ﻛﻪ ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺳﺨﺘﻲ ﻣﺆﺛﺮ ﻣﻘﻄﻊ ﻛﺎﻫﺶ ﻳﺎﻓﺘﻪ‬ ‫ﺍﺳﺖ‪.‬‬ ‫‪27‬‬ ‫ﻛﻤﺎﻧﺶ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Inelastic Buckling‬‬ ‫ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﭼﻨﻴﻦ ﻣﻘﺪﺍﺭ ﻛﺎﻫﺶ ﻳﺎﻓﺘﻪ ﺍﻱ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ ﻣﻨﺠﺮ ﺑﻪ ﻣﻨﺤﻨﻲ ﺍﺻﻼﺡ‬ ‫ﺷﺪﻩ ﺍﻱ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺯﻳﺮ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﺷﻜﻞ ﻋﻤﻠﻲ ﻣﻨﺤﻨﻲ ﻃﺮﺍﺣﻲ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﺑﺮﺍﻱ ﺑﺨﺶ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ ﻣﻨﺤﻨﻲ‬ ‫ﻛﻤﺎﻧﺶ ﺳﺘﻮﻥ‪ ،‬ﺑﺼﻮﺭﺕ ﺗﺠﺮﺑﻲ ﺗﻮﺳﻂ ﺁﺯﻣﺎﻳﺸﻬﺎﻱ ﮔﺴﺘﺮﺩﻩﺍﻱ ﺗﻌﻴﻴﻦ ﮔﺮﺩﻳﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪28‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻣﻨﺤﻨﻲ ﻇﺮﻓﻴﺖ ﺳﺘﻮﻥ‬ Column Strength Curve Urmia University - M.Sheidaii 29 ‫ﻧﺎﻛﺎﻣﻠﻲ ﻫﺎﻱ ﻫﻨﺪﺳﻲ ﺍﻭﻟﻴﻪ‬ ‫‪Initial Geometric Imperfections‬‬ ‫ ﻋﺎﻣﻞ ﻣﻬﻢ ﺗﺎٴﺛﻴﺮﮔﺬﺍﺭ ﺩﻳﮕﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺳﺘﻮﻧﻬﺎ‪ ،‬ﺍﺛﺮ ﺍﻧﺤﻨﺎﻱ ﺍﻭﻟﻴﻪ ﻋﻀﻮ ﺍﺳﺖ‪.‬‬ ‫ ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﺭﻓﺘﺎﺭ ﻳﻚ ﺳﺘﻮﻥ ﺍﻳﺪﻩ ﺁﻝ ﺑﺎ ﺳﺘﻮﻧﻲ ﻛﻪ ﺍﺯ ﺍﻭﻝ ﻛﺎﻣﻼً ﻣﺴﺘﻘﻴﻢ ﻧﻴﺴﺖ‬ ‫ﻣﻘﺎﻳﺴﻪ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬ ‫ ﺍﻳﻦ ﻋﺎﻣﻞ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ﻇﺮﻓﻴﺖ ﻭ ﺳﺨﺘﻲ ﺳﺘﻮﻥ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪30‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺿﻮﺍﺑﻂ ‪ AISC‬ﺑﺮﺍﻱ ﻣﻘﺎﻭﻣﺖ ﻋﻀﻮ ﻓﺸﺎﺭﻱ‬ ‫‪AISC Specifications for Compression Member Strength‬‬ ‫‬ ‫ﺿﻮﺍﺑﻂ ‪ AISC‬ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‪،‬ﻛﻪ ﺩﺭ ‪AISC Chapter E p.16.1-32‬‬ ‫ﺑﻴﺎﻥ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺑﺮ ﻣﺒﻨﺎﻱ ﺗﺤﻘﻴﻘﺎﺕ ﺍﻧﺠﺎﻡ ﺷﺪﻩ ﺩﺭ ﺳﺎﻟﻬﺎﻱ ﻣﺘﻤﺎﺩﻱ ﺍﺳﺘﻮﺍﺭ ﮔﺮﺩﻳﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‬ ‫ﺿﻮﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻓﻮﻻﺩ ﺍﻳﺮﺍﻥ ﺩﺭ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻧﻴﺰ ﺍﺯ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC‬ﺍﻗﺘﺒﺎﺱ‬ ‫ﮔﺮﺩﻳﺪﻩ ﻭ ﺩﺭ ﺑﺨﺶ ‪ 4-2-10‬ﺻﻔﺤﻪ ‪ 173‬ﻣﺒﺤﺚ ﺷﺸﻢ ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‬ ‫ﺩﺭ ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ ﻫﺎﻱ ‪ AISC‬ﺑﺮﺍﻱ ﻃﺮﺡ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﺳﻪ ﺣﺎﻟﺖ ﺣﺪﻱ ﺍﺻﻠﻲ ﺩﺭ ﻧﻈﺮ‬ ‫ﮔﺮﻓﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬ ‫)‪(i‬‬ ‫)‪(ii‬‬ ‫)‪(iii‬‬ ‫‬ ‫‪31‬‬ ‫ﻛﻤﺎﻧﺶ ﻛﻠﻲ ﺍﺭﺗﺠﺎﻋﻲ )ﺑﺮ ﺍﺳﺎﺱ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻟﺮ(‬ ‫ﻛﻤﺎﻧﺶ ﻛﻠﻲ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ )ﺷﺎﻣﻞ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻓﺸﺎﺭﻱ ﻧﻴﺰ ﻫﺴﺖ(‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ‬ ‫ﺍﻳﻦ ﺿﻮﺍﺑﻂ ﺩﺭ ﺑﺮ ﮔﻴﺮﻧﺪﻩ ﻛﻤﺎﻧﺶ ﺍﺭﺗﺠﺎﻋﻲ ﻭ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ ﺳﺘﻮﻧﻬﺎ ﺷﺎﻣﻞ ﻛﻠﻴﻪ ﻣﺒﺎﺣﺚ ﻓﻮﻕ‬ ‫ﺍﻟﺬﻛﺮ )ﺍﻧﺤﻨﺎﻱ ﻋﻀﻮ‪ ،‬ﺗﻨﺶ ﻫﺎﻱ ﭘﺴﻤﺎﻧﺪ‪ ،‬ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺗﺼﺎﺩﻓﻲ ﻭ ‪ (...‬ﻣﻲ ﺑﺎﺷﻨﺪ‪.‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ ﺑﺮﺍﻱ ﻣﻘﺎﻭﻣﺖ ﻋﻀﻮ ﻓﺸﺎﺭﻱ‬AISC ‫ﺿﻮﺍﺑﻂ‬ AISC Specifications for Compression Member Strength ‫ ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﺍﺳﻤﻲ‬Pn = Fcr Ag LRFD ‫ = ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻃﺮﺍﺣﻲ‬IcPn ASD ‫ = ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻣﺠﺎﺯ‬Pn /:c : ‫ﻛﻪ‬ Fcr= ‫ﺗﻨﺶ ﺑﺤﺮﺍﻧﻲ ﺧﻤﺸﻲ‬ Ag= ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ‬ Urmia University - M.Sheidaii 32 ‫ﺭﻭﺵ ‪LRFD‬‬ ‫‪LRFD Method‬‬ ‫‪Pu d Ic Pn‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻧﻬﺎﻳﻲ ﻻﺯﻡ ﺍﺯ ﺗﺤﻠﻴﻞ ﺳﺎﺯﻩ ﺗﺤﺖ ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭ ﺿﺮﻳﺐ ﺩﺍﺭ= ‪Pu‬‬ ‫ﺿﺮﻳﺐ ﻣﻘﺎﻭﻣﺖ ﺑﺮﺍﻱ ﻓﺸﺎﺭ ‪Ic = 0.90‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻃﺮﺍﺣﻲ= ‪Ic Pn‬‬ ‫‪33‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ASD‫ﺭﻭﺵ‬ ASD Method Pn Pa d :c Pa =‫ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ ﺍﺯ ﺗﺤﻠﻴﻞ ﺳﺎﺯﻩ ﺗﺤﺖ ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ‬ :c = safety factor compression =1.67 ‫ﺿﺮﻳﺐ ﺍﻳﻤﻨﻲ ﺑﺮﺍﻱ ﻓﺸﺎﺭ‬ Pn / :c=allowable compressive strength ‫ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻣﺠﺎﺯ‬ Urmia University - M.Sheidaii 34 ASD – Allowable Stress ASD – ‫ﺭﻭﺵ ﺗﻨﺶ ﻣﺠﺎﺯ‬ f a d Fa fa= computed axial compressive stress = Pa /Ag ‫ﺗﻨﺶ ﻓﺸﺎﺭﻱ ﻣﺤﻮﺭﻱ ﻣﺤﺎﺳﺒﻪ ﺷﺪﻩ‬ Fa = allowable axial compressive stress ‫ﺗﻨﺶ ﻓﺸﺎﺭﻱ ﻣﺤﻮﺭﻱ ﻣﺠﺎﺯ‬ Fcr :c Urmia University - M.Sheidaii Fcr 1.67 0.6 Fcr 35 ‫ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﺑﺮﺍﻱ ﻛﻤﺎﻧﺶ ﺧﻤﺸﻲ ﺩﺭ ﺍﻋﻀﺎﻱ ﺑﺪﻭﻥ ﺍﺟﺰﺍﻱ ﻻﻏﺮ‬ Compressive Strength for Flexural Buckling of Members without Slender Elements The design compressive strength is IcPn with Øc = 0.90 Design Stress for Compression Member of Fy=2333 kgf/cm2 , Øc=0.9 KL/r 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 c Fcr 2100 2099 2099 2098 2097 2096 2095 2093 2091 2089 2087 2085 2082 2079 2076 2073 2070 2066 2062 2059 2054 2050 2045 2041 2036 2031 2025 2020 2014 2008 2002 1996 1989 1983 1976 1969 1962 1955 1947 1940 KL/r 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 c Fcr 1932 1924 1916 1908 1899 1891 1882 1873 1864 1855 1846 1837 1827 1817 1808 1798 1788 1778 1767 1757 1746 1736 1725 1714 1703 1692 1681 1670 1659 1647 1636 1624 1613 1601 1589 1577 1565 1554 1541 1529 KL/r 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 c Fcr 1517 1505 1493 1481 1468 1456 1443 1431 1418 1406 1393 1381 1368 1356 1343 1330 1318 1305 1292 1280 1267 1254 1242 1229 1216 1204 1191 1178 1166 1153 1141 1128 1116 1103 1091 1078 1066 1054 1041 1029 KL/r 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 c Fcr 1017 1005 993 981 969 957 945 933 921 909 898 886 874 863 852 840 829 818 806 794 783 772 761 751 740 730 720 711 701 692 683 674 665 656 648 640 631 623 616 608 KL/r 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 c Fcr 600 593 586 579 572 565 558 551 545 539 532 526 520 514 508 502 497 491 486 480 475 470 465 460 455 450 445 440 436 431 427 422 418 414 409 405 401 397 393 389 37 40 ‫ﻣﺮﺍﺣﻞ ﺗﺤﻠﻴﻞ ﺳﺘﻮﻥ‬ Steps for Column Analysis K ‫ ﺍﺭﺯﻳﺎﺑﻲ ﺷﺮﺍﻳﻂ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﻭ ﺍﻧﺘﻬﺎﻳﻲ ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ‬ KL/r ‫ ﻣﺤﺎﺳﺒﻪ‬ Pn=FcrAg ‫ ﺗﻌﻴﻴﻦ‬ LRFD ‫ ﺑﻪ ﺭﻭﺵ‬: Pu≤ IcPn= IcFcr Ag ‫ ﻣﺤﺎﺳﺒﻪ ﻇﺮﻓﻴﺖ ﻋﻀﻮ‬ ASD‫ ﺑﻪ ﺭﻭﺵ‬: Pa≤ Pn/Ωc= FcrAg/ Ωc Urmia University - M.Sheidaii 41 ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﻣﻄﻠﻮﺑﺴﺖ ﺗﻌﻴﻴﻦ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺳﺘﻮﻥ ﺷﻜﻞ ﺯﻳﺮ‪ .‬ﺳﺘﻮﻥ ﺍﺯ ﻓﻮﻻﺩ‬ ‫ﺳﺎﺧﺘﻤﺎﻧﻲ ﻣﻌﻤﻮﻟﻲ)‪ (Fy= 2333 kgf/cm2‬ﻭ ﺑﺎﺭ ﻭﺍﺭﺩﻩ ﻣﺤﻮﺭﻱ ﺍﺳﺖ‪.‬‬ ‫ﺣﻞ‪:‬‬ ‫)‪IPB 28 (Ag=131.4 cm2, ry= 7.09 cm‬‬ ‫‪K=0.8‬‬ ‫‪KL/r=0.8(450)/7.09=50.8‬‬ ‫‪IcFcr = 1849 kgf/cm2‬‬ ‫‪IcPn = IcFcrAg = 1.849(131.4) =243 ton‬‬ ‫‪42‬‬ ‫‪Urmia University - M.Sheidaii‬‬ Example: Determine the design strength IcPn of the axially loaded column shown in the figure if KL=5.7 m and ST37 steel is used. U30 ( A=58.8 cm2, d=300mm, Ix=8030cm4, Iy=495cm4, e=2.7cm from back of U) Solution: A=(50)(1.2)+2(58.8)=177.6 cm2 Ў from top = [2(58.8)(16.2)+(50)(1.2)(0.6)]/177.6= 10.93 cm Ix = 2(8030)+2(58.8)(5.27)2+(50)(1.2)(10.33)2+(50)(1.2)3/12= 25736 cm4 Iy = 2(492)+2(58.8)(14.7)2 + (1.2) (50)3/12= 38902 cm4 Min r=(25736/177.6)= 12.04 cm KL/r=570/12.04=47.3 IcFcr = 1879.4 kgf/cm2 IcPn = IcFcrAg = 1.879(177.6)=333.8 ton 43 1802 1802 Urmia University - M.Sheidaii 291 44 ‫ﺟﺰﺋﻴﺎﺕ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﺑﻪ ﺭﻭﺵ ‪LRFD‬‬ ‫‪Design of Compression Members‬‬ ‫ﻃﺮﺍﺣﻲ ﺳﺘﻮﻧﻬﺎ ﺩﺍﺭﺍﻱ ﻳﻚ ﺭﻭﻧﺪ ﺳﻌﻲ ﻭ ﺧﻄﺎ ﺑﻮﺩﻩ ﻭ ﺷﺎﻣﻞ ﮔﺎﻣﻬﺎﻱ ﺯﻳﺮ ﺍﺳﺖ‪:‬‬ ‫‪ .1‬ﺍﺭﺯﻳﺎﺑﻲ ﺷﺮﺍﻳﻂ ﺍﻧﺘﻬﺎﻳﻲ ﻭ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ‪K‬‬ ‫‪ .2‬ﻓﺮﺽ ﻣﻘﺪﺍﺭﻱ ﺑﺮﺍﻱ ﺗﻨﺶ ﻃﺮﺍﺣﻲ ‪IcFcr‬‬ ‫• ﺑﺎ ﻓﺮﺽ ‪ KL/r‬ﺑﻴﻦ ‪ 40‬ﺗﺎ ‪ 60‬ﺑﺮﺍﻱ ﺳﺘﻮﻥ ﻫﺎﻱ ﻣﺘﻮﺳﻄﻲ ﺑﻄﻮﻝ ‪ 3‬ﺗﺎ ‪ 4.5‬ﻣﺘﺮ‬ ‫• ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﻣﻬﺎﺭﺑﻨﺪﻱ ﺑﺎ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺳﺒﻚ‪ ،‬ﺿﺮﺍﻳﺐ ﻻﻏﺮﻱ ﺑﺰﺭﮔﻲ ﺩﺭ ﺣﺪ ‪ 100‬ﻳﺎ ﺑﻴﺸﺘﺮ ﻣﻤﻜﻦ‬ ‫ﺍﺳﺖ ﺗﺨﻤﻴﻦ ﺯﺩﻩ ﺷﻮﺩ‪.‬‬ ‫‪.3‬‬ ‫‪.4‬‬ ‫‪.5‬‬ ‫‪.6‬‬ ‫‪.7‬‬ ‫‪45‬‬ ‫ﺗﻌﻴﻴﻦ ‪Ag ≥ Pu / IcFcr‬‬ ‫ﻣﻘﻄﻌﻲ ﺍﻧﺘﺨﺎﺏ ﺷﻮﺩ ﻛﻪ ﻧﻴﺎﺯ ﺳﻄﺢ ﻣﻘﻄﻊ ﺭﺍ ﺍﺭﺿﺎ ﻧﻤﺎﻳﺪ‪،‬‬ ‫ﻣﺤﺎﺳﺒﻪ ﻣﻘﺪﺍﺭ ‪ IcFcr Ag‬ﺑﺮﺍﻱ ﻣﻘﻄﻊ ﺍﻧﺘﺨﺎﺑﻲ‪،‬‬ ‫ﺗﺠﺪﻳﺪﻧﻈﺮ ﺩﺭ ﻣﻘﻄﻊ ﺍﻧﺘﺨﺎﺑﻲ ﺩﺭ ﺻﻮﺭﺕ ﻟﺰﻭﻡ )ﺗﻜﺮﺍﺭ ﺭﻭﻧﺪ ﻓﻮﻕ ﺑﺎ ﻣﻘﺪﺍﺭ ‪ Fcr‬ﻣﻘﻄﻊ ﺍﻧﺘﺨﺎﺑﻲ‬ ‫ﻣﺰﺑﻮﺭ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻘﺪﺍﺭﻱ ﺑﺮﺍﻱ ﮔﺎﻡ ‪،( 2‬‬ ‫ﻛﻨﺘﺮﻝ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‪.‬‬ Example: Using Fy=2333kgf/cm2, select the lightest IPB available for the service column loads PD=45ton and PL=70ton. KL=3 m. Solution: Pu = 1.2(45) + 1.6(70) = 166 ton Assume KL/r = 50 → IcFcr = 1855 kgf/cm2 Required A = 166000 / 1855 =89.49 cm2 Try IPB24 (A=106 cm2 , ry=6.08 cm) KL/r = 300 / 6.08 = 49.34 → IcFcr = 1861 kgf/cm2 IcPn = IcFcrAg = 1.861 (106) = 197.3 > 166 ton OK. Urmia University - M.Sheidaii 46 Example: Select the lightest satisfactory IPB for the following conditions: Fy=2333kgf/cm2, Pu=200 ton , KxLx=6.0 m , KyLy=3.0 m Solution: Assume KL/r = 50 → IcFcr = 1855 kgf/cm2 Required A = 200000 / 1855 =107.82 cm2 Try IPB26 (A=118.4 cm2 , rx=11.2 cm, ry=6.58 cm) (KL/r)x = 600 / 11.2 = 53.6 ← (KL/r)y = 300 / 6.58 = 45.6 IcFcr = 1822 kgf/cm2 IcPn = IcFcrAg = 1.822 (1118.4) = 215.7 > 200 ton OK. Urmia University - M.Sheidaii 47 ‫ﻣﻼﺣﻈﺎﺕ ﻣﺮﺑﻮﻁ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻭ ﭘﻴﭽﺸﻲ‪-‬ﺧﻤﺸﻲ‬ ‫‪Remarks on Torsional and Flexural-torsional Buckling‬‬ ‫ ﺳﺘﻮﻧﻬﺎ ﺑﻪ ﻃﺮﻕ ﻣﺨﺘﻠﻔﻲ ﻣﻲ ﺗﻮﺍﻧﻨﺪ ﻛﻤﺎﻧﻪ ﻛﻨﻨﺪ ‪:‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ‬ ‫ﻛﻤﺎﻧﺶ ﺧﻤﺸﻲ‬ ‫ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ‬ ‫ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ‪ -‬ﺧﻤﺸﻲ‬ ‫ ﺗﺎ ﺑﻪ ﺍﻳﻨﺠﺎ ﻓﻘﻂ ﻛﻤﺎﻧﺶ ﺧﻤﺸﻲ ﻣﻮﺭﺩ ﺗﻮﺟﻪ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﺍﺳﺖ‪.‬‬ ‫ ﺑﺎ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩﻥ ﻣﻘﺎﻃﻌﻲ ﻛﻪ ﻓﺎﻗﺪ ﺍﺟﺰﺍﻱ ﻻﻏﺮ ﻫﺴﺘﻨﺪ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﺣﺎﻟﺖ ﺣﺪﻱ ﻛﻤﺎﻧﺶ‬ ‫ﻣﻮﺿﻌﻲ ﺩﺭ ﺍﺭﺯﻳﺎﺑﻲ ﻣﻘﺎﻭﻣﺖ ﺳﺘﻮﻥ ﺻﺮﻓﻨﻈﺮ ﻛﺮﺩ‪.‬‬ ‫‪48‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻣﻼﺣﻈﺎﺕ ﻣﺮﺑﻮﻁ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻭ ﭘﻴﭽﺸﻲ‪-‬ﺧﻤﺸﻲ‬ ‫‪Remarks on Torsional and Flexural-torsional Buckling‬‬ ‫ ﭼﻮﻥ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻣﻲ ﺗﻮﺍﻧﺪ ﺧﻴﻠﻲ ﭘﻴﭽﻴﺪﻩ ﺑﺎﺷﺪ ‪ ،‬ﺍﺯ ﺍﻳﻨﺮﻭ ﻣﻄﻠﻮﺑﺘﺮ ﺍﺳﺖ ﻛﻪ ﺍﺯ‬ ‫ﻭﻗﻮﻉ ﺁﻥ ﺟﻠﻮﮔﻴﺮﻱ ﺷﻮﺩ ‪ .‬ﺍﻳﻦ ﻛﺎﺭ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﻪ ﻃﺮﻕ ﺯﻳﺮ ﺍﻧﺠﺎﻡ ﭘﺬﻳﺮﺩ ‪:‬‬ ‫ ﭼﻴﻨﺶ )‪ ( arrangement‬ﺩﻗﻴﻖ ﺍﻋﻀﺎ‬ ‫ ﻧﺼﺐ ﻣﻬﺎﺭ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻭ ﭘﻴﭽﺸﻲ‬ ‫ ﺗﺎﻣﻴﻦ ﺗﻜﻴﻪ ﮔﺎﻩ ﻫﺎﻱ ﺍﻧﺘﻬﺎﻳﻲ ﻭ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻣﻴﺎﻧﻲ ﺑﻪ ﻣﻘﺪﺍﺭ ﻛﺎﻓﻲ‬ ‫ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻘﺎﻃﻊ ﻗﻮﻃﻲ ﺷﻜﻞ ‪ ،‬ﻳﺎ ﺑﻪ ﺻﻮﺭﺕ ﻗﻮﻃﻲ ﺷﻜﻞ ﺩﺭ ﺁﻭﺭﺩﻥ ﻣﻘﺎﻃﻊ ‪I‬‬ ‫ﺷﻜﻞ ﺑﺎ ﺍﻓﺰﻭﺩﻥ ﻭﺭﻗﻬﺎﻱ ﺟﺎﻧﺒﻲ ﺟﻮﺵ ﺷﺪﻩ ) ‪( ৗIৗ‬‬ ‫ ﻛﻮﺗﺎﻩ ﮔﺮﻓﺘﻦ ﻃﻮﻝ ﺍﻋﻀﺎﻳﻲ ﻛﻪ ﺗﺤﺖ ﭘﻴﭽﺶ ﻫﺴﺘﻨﺪ‪.‬‬ ‫‪49‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻣﻼﺣﻈﺎﺕ ﻣﺮﺑﻮﻁ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻭ ﭘﻴﭽﺸﻲ‪-‬ﺧﻤﺸﻲ‬ ‫‪Remarks on Torsional and Flexural-torsional Buckling‬‬ ‫ﺍﻣﺎ ﺑﻪ ﻫﺮ ﺣﺎﻝ‪:‬‬ ‫ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ – ﺧﻤﺸﻲ ﻗﻄﻌﺎ ﺍﻣﻜﺎﻥ ﻭﻗﻮﻉ ﺩﺍﺷﺘﻪ ﻭ ﻣﻤﻜﻦ ﺍﺳﺖ ﻛﻨﺘﺮﻝ ﻛﻨﻨﺪﻩ‬ ‫ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻃﺮﺍﺣﻲ ﺩﺭ ﺍﻋﻀﺎﻱ ﺯﻳﺮ ﺑﺎﺷﺪ‪:‬‬ ‫ ﻣﻘﺎﻃﻊ ﺩﺍﺭﺍﻱ ﻳﻚ ﻣﺤﻮﺭ ﺗﻘﺎﺭﻥ ﻧﻈﻴﺮ ﺳﭙﺮﻱ ﻫﺎ ) ﻣﻘﺎﻃﻊ ‪ T‬ﺷﻜﻞ ( ﻳﺎ ﺯﻭﺝ‬ ‫ﻧﺒﺸﻲ ﻫﺎ‬ ‫ ﺗﻚ ﻧﺒﺸﻲ ﺑﺎ ﻃﻮﻝ ﺑﺎﻟﻬﺎﻱ ﺑﺮﺍﺑﺮ‬ ‫ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ – ﺧﻤﺸﻲ ﺍﻏﻠﺐ ﻛﻨﺘﺮﻝ ﻛﻨﻨﺪﻩ ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻃﺮﺍﺣﻲ ﻧﺒﺸﻲ ﻫﺎﻱ‬ ‫ﺗﻜﻲ ﺑﺎ ﻃﻮﻝ ﺑﺎﻝ ﻫﺎﻱ ﻧﺎﻣﺴﺎﻭﻱ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ‪.‬‬ ‫ ﻣﻘﺎﻃﻊ ﺑﺎﺯ ﻧﻈﻴﺮ ﻣﻘﺎﻃﻊ ‪ I‬ﺷﻜﻞ ﻭ ﻧﺎﻭﺩﺍﻧﻲ ﻫﺎ ﻣﻘﺎﻭﻣﺖ ﭘﻴﭽﺸﻲ ﻛﻤﻲ ﺩﺍﺭﻧﺪ‪.‬‬ ‫‪50‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻣﻼﺣﻈﺎﺕ ﻣﺮﺑﻮﻁ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻭ ﭘﻴﭽﺸﻲ‪-‬ﺧﻤﺸﻲ‬ ‫‪Remarks on Torsional and Flexural-torsional Buckling‬‬ ‫ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻳﺎ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ – ﺧﻤﺸﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪ‬ ‫ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﺳﺘﻮﻧﻬﺎﻱ ﻧﺎﻣﺘﻘﺎﺭﻥ ﻳﺎ ﺳﺘﻮﻧﻬﺎﻱ ﻣﺘﻘﺎﺭﻥ ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺍﺯ ﻭﺭﻗﻬﺎﻱ ﻧﺎﺯﻙ‬ ‫ﺷﻮﺩ‪.‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﺑﺮﺍﻱ ﻛﻤﺎﻧﺶ ﭘﻴﭽﺸﻲ ﻭ ﭘﻴﭽﺸﻲ‪-‬ﺧﻤﺸﻲ ﺍﻋﻀﺎﻱ ﺑﺪﻭﻥ ﺍﺟﺰﺍﻱ ﻻﻏﺮ‬ ‫ﺑﺎﻳﺪ ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪) AISC05 E4 p.16.1-34‬ﻳﺎ ﻋﻴﻨﺎ ﺑﻄﻮﺭ ﻣﺸﺎﺑﻪ ﺑﺮﺍﺳﺎﺱ ﺑﻨﺪ‬ ‫‪ 2-2-4-2-10‬ﺻﻔﺤﻪ ‪ 178‬ﻣﺒﺤﺚ ﺩﻫﻢ( ﺗﻌﻴﻴﻦ ﺷﻮﺩ ) ﺍﺳﻼﻳﺪ ﺑﻌﺪﻱ(‪.‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﺑﺎﺍﺟﺰﺍﻱ ﻻﻏﺮ ﺑﺎﻳﺪ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻻﺕ ﺣﺪﻱ ﻛﻤﺎﻧﺶ ﺧﻤﺸﻲ‪،‬‬ ‫ﭘﻴﭽﺸﻲ ﻭ ﭘﻴﭽﺸﻲ‪ -‬ﺧﻤﺸﻲ ﺑﺮﻃﺒﻖ ﺿﻮﺍﺑﻂ ‪ AISC05 E7 p.16.1-39‬ﺗﻌﻴﻴﻦ ﺷﻮﺩ‪.‬‬ ‫‪51‬‬ ‫‪Urmia University - M.Sheidaii‬‬ Urmia University - M.Sheidaii 52 Urmia University - M.Sheidaii 53 , based on the amount of end restraints against twisting 54 Urmia University - M.Sheidaii 58 Check torsional and flexural-torsional buckling about y-y axis: Urmia University - M.Sheidaii 59 Urmia University - M.Sheidaii 60 ‫ﺍﺟﺘﻨﺎﺏ ﺍﺯ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‬ ‫‪Avoiding Local Buckling‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪61‬‬ ‫ﺍﻏﻠﺐ ﻣﻘﺎﻃﻊ ﻓﻮﻻﺩﻱ ﺍﺯ ﺍﺟﺰﺍﻱ ﻧﺎﺯﻙ ﺟﺪﺍﮔﺎﻧﻪﺍﻱ ﺳﺎﺧﺘﻪ ﺷﺪﻩﺍﻧﺪ‪.‬‬ ‫ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﺎ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺣﻮﻝ ﻣﺤﻮﺭ ﺿﻌﻴﻒ ﺗﺮ ﻋﻀﻮ‬ ‫ﻧﺎﺯﻙ ﺩﭼﺎﺭ ﺧﺮﺍﺑﻲ ﺷﻮﻧﺪ‪.‬‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‪ ،‬ﺍﺯ ﺭﺳﻴﺪﻥ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺑﻪ ﺗﺴﻠﻴﻢ ﻓﺸﺎﺭﻱ ﻳﺎ ﻇﺮﻓﻴﺖ ﺧﻤﻴﺮﻱ‬ ‫ﻛﺎﻣﻞ ﺟﻠﻮﮔﻴﺮﻱ ﺧﻮﺍﻫﺪ ﻛﺮﺩ ‪.‬‬ ‫ﺑﺮﺍﻱ ﺍﺟﺘﻨﺎﺏ ﺍﺯ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‪ ،‬ﻧﺴﺒﺖ ﻫﺎﻱ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ‬ ‫ﻋﺮﺿﻲ ﻣﻄﺎﺑﻖ ﺿﻮﺍﺑﻂ ﺑﻨﺪ ‪) AISC05 B4 p. 16.1-14‬ﻳﺎ ﻣﺘﻨﺎﻇﺮ ﺑﺎ ﺁﻥ ﺑﻨﺪ‬ ‫‪ 5-2-2-10‬ﺻﻔﺤﻪ ‪ 154‬ﻣﺒﺤﺚ ﺩﻫﻢ( ﻣﺤﺪﻭﺩ ﻣﻲﮔﺮﺩﺩ‪.‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺍﺟﺰﺍﻱ ﺗﻘﻮﻳﺖ ﺷﺪﻩ ﻭ ﺗﻘﻮﻳﺖ ﻧﺸﺪﻩ‬ ‫‪Stiffened and Unstiffened Elements‬‬ ‫‪Unstiffened Elements‬‬ ‫ ﺍﺟﺰﺍﻱ ﺗﻘﻮﻳﺖ ﻧﺸﺪﻩ)ﺑﺎ ﻳﻚ ﻟﺒﻪ ﻣﺘﻜﻲ(‬ ‫ﻓﻘﻂ ﺩﺭ ﻃﻮﻝ ﻳﻚ ﻟﺒﻪ ﻣﻮﺍﺯﻱ ﺑﺎ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭﻱ ﻓﺸﺎﺭﻱ ﺗﻜﻴﻪ ﺩﺍﺭﻧﺪ ‪.‬‬ ‫ ﺍﺟﺰﺍﻱ ﺗﻘﻮﻳﺖ ﺷﺪﻩ)ﺑﺎ ﺩﻭ ﻟﺒﻪ ﻣﺘﻜﻲ(‬ ‫‪Stiffened Elements‬‬ ‫ﺩﺭ ﻃﻮﻝ ﻫﺮﺩﻭ ﻟﺒﻪ ﻣﻮﺍﺯﻱ ﺑﺎ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭﻱ ﻓﺸﺎﺭﻱ ﺗﻜﻴﻪ ﺩﺍﺭﻧﺪ ‪.‬‬ ‫‪62‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﺍﺟﺰﺍﻱ ﺗﻘﻮﻳﺖ ﺷﺪﻩ ﻭ ﺗﻘﻮﻳﺖ ﻧﺸﺪﻩ‬ ‫‪Stiffened and Unstiffened Elements‬‬ ‫ﻋﺮﺽ ﺍﺟﺰﺍ ﺑﺎﻳﺪ ﺑﻪ ﺻﻮﺭﺕ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﺷﻜﻞ ﻫﺎﻱ ﺯﻳﺮ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﻮﺩ ‪:‬‬ ‫‪63‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻃﺒﻘﻪ ﺑﻨﺪﻱ ﻣﻘﺎﻃﻊ ﻋﺮﺿﻲ ﺑﺮﺍﻱ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‬ ‫‪Classifications Cross-section for Local Buckling‬‬ ‫ ﻣﻘﺎﻃﻊ ﻓﺸﺮﺩﻩ )‪ :(Compact‬ﺗﺴﻠﻴﻢ ﻛﻞ ﻣﻘﻄﻊ ﻗﺒﻞ ﺍﺯ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ‪.‬‬ ‫ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﻣﻘﻄﻌﻲ ﻓﺸﺮﺩﻩ ﻣﺤﺴﻮﺏ ﺷﻮﺩ ﺑﺎﻟﻬﺎﻱ ﺁﻥ ﺑﺎﻳﺪ ﺑﻄﻮﺭ ﻣﻤﺘﺪ ﺑﻪ ﺟﺎﻥ ﻳﺎ ﺟﺎﻧﻬﺎ‬ ‫ﻣﺘﺼﻞ ﺷﺪﻩ ﻭ ﻧﺴﺒﺖ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﺍﺟﺰﺍﻱ ﻓﺸﺎﺭﻱ ﺁﻥ ﻧﺒﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ﻧﺴﺒﺖ ﺣﺪﻱ‬ ‫ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ‪ λp‬ﺑﺎﺷﺪ ‪ . b/t d λp‬ﺑﺮﺍﻱ ﻣﻘﺎﺩﻳﺮ ﺣﺪﻱ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ‪ λp‬ﻭ ‪λr‬‬ ‫ﺑﻪ ‪) AISC05 Table B4.1 p.16.1-16‬ﻳﺎ ﻣﺘﻨﺎﻇﺮ ﺑﺎ ﺁﻥ ﺟﺪﻭﻝ ‪ 1-2-2-10‬ﺹ‪155‬‬ ‫ﻣﺒﺤﺚ ﺩﻫﻢ( ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ‪.‬‬ ‫ ﻣﻘﺎﻃﻊ ﻏﻴﺮﻓﺸﺮﺩﻩ )‪ :(Non-Compact‬ﺗﺴﻠﻴﻢ ﺑﺨﺸﻲ ﺍﺯ ﻣﻘﻄﻊ ﻗﺒﻞ ﺍﺯ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‪.‬‬ ‫ﺍﮔﺮﻧﺴﺒﺖ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﻳﻚ ﻳﺎ ﭼﻨﺪ ﺟﺰء ﻓﺸﺎﺭﻱ ﻣﻘﻄﻊ ﺍﺯ‪ λp‬ﺑﻴﺸﺘﺮ ﺑﻮﺩﻩ ﺍﻣﺎ ﺩﺭ‬ ‫ﻋﻴﻦ ﺣﺎﻝ ﻛﻤﺘﺮ ﺍﺯ ‪ λr‬ﻗﻴﺪ ﺷﺪﻩ ﺩﺭ ﺟﺪﻭﻝ ‪ B4.1‬ﺑﺎﺷﺪ‪ ،‬ﻣﻘﻄﻊ ﻏﻴﺮﻓﺸﺮﺩﻩ ﺍﺳﺖ‬ ‫) ‪.(λp < b/t d λr‬‬ ‫ ﻣﻘﺎﻃﻊ ﻻﻏﺮ) ‪ :(Slender‬ﺑﺪﻭﻥ ﻫﺮﮔﻮﻧﻪ ﺗﺴﻠﻴﻢ ﺷﺪﮔﻲ ﺩﺭ ﻣﻘﻄﻊ ﻗﺒﻞ ﺍﺯ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‪.‬‬ ‫ﺍﮔﺮ ﻧﺴﺒﺖ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﺣﺘﻲ ﺗﻨﻬﺎ ﻳﻚ ﺟﺰء ﻣﻘﻄﻊ ﺍﺯ ‪ λr‬ﺗﺠﺎﻭﺯ ﻧﻤﺎﻳﺪ‪ ،‬ﻣﻘﻄﻊ‬ ‫ﻣﺰﺑﻮﺭ ﻣﻘﻄﻌﻲ ﺑﺎ ﺍﺟﺰﺍﻱ ﻻﻏﺮ ﻣﺤﺴﻮﺏ ﻣﻲ ﺷﻮﺩ ) ‪(λr < b/t‬‬ ‫‪64‬‬ ‫‪Urmia University - M.Sheidaii‬‬ Urmia University - M.Sheidaii 66 Urmia University - M.Sheidaii 67 Urmia University - M.Sheidaii 68 Urmia University - M.Sheidaii 69 Urmia University - M.Sheidaii 70 ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‬ ‫‪Local Buckling‬‬ ‫ ﺩﺭ ﺍﻏﻠﺐ ﻣﻮﺍﺭﺩ ﻋﻤﻠﻲ‪ ،‬ﻧﺴﺒﺖﻫﺎﻱ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﺩﺭ ﺣﺪﻱ ﻫﺴﺘﻨﺪ‬ ‫ﻛﻪ ﻣﻘﻄﻊ ﺑﺎ ﺍﺟﺰﺍﻱ ﻻﻏﺮ ﻧﺒﻮﺩﻩ ﻭ ﺣﺎﻟﺖ ﺣﺪﻱ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺗﻌﻴﻴﻦﻛﻨﻨﺪﻩ ﻧﻴﺴﺖ‪.‬‬ ‫ﻟﺬﺍ ﻏﺎﻟﺒﺎ ﻧﻴﺎﺯﻱ ﺑﻪ ﺭﻭﻧﺪ ﻣﺤﺎﺳﺒﺎﺗﻲ ﺫﻳﻞ ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ ﺍﮔﺮ ﺩﺭ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ‪ ،‬ﺟﺰﺋﻲ ﻻﻏﺮ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ‪ ،‬ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻓﺸﺎﺭﻱ‬ ‫ﺑﺎﻳﺪ ﻛﺎﻫﺶ ﻳﺎﺑﺪ)ﺑﻪ ﻛﻤﻚ ﺿﺮﻳﺐ ﻛﺎﻫﺶ ‪ (Q‬ﺿﻮﺍﺑﻂ ﻣﺮﺑﻮﻃﻪ ﺩﺭ ‪AISC05 E7‬‬ ‫‪ p.16.1-39‬ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﺳﺖ)ﺍﺳﻼﻳﺪ ﺑﻌﺪﻱ(‪.‬‬ ‫‪71‬‬ ‫‪Urmia University - M.Sheidaii‬‬ 72 Urmia University - M.Sheidaii 73 74 Urmia University - M.Sheidaii 75 76 ‫ﺟﺪﺍﻭﻝ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬ ‫)‪Design Tables ( Manual of Steel Construction‬‬ ‫ ﺟﺪﺍﻭﻝ ﺗﻨﺶ ﺑﺤﺮﺍﻧﻲ ﻣﻮﺟﻮﺩ ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬ ‫)‪(Available critical stress for compression members-table 4-22‬‬ ‫ﺑﺎﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻳﻦ ﻧﻮﻉ ﺟﺪﺍﻭﻝ ﻣﻘﺪﺍﺭ ‪) IcFcr‬ﺑﺮﺍﻱ ﺭﻭﺵ‪ (LRFD‬ﻭ ‪) Fcr/:c‬ﺑﺮﺍﻱ‬ ‫ﺭﻭﺵ‪ (ASD‬ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ ‪ KL/r‬ﺑﻪ ﺍﺯﺍﻱ ﻣﻘﺎﺩﻳﺮ ﻣﺨﺘﻠﻒ ‪ Fy‬ﻗﺎﺑﻞ ﺗﻌﻴﻴﻦ ﺍﺳﺖ ‪.‬‬ ‫ ﺟﺪﺍﻭﻝ ﻣﻘﺎﻭﻣﺖ ﻣﻮﺟﻮﺩ ﺩﺭ ﻓﺸﺎﺭ ﻣﺤﻮﺭﻱ )ﺟﺪﺍﻭﻝ ﺑﺎﺭ ﺳﺘﻮﻥ(‬ ‫)‪(Available strength in axial compression (column load tables)-Table 4-1‬‬ ‫ﺑﺎﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻳﻦ ﻧﻮﻉ ﺟﺪﺍﻭﻝ ﻣﻘﺪﺍﺭ ‪) IcPn‬ﺑﺮﺍﻱ ﺭﻭﺵ‪ (LRFD‬ﻭ ‪) Pn/:c‬ﺑﺮﺍﻱ‬ ‫ﺭﻭﺵ‪ (ASD‬ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ‪ KL‬ﺑﻪ ﺍﺯﺍﻱ ﻣﻘﺎﺩﻳﺮ ﻣﺨﺘﻠﻒ ‪ Fy‬ﻗﺎﺑﻞ ﺗﻌﻴﻴﻦ ﺍﺳﺖ ‪.‬‬ ‫‪77‬‬ ‫ﺟﺪﺍﻭﻝ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‬ Design Tables ( Manual of Steel Construction) 78 ‫ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺩﺭ ﻗﺎﺑﻬﺎ‬ ‫‪Effective Length of Columns in Frames‬‬ ‫ ﺗﺎ ﺑﻪ ﺍﻳﻨﺠﺎ‪ ،‬ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺸﻲ ﺳﺘﻮﻥﻫﺎﻱ ﻣﻨﻔﺮﺩ )ﺗﻜﻲ( ﻣﻮﺭﺩ ﺑﺮﺭﺳﻲ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ‬ ‫ﺍﺳﺖ‪ .‬ﺍﻳﻦ ﺳﺘﻮﻧﻬﺎ ﺍﺯ ﺷﺮﺍﻳﻂ ﺍﻧﺘﻬﺎﻳﻲ ﻣﺨﺘﻠﻔﻲ ﺑﺮﺧﻮﺭﺩﺍﺭ ﺑﻮﺩﻧﺪ‪ ،‬ﺍﻣﺎ ﺑﻪ ﺍﻋﻀﺎﻱ ﺩﻳﮕﺮ‬ ‫ﺗﻮﺳﻂ ﺍﺗﺼﺎﻻﺕ ﺧﻤﺸﻲ )ﮔﻴﺮﺩﺍﺭ( ﻣﺘﺼﻞ ﻧﺒﻮﺩﻧﺪ‪.‬‬ ‫ ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﻮﺛﺮ ‪ K‬ﺑﺮﺍﻱ ﻛﻤﺎﻧﺶ ﻳﻚ ﺳﺘﻮﻥ ﻣﻨﻔﺮﺩ ﺭﺍ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ‪AISC Table‬‬ ‫‪ C-C2.2 p. 16.1-240‬ﺗﻌﻴﻴﻦ ﻛﺮﺩ‪.‬‬ ‫‬ ‫‬ ‫‪79‬‬ ‫ﺍﻳﻦ ﺿﺮﺍﻳﺐ ﺑﺮﺍﻱ ﺳﺘﻮﻥﻫﺎﻱ ﺑﺎ ﻗﻴﺪﻫﺎﻱ ﺍﻧﺘﻬﺎﻳﻲ ﺍﻳﺪﻩﺁﻝ ﻣﺸﺨﺺ ﺑﺴﻂ ﺩﺍﺩﻩ‬ ‫ﺷﺪﻩﺍﻧﺪ‪.‬‬ ‫ﺍﻳﻦ ﺿﺮﺍﻳﺐ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲﻫﺎﻱ ﺍﻭﻟﻴﻪ ﻭ ﺑﺮﺍﻱ ﺷﺮﺍﻳﻄﻲ ﻛﻪ ﺍﺯ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ‬ ‫ﺟﻠﻮﮔﻴﺮﻱ ﺷﺪﻩ ﺑﺎﺷﺪ‪ ،‬ﻣﻌﻤﻮﻻ ﺑﺴﻴﺎﺭ ﺭﺿﺎﻳﺘﺒﺨﺶ ﻣﻲﺑﺎﺷﻨﺪ ‪.‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻗﺎﺑﻬﺎﻱ ﺑﺪﻭﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ‬ Sidesway Inhibited Frames Urmia University - M.Sheidaii 80 ‫ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺩﺭ ﻗﺎﺑﻬﺎ‬ ‫‪Effective Length of Columns in Frames‬‬ ‫ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﻮﺛﺮ ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ ﻗﺎﺑﻬﺎ ﺑﺎﻳﺪ ﺑﻪ ﺭﻭﺍﻝ ﺯﻳﺮ ﺗﻌﻴﻴﻦ ﺷﻮﺩ‬ ‫‪81‬‬ ‫‬ ‫ﺍﮔﺮ ﺳﺘﻮﻥ ﻋﻀﻮﻱ ﺍﺯ ﻳﻚ ﻗﺎﺏ ﻣﻬﺎﺭ ﺷﺪﻩ ﺑﺎﺷﺪ ﻃﻮﻝ ﻣﻮﺛﺮ ﺁﻥ‪0 < K ≤ 1 :‬‬ ‫ﺑﺪﻳﻬﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺣﺎﻟﺖ ﻗﺎﺑﻬﺎﻱ ﻣﻬﺎﺭﺷﺪﻩ ‪ ،‬ﻳﻚ ﺗﻘﺮﻳﺐ ﺩﺳﺖ ﺑﺎﻻ ﺍﺯ ‪ K‬ﻋﻤﻮﻣﺎ‬ ‫ﻛﺎﻓﻲ ﺑﻮﺩﻩ‪ ،‬ﺑﺪﻭﻥ ﺍﻳﻨﻜﻪ ﻧﻴﺎﺯﻱ ﺑﻪ ﻣﺤﺎﺳﺒﻪ ﻧﻤﻮﺩﻥ ﺟﺰﺋﻴﺎﺕ ﺳﺨﺘﻲ ﻗﺎﺏ ﺑﺎﺷﺪ ‪.‬‬ ‫‬ ‫ﺍﮔﺮﺳﺘﻮﻥ ﻋﻀﻮﻱ ﺍﺯﻳﻚ ﻗﺎﺏ ﻣﻬﺎﺭﻧﺸﺪﻩ ﺑﺎﺷﺪ ﺩﺭﺍﻳﻦ ﺻﻮﺭﺕ‪1 ≤ K ≤ ∞ :‬‬ ‫ﺑﻌﻀﻲ ﺭﻭﺷﻬﺎﻱ ﻣﺮﺳﻮﻡ ﺑﺮﺍﻱ ﺗﺨﻤﻴﻦ ‪ K‬ﺑﺎﻳﺪ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ‪ ،‬ﺯﻳﺮﺍ‬ ‫ﻣﺤﺪﻭﺩﻩ ﻣﻘﺎﺩﻳﺮ ﻣﻤﻜﻦ ﺗﺎ ﺑﻲ ﻧﻬﺎﻳﺖ ﺍﺳﺖ ‪.‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺩﺭ ﻗﺎﺑﻬﺎ‬ ‫‪Effective Length of Columns in Frames‬‬ ‫ﺭﻭﺷﻬﺎﻳﻲ ﻛﻪ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ﻃﻮﻝ ﻣﻮﺛﺮ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ‪:‬‬ ‫‪82‬‬ ‫‬ ‫ﺗﺤﻠﻴﻞ ﻫﺎﻱ ﻛﻤﺎﻧﺶ ﻛﺎﻣﻞ ﺑﻪ ﺍﺯﺍﻱ ﻭﺿﻌﻴﺖ ﻫﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺧﺎﺹ ﻭ ﻣﻌﻴﻦ‬ ‫)ﺭﻭﺷﻲ ﺧﻴﻠﻲ ﻃﻮﻻﻧﻲ ﻭ ﺷﺎﻳﺪ ﺧﻴﻠﻲ ﻣﺸﻜﻞ(‪.‬‬ ‫‬ ‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ )ﻧﻤﻮﮔﺮﺍﻓﻬﺎ(‪ .‬ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ ﺑﻪ‪:‬‬ ‫‪AISC05 p. 16.1-240‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ‬ ‫‪Alignment Charts‬‬ ‫ﺭﻭﺵ ﺑﺴﻴﺎﺭ ﻣﺘﺪﺍﻭﻝ ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ‪ K‬ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺍﺳﺖ ﻛﻪ ﺑﻪ‬ ‫ﺁﻧﻬﺎ ﻧﻤﻮﮔﺮﺍﻑ ﻧﻴﺰ ﮔﻔﺘﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﺍﻳﻦ ﻧﻤﻮﺩﺍﺭﻫﺎ ﺩﺭﺷﻜﻞ ﻫﺎﻱ ﺯﻳﺮ ﺍﺯ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC05‬ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﻧﺪ‪:‬‬ ‫ ﺷﻜﻞ ‪ C-C2.3‬ﺑﺮﺍﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺪﻭﻥ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ)‪ ،(sidesway inhibited‬ﻭ‬ ‫ ﺷﻜﻞ ‪ C-C2.4‬ﺑﺮﺍﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺎ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ)‪. (sidesway uninhibited‬‬ ‫‪83‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ‬ ‫‪Alignment Charts‬‬ ‫‪84‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ‬ ‫‪Alignment Charts‬‬ ‫‪85‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ‬ ‫‪Alignment Charts‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺑﺮﺍﺳﺎﺱ ﻓﺮﺿﻴﺎﺕ ﺷﺮﺍﻳﻂ ﺍﻳﺪﻩ ﺁﻟﻲ ﺗﻬﻴﻪ ﺷﺪﻩ ﺍﻧﺪ ﻛﻪ ﺩﺭ ﺳﺎﺯﻩ ﻫﺎﻱ ﻭﺍﻗﻌﻲ‬ ‫ﺑﻪ ﻧﺪﺭﺕ ﻣﺤﻘﻖ ﻣﻲ ﺷﻮﻧﺪ ‪ .‬ﺍﻳﻦ ﻓﺮﺿﻴﺎﺕ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪:‬‬ ‫‪ .1‬ﺭﻓﺘﺎﺭ ﻛﺎﻣﻼ ﺍﺭﺗﺠﺎﻋﻲ ﺍﺳﺖ‪.‬‬ ‫‪ .2‬ﻫﻤﻪ ﺍﻋﻀﺎ ﺳﻄﺢ ﻣﻘﻄﻊ ﺷﺎﻥ ﺛﺎﺑﺖ ﺍﺳﺖ‪.‬‬ ‫‪ .3‬ﻫﻤﻪ ﮔﺮﻩ ﻫﺎ ﺻﻠﺐ ﻫﺴﺘﻨﺪ‪.‬‬ ‫‪ .4‬ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺪﻭﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ‪ ،‬ﺩﻭﺭﺍﻧﻬﺎ ﺩﺭ ﺩﻭ ﺍﻧﺘﻬﺎﻱ ﺗﻴﺮﻫﺎﻱ ﻣﻘﻴﺪﻛﻨﻨﺪﻩ‬ ‫ﺳﺘﻮﻥ‪ ،‬ﻣﻘﺪﺍﺭﺷﺎﻥ ﺑﺮﺍﺑﺮ ﻭ ﺟﻬﺖ ﺷﺎﻥ ﻣﺨﺎﻟﻒ ﻫﻢ ﺑﻮﺩﻩ؛ ﻟﺬﺍ ﺩﺍﺭﺍﻱ ﺧﻤﺶ ﺑﺎ ﺍﻧﺤﻨﺎﻱ ﺳﺎﺩﻩ‬ ‫)ﺗﻜﻲ( ﻫﺴﺘﻨﺪ‪.‬‬ ‫‪ .5‬ﺑﺮﺍﻱ ﺳﺘﻮﻥ ﻫﺎﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺎ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ‪ ،‬ﺩﻭﺭﺍﻧﻬﺎﻱ ﺩﻭ ﺍﻧﺘﻬﺎﻱ ﺗﻴﺮ ﻣﻘﻴﺪﻛﻨﻨﺪﻩ‬ ‫ﺳﺘﻮﻥ‪ ،‬ﻣﻘﺪﺍﺭ ﻭ ﺟﻬﺖ ﺷﺎﻥ ﺑﺮﺍﺑﺮ ﺑﻮﺩﻩ‪ ،‬ﻟﺬﺍ ﺩﺍﺭﺍﻱ ﺧﻤﺶ ﺑﺎ ﺍﻧﺤﻨﺎﻱ ﻣﻀﺎﻋﻒ )ﻣﻌﻜﻮﺱ(‬ ‫ﻫﺴﺘﻨﺪ‪.‬‬ ‫‪ L√P/EI .6‬ﭘﺎﺭﺍﻣﺘﺮ ﺳﺨﺘﻲ ﻫﻤﻪ ﺳﺘﻮﻧﻬﺎ ﺑﺮﺍﺑﺮ ﺍﺳﺖ‪.‬‬ ‫‪ .7‬ﺗﻮﺯﻳﻊ ﮔﻴﺮﺩﺍﺭﻱ ﮔﺮﻫﻲ ﺑﻴﻦ ﺳﺘﻮﻥ ﺑﺎﻻ ﻭ ﭘﺎﻳﻴﻦ ﮔﺮﻩ ﺑﻪ ﻧﺴﺒﺖ ‪ EI/L‬ﺩﻭ ﺳﺘﻮﻥ ﺑﻮﺩﻩ ﺍﺳﺖ‪.‬‬ ‫‪ .8‬ﻫﻤﻪ ﺳﺘﻮﻥ ﻫﺎ ﻫﻤﺰﻣﺎﻥ ﻛﻤﺎﻧﻪ ﻣﻲ ﻛﻨﻨﺪ‪.‬‬ ‫‪ .9‬ﻫﻴﭻ ﻧﻴﺮﻭﻱ ﻓﺸﺎﺭﻱ ﻣﺤﻮﺭﻱ ﻗﺎﺑﻞ ﺗﻮﺟﻬﻲ ﺩﺭ ﺗﻴﺮﻫﺎ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‪.‬‬ ‫‪86‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ‬ ‫‪Alignment Charts‬‬ ‫ ﻧﻤﻮﺩﺍﺭ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺑﺮﺍﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺪﻭﻥ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻛﻪ ﺩﺭ ﺷﻜﻞ ‪C-C2.3‬‬ ‫ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ‪ ،‬ﺑﺮﺍﺳﺎﺱ ﻣﻌﺎﺩﻟﻪ ﺯﻳﺮ ﺍﺳﺖ‪:‬‬ ‫ ﻧﻤﻮﺩﺍﺭ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺑﺮﺍﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺎ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻛﻪ ﺩﺭ ﺷﻜﻞ ‪C-C2.4‬‬ ‫ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ‪ ،‬ﺑﺮﺍﺳﺎﺱ ﻣﻌﺎﺩﻟﻪ ﺯﻳﺮ ﺍﺳﺖ‪:‬‬ ‫ﻛﻪ ﺿﺮﻳﺐ ﺻﻠﺒﻴﺖ ﻧﺴﺒﻲ ‪ G‬ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ ‪:‬‬ ‫‪87‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ‬ ‫‪Alignment Charts‬‬ ‫ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ‪:K‬‬ ‫‪ .1‬ﻧﻤﻮﺩﺍﺭ ﻣﻨﺎﺳﺐ ﺍﻧﺘﺨﺎﺏ ﺷﻮﺩ‬ ‫‪ G .2‬ﺑﺮﺍﻱ ﻫﺮﺩﻭ ﺍﻧﺘﻬﺎﻱ ﺳﺘﻮﻥ ﻣﺤﺎﺳﺒﻪ ﺷﻮﺩ)‪(GA ,GB‬‬ ‫‪ .3‬ﺧﻂ ﻣﺴﺘﻘﻴﻤﻲ ﺑﻴﻦ ‪ GA‬ﻭ‪ GB‬ﺭﺳﻢ ﺷﺪﻩ‪ ،‬ﻣﻘﺪﺍﺭ ‪K‬‬ ‫ﻗﺮﺍﺋﺖ ﺷﻮﺩ ‪.‬‬ ‫‪88‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ – ﺗﺼﺤﻴﺤﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﺩﻭﺭﺍﻧﻬﺎﻱ ﮔﺮﻫﻲ‬ ‫‪Modifications on Alignment Charts – Joint Rotations‬‬ ‫‬ ‫ﺗﺼﺤﻴﺤﺎﺕ ﺑﺮﺍﻱ ﺷﺮﺍﻳﻂ ﺍﻧﺘﻬﺎﻳﻲ ﻣﺨﺘﻠﻒ ﺳﺘﻮﻥ‪:‬‬ ‫‪.1‬‬ ‫‪.2‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ ﻣﻔﺼﻠﻲ‪ :‬ﺍﺯ ﻧﻈﺮ ﺗﺌﻮﺭﻱ ‪ G‬ﺑﻲ ﻧﻬﺎﻳﺖ ﺑﻮﺩﻩ ﺍﻣﺎ ﺑﻜﺎﺭﮔﻴﺮﻱ ﻣﻘﺪﺍﺭ ‪10‬‬ ‫ﭘﻴﺸﻨﻬﺎﺩ ﺷﺪﻩ ﺍﺳﺖ )ﺍﻟﺒﺘﻪ ﻣﮕﺮ ﺩﺭ ﻣﻮﺍﺭﺩﻳﻜﻪ ﻳﻚ ﻣﻔﺼﻞ ﺑﺪﻭﻥ ﺍﺻﻄﻜﺎﻙ ﻭﺍﻗﻌﻲ ﻃﺮﺡ‬ ‫ﺷﺪﻩ ﺑﺎﺷﺪ(‪.‬‬ ‫ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﺻﻠﺐ ﺳﺘﻮﻧﻬﺎ ﺑﻪ ﺷﺎﻟﻮﺩﻩ ﻫﺎ‪ :‬ﺍﺯ ﻧﻈﺮ ﺗﺌﻮﺭﻱ ‪ G‬ﺻﻔﺮ ﺑﻮﺩﻩ ﺍﻣﺎ ﺑﻜﺎﺭﮔﻴﺮﻱ‬ ‫ﻣﻘﺪﺍﺭ ‪ 1.0‬ﭘﻴﺸﻨﻬﺎﺩ ﺷﺪﻩ ﺍﺳﺖ ﭼﻮﻥ ﻋﻤﻼ ﻫﻴﭻ ﺍﺗﺼﺎﻝ ﻛﺎﻣﻼ ﺻﻠﺒﻲ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‪.‬‬ ‫ﺗﺼﺤﻴﺤﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﺷﺮﺍﻳﻂ ﺍﻧﺘﻬﺎﻳﻲ ﻣﺨﺘﻠﻒ ﺗﻴﺮ ‪) :‬ﺿﺮﺍﻳﺐ ﺟﺪﻭﻝ ﺯﻳﺮ ﺩﺭ ‪(EI/L)g‬‬ ‫ﻋﻀﻮ ﺗﻴﺮ ﺿﺮﺏ ﺷﻮﺩ(‬ EXAMPLE Calculate the effective length factor for the W12 x 53 column AB of the frame shown below. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane buckling. Assume that the same column section is used for the stories above and below. Urmia University - M.Sheidaii 90 Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible. • It is an unbraced (sidesway uninhibited) frame. • Lx = Ly = 12 ft. • Ky = 1.0 • Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB. • Need to calculate Kx using alignment charts. Step II - Calculate Kx • W 12 x 53 ( Ixx = 425 in4 , rx = 5.23 in, W 14 x 68 ( Ixx = 723 in4 ) Using GA and GB: Kx = 1.3 ry = 2.48 in ), from Alignment Chart Step III – Design strength of the column • KyLy = 1.0 x 12 = 12 ft =144 in. • Kx Lx = 1.3 x 12 = 15.6 ft = 187.2 in. Therefore, y-axis buckling governs. Therefore ØcPn = 548 kips (KL/r)y=144 / 2.48 =58.06 ← (KL/r)x=187.2 / 5.23 =35.79 91 ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ – ﺗﺼﺤﻴﺤﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Modifications on Alignment Charts –Inelastic Buckling‬‬ ‫‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺑﺮ ﻣﺒﻨﺎﻱ ﻓﺮﺽ ﺧﺮﺍﺑﻲ ﻛﺎﻣﻼ ﺍﺭﺗﺠﺎﻋﻲ ﺳﺘﻮﻥ ﺍﺳﺘﻮﺍﺭ‬ ‫ﻫﺴﺘﻨﺪ‪ ،‬ﻛﻪ ﺍﻳﻦ ﻓﺮﺽ ﺩﺭ ﻋﻤﻞ ﺑﻨﺪﺭﺕ ﻣﺤﻘﻖ ﻣﻲ ﮔﺮﺩﺩ‪.‬‬ ‫‬ ‫ﺧﺮﺍﺑﻲ ﺩﺭﺻﺪ ﺑﺎﻻﻳﻲ ﺍﺯ ﺳﺘﻮﻧﻬﺎ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮ ﺍﺭﺗﺠﺎﻋﻲ ﺍﺳﺖ ‪ .‬ﺩﺭ ﺍﻳﻦ ﮔﻮﻧﻪ ﻣﻮﺍﺭﺩ‪،‬‬ ‫ﻣﻘﺎﺩﻳﺮ ‪ K‬ﻧﻤﻮﺩﺍﺭ ﺧﻴﻠﻲ ﻣﺤﺎﻓﻈﻪ ﻛﺎﺭﺍﻧﻪ ﺑﻮﺩﻩ ﻭ ﺑﺎﻳﺪ ﺗﺼﺤﻴﺢ ﺷﻮﺩ‪.‬‬ ‫‪92‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ – ﺗﺼﺤﻴﺤﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Modifications on Alignment Charts –Inelastic Buckling‬‬ ‫‬ ‫ﻫﻤﺎﻧﻄﻮﺭ ﻛﻪ ﻗﺒﻼ ﺫﻛﺮ ﺷﺪ ‪ ،‬ﺿﺮﻳﺐ ﺻﻠﺒﻴﺖ ﻧﺴﺒﻲ ‪ G‬ﻧﺸﺎﻥ ﺩﻫﻨﺪﻩ ﻧﺴﺒﺖ ﺻﻠﺒﻴﺖ‬ ‫ﺧﻤﺸﻲ ﺳﺘﻮﻧﻬﺎ )‪ (EIc/Lc‬ﺑﻪ ﺗﻴﺮﻫﺎ )‪ (EIg/Lg‬ﺍﺳﺖ ‪.‬‬ ‫‬ ‫ﺍﻣﺎ‪ ،‬ﺍﮔﺮ ﻛﻤﺎﻧﺶ ﺳﺘﻮﻥ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ ﺍﺗﻔﺎﻕ ﺑﻴﺎﻓﺘﺪ‬ ‫) )‪ ، ( KL/r d 4.71√(E/Fy‬ﺻﻠﺒﻴﺖ ﺧﻤﺸﻲ ﺳﺘﻮﻥ ﻛﺎﻫﺶ ﺧﻮﺍﻫﺪ ﻳﺎﻓﺖ ﭼﻮﻥ‬ ‫‪ Ic‬ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﻓﻘﻂ ﻫﺴﺘﻪ ﺍﺭﺗﺠﺎﻋﻲ ﻣﻘﻄﻊ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ‪.‬‬ ‫‪93‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ – ﺗﺼﺤﻴﺤﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Modifications on Alignment Charts –Inelastic Buckling‬‬ ‫‬ ‫ﺗﻴﺮﻫﺎ ﺻﻠﺒﻴﺖ ﺧﻤﺸﻲ ﺑﻴﺸﺘﺮﻱ ﺩﺭ ﻣﻘﺎﻳﺴﻪ ﺑﺎ ﺻﻠﺒﻴﺖ ﻛﺎﻫﺶ ﻳﺎﻓﺘﻪ )‪ (EIc‬ﺳﺘﻮﻧﻬﺎﻱ‬ ‫ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ ﺧﻮﺍﻫﻨﺪ ﺩﺍﺷﺖ‪ .‬ﺩﺭ ﻧﺘﻴﺠﻪ ﺗﻴﺮﻫﺎ ﺧﻮﺍﻫﻨﺪ ﺗﻮﺍﻧﺴﺖ ﺳﺘﻮﻧﻬﺎ ﺭﺍ ﺑﻬﺘﺮ ﻣﻘﻴﺪ‬ ‫ﻛﻨﻨﺪ ﻛﻪ ﺍﻳﻦ ﺍﻣﺮ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺳﺘﻮﻥ ﻣﻄﻠﻮﺑﺘﺮ ﺍﺳﺖ ‪.‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻥ ﺍﻳﻦ ﻣﻮﺿﻮﻉ‪ ،‬ﺿﺮﻳﺐ ﺻﻠﺒﻴﺖ ﻧﺴﺒﻲ ‪ G‬ﺗﺼﺤﻴﺢ ﻣﻲ ﺷﻮﺩ ﺑﺪﻳﻦ‬ ‫ﺗﺮﺗﻴﺐ ﻛﻪ ﺩﺭ ﻋﺒﺎﺭﺕ ﺳﺨﺘﻲ ﺳﺘﻮﻥ ﺑﻪ ﺟﺎﻱ ﻣﺪﻭﻝ ﺍﺭﺗﺠﺎﻋﻲ ‪ E‬ﺍﺯ ﻣﺪﻭﻝ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫ﻣﻤﺎﺳﻲ ‪ Et‬ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ ‪:‬‬ ‫‪94‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ – ﺗﺼﺤﻴﺤﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪Modifications on Alignment Charts –Inelastic Buckling‬‬ ‫‬ ‫ﻣﻌﺎﺩﻟﻪ ﻓﻮﻕ ﺍﻟﺬﻛﺮ ﺭﻭﻧﺪ ﺗﺼﺤﻴﺢ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ‪ G :‬ﺑﻪ ﺭﻭﺍﻝ ﭘﻴﺸﻴﻦ ﻣﺤﺎﺳﺒﻪ‬ ‫ﻣﻲﺷﻮﺩ‪ ،‬ﻭ ﺳﭙﺲ ﺑﻪ ﺿﺮﻳﺐ ﺗﺼﺤﻴﺢ ‪ W = Et/E‬ﻛﻪ ﺿﺮﻳﺐ ﻛﺎﻫﺶ ﺳﺨﺘﻲ )‪(SRF‬‬ ‫ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ﺿﺮﺏ ﻣﻲ ﮔﺮﺩﺩ‪.‬‬ ‫‪Et‬‬ ‫‪E‬‬ ‫‬ ‫ﻧﺴﺒﺖ ‪ Et/E‬ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﻣﻌﺎﺩﻝ ﺑﺎﺷﺪ ﺑﺎ ﻧﺴﺒﺖ ‪:‬‬ ‫‪Pu / A‬‬ ‫‪c iF‬‬ ‫‪t scr‬‬ ‫‪al e‬‬ ‫‬ ‫‪95‬‬ ‫‪Et Fcr inelastic‬‬ ‫|‬ ‫‪E Fcr elastic‬‬ ‫‪W‬‬ ‫ﺟﺪﻭﻝ ﺯﻳﺮ ﺿﺮﻳﺐ ﻛﺎﻫﺶ ﺳﺨﺘﻲ )‪ (W‬ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ‪ Fy‬ﻭ‬ ‫ﺗﻨﺶ ﺗﺴﻠﻴﻢ ‪ Pu/Ag‬ﻣﻲ ﺩﻫﺪ ‪ ،‬ﻛﻪ ‪ Pu‬ﺑﺎﺭ ﻃﺮﺍﺣﻲ ﺿﺮﻳﺐ ﺩﺍﺭ )ﺍﺯ ﺗﺤﻠﻴﻞ(‬ ‫ﻣﻲ ﺑﺎﺷﺪ ‪.‬‬ Urmia University - M.Sheidaii 96 ● KxLx/rx = (1.75)(15×12)/(4.39) = 71.75 ● Øc Fcr = 30.92 ksi ● Øc Pn = Øc Fcr Ag= 30.92 × 17.6 = 544.2 kips 97 ‫ﺳﺘﻮﻥﻫﺎﻱ ﻣﺘﻜﻲ ﺑﻪ ﻳﻜﺪﻳﮕﺮ‬ ‫‪Columns Leaning on Each Other‬‬ ‫ ﺩﺭ ﺟﻬﺖ ﺍﻳﻤﻨﻲ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﺍﮔﺮ ﻫﺮ ﺳﺘﻮﻥ ﺍﺯ ﻗﺎﺏ ﻣﻬﺎﺭ ﻧﺸﺪﻩ‪ ،‬ﻣﺴﺘﻘﻼ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﻗﺎﺑﻬﺎﻱ ﺑﺎ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻃﺮﺍﺣﻲ ﺷﻮﺩ ‪.‬‬ ‫ ﺿﺮﻳﺐ ‪ K‬ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺍﺣﺘﻤﺎﻻ ﺑﻪ ﻣﻴﺰﺍﻥ ﻗﺎﺑﻞ ﺗﻮﺟﻬﻲ ﺑﻴﺶ ﺍﺯ ‪ 1.0‬ﺧﻮﺍﻫﺪ ﺑﻮﺩ ‪.‬‬ ‫ ﺍﻣﺎ‪ ،‬ﻋﻤﻼ ﻳﻚ ﺳﺘﻮﻥ ﻧﻤﻲ ﺗﻮﺍﻧﺪ ﻛﻤﺎﻧﺶ ﺑﺎ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻧﻤﺎﻳﺪ ﻣﮕﺮ ﺁﻧﻜﻪ ﻫﻤﻪ‬ ‫ﺳﺘﻮﻧﻬﺎﻱ ﺁﻥ ﻃﺒﻘﻪ ﻛﻤﺎﻧﺶ ﺑﺎ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻧﻤﺎﻳﻨﺪ ‪.‬‬ ‫‪98‬‬ ‫ﺳﺘﻮﻥﻫﺎﻱ ﻣﺘﻜﻲ ﺑﻪ ﻳﻜﺪﻳﮕﺮ‬ ‫‪Columns Leaning on Each Other‬‬ ‫ ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﻮﻝ ﻣﻮﺛﺮ ﺳﺘﻮﻥ ﺑﺮﺍﻱ ﻗﺎﺑﻬﺎﻱ ﺑﺎ ﺍﻣﻜﺎﻥ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﺑﺮﺍﺳﺎﺱ ﺍﻳﻦ ﻓﺮﺽ‬ ‫ﺗﻬﻴﻪ ﺷﺪﻩ ﺍﻧﺪ ﻛﻪ ﻫﻤﻪ ﺳﺘﻮﻥﻫﺎﻱ ﻃﺒﻘﻪ ﺑﻪ ﺻﻮﺭﺕ ﻫﻤﺰﻣﺎﻥ ﻛﻤﺎﻧﻪ ﻛﻨﻨﺪ‪ ،‬ﻳﻌﻨﻲ‬ ‫ﺳﺘﻮﻥﻫﺎ ﻧﺘﻮﺍﻧﻨﺪ ﻫﻤﺪﻳﮕﺮ ﺭﺍ ﻣﻬﺎﺭ ﻛﺮﺩﻩ ﻳﺎ ﺑﻪ ﻫﻢ ﺗﻜﻴﻪ ﻧﻤﺎﻳﻨﺪ‪.‬‬ ‫ ﺑﻌﻀﻲ ﻣﻮﺍﻗﻊ‪ ،‬ﺳﺘﻮﻥﻫﺎﻱ ﺧﺎﺻﻲ ﺍﺯ ﻳﻚ ﻗﺎﺏ ﺩﺍﺭﺍﻱ ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺸﻲ ﺍﺿﺎﻓﻪﺗﺮﻱ‬ ‫ﻫﺴﺘﻨﺪ‪،‬‬ ‫ ﻫﻨﮕﺎﻣﻴﻜﻪ ﺳﺎﻳﺮ ﺳﺘﻮﻥﻫﺎ ﺑﻪ ﺑﺎﺭ ﻛﻤﺎﻧﺶﺷﺎﻥ ﻣﻲﺭﺳﻨﺪ‪ ،‬ﭼﻨﻴﻦ ﺳﺘﻮﻧﻲ ﺑﻪ ﺑﺎﺭ ﻛﻤﺎﻧﺶ‬ ‫ﻧﺨﻮﺍﻫﺪ ﺭﺳﻴﺪ‪ ،‬ﻟﺬﺍ ﻗﺎﺏ ﻛﻤﺎﻧﻪ ﻧﺨﻮﺍﻫﺪ ﻛﺮﺩ‪.‬‬ ‫‪99‬‬ ‫ﺳﺘﻮﻥﻫﺎﻱ ﻣﺘﻜﻲ ﺑﻪ ﻳﻜﺪﻳﮕﺮ‬ ‫‪Columns Leaning on Each Other‬‬ ‫ ﺳﺘﻮﻥﻫﺎﻱ ﺑﺎ ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺸﻲ ﺍﺿﺎﻓﻲ‪ ،‬ﺳﺘﻮﻥﻫﺎﻱ ﺩﻳﮕﺮ ﺭﺍ ﻣﻬﺎﺭ ﺧﻮﺍﻫﻨﺪ ﻛﺮﺩ‪.‬‬ ‫ ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ‪ ،‬ﺳﺘﻮﻥﻫﺎﻱ ﺩﻳﮕﺮ ﺑﺮ ﺳﺘﻮﻥﻫﺎﻱ ﺑﺎ ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺸﻲ ﺍﺿﺎﻓﻲ ﺗﻜﻴﻪ‬ ‫ﺧﻮﺍﻫﻨﺪ ﻛﺮﺩ‪ .‬ﻟﺬﺍ ﺿﺮﻳﺐ ﻃﻮﻝ ﻣﻮﺛﺮ ‪ K‬ﺍﻳﻦ ﺳﺘﻮﻥﻫﺎ ﺑﻪ ‪ 1.0‬ﺧﻮﺍﻫﺪ ﺭﺳﻴﺪ‪.‬‬ ‫ ﻭﺿﻌﻴﺖﻫﺎﻱ ﺑﺴﻴﺎﺭﻱ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻛﻪ ﺩﺭ ﺁﻧﻬﺎ ﺑﻌﻀﻲ ﺳﺘﻮﻥﻫﺎ ﺩﺍﺭﺍﻱ ﻣﻘﺎﻭﻣﺖ ﻛﻤﺎﻧﺸﻲ‬ ‫ﺍﺿﺎﻓﻲ ﻫﺴﺘﻨﺪ‪ ،‬ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﻣﻲﺗﻮﺍﻥ ﺑﻪ ﻣﻮﺍﺭﺩﻱ ﺍﺷﺎﺭﻩ ﻛﺮﺩ ﻛﻪ ﺩﺭ ﺁﻥ ﺩﺭ ﻃﺮﺍﺣﻲ‬ ‫ﺳﺘﻮﻧﻬﺎﻱ ﻣﺨﺘﻠﻒ ﻳﻚ ﻃﺒﻘﻪ ﺳﺎﺧﺘﻤﺎﻥ‪ ،‬ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭﮔﺬﺍﺭﻱ ﻣﺨﺘﻠﻔﻲ ﻛﻨﺘﺮﻝ ﻛﻨﻨﺪﻩ‬ ‫ﺑﺎﺷﺪ‪.‬‬ ‫ ﻛﻞ ﺑﺎﺭ ﺛﻘﻠﻲ ﻛﻪ ﻳﻚ ﻗﺎﺏ ﻣﻬﺎﺭ ﺷﺪﻩ ﻗﺎﺩﺭ ﺑﻪ ﺗﺤﻤﻞ ﺁﻥ ﺍﺳﺖ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﺟﻤﻊ ﻣﻘﺎﻭﻣﺖ‬ ‫ﺗﻚ ﺗﻚ ﺳﺘﻮﻧﻬﺎ‪.‬‬ ‫‪100‬‬ ‫ﻛﻞ ﺑﺎﺭ ﺛﻘﻠﻲ ﻛﻪ ﺳﺒﺐ ﻛﻤﺎﻧﺶ ﺑﺎ ﺣﺮﻛﺖ ﺟﺎﻧﺒﻲ ﻳﻚ ﻗﺎﺏ ﻣﻲﺷﻮﺩ ﺭﺍ ﻣﻲﺗﻮﺍﻥ ﺑﻴﻦ ﺳﺘﻮﻧﻬﺎ‬ ‫ﺑﻪ ﻧﺴﺒﺖ ﻫﺎﻱ ﻣﺨﺘﻠﻒ ﺗﻘﺴﻴﻢ ﻛﺮﺩ ﺍﻟﺒﺘﻪ ﺑﺸﺮﻃﻲ ﻛﻪ ﺑﺎﺭ ﺣﺪﺍﻛﺜﺮ ﻭﺍﺭﺩ ﺑﺮ ﻫﺮ ﺳﺘﻮﻥ ﺍﺯ ﺑﺎﺭﻱ‬ ‫ﻛﻪ ﺳﺘﻮﻥ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﻣﻬﺎﺭﻧﺸﺪﻩ)ﻳﻌﻨﻲ ﺑﺎ ‪ ( K=1‬ﺑﻪ ﻛﻤﺎﻧﺶ ﻣﻲﺭﺳﺎﻧﺪ ﺑﻴﺸﺘﺮ ﻧﺒﺎﺷﺪ‪.‬‬ ‫ﻭﺻﻠﻪ ﺳﺘﻮﻥﻫﺎ‬ ‫‪Column Splices‬‬ ‫ ﺩﺭ ﺳﺎﺧﺘﻤﺎﻧﻬﺎﻱ ﭼﻨﺪ ﻃﺒﻘﻪ ﻧﻴﺎﺯ ﺑﻪ ﺍﺟﺮﺍﻱ ﻭﺻﻠﻪ ﺳﺘﻮﻥ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ ﻣﺤﻞ ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﺍﺟﺮﺍﻱ ﻭﺻﻠﻪ ﺣﺪﻭﺩ ﻳﻚ ﻣﺘﺮ ﺑﺎﻻﺗﺮ ﺍﺯ ﻛﻒ ﻃﺒﻘﻪ ﺍﺳـﺖ ﺩﺭ ﺍﻳـﻦ‬ ‫ﺣﺎﻟﺖ ﻭﺻﻠﻪﻫﺎ ﺗﺪﺍﺧﻠﻲ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﺗﻴﺮ ﻭ ﺳﺘﻮﻥ‪ ،‬ﻭ ﻣﻬﺎﺭﺑﻨﺪﻫﺎ ﻧﺨﻮﺍﻫﻨﺪ ﺩﺍﺷﺖ‪.‬‬ ‫‪101‬‬ ‫ﻭﺻﻠﻪ ﺳﺘﻮﻥﻫﺎ‬ ‫‪Column Splices‬‬ ‫ ﺗﻔﺎﻭﺕ ﻭﺻﻠﻪﻫﺎﻱ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻭ ﻛﺸﺸﻲ‪:‬‬ ‫ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻛﻞ ﺑﺎﺭ ﺗﻮﺳﻂ ﻭﺻﻠﻪ ﺍﻧﺘﻘﺎﻝ ﻣﻲ ﻳﺎﺑﺪ‬ ‫ ﺩﺭ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﺑﺨﺶ ﻋﻤﺪﻩﺍﻱ ﺍﺯ ﺑﺎﺭ ﻣﺴﺘﻘﻴﻤﺎ ﺑﺎ ﺗﻤﺎﺱ ﺩﻭ ﺳـﺘﻮﻥ ﻭ ﺑﻘﻴـﻪ‬ ‫ﺗﻮﺳﻂ ﻭﺻﻠﻪﻫﺎ ﺍﻧﺘﻘﺎﻝ ﻣﻲﻳﺎﺑﺪ‪.‬‬ ‫ ﺳﻨﮓﺯﻧﻲ ﺍﻧﺘﻬﺎﻱ ﺳﺘﻮﻥ ﺑﺎﻋﺚ ﺻﺎﻑ ﺷﺪﻥ ﻣﻘﻄﻊ ﻭ ﺗﻤﺎﺱ ﺑﻬﺘـﺮ ﺩﺭ ﻣﺤـﻞ ﺍﺗﻜـﺎ ﻭ‬ ‫ﺍﻧﺘﻘﺎﻝ ﻣﺴﺘﻘﻴﻢ ﻗﺴﻤﺖ ﻋﻤﺪﻩﺍﻱ ﺍﺯ ﻧﻴﺮﻭ ﺍﺯ ﻃﺮﻳﻖ ﺳﻄﺢ ﺗﻤﺎﺱ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺍﻣﺎ ﺑﻬﺮ ﺣﺎﻝ ﺣﺘﻲ ﺍﮔﺮ ﺗﻤﺎﺱ ﻛﺎﻣﻞ ﻧﻴﺰ ﺩﺭ ﻣﺤﻞ ﻭﺻﻠﻪ ﺑﺮﻗﺮﺍﺭ ﮔﺮﺩﺩ ﺑﺎﺯ ﻭﺟـﻮﺩ ﻭﺭﻕ‬ ‫ﻭﺻﻠﻪ ﺑﻪ ﺩﻻﻳﻞ ﺯﻳﺮ ﺿﺮﻭﺭﺕ ﺩﺍﺭﺩ‪:‬‬ ‫ ﺑﺮﺍﻱ ﺟﻔﺖ ﻛﺮﺩﻥ)ﺑﺎ ﻫﻢ ﻧﮕﻬﺪﺍﺷﺘﻦ( ﺩﻭ ﺳﺘﻮﻥ ﺩﺭ ﺣﻴﻦ ﺍﺟﺮﺍ‬ ‫ ﺑﺮﺍﻱ ﺗﺤﻤﻞ ﻧﻴﺮﻭﻫﺎﻱ ﺑﺮﺷﻲ ﻭ ﺧﻤﺸﻲ ﻣﻮﺟﻮﺩ ﺩﺭ ﺳﺘﻮﻧﻬﺎ‬ ‫‪102‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫‪Typical Bolted or Riveted Splices‬‬ ‫ﺷﻜﻞ)‪ :(c‬ﻭﺻﻠﻪ ﺑﺎ ﻭﺭﻕ ﺯﻳﺮﺳﺮﻱ ‪-‬‬ ‫ﻋﺪﻡ ﺿﺮﻭﺭﺕ ﺍﺗﻜﺎﻱ ﻭﺭﻗﻬﺎﻱ ﭘﺮﻛﻨﻨﺪﻩ‬ ‫‪103‬‬ ‫ﺷﻜﻞ)‪ :(b‬ﻭﺻﻠﻪ ﺑﺎ ﻭﺭﻗﻬﺎﻱ ﭘﺮﻛﻨﻨﺪﻩ‪-‬‬ ‫ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﻭﺻﻠﻪ ﺩﻭﺳﺘﻮﻥ ﺑﺎ ﻋﻤﻖ‬ ‫ﻧﺎﺑﺮﺍﺑﺮ‬ ‫ﺷﻜﻞ)‪ :(a‬ﻭﺻﻠﻪ ﭘﻴﭽﻲ ﺳﺎﺩﻩ ‪-‬‬ ‫ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﻭﺻﻠﻪ ﺩﻭﺳﺘﻮﻥ ﺑﺎ ﻋﻤﻖ‬ ‫ﺑﺮﺍﺑﺮ‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫‪Typical Welded Splices‬‬ ‫ﺷﻜﻞ)‪ :(c‬ﻭﺻﻠﻪ ﺟﻮﺷﻲ ﺑﺎ ﻭﺭﻕ ﺯﻳﺮﺳﺮﻱ‬ ‫ ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﻭﺻﻠﻪ ﺩﻭﺳﺘﻮﻥ ﺑﺎ ﻋﻤﻖ‬‫ﻧﺎﺑﺮﺍﺑﺮ‬ ‫‪104‬‬ ‫ﺷﻜﻞ)‪ :(b‬ﻭﺻﻠﻪ ﺑﺎ ﻭﺭﻗﻬﺎﻱ ﻭﺻﻠﻪ‬ ‫ﺟﻮﺵ ﺷﺪﻩ ‪ -‬ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﻭﺻﻠﻪ‬ ‫ﺩﻭﺳﺘﻮﻥ ﺑﺎ ﻋﻤﻖ ﺍﺳﻤﻲ ﺑﺮﺍﺑﺮ‬ ‫ﺷﻜﻞ)‪ :(a‬ﻭﺻﻠﻪ ﺟﻮﺷﻲ ﺑﺎ ﺟﻮﺵ‬ ‫ﺷﻴﺎﺭﻱ ﻧﻔﻮﺫﻱ‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻭﺻﻠﻪ ﺳﺘﻮﻧﻬﺎ ‪ -‬ﺗﻮﺻﻴﻪﻫﺎ‬ ‫ ﺑﻬﺘﺮ ﺍﺳﺖ ﺍﺯ ﻧﻴﻤﺮﺧﻬﺎﻱ ﻫﻢﻋﻤﻖ ﺩﺭ ﻃﺒﻘﺎﺕ ﻣﺨﺘﻠـﻒ ﺳـﺎﺧﺘﻤﺎﻥ ﺍﺳـﺘﻔﺎﺩﻩ ﺷـﻮﺩ‪ .‬ﺩﺭ‬ ‫ﺻﻮﺭﺗﻲ ﻛﻪ ﻋﻤﻖ ﺳﺘﻮﻥ ﺑﺎﻻﻳﻲ ﺧﻴﻠﻲ ﻛﻤﺘـﺮ ﺍﺯ ﻋﻤـﻖ ﺳـﺘﻮﻥ ﭘـﺎﻳﻴﻨﻲ ﺑﺎﺷـﺪ ﺑﺎﻳـﺪ ﺍﺯ‬ ‫ﻭﺭﻗﻬﺎﻱ ﭘﺮﻛﻨﻨﺪﻩ)ﻟﻘﻤﻪ‪ ،‬ﻻﻳﻲ‪ (Filler‬ﺑﻴﻦ ﺳﺘﻮﻥ ﺑﺎﻻﻳﻲ ﻭ ﻭﺭﻕ ﻭﺻﻠﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪.‬‬ ‫ ﺩﺭ ﺳﺎﺧﺘﻤﺎﻧﻬﺎﻱ ﭼﻨﺪ ﻃﺒﻘﻪ ﺍﺯ ﺟﻨﺒﻪ ﺗﺌﻮﺭﻳﻚ ﺑﺎﻳﺪ ﺍﻧﺪﺍﺯﻩ ﺳﺘﻮﻥ ﺩﺭ ﻫﺮ ﻃﺒﻘـﻪ ﺗﻐﻴﻴـﺮ‬ ‫ﻳﺎﺑﺪ ﺍﻣﺎ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻫﺰﻳﻨﻪ ﻧﺴﺒﺘﺎ ﺑﺎﻻﻱ ﺍﺟﺮﺍﻱ ﻭﺻﻠﻪ‪ ،‬ﭘﻴﺸﻨﻬﺎﺩ ﻣﻲﺷﻮﺩ ﺣﺪﺍﻗﻞ ﺩﺭ ﻫـﺮ‬ ‫ﺩﻭ ﻃﺒﻘﻪ‪ ،‬ﺍﻧﺪﺍﺯﻩ ﺳﺘﻮﻥ ﺛﺎﺑﺖ ﺑﺎﻗﻲ ﺑﻤﺎﻧﺪ )ﻋﻠﻴﺮﻏﻢ ﺍﻳﻨﻜﻪ ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﻭﺯﻥ ﻛﻞ ﻓﻮﻻﺩ‬ ‫ﻣﺼﺮﻓﻲ ﺑﺎﻻ ﺧﻮﺍﻫﺪ ﺭﻓﺖ(‪.‬‬ ‫ ﺍﻟﺒﺘﻪ ﻭﺻﻠﻪ ﺳﺘﻮﻧﻬﺎ ﺩﺭ ﻫﺮ ﺳﻪ ﻃﺒﻘﻪ ﻭ ﺑﻴﺸﺘﺮ ﺍﺯ ﺁﻥ ﻧﻴﺰ‪ ،‬ﻣﺸﻜﻼﺕ ﺍﺟﺮﺍﻳﻲ ﺯﻳﺎﺩﻱ ﺩﺍﺭﺩ‬ ‫ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺑﻜﺎﺭﮔﻴﺮﻱ ﺳﺘﻮﻥ ﺑﺎ ﺍﻧﺪﺍﺯﻩ ﺛﺎﺑﺖ ﺩﺭ ﺳﻪ ﻃﺒﻘﻪ ﺗﻮﺻﻴﻪ ﻧﻤﻲﺷﻮﺩ‪.‬‬ ‫‪105‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻭﺻﻠﻪ ﺳﺘﻮﻧﻬﺎ ‪ -‬ﺗﻮﺻﻴﻪﻫﺎ‬ ‫ ﺑﻜﺎﺭﮔﻴﺮﻱ ﺳـﺘﻮﻧﻲ ﺑـﺎ ﺷـﻤﺎﺭﻩ ﺛﺎﺑـﺖ ﺩﺭ‬ ‫ﻃﺒﻘﺎﺕ ﻣﺨﺘﻠـﻒ ﻭ ﺍﺳـﺘﻔﺎﺩﻩ ﺍﺯ ﻭﺭﻗﻬـﺎﻱ‬ ‫ﺗﻘﻮﻳﺘﻲ ﺩﺭ ﻃﺒﻘﺎﺕ ﭘﺎﻳﻴﻦﺗﺮ ﺑﻨﺤـﻮﻱ ﻛـﻪ‬ ‫ﺩﺭ ﻃﺒﻘــﺎﺕ ﺑــﺎﻻﺗﺮ ﺑﺘــﺪﺭﻳﺞ ﺍﺯ ﻣﻘــﺪﺍﺭ‬ ‫ﻭﺭﻗﻬــﺎﻱ ﺗﻘــﻮﻳﺘﻲ ﻛﺎﺳــﺘﻪ ﺷــﻮﺩ ﺗﻮﺻــﻴﻪ‬ ‫ﻣﻲﮔﺮﺩﺩ‪.‬‬ ‫ ﺍﻟﺒﺘﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﭼﻨﻴﻦ ﻣﻘﺎﻃﻊ ﻣﺮﻛﺒـﻲ‬ ‫ﺑﺪﻭﻥ ﺍﺟﺮﺍﻱ ﻭﺻﻠﻪ ﻓﻘﻂ ﺗﺎ ﺣﺪﺍﻛﺜﺮ ﭼﻬﺎﺭ‬ ‫ﻃﺒﻘﻪ ﺍﻣﻜﺎﻥﭘﺬﻳﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﭼﺮﺍ ﻛﻪ ﺍﮔﺮ‬ ‫ﻃــﻮﻝ ﺳــﺘﻮﻥ ﺑﻴﺸــﺘﺮ ﺍﺯ ‪ 12‬ﻣﺘــﺮ ﺷــﻮﺩ‬ ‫ﺍﺟﺮﺍﻱ ﻭﺻﻠﻪ ﺍﺟﺘﻨﺎﺏﻧﺎﭘﺬﻳﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫‪106‬‬ ‫‪Urmia University - M.Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﻭﺻﻠﻪ ﺳﺘﻮﻧﻬﺎ‬ ‫ ﺗﺨﻤﻴﻦ ﺑﺎﺭ ﻭﺻﻠﻪﻫﺎ‪:‬‬ ‫‪ (1‬ﺍﮔﺮ ﺍﻧﺘﻬﺎﻱ ﺳﺘﻮﻥ ﺳﻨﮓﺧﻮﺭﺩﻩ ﻧﺒﺎﺷﺪ ← ‪ %100‬ﻛﻞ ﺑﺎﺭ ﺳﺘﻮﻥ‬ ‫‪ (2‬ﺍﮔﺮ ﺳﻄﻮﺡ ﺳﻨﮓﺧﻮﺭﺩﻩ ﺑﻮﺩﻩ ﻭﺳﺘﻮﻥ ﻓﻘﻂ ﺗﺤﺖ ﺑﺎﺭ ﻣﺤﻮﺭﻱ ﺑﺎﺷﺪ ←‬ ‫‪ 25‬ﺍﻟﻲ ‪ %50‬ﻛﻞ ﺑﺎﺭ ﺳﺘﻮﻥ‬ ‫‪ (3‬ﺍﮔﺮ ﺳﺘﻮﻥ ﺗﺤﺖ ﺧﻤﺶ ﻧﻴﺰ ﺑﺎﺷﺪ ← ‪ 50‬ﺍﻟﻲ ‪ %75‬ﺑﺎﺭ ﺳﺘﻮﻥ‬ ‫‬ ‫‬ ‫‪107‬‬ ‫ﺍﻧﺘﻘﺎﻝ ﺧﻤﺶ ﺗﻮﺳﻂ ﻭﺭﻗﻬﺎﻱ ﻭﺻﻠﻪ ﺑﺎﻟﻬﺎ‪ ،‬ﺍﻧﺘﻘﺎﻝ ﺑﺮﺵ ﺗﻮﺳﻂ ﻭﺭﻗﻬﺎﻱ ﻭﺻﻠﻪ ﺟﺎﻥ‪ ،‬ﻭ ﻧﻴﺮﻭﻱ‬ ‫ﻣﺤﻮﺭﻱ ﻧﻴﺰ ﺑﻴﻦ ﻫﻤﻪ ﻭﺭﻗﻬﺎ ﺗﻮﺯﻳﻊ ﻣﻲﺷﻮﺩ‬ ‫ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﻣﺴﺎﺣﺖ ﻭﺭﻕ ﻭﺻﻠﻪ‪،‬‬ ‫ﻣﺠﻤﻮﻉ ﻣﺴﺎﺣﺖ ﻣﻮﺭﺩﻧﻴﺎﺯ ﺑﺮﺍﻱ‬ ‫ﺗﺤﻤﻞ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ‪ ،‬ﻟﻨﮕﺮ ﺧﻤﺸﻲ‬ ‫ﻭ ﻧﻴﺮﻭﻱ ﺑﺮﺷﻲ ﻣﺤﺎﺳﺒﻪ ﻭ ﻭﺭﻕ ﻳﻜﺠﺎ‬ ‫ﻃﺮﺡ ﺷﻮﺩ‪.‬‬ ‫‪M/d‬ﻧﻴﺮﻭﻱ ﻫﺮ ﺑﺎﻝ‬ ‫‪M/‬‬ ‫‪d‬‬ ‫‪M‬‬ ‫‪Wmax= 1.5 Wm‬‬ ‫‪= 1.5 V/A‬‬ ‫‪V‬‬ ‫‪d‬‬ Built-up Compression Members ‫ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ‬:‫ ﻗﺴﻤﺖ ﺩﻭﻡ‬-‫ﻓﺼﻞ ﭼﻬﺎﺭﻡ‬ Urmia University, M. Sheidaii 1 ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ‬ ‫ﻣﻤﻜﻦ ﺍﺳﺖ ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﺍﺯ ﺗﺮﻛﻴﺐ ﺩﻭ ﻳﺎ ﭼﻨﺪ ﺟﺰء ﺑﻪ ﺻﻮﺭﺗﻬﺎﻱ ﻣﺨﺘﻠﻒ ﺳﺎﺧﺘﻪ ﺷﻮﺩ‪:‬‬ ‫‪ (1‬ﺳﺘﻮﻧﻬﺎﻱ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭ ﺗﻤﺎﺱ ﻣﺴﺘﻘﻴﻢ ﺑﺎ ﻫﻢ‪،‬‬ ‫ﻧﻈﻴﺮ ﻧﻴﻤﺮﺧﻬﺎﻱ ﺑﺎ ﻭﺭﻗﻬﺎﻱ ﺗﻘﻮﻳﺘﻲ‬ ‫‪ (2‬ﺳﺘﻮﻧﻬﺎﻱ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭ ﺗﻤﺎﺱ ﻧﺰﺩﻳﻚ ﺑﺎ ﻫﻢ‪،‬‬ ‫ﻧﻈﻴﺮ ﺯﻭﺝ ﻧﺒﺸﻲ ﺑﺎ ﻓﺎﺻﻠﻪ ﺟﺰﺋﻲ ﺍﺯ ﻫﻢ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﺿﺨﺎﻣﺖ ﻭﺭﻕ ﻟﭽﻜﻲ‬ ‫‪ (3‬ﺳﺘﻮﻧﻬﺎﻱ ﺑﺎ ﺍﺟﺰﺍﻱ ﺑﺪﻭﻥ ﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‪،‬‬ ‫ﻧﻈﻴﺮ ﺯﻭﺝ ﻧﺎﻭﺩﺍﻧﻲ ﻳﺎ ﭼﻬﺎﺭ ﻧﺒﺸﻲ‪ ،‬ﻣﻨﺎﺳﺐ ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ ﺑﻠﻨﺪ‪،‬‬ ‫ﺿﺮﻭﺭﺕ ﺍﺗﺼﺎﻝ ﺍﺟﺰﺍ ﺑﺎ ﻭﺭﻗﻬﺎﻱ ﺑﺴﺖ‬ ‫‪2‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ﺑﺤﺚ ﻣﻘﺪﻣﺎﺗﻲ‪ :‬ﺑﺮﺭﺳﻲ ﺭﻓﺘﺎﺭ ﺳﺘﻮﻧﻲ ﻣﺮﻛﺐ ﺍﺯ ﺩﻭ ﻭﺭﻕ ﻣﺸﺎﺑﻪ ﺩﺭ ﺣﺎﻻﺕ ﻣﺨﺘﻠﻒ ﺍﺗﺼﺎﻝ‬ ‫‪ (1‬ﺩﺭ ﺻﻮﺭﺕ ﻋﺪﻡ ﺍﺗﺼﺎﻝ ﻭﺭﻗﻬﺎ ﺑﻬﻢ‪:‬‬ ‫‬ ‫ﻋﻤﻠﻜﺮﺩ ﻣﺸﺎﺑﻪ ﻫﺮ ﺩﻭﺟﺰء ﻭ ﺗﻐﻴﻴﺮﺷﻜﻞﻫﺎﻱ ﻣﺸﺎﺑﻪ‬ ‫‬ ‫ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺳﺘﻮﻥ ﺩﻭ ﺑﺮﺍﺑﺮ ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﻳﻚ ﻭﺭﻕ‬ ‫ﺑﺎﺭ ﻫﺮ ﺟﺰء ﻧﺼﻒ ﺑﺎﺭ ﺳﺘﻮﻥ‬ ‫‪K=1‬‬ ‫‬ ‫‬ ‫‪3‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫‪(2‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﺗﺎﻣﻴﻦ ﺍﺗﺼﺎﻻﺕ ﻛﺎﻓﻲ ﺩﺭ ﺑﺮﺍﺑﺮ ﻟﻐﺰﺵ‪:‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪4‬‬ ‫ﻋﻤﻠﻜﺮﺩ ﺑﻪ ﺻﻮﺭﺕ ﻳﻚ ﻗﻄﻌﻪ ﻭﺍﺣﺪ‬ ‫ﺗﻐﻴﻴﺮﺷﻜﻞ ﻣﺘﻔﺎﻭﺕ ﻭﺭﻗﻬﺎ‬ ‫ﻣﺤﺎﺳﺒﻪ ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺑﺮﺍﻱ ﻛﻞ ﻣﻘﻄﻊ ﻣﺮﻛﺐ‬ ‫)‪ 4‬ﺑﺮﺍﺑﺮ ﺣﺎﻟﺖ ﻗﺒﻞ(‪.‬‬ ‫‪K=1‬‬ ‫‪1.732L / d.‬‬ ‫)‪(1)(L‬‬ ‫‪3‬‬ ‫‪4bd‬‬ ‫‪/ 2bd‬‬ ‫‪6‬‬ ‫‪KL‬‬ ‫‪r‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫‪(3‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﺗﺎﻣﻴﻦ ﺍﺗﺼﺎﻻﺕ ﻣﺤﺪﻭﺩ ﻓﻘﻂ ﺩﺭﭼﻨﺪ ﺟﺎ‪:‬‬ ‫‬ ‫‬ ‫‬ ‫ﻣﻘﺎﻭﻣﺖ ﻗﻄﻌﻪ ﺑﻴﻦ ﺩﻭﺣﺎﻟﺖ ﻗﺒﻞ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‬ ‫ﭼﻮﻥ ﻛﻤﺘﺮﻳﻦ ﺟﺎﺑﺠﺎﻳﻲ ﻭﺭﻗﻬﺎ ﻧﺴﺒﺖ ﺑﻬﻢ ﺩﺭ ﻣﺮﻛﺰ ﺳﺘﻮﻥ ﺍﺳﺖ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺍﺗﺼﺎﻻﺕ‬ ‫ﻟﻐﺰﺵﮔﻴﺮ ﺩﺭ ﻭﺳﻂ ﻛﻤﺘﺮﻳﻦ ﺍﺛﺮ ﺭﺍ ﺩﺍﺭﺩ‬ ‫ﭼﻮﻥ ﺑﻴﺸﺘﺮﻳﻦ ﺟﺎﺑﺠﺎﻳﻲ ﻭﺭﻗﻬﺎ ﻧﺴﺒﺖ ﺑﻬﻢ ﺩﺭ ﺩﻭ ﺍﻧﺘﻬﺎﺳﺖ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺍﺗﺼﺎﻻﺕ ﻟﻐﺰﺵﮔﻴﺮ‬ ‫ﺩﺭ ﺩﻭ ﺍﻧﺘﻬﺎ ﺑﻴﺸﺘﺮﻳﻦ ﺍﺛﺮ ﺭﺍ ﺩﺍﺭﺩ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ‪:‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪5‬‬ ‫ﺗﻐﻴﻴﺮﺷﻜﻞ ﺳﺘﻮﻥ ﺑﻪ ﺷﻜﻞ ﺣﺮﻑ ‪S‬‬ ‫‪K=0.5‬‬ ‫‪1.732L / d.‬‬ ‫)‪(0.5)(L‬‬ ‫‪3‬‬ ‫‪KL‬‬ ‫‪r‬‬ ‫‪bd‬‬ ‫‪/ 2bd‬‬ ‫‪6‬‬ ‫ﺗﻨﺶ ﻃﺮﺍﺣﻲ ﻭ ﻇﺮﻓﻴﺖ ﺑﺎﺭﺑﺮﻱ ﺍﺯ ﻟﺤﺎﻅ ﻧﻈﺮﻱ ﺑﺮﺍﺑﺮ ﺑﺎ ﺣﺎﻟﺖ‬ ‫ﻗﺒﻞ ﺍﺳﺖ‪.‬‬ ‫ﺍﻟﺒﺘﻪ ﺩﺭ ﺣﺎﻻﺕ ﻣﺘﻌﺎﺭﻑ ﻣﻮﺟﻮﺩ‪ ،‬ﻗﻄﻌﺎﺕ ﺗﻤﺎﻳﻞ ﺑﻪ ﻛﻨﺪﻩ‬ ‫ﺷﺪﻥ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺧﻮﺍﻫﻨﺪ ﺩﺍﺷﺖ‪.‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ ﺑﺮﺍﻱ ﺿﻮﺍﺑﻂ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ ﺑﻪ ‪AISC E6 p. 16.1-37‬‬ ‫ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﻭﺍﻗﻊ ﺑﺮ ﺭﻭﻱ ﻛﻒ ﺳﺘﻮﻧﻬﺎ ﺑﺎ ﺳﻄﻮﺡ ﺻﺎﻑ ﻭ ﻳﺎ ﺩﺭ‬ ‫ﻣﺤﻞ ﻭﺻﻠﻪ ﻫﺎ‪ ،‬ﺑﺎﻳﺪ ﺍﺟﺰﺍ ﺗﻮﺳﻂ ﺟﻮﺵ ﻳﺎ ﭘﻴﭻ ﺍﺻﻄﻜﺎﻛﻲ ﺑﻪ ﻫﻢ ﺍﺗﺼﺎﻝ ﻳﺎﺑﻨﺪ‪.‬‬ ‫ ﺍﺗﺼﺎﻝ ﺍﺟﺰﺍ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﻣﻲ ﺗﻮﺍﻧﺪ ﺑﻪ ﺻﻮﺭﺗﻬﺎﻱ ﻣﺨﺘﻠﻒ ﺯﻳﺮ ﺍﻧﺠﺎﻡ ﮔﻴﺮﺩ‪:‬‬ ‫)‪(i‬ﻣﺴﺘﻘﻴﻢ‪(ii) ،‬ﺑﻪ ﻛﻤﻚ ﻟﻘﻤﻪ)ﻻﻳﻲ(‪(iii) ،‬ﺑﻪ ﻛﻤﻚ ﻭﺭﻕ ﺧﺎﺭﺟﻲ‬ ‫ ﺩﺭ ﺻﻮﺭﺕ ﺟﻮﺷﻜﺎﺭﻱ‪ ،‬ﺑﺎﻳﺪ ﺍﺟﺰﺍ ﺩﺭ ﻣﺤﻠﻬﺎﻱ ﻳﺎﺩﺷﺪﻩ ﺑﺎ ﺟﻮﺵ ﭘﻴﻮﺳﺘﻪ‬ ‫ﺩﺭ ﻃﻮﻟﻲ ﺣﺪﺍﻗﻞ ﺑﺮﺍﺑﺮ ﺑﺎ ﻋﺮﺽ ﺣﺪﺍﻛﺜﺮ ﻋﻀﻮ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﻣﺘﺼﻞ ﺷﻮﻧﺪ‪.‬‬ ‫ ﺩﺭ ﺻﻮﺭﺕ ﭘﻴﭽﻜﺎﺭﻱ‪:‬‬ ‫‬ ‫‬ ‫‪6‬‬ ‫ﻓﺎﺻﻠﻪ ﻣﺮﻛﺰ ﺑﻪ ﻣﺮﻛﺰ ﭘﻴﭽﻬﺎ ﺩﺭ ﺟﻬﺖ ﻃﻮﻟﻲ ﻋﻀﻮ‬ ‫ﻧﺒﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ‪ 4‬ﺑﺮﺍﺑﺮ ﻗﻄﺮ ﭘﻴﭽﻬﺎ ﺑﺎﺷﺪ‬ ‫ﻃﻮﻝ ﻛﻞ ﺍﺗﺼﺎﻝ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ‪ 1/5‬ﺑﺮﺍﺑﺮ ﺑﺰﺭﮔﺘﺮﻳﻦ‬ ‫ﻋﺮﺽ ﻋﻀﻮ ﺑﺎﺷﺪ‪.‬‬ ‫‪t Wmax‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ‬ ‫ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ ﻻﻏﺮﻱ ﻣﻮﺛﺮ ﺗﻚ ﺗﻚ ﻧﻴﻤﺮﺧﻬﺎ )‪ (a/ri‬ﺩﺭ ﺣﺪ ﻓﺎﺻﻞ ﺍﺗﺼﺎﻻﺕ ﺑﺎﻳﺪ ﺍﺯ ¾ ﺿﺮﻳﺐ‬ ‫ﻻﻏﺮﻱ ﺣﺎﻛﻢ ﻋﻀﻮ ﻣﺮﻛﺐ ﺑﺰﺭﮔﺘﺮ ﻧﺒﺎﺷﺪ ) ﺩﺭ ﻣﺤﺎﺳﺒﻪ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﻫﺮ ﻧﻴﻤﺮﺥ‬ ‫ﺑﺎﻳﺪ ﻛﻮﭼﻜﺘﺮﻳﻦ ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﻣﻨﻈﻮﺭ ﮔﺮﺩﺩ(‪.‬‬ ‫ ﺩﺭ ﻃﻮﻝ ﻳﻚ ﻋﻀﻮ ﻣﺮﻛﺐ‪ ،‬ﺣﺪﺍﻗﻞ ﺩﻭ ﻧﻘﻄﻪ ﺍﺗﺼﺎﻝ ﻣﻴﺎﻧﻲ ﺩﺭ ﻧﻘﺎﻁ ﻳﻚ ﺳﻮﻡ ﻃﻮﻝ‬ ‫ﺑﻴﻦ ﺩﻭ ﺳﺮ ﻋﻀﻮ ﻣﻮﺟﻮﺩ ﺑﺎﺷﺪ‪.‬‬ ‫‪a‬‬ ‫‪7‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ ﺩﺭ ﺻﻮﺭﺕ ﺍﺗﺼﺎﻝ ﺍﺟﺰﺍ ﺗﻮﺳﻂ ﻭﺭﻕ ﺧﺎﺭﺟﻲ)‪ (outside plate‬ﻓﻮﺍﺻﻞ ﭘﻴﭽﻬﺎ ﻭ‬ ‫ﺟﻮﺷﻬﺎﻱ ﻣﻨﻘﻄﻌﻲ ﻛﻪ ﻭﺭﻕ ﺭﺍ ﺑﻪ ﻧﻴﻤﺮﺧﻬﺎ ﻣﺘﺼﻞ ﻣﻲ ﻧﻤﺎﻳﻨﺪ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﻣﻘﺎﺩﻳﺮ ﺯﻳﺮ‬ ‫ﺑﺎﺷﺪ‪:‬‬ ‫• ﺍﮔﺮ ﻋﻮﺍﻣﻞ ﺍﺗﺼﺎﻝ ﺭﻭﺑﺮﻭﻱ ﻫﻢ ﺑﺎﺷﻨﺪ )‪Min (0.75t√(E/Fy )=1065t /√Fy , 30 cm‬‬ ‫• ﺍﮔﺮ ﻋﻮﺍﻣﻞ ﺍﺗﺼﺎﻝ ﺭﻭﺑﺮﻭﻱ ﻫﻢ ﻧﺒﺎﺷﻨﺪ )‪Min (1.12t√(E/Fy )=1595t /√Fy , 45 cm‬‬ ‫‪ = t‬ﺿﺨﺎﻣﺖ ﻧﺎﺯﻛﺘﺮﻳﻦ ﻭﺭﻕ ﭘﻮﺷﺸﻲ ﺧﺎﺭﺟﻲ‬ ‫‪45 cm‬‬ ‫‪8‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ‬ ‫ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻳﻦ ﻧﻮﻉ ﺳﺘﻮﻧﻬﺎ ﺑﺮﺍﺳﺎﺱ ﻣﻄﺎﻟﺐ ﻓﺼﻞ ﭘﻴﺸﻴﻦ‪ ،‬ﻭ ﺑﺎ ﻣﻨﻈﻮﺭ ﻧﻤﻮﺩﻥ ﺍﺻﻼﺡ ﺯﻳﺮ‬ ‫ﺗﻌﻴﻴﻦ ﻣﻲﮔﺮﺩﺩ‪:‬‬ ‫ﺍﮔﺮ ﻣﺪ ﻛﻤﺎﻧﺶ ﻫﻤﺮﺍﻩ ﺑﺎ ﺗﻐﻴﻴﺮﺷﻜﻞ ﻧﺴﺒﻲ )ﻟﻐﺰﺵ( ﺑﺎﺷﺪ ﺗﻨﺶﻫﺎﻱ ﺑﺮﺷﻲ ﺩﺭ ﺍﺗﺼﺎﻻﺕ ﺑﻴﻦ‬ ‫‬ ‫ﻧﻴﻤﺮﺧﻬﺎ ﺍﻳﺠﺎﺩ ﺷﺪﻩ ﻭ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ‪ KL/r‬ﺑﺎ ‪ (KL/r)m‬ﺟﺎﻳﮕﺰﻳﻦ ﻣﻲﺷﻮﺩ‪:‬‬ ‫‪.i‬‬ ‫ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﻣﻴﺎﻧﻲ ﭘﻴﭽﻲ ﻣﻌﻤﻮﻟﻲ)‪(snug-tight bolted‬‬ ‫‪y‬‬ ‫ﺩﺭ ﮐﻤﺎﻧﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ‪ x‬ﺑﻴﻦ ﻧﻴﻤﺮﺧﻬﺎ ﻟﻐﺰﺵ ﺍﻳﺠﺎﺩ ﻧﻤﯽﺷﻮﺩ‬ ‫‪.ii‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫ﻣﯽﺷﻮﺩ‬ ‫ﻟﻐﺰﺵ ﺍﻳﺠﺎﺩ‬ ‫ﻧﻴﻤﺮﺧﻬﺎ‬ ‫ﻣﻴﺎﻧﻲ ‪ y‬ﺑﻴﻦ‬ ‫ﺍﺗﺼﺎﻻﺕ ﻣﺤﻮﺭ‬ ‫ﮐﻤﺎﻧﺶ ﺣﻮﻝ‬ ‫‪(welded or pretensioned bolted‬‬ ‫ﺍﺻﻄﻜﺎﻛﻲ)‬ ‫ﭘﻴﭽﻲ‬ ‫ﺟﻮﺷﻲ ﻭ‬ ‫ﺩﺭﺑﺮﺍﻱ‬ ‫‪y‬‬ ‫‪9‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ‬ ‫ﺩﺭﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺭﻭﺍﺑﻂ‪:‬‬ ‫‪ =(KL/r)o‬ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﻋﻀﻮ ﺳﺘﻮﻥ ﻣﺮﻛﺐ ﺑﻪ ﻋﻨﻮﺍﻥ ﻳﻚ ﻋﻀﻮ ﻳﻜﭙﺎﺭﭼﻪ‬ ‫‪ =(KL/r)m‬ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺍﺻﻼﺡ ﺷﺪﻩ ﻋﻀﻮ ﻣﺮﻛﺐ‬ ‫‪a‬‬ ‫= ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺍﺗﺼﺎﻻﺕ‬ ‫‪ = ri‬ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﺣﺪﺍﻗﻞ ﺍﺟﺰﺍﻱ ﺗﺸﻜﻴﻞ ﺩﻫﻨﺪﻩ ﺳﺘﻮﻥ‬ ‫‪ = rib‬ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﺟﺰء ﺗﻜﻲ ﻧﺴﺒﺖ ﺑﻪ ﻣﺤﻮﺭ ﻣﺮﻛﺰﻱ ﺁﻥ ﻣﻮﺍﺯﻱ ﺑﺎ ﻣﺤﻮﺭ ﻛﻤﺎﻧﺶ‬ ‫‪ = h‬ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﻣﺮﺍﻛﺰ ﺍﺟﺰﺍ )ﻧﻴﻤﺮﺧﻬﺎ( ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻋﻤﻮﺩ ﺑﺮ ﻣﺤﻮﺭ ﻛﻤﺎﻧﺶ ﻋﻀﻮ‬ ‫‪ = D‬ﺿﺮﻳﺐ ﻃﺒﻠﻪ )ﺟﺪﺍﺷﺪﮔﻲ( = ‪h/2rib‬‬ ‫‪10‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺑﺪﻭﻥ ﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫‬ ‫ﺑﺮﺍﻱ ﺿﻮﺍﺑﻂ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ ﺑﻪ ‪AISC E6 p. 16.1-37‬‬ ‫‬ ‫ﺳﻄﻮﺡ ﺟﺎﻧﺒﻲ ﺑﺎﺯ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺍﺯ ﭼﻨﺪ ﻧﻴﻤﺮﺥ ﺭﺍ ﺑﺎﻳﺪ ﺑﻪ ﺭﻭﺷﻬﺎﻱ ﺯﻳﺮ‬ ‫ﭘﻮﺷﺎﻧﺪ‪:‬‬ ‫‪.1‬‬ ‫‪.2‬‬ ‫‪11‬‬ ‫ﺗﻮﺳﻂ ﻭﺭﻕ ﭘﻮﺷﺸﻲ ﻣﻤﺘﺪ ﺑﺎ ﺳﻮﺭﺍﺧﻬﺎﻱ ﺩﺳﺘﺮﺳﻲ ‪(continuous‬‬ ‫)‪perforated cover plate‬‬ ‫ﺗﻮﺳﻂ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ)‪ (lacing‬ﻭ ﺑﺎ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ )‪(tie plates‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺍﺟﺰﺍﻱ ﺑﺪﻭﻥ ﺗﻤﺎﺱ ﺑﺎ ﻫﻢ‬ ‫ﻧﻘﺶ ﺍﻳﻦ ﺍﺗﺼﺎﻻﺕ‪:‬‬ ‫ ﺣﻔﻆ ﻧﻴﻤﺮﺧﻬﺎﻱ ﻣﺨﺘﻠﻒ ﺳﺘﻮﻥ ﺑﻪ ﻣﻮﺍﺯﺍﺕ ﻫﻢ ﻭ ﺩﺭ ﻓﺎﺻﻠﻪ ﻳﻜﻨﻮﺍﺧﺖ ﺍﺯ ﻫﻢ ﻭ‬ ‫ﺗﻮﺯﻳﻊ ﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﺑﻴﻦ ﺁﻧﻬﺎ‪،‬‬ ‫ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﻛﻤﺎﻧﺶ ﺍﻧﻔﺮﺍﺩﻱ ﻫﺮ ﻳﻚ ﺍﺯ ﺍﺟﺰﺍ‬ ‫ ﻋﻤﻠﻜﺮﺩ ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﻪ ﺻﻮﺭﺕ ﻳﻚ ﻗﻄﻌﻪ ﻭﺍﺣﺪ ﺗﺤﺖ ﺑﺎﺭ ﻭﺍﺭﺩﻩ‬ ‫‪12‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﻭﺭﻕ ﺳﻮﺭﺍﺧﺪﺍﺭ ﻣﻤﺘﺪ‬ ‫ﻣﺰﺍﻳﺎﻱ ﻭﺭﻗﻬﺎﻱ ﭘﻮﺷﺸﻲ ﺳﻮﺭﺍﺧﺪﺍﺭ‪:‬‬ ‫‪ (1‬ﺳﺎﺧﺖ ﺁﻧﻬﺎ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺩﺳﺘﮕﺎﻩ ﺑﺮﺵ ﺁﺳﺎﻥ ﺍﺳﺖ‪،‬‬ ‫‪ (2‬ﺑﺮﺧﻲ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻫﺎ ﺳﻄﺢ ﺧﺎﻟﺺ ﻭﺭﻕ ﺭﺍ ﺩﺭ ﺑﺎﺭﺑﺮﻱ ﻋﻀﻮ ﻣﻮﺛﺮ ﻣﻲ ﺩﺍﻧﻨﺪ ﺍﻟﺒﺘﻪ‬ ‫ﺑﺸﺮﻃﻲ ﻛﻪ ﺍﻳﺠﺎﺩ ﺳﻮﺭﺍﺧﻬﺎ ﺑﺮ ﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ﺗﺠﺮﺑﻲ ﻣﻘﺮﺭ ﺷﺪﻩ)ﺍﺳﻼﻳﺪ ﺑﻌﺪﻱ(‪،‬‬ ‫ﺻﻮﺭﺕ ﮔﻴﺮﺩ‪،‬‬ ‫‪ (3‬ﺭﻧﮓ ﺁﻣﻴﺰﻱ ﺍﻋﻀﺎ ﺳﺎﺩﻩ ﺗﺮ ﺍﺳﺖ‪.‬‬ ‫‪13‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﻭﺭﻕ ﺳﻮﺭﺍﺧﺪﺍﺭ ﻣﻤﺘﺪ‬ ‫ﺍﮔﺮ ﻋﺮﺽ ﺑﺎﻗﻴﻤﺎﻧﺪﻩ )‪ (unsupported width‬ﻭﺭﻕ ﺩﺭ ﻣﺤﻞ ﺳﻮﺭﺍﺥ ﺣﺎﺋﺰ ﺷﺮﺍﻳﻂ ﺯﻳﺮ ﺑﺎﺷﺪ ﺩﺭ‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ﻣﺸﺎﺭﻛﺖ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ‪:‬‬ ‫‪ (1‬ﻧﺴﺒﺖ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﺁﻥ ﻣﺤﺪﻭﺩﻳﺘﻬﺎﻱ ﺑﺨﺶ ‪ B4‬ﺁﻳﻴﻦﻧﺎﻣﻪ ﺭﺍ ﺑﺮﺁﻭﺭﺩﻩ ﻧﻤﺎﻳﺪ‬ ‫)‪،(AISC05 Table B4.1 case 14‬‬ ‫‪ (2‬ﻧﺴﺒﺖ ﻃﻮﻝ ﺳﻮﺭﺍﺥ)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺗﻨﺶ( ﺑﻪ ﻋﺮﺽ ﺁﻥ ﺍﺯ ﺩﻭ ﺗﺠﺎﻭﺯ ﻧﻜﻨﺪ‪،‬‬ ‫‪ (3‬ﻓﺎﺻﻠﻪ ﺧﺎﻟﺺ ﺑﻴﻦ ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺗﻨﺶ‪ ،‬ﻛﻤﺘﺮ ﺍﺯ ﻓﺎﺻﻠﻪ ﻋﺮﺿﻲ ﺑﻴﻦ ﻧﺰﺩﻳﻜﺘﺮﻳﻦ‬ ‫ﺧﻄﻮﻁ ﺍﺗﺼﺎﻝ ﺟﻮﺵ ﻳﺎ ﭘﻴﭻ ﻧﺒﺎﺷﺪ‪،‬‬ ‫‪ (4‬ﻣﺤﻴﻂ ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﻫﻴﭻ ﺟﺎﻳﻲ‪ ،‬ﺍﻧﺤﻨﺎﻳﻲ ﺑﺎ ﺷﻌﺎﻉ ﻛﻤﺘﺮ ﺍﺯ ‪ 4‬ﺳﺎﻧﺘﻴﻤﺘﺮ ﻧﺪﺍﺷﺘﻪ ﺑﺎﺷﺪ‪.‬‬ ‫ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ ﺑﺎ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻭ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ‬ ‫‪15‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﻮﺍﺑﻂ ﻃﺮﺍﺣﻲ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ )ﻗﻴﺪﻫﺎﻱ ﺍﻧﺘﻬﺎﻳﻲ‪(Tie plates ،‬‬ ‫‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﺑﻜﺎﺭﮔﻴﺮﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ‪ ،‬ﺑﺎﻳﺪ ﺩﺭ ﺩﻭ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﻭ ﻧﻴﺰ ﺩﺭ ﻣﺤﻠﻲ ﻛﻪ‬ ‫ﭼﭗ ﻭ ﺭﺍﺳﺖﻫﺎ ﻗﻄﻊ ﻣﻲﺷﻮﻧﺪ ﺍﺯ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪.‬‬ ‫‬ ‫ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ ﺑﺎﻳﺪ ﺗﺎﺣﺪﺍﻣﻜﺎﻥ ﺑﻪ ﺩﻭ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ﻧﺰﺩﻳﻚ ﺑﺎﺷﻨﺪ‪،‬‬ ‫‬ ‫ﻃﻮﻝ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ ﺍﻧﺘﻬﺎﻳﻲ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﻓﺎﺻﻠﻪ ﺧﻄﻮﻁ ﺟﻮﺵ ﻳﺎ ﭘﻴﭻ ﺍﺗﺼﺎﻝ ﺩﻫﻨﺪﻩ ﺁﻧﻬﺎ ﺑﻪ‬ ‫ﺍﺟﺰﺍﻱ )ﻧﻴﻤﺮﺧﻬﺎ( ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﺑﺎﺷﺪ ﻭ ﻃﻮﻝ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ ﻣﻴﺎﻧﻲ ﻧﻴﺰ ﻧﺒﺎﻳﺪ ﺍﺯ ﻧﺼﻒ ﺍﻳﻦ ﻣﻘﺪﺍﺭ‬ ‫ﻛﻤﺘﺮ ﺑﺎﺷﺪ‪.‬‬ ‫ﺿﺨﺎﻣﺖ ﺑﺴﺖﻫﺎﻱ ﺍﻓﻘﻲ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ‪ 1 ⁄ 50‬ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺩﻭ ﺧﻂ ﺍﺗﺼﺎﻝ ﺩﻭ ﻃﺮﻑ ﺁﻥ ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﺟﻮﺷﻜﺎﺭﻱ‪ ،‬ﻛﻞ ﻃﻮﻝ ﺟﻮﺵ ﻫﺮ ﺧﻂ ﺍﺗﺼﺎﻝ ﺑﺴﺖ ﺍﻓﻘﻲ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ‪ 1 ⁄ 3‬ﻃﻮﻝ ﺑﺴﺖ‬ ‫ﺑﺎﺷﺪ‪.‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﭘﻴﭽﻜﺎﺭﻱ‬ ‫‬ ‫‬ ‫‪16‬‬ ‫‪ 9‬ﻓﺎﺻﻠﻪ ﭘﻴﭽﻬﺎﻱ ﺑﺴﺖ ﺍﻓﻘﻲ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺩﺭ ﺟﻬﺖ ﺗﻨﺶ‪ ،‬ﻧﺒﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ‪ 6‬ﺑﺮﺍﺑﺮ ﻗﻄﺮ ﭘﻴﭻ‬ ‫ﺑﺎﺷﺪ‬ ‫‪ 9‬ﺑﺴﺖ ﺍﻓﻘﻲ ﺑﺎﻳﺪ ﺩﺭ ﻫﺮ ﻃﺮﻑ ﺧﻮﺩ ﺣﺪﺍﻗﻞ ﺑﺎ ‪ 3‬ﭘﻴﭻ ﺑﻪ ﻧﻴﻤﺮﺥ ﺍﺗﺼﺎﻝ ﻳﺎﺑﺪ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﻮﺍﺑﻂ ﻃﺮﺍﺣﻲ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ )ﻗﻴﺪﻫﺎﻱ ﻣﻮﺭﺏ‪(Lacing ،‬‬ ‫‬ ‫ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻣﻌﻤﻮﻻ ﺍﺯ ﻧﻮﻉ ﺗﺴﻤﻪ ﺍﻧﺘﺨﺎﺏ ﻣﻲﺷﻮﻧﺪ ﻭﻟﻲ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﻧﺒﺸﻲ‪ ،‬ﻧﺎﻭﺩﺍﻧﻲ ﻭ ﻳﺎ‬ ‫ﺳﺎﻳﺮ ﻧﻴﻤﺮﺥﻫﺎﻱ ﻧﻮﺭﺩ ﺷﺪﻩ ﻧﻴﺰ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫‬ ‫ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺭﺍ ﺑﺎﻳﺪ ﻃﻮﺭﻱ ﻗﺮﺍﺭ ﺩﺍﺩ ﻛﻪ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺣﺪﺍﻛﺜﺮ ‪ L1/r‬ﻫﺮ ﻧﻴﻤﺮﺥ ﺩﺭ ﺑﻴﻦ‬ ‫ﺍﺗﺼﺎﻻﺕ ﻣﺮﺑﻮﻃﻪ ﺍﺯ ‪ 3 / 4‬ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺣﺎﻛﻢ ﻋﻀﻮ ﺑﻴﺸﺘﺮ ﻧﺸﻮﺩ‪،‬‬ ‫‬ ‫ﻣﻘﺎﻭﻣﺖ ﺑﺮﺷﻲ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﻋﻤﻮﺩ ﺑﺮ ﻣﺤﻮﺭ ﻋﻀﻮ ﻓﺸﺎﺭﻱ ﻣﺮﻛﺐ‪ ،‬ﺣﺪﺍﻗﻞ‬ ‫ﺑﺎﻳﺪ ‪ %2‬ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻃﺮﺍﺣﻲ ﻋﻀﻮ )‪ (ØcPn‬ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﻧﺴﺒﺖ ﻻﻏﺮﻱ ‪ L/r‬ﺑﺮﺍﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺳﺎﺩﻩ )ﺗﻜﻲ( ﻧﺒﺎﻳﺪ ﺑﻴﺸﺘﺮ ﺍﺯ ‪ 140‬ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﻧﺴﺒﺖ ﻻﻏﺮﻱ ‪ L/r‬ﺑﺮﺍﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺯﻭﺝ ﻧﺒﺎﻳﺪ ﺑﻴﺸﺘﺮ ﺍﺯ ‪ 200‬ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺯﻭﺝ ﺑﺎﻳﺪ ﺩﺭ ﻣﺤﻞ ﺗﻘﺎﻃﻊ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﺍﺗﺼﺎﻝ ﻳﺎﺑﻨﺪ‪.‬‬ ‫‪17‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﻮﺍﺑﻂ ﻃﺮﺍﺣﻲ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ )ﻗﻴﺪﻫﺎﻱ ﻣﻮﺭﺏ‪(Lacing ،‬‬ ‫‬ ‫ﻃﻮﻝ ‪ L‬ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺘﻲ ﻛﻪ ﺩﺭ ﻓﺸﺎﺭ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ﺭﺍ ﻣﻲﺗﻮﺍﻥ‬ ‫ﺑﺮﺍﺑﺮ ﺑﺎ ﻣﻘﺎﺩﻳﺮ ﺯﻳﺮ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺖ‪:‬‬ ‫¾ ﻃﻮﻝ ﺁﺯﺍﺩ ﺗﺴﻤﻪ ﺑﻴﻦ ﺟﻮﺷﻬﺎ ﻳﺎ ﭘﻴﭻﻫﺎﻱ ﺍﺗﺼﺎﻝ ﺑﻪ ﻧﻴﻤﺮﺥﻫﺎﻱ ﻋﻀﻮ ﻣﺮﻛﺐ )‪ ،(ℓ‬ﺑﺮﺍﻱ‬ ‫ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺳﺎﺩﻩ‪،‬‬ ‫¾ ‪ 70‬ﺩﺭﺻﺪ ﻣﻘﺪﺍﺭ ﻓﻮﻕ )‪ ،(0.70ℓ‬ﺑﺮﺍﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺯﻭﺝ‪.‬‬ ‫‬ ‫ﺯﺍﻭﻳﻪ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺑﺎ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ﻋﻀﻮ ﺍﺭﺟﺢ ﺍﺳﺖ ﻛﻤﺘﺮ ﺍﺯ ﻣﻘﺎﺩﻳﺮ ﺯﻳﺮ ﻧﺒﺎﺷﺪ‪:‬‬ ‫ ‪ 60‬ﺩﺭﺟﻪ‪ ،‬ﺑﺮﺍﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺳﺎﺩﻩ‪،‬‬ ‫ ‪ 45‬ﺩﺭﺟﻪ‪ ،‬ﺑﺮﺍﻱ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺯﻭﺝ‪.‬‬ ‫‬ ‫ﺍﮔﺮ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺧﻄﻮﻁ ﺟﻮﺵ ﻳﺎ ﭘﻴﭻ ﺩﻭ ﺳﺮ ﺑﺴﺖ ﺑﻴﺶ ﺍﺯ ‪ 380 mm‬ﺑﺎﺷﺪ ﺍﺭﺟﺢ ﺍﺳﺖ ﻛﻪ‬ ‫ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺯﻭﺝ ﺑﻜﺎﺭ ﺭﻭﻧﺪ ﻭ ﻳﺎ ﺍﻳﻨﻜﻪ ﺍﺯ ﻧﻴﻤﺮﺥ ﻧﺒﺸﻲ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪.‬‬ ‫‪18‬‬ ‫‪Urmia University, M. Sheidaii‬‬ :‫ﻣﺜﺎﻝ‬ .‫ ﻃﺮﺍﺣﻲ ﻧﻤﺎﻳﻴﺪ‬Pu=106 ton ‫ ﺑﺮﺍﻱ ﺗﺤﻤﻞ ﺑﺎﺭ ﺿﺮﻳﺒﺪﺍﺭ‬25 cm ‫ﺳﺘﻮﻥ ﻣﺮﻛﺒﻲ ﺍﺯ ﺩﻭﻧﺎﻭﺩﺍﻧﻲ ﺑﺎ ﻓﺎﺻﻠﻪ ﭘﺸﺖ ﺑﻪ ﭘﺸﺖ‬ :‫ﺣﻞ‬ :‫ﺍﻟﻒ( ﻃﺮﺍﺣﻲ ﻣﻘﻄﻊ‬ KL/r = 50 Pu=106 ton → ØcFcr = 1855 kgf/cm2 A t Pu/ ØcFcr= 106000 / 1855 = 53.91 cm2‫ﻻﺯﻡ‬ USE 2UNP200 A = 2(32.2) = 64.4 cm2 4.5 m Ix = 2(1910) = 3820 cm4 ← Iy = 2(148) + 2(32.2) (25/2-2.01)2 = 7383 cm4 rx = √Ix/A = √(3820/(2×32.2)) = 7.70 cm KL/r = 1× 450 / 7.7 = 58.43 → ØcFcr = 1773 kgf/cm2 ØcPn = 1773 × (2×32.2) = 114.2 ton > 106 ton O.K. Pu=106 ton Urmia University, M. Sheidaii 19 ‫ﺍﺩﺍﻣﻪ ﺣﻞ‪:‬‬ ‫ﺏ( ﻃﺮﺍﺣﻲ ﺑﺴﺖﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﭘﻴﭽﻲ)ﻗﻄﺮ ﭘﻴﭽﻬﺎ ‪:( ¾ in‬‬ ‫ﺑﺴﺖ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﺳﺎﺩﻩ ﻛﺎﻓﻲ ﺍﺳﺖ → ‪ 17 cm < 38 cm‬ﻓﺎﺻﻠﻪ ﺧﻄﻮﻁ ﭘﻴﭻ‬ ‫ﺯﺍﻭﻳﻪ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻫﺎ ﺑﺎ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ﻋﻀﻮ ‪ 60‬ﺩﺭﺟﻪ ﺍﻧﺘﺨﺎﺏ ﻣﻲ ﺷﻮﺩ‬ ‫ﻛﻨﺘﺮﻝ ﻻﻏﺮﻱ ﻧﻴﻤﺮﺥ ﺑﻴﻦ ﺍﺗﺼﺎﻻﺕ‪:‬‬ ‫‪ L = 17 / cos 30 = 19.6 cm‬ﻃﻮﻝ ﻧﺎﻭﺩﺍﻧﻲ ﺑﻴﻦ ﺑﺴﺖ ﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ‬ ‫‪L/r = 19.6 / 2.14 = 9.16 (3/4) (58.43)=43.82‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﺑﻌﺎﺩ ﻣﻘﻄﻊ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻫﺎ‪:‬‬ ‫‪ Vu= 0.02 ØcPn‬ﻣﻘﺎﻭﻣﺖ ﺑﺮﺷﻲ ﺣﺪﺍﻗﻞ ﺑﺴﺖ ﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ‬ ‫‪= 0.02(114.2) = 2282 kgf‬‬ ‫‪ Vu/2 = 1141 kgf‬ﻧﻴﺮﻭﻱ ﺑﺮﺷﻲ ﻣﻮﺛﺮ ﺑﺮ ﻫﺮ ﺻﻔﺤﻪ ﭼﭗ ﻭ ﺭﺍﺳﺖ‬ ‫)‪F= Vu/2sinD =1141/(17/19.6‬‬ ‫ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﻣﻮﺛﺮ ﺑﺮ ﺑﺴﺖ ﭼﭗ ﻭ ﺭﺍﺳﺖ‬ ‫‪=1315 kgf‬‬ ‫‪20‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﺩﺍﻣﻪ ﻃﺮﺍﺣﻲ ﺍﺑﻌﺎﺩ ﻣﻘﻄﻊ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻫﺎ‪:‬‬ ‫‪A=bt‬‬ ‫‪r=√(I/A)=0.289t‬‬ ‫‪ I = bt3/12‬ﻣﺸﺨﺼﺎﺕ ﺍﺑﻌﺎﺩﻱ ﺑﺴﺖ ﻫﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ‬ ‫ﺑﺎ ﻓﺮﺽ ﺍﻳﻨﻜﻪ ﺣﺪﺍﻛﺜﺮ ﻻﻏﺮﻱ ﺑﺴﺘﻬﺎﻱ ﭼﭗ ﻭ ﺭﺍﺳﺖ ‪ 140‬ﺑﺎﺷﺪ‪:‬‬ ‫‪t=0.484 cm‬‬ ‫‪19.6/0.289t = 140‬‬ ‫‪L/r=140‬‬ ‫‪ t=0.5 cm‬ﺍﻧﺘﺨﺎﺏ ﺷﺪ‬ ‫‪L/r=19.6/(0.289(0.5))=136 → ØcFcr = 841 kgf/cm2‬‬ ‫)‪ A t F/ ØcFcr= 1315 / 841 = 1.56 cm2 → (32 × 5 mm‬ﻻﺯﻡ‬ ‫ﻓﺎﺻﻠﻪ ﻻﺯﻡ ﺑﺮﺍﻱ ﭘﻴﭻ ¾ ﺍﻳﻨﭻ ﺍﺯ ﻟﺒﻪ ﻭﺭﻕ ‪ 32 mm‬ﺍﺳﺖ ﻟﺬﺍ‪:‬‬ ‫‪ t 19.6 + 2 (3.2) = 26 cm‬ﻃﻮﻝ ﻫﺮ ﺗﺴﻤﻪ ﭼﭗ ﻭ ﺭﺍﺳﺖ‬ ‫‪ PL260 × 32 × 5 mm‬ﺍﻧﺘﺨﺎﺏ ﺷﺪ‪.‬‬ ‫ﻃﺮﺍﺣﻲ ﻭﺭﻗﻬﺎﻱ ﺍﻧﺘﻬﺎﻳﻲ ﻗﻄﻌﻪ )ﺑﺴﺖ ﻫﺎﻱ ﺍﻓﻘﻲ( ‪:‬‬ ‫‪= 17 cm‬‬ ‫ﺣﺪﺍﻗﻞ ﻃﻮﻝ‬ ‫‪t = 17 /50 =3.4 cm‬ﺣﺪﺍﻗﻞ ﺿﺨﺎﻣﺖ‬ ‫‪= 17 + 2(3.2) = 23.4 cm‬‬ ‫‪ PL240 × 170 × 4 mm‬ﺍﻧﺘﺨﺎﺏ ﺷﺪ‪.‬‬ ‫ﺣﺪﺍﻗﻞ ﻋﺮﺽ‬ ‫ﺳﺘﻮﻥ ﺑﺎ ﺑﺴﺖ ﻫﺎﻱ ﻣﻮﺍﺯﻱ‬ Urmia University, M. Sheidaii 22 Urmia University, M. Sheidaii 23 Urmia University, M. Sheidaii 24 25 Introduction to B E A M S ‫ ﻣﻘﺪﻣﻪﺍﻱ ﺑﺮ ﺗﻴﺮﻫﺎ‬:‫ ﻗﺴﻤﺖ ﺍﻭﻝ‬-‫ﻓﺼﻞ ﭘﻨﺠﻢ‬ Urmia University, M. Sheidaii 1 ‫ﻣﻘﺪﻣﻪﺍﻱ ﺑﺮ ﺗﻴﺮﻫﺎ‬ ‫ ﻣﻌﻤﻮﻻ ﺑﻪ ﻋﻀﻮﻱ ﻛﻪ ﺗﺤﺖ ﺍﺛﺮ ﺑﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻗﺮﺍﺭ ﮔﻴﺮﺩ ﺗﻴﺮ ﻣﻲﮔﻮﻳﻨﺪ‪.‬‬ ‫ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﻋﻼﻭﻩ ﺑﺮ ﺑﺎﺭ ﺟﺎﻧﺒﻲ‪ ،‬ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﻧﻴﺰ ﻣﻮﺟﻮﺩ ﺑﺎﺷﺪ ﻋﻀﻮ‬ ‫ﻣﺰﺑﻮﺭ ﺭﺍ ﺗﻴﺮﺳﺘﻮﻥ ﻧﺎﻣﻨﺪ‪.‬‬ ‫ ﺗﻴﺮﻫﺎ ﺭﺍ ﺍﻋﻀﺎﻱ ﺧﻤﺸﻲ ‪ flexure members‬ﻧﻴﺰ ﻣﻲﻧﺎﻣﻨﺪ‪ ،‬ﻃﺮﺍﺣﻲ‬ ‫ﺗﻴﺮﻫﺎ ﺑﺮ ﺍﺳﺎﺱ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺻﻮﺭﺕ ﻣﻲﮔﻴﺮﺩ‪.‬‬ ‫‪2‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻧﻮﺍﻉ ﺗﻴﺮﻫﺎ‬ ‫ ﺗﻴﺮﻫﺎﻱ ﻓﺮﻋﻲ)‪ : (Joists‬ﺗﻴﺮﻫﺎﻳﻲ ﺑﺎ ﻓﻮﺍﺻﻞ ﻛﻢ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﻛﻪ ﺑﺎﺭ ﻛﻒﻫﺎ ﻳﺎ ﺳﻘﻒﻫﺎﻱ‬ ‫ﺳﺎﺧﺘﻤﺎﻥ ﺭﺍ ﺑﻪ ﺷﺎﻫﺘﻴﺮﻫﺎ ﺣﻤﻞ ﻣﻲﻛﻨﻨﺪ‪.‬‬ ‫ ﺗﻴﺮ ﺍﺻﻠﻲ)‪ :(Spandrel‬ﺗﻴﺮﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﻟﺒﻪﻫﺎﻱ ﺧﺎﺭﺟﻲ ﺳﺎﺧﺘﻤﺎﻥ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﻭ ﺑﺎﺭ‬ ‫ﺩﻳﻮﺍﺭﻫﺎﻱ ﺧﺎﺭﺟﻲ‪ ،‬ﺑﺨﺸﻲ ﺍﺯ ﺑﺎﺭ ﻛﻒ ﻭ ﻳﺎ ﻭﺭﻭﺩﻱ ﺳﺎﺧﺘﻤﺎﻥ ﺭﺍ ﺗﺤﻤﻞ ﻣﻲﻛﻨﻨﺪ‪ .‬ﭼﻨﻴﻦ ﻋﻀﻮﻱ‬ ‫ﻣﻲﺗﻮﺍﻧﺪ ﻳﻚ ﺗﻴﺮ ﺍﺻﻠﻲ‪ ،‬ﻭ ﻳﺎ ﺗﻴﺮﻱ ﺍﺻﻠﻲ ﺍﺯ ﻧﻮﻉ ﺷﺎﻫﺘﻴﺮ ﺑﺎﺷﺪ‪.‬‬ ‫ ﺷﺎﻫﺘﻴﺮ)‪ :(Girder‬ﻋﻀﻮﻱ ﻛﻪ ﺑﺎﺭ ﺗﻴﺮﻫﺎﻱ ﺩﻳﮕﺮ ﺭﺍ ﺗﺤﻤﻞ ﻛﺮﺩﻩ ﻭ ﺑﻪ ﺳﺘﻮﻧﻬﺎ ﻣﻨﺘﻘﻞ ﻣﻲﻧﻤﺎﻳﺪ‪.‬‬ ‫‪3‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻧﻮﺍﻉ ﺗﻴﺮﻫﺎ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪4‬‬ ‫ﻧﻌﻞ ﺩﺭﮔﺎﻩ )‪ :(Lintel‬ﺗﻴﺮﻱ ﻛﻪ ﺭﻭﻱ ﺑﺎﺯﺷﻮ ﺩﻳﻮﺍﺭ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﻭ ﺑﺎﺭ ﺩﻳﻮﺍﺭ ﺭﻭﻱ ﺑﺎﺯﺷﻮ ﺭﺍ‬ ‫ﺗﺤﻤﻞ ﻣﻲﻛﻨﺪ‪.‬‬ ‫ﻻﭘﻪ )‪ : (Purlin‬ﺗﻴﺮﻱ ﻛﻪ ﺑﺎﺭ ﭘﻮﺷﺶ ﺳﻘﻒ ﺩﺭ ﺳﺎﺧﺘﻤﺎﻧﻬﺎﻱ ﺻﻨﻌﺘﻲ ﺭﺍ ﺗﺤﻤﻞ ﻛﺮﺩﻩ‬ ‫ﻭ ﺑﻪ ﺧﺮﭘﺎ ﻳﺎ ﻗﺎﺏ ﺍﺻﻠﻲ ﺳﺎﺧﺘﻤﺎﻥ ﺻﻨﻌﺘﻲ ﺍﻧﺘﻘﺎﻝ ﻣﻲﺩﻫﺪ‪.‬‬ ‫ﺗﻴﺮ ﺷﻴﺮﻭﺍﻧﻲ‪ ،‬ﺗﻴﺮ ﺷﻴﺐ)‪ :(Rafter‬ﻋﻀﻮﻱ ﻏﺎﻟﺒﺎ ﻣﻮﺭﺏ ﻛﻪ ﺑﺎﺭ ﻻﭘﻪﻫﺎ ﻭ ﺗﻴﺮﻫﺎﻱ ﻓﺮﻋﻲ‬ ‫ﺭﺍ ﺗﺤﻤﻞ ﻣﻲﻛﻨﺪ‪.‬‬ ‫ﺗﻴﺮ ﻃﻮﻟﻲ)‪ :(Stinger‬ﺗﻴﺮﻫﺎﻱ ﻃﻮﻟﻲ ﺩﺭ ﻛﻪ ﺑﻪ ﻣﻮﺍﺯﺍﺕ ﻃﻮﻝ ﭘﻞ ﻗﺮﺍﺭ ﺩﺍﺷﺘﻪ ﻭ ﺑﺎﺭ‬ ‫ﻋﺒﻮﺭﮔﺎﻩ ﭘﻞ ﺭﺍ ﺗﺤﻤﻞ ﻣﻲﻛﻨﻨﺪ‪ ،‬ﻋﻤﻮﺩ ﺑﺮ ﺍﻳﻦ ﺗﻴﺮﻫﺎ ‪ ،‬ﺗﻴﺮﻫﺎﻱ ﻛﻒ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ﻛﻪ ﺑﺎﺭ‬ ‫ﻋﺒﻮﺭﮔﺎﻩ ﺭﺍ ﺍﺯ ﺗﻴﺮﻫﺎﻱ ﻃﻮﻟﻲ ﺑﻪ ﺷﺎﻫﺘﻴﺮﻫﺎﻱ ﭘﻞ ﻳﺎ ﺧﺮﭘﺎﻫﺎ ﻣﻨﺘﻘﻞ ﻣﻲﻧﻤﺎﻳﻨﺪ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺭﮔﺬﺍﺭﻱ ﺗﻴﺮﻫﺎ‬ Urmia University, M. Sheidaii 5 Urmia University, M. Sheidaii 6 Urmia University, M. Sheidaii 7 ‫ﻣﻘﺎﻃﻊ ﻣﺘﺪﺍﻭﻝ ﺑﺮﺍﻱ ﺗﻴﺮﻫﺎ‬ ‫‪Sections Used as Beams‬‬ ‫ ﻣﻘﺎﻃﻊ ‪ I‬ﺷﻜﻞ ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ ﻭ ﺍﻗﺘﺼﺎﺩﻳﺘﺮﻳﻦ ﻣﻘﺎﻃﻊ ﺗﻴﺮ ﻫﺴﺘﻨﺪ‪.‬‬ ‫ ﻣﻘﺎﻃﻊ ‪ IPE‬ﺩﺍﺭﺍﻱ ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺑﺎﻻﻳﻲ ﺑﻮﺩﻩ ﻭ ﻧﺴﺒﺖ ﺑﻪ ﺳﺎﻳﺮ ﻧﻴﻤﺮﺧﻬﺎ ﺩﺍﺭﺍﻱ‬ ‫ﺭﺍﻧﺪﻣﺎﻥ ﻣﻘﺎﻭﻣﺘﻲ )ﻧﺴﺒﺖ ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺑﻪ ﻭﺯﻥ ﻭﺍﺣﺪ ﻃﻮﻝ( ﺑﻴﺸﺘﺮﻱ ﻫﺴﺘﻨﺪ‪.‬‬ ‫ ﻣﻘﺎﻃﻊ ‪ INP‬ﻛﻪ ﺩﺍﺭﺍﻱ ﺷﻴﺐ ﺩﺭ ﺳﻄﺢ ﺩﺍﺧﻠﻲ ﺑﺎﻝ ﻫﺴﺘﻨﺪ )ﻧﻈﻴﺮ ﻣﻘﺎﻃﻊ ‪S‬‬ ‫ﺍﻣﺮﻳﻜﺎﻳﻲ( ﺩﺍﺭﺍﻱ ﺭﺍﻧﺪﻣﺎﻥ ﻣﻘﺎﻭﻣﺘﻲ ﻛﻤﺘﺮ ﻭ ﺍﺳﺘﻔﺎﺩﻩ ﻣﺤﺪﻭﺩﻱ ﻣﻲﺑﺎﺷﻨﺪ‪.‬‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻳﻦ ﻣﻘﺎﻃﻊ ﺯﻣﺎﻧﻲ ﻣﻄﻠﻮﺑﻴﺖ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ ﻛﻪ‪:‬‬ ‫ ﻋﺮﺽ ﻛﻢ ﺑﺎﻝ ﻣﻄﻠﻮﺏ ﺑﺎﺷﺪ‪،‬‬ ‫ ﻭ ﻳﺎ ﻧﻴﺮﻭﻱ ﺑﺮﺷﻲ ﺑﺎﻻﻳﻲ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ‪،‬‬ ‫ ﻭ ﻳﺎ ﻧﻴﺎﺯ ﺑﻪ ﺿﺨﺎﻣﺖ ﺯﻳﺎﺩ ﺑﺎﻝ ﺩﺭ ﻣﺠﺎﻭﺭﺕ ﺟﺎﻥ ﺑﺎﺷﺪ )ﻣﺜﻼ ﺩﺭ ﺭﻳﻠﻬﺎﻱ‬ ‫ﺟﺮﺛﻘﻴﻞ(‬ ‫‪8‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻃﻊ ﻣﺘﺪﺍﻭﻝ ﺑﺮﺍﻱ ﺗﻴﺮﻫﺎ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪9‬‬ ‫‪Sections Used as Beams‬‬ ‫ﻣﻘﺎﻃﻊ ﻧﺎﻭﺩﺍﻧﻲ ﺩﺍﺭﺍﻱ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻤﻲ ﺑﻮﺩﻩ ﻟﻴﻜﻦ ﺩﺭ ﻣﻮﺍﺭﺩ ﺯﻳﺮ ﻛﺎﺭﺑﺮﺩ ﺩﺍﺭﻧﺪ‪:‬‬ ‫ ﺗﻴﺮﻫﺎﻳﻲ ﻛﻪ ﺑﺎﺭ ﺳﺒﻜﻲ ﺗﺤﻤﻞ ﻣﻲﻛﻨﻨﺪ ﻧﻈﻴﺮ ﻻﭘﻪﻫﺎ‪،‬‬ ‫ ﺯﻣﺎﻧﻲ ﻛﻪ ﺑﺎﺭ ﺗﻴﺮ ﺑﺎﻳﺪ ﻛﻢ ﻋﺮﺽ ﺑﺎﺷﺪ‪.‬‬ ‫ﻣﻘﺎﻃﻊ ‪ Z‬ﺷﻜﻞ ﺩﺭ ﺑﺎﻣﻬﺎﻱ ﺷﻴﺒﺪﺍﺭ ﺑﻪ ﻋﻨﻮﺍﻥ ﻻﭘﻪ ﺳﻘﻒ ﺑﺮ ﻣﻘﺎﻃﻊ ‪ I‬ﺷﻜﻞ‬ ‫ﺍﺭﺟﺤﻴﺖ ﺩﺍﺭﻧﺪ ﺯﻳﺮﺍ ﻧﻴﺮﻭﻱ ﻭﺍﺭﺩ ﺑﺮ ﻻﭘﻪ ‪ Z‬ﺷﻜﻞ ﺗﻘﺮﻳﺒﺎ ﺍﺯ ﻣﺮﻛﺰ ﺑﺮﺵ ﻋﺒﻮﺭ‬ ‫ﻣﻲﻧﻤﺎﻳﺪ‪.‬‬ ‫ﻣﻌﻤﻮﻻ ﻧﺎﻭﺩﺍﻧﻲﻫﺎ ﻭ ﻧﻴﻤﺮﺧﻬﺎﻱ ‪ Z‬ﺷﻜﻞ ﻣﻘﺎﻭﻣﺖ ﻛﻤﻲ ﺩﺭ ﺑﺮﺍﺑﺮ ﺑﺎﺭﻫﺎﻱ ﻋﻤﻮﺩ ﺑﺮ‬ ‫ﺟﺎﻥ ﺧﻮﺩ ﺩﺍﺭﻧﺪ ﻟﺬﺍ ﺑﺮﺍﻱ ﭘﺎﻳﺪﺍﺭﺳﺎﺯﻱ ﺁﻧﻬﺎ ﺍﺯ ﻣﻴﻞ ﻣﻬﺎﺭ )‪ (sag rod‬ﺍﺳﺘﻔﺎﺩﻩ‬ ‫ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﻣﻘﺎﻃﻊ ﻣﺮﻛﺐ‪ ،‬ﺗﻴﺮﻫﺎﻱ ﻻﻧﻪ ﺯﻧﺒﻮﺭﻱ ﻭ ﺗﻴﺮﻫﺎﻱ ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺍﺯ ﻣﻴﻠﮕﺮﺩ ﺍﺯ ﺟﻤﻠﻪ‬ ‫ﺳﺎﻳﺮ ﺍﻧﻮﺍﻉ ﻣﻘﺎﻃﻊ ﻣﺘﺪﺍﻭﻝ ﺑﺮﺍﻱ ﺗﻴﺮﻫﺎ ﻫﺴﺘﻨﺪ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪Design Beams‬‬ ‫ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺗﻴﺮ ﻣﺎﻧﻨﺪ ﻳﻚ ﺳﺘﻮﻥ ﻋﻤﻞ ﻣﻲﻛﻨﺪ ﻭ ﺩﺭ ﺻﻮﺭﺕ ﻋﺪﻡ ﺗﺎﻣﻴﻦ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ‬ ‫ﻣﻤﻜﻦ ﺍﺳﺖ ﻛﻤﺎﻧﻪ ﻛﻨﺪ‪.‬‬ ‫ﭘﺪﻳﺪﻩ ﻛﻤﺎﻧﺶ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺗﻴﺮ ﺭﺍ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ )‪(lateral-torsional buckling‬‬ ‫ﻣﻲﻧﺎﻣﻨﺪ ﺑﺮﻭﺯ ﺍﻳﻦ ﭘﺪﻳﺪﻩ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻋﻀﻮ ﺗﻴﺮ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﺗﻌﺒﻴﻪ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﺑﺎﻋﺚ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﻣﻲﺷﻮﺩ)ﺑﻪ ﻭﺍﺳﻄﻪ ﻛﺎﻫﺶ‬ ‫ﻃﻮﻝ ﻛﻤﺎﻧﺶ(‪.‬‬ ‫ﻳﻚ ﺭﻭﺵ ﺑﺮﺍﻱ ﻣﻬﺎﺭ ﻧﻤﻮﺩﻥ ﺗﻴﺮ‪ ،‬ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺁﻥ ﺩﺭ ﺗﻮﺩﻩ ﺗﻮﭘﺮ ﻣﻲﺑﺎﺷﺪ )ﻣﺜﻼ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺗﻴﺮﭼﻪ ﻭ‬ ‫ﺑﻠﻮﻙ ﺑﻴﻦ ﺩﻭ ﺗﻴﺮ ﻭ ﺍﻧﺠﺎﻡ ﺑﺘﻦﺭﻳﺰﻱ(‬ ‫ﺑﺮﺍﻱ ﻣﻘﺎﻭﻡ ﻧﻤﻮﺩﻥ ﺗﻴﺮ ﺩﺭ ﺑﺮﺍﺑﺮ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﻣﻘﺎﻃﻊ ﺑﺎ ﺑﺎﻟﻬﺎﻱ ﭘﻬﻦﺗﺮ ﻭ ﻳﺎ‬ ‫ﻣﻘﺎﻃﻊ ﺑﺴﺘﻪ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ‪.‬‬ ‫‪10‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫ﺭﺍﺑﻄﻪ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻭ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ‬ ‫ﺗﻴﺮ ﺩﺭ ﺷﻜﻞ ﻣﻘﺎﺑﻞ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﺗﺎﻣﻴﻦ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﺧﺮﺍﺑﻲ ﺗﻴﺮ ﺑﺎ ﺗﺴﻠﻴﻢ‬ ‫ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﻫﻤﺮﺍﻩ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﻋﺪﻡ ﺗﺎﻣﻴﻦ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﺧﺮﺍﺑﻲ ﺗﻴﺮ ﺑﺎ‬ ‫ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﻫﻤﺮﺍﻩ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎ ﺑﺮ ﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ‬ ‫‪ AISC Chapter F p.16.1-44‬ﺍﻧﺠﺎﻡ ﻣﻲﮔﻴﺮﺩ‪.‬‬ ‫ﻣﻼﺣﻈﺎﺕ ﻃﺮﺍﺣﻲ ﺩﺭ ﺁﻳﻴﻦﻧﺎﻣﻪ ‪AISC‬‬ ‫‬ ‫‬ ‫‬ ‫‪11‬‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮﺍﻱ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ‬ ‫ﺍﻟﺒﺘﻪ ﺣﺎﻻﺕ ﺣﺪﻱ ﺩﻳﮕﺮﻱ ﻧﻈﻴﺮﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺑﺎﻝ‪،‬‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺟﺎﻥ ﻭ ﺗﺴﻠﻴﻢ ﻛﺸﺸﻲ ﺑﺎﻝ ﺍﺯ ﺩﻳﮕﺮ‬ ‫ﻣﻼﺣﻈﺎﺕ ﻣﻄﺮﺡ ﺩﺭ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎ ﻣﻲ ﺑﺎﺷﻨﺪ‪.‬‬ ‫‪Design Beams‬‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫ ﺭﺍﻫﻨﻤﺎﻱ ﻛﺎﺭﺑﺮﺩ ﺑﺨﺶ ﻫﺎﻱ ﻣﺨﺘﻠﻒ ﻓﺼﻞ‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﺧﻤﺸﻲ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC‬ﺩﺭ‬ ‫‪AISC Chapter F Table F1. p.16.1-45‬‬ ‫ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪12‬‬ ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ‬ ‫ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺗﻴﺮ ﺩﺍﺭﺍﻱ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﺑﺎﺷﺪ ﺑﺎ ﺗﺴﻠﻴﻢ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺗﻴﺮ ﺧﺮﺍﺏ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮ ﺍﺳﺎﺱ ﻟﻨﮕﺮ ﺧﻤﻴﺮﻱ ﻣﻘﻄﻊ ﺗﻌﻴﻴﻦ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﻣﻮﺟﻮﺩ ﻃﺒﻖ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC‬ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﺤﺎﺳﺒﻪ ﻣﻲﺷﻮﺩ‪:‬‬ ‫)‪(LRFD method‬‬ ‫)‪(ASD method‬‬ ‫‪Available Design Flexural Strength = Ib Mn‬‬ ‫‪Allowable Design Flexural Strength = Mn / :b‬‬ ‫‪:b = 1.67‬‬ ‫;‬ ‫‪Ib = 0.9‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ = ‪Mn = nominal flexural moment‬‬ ‫ﻟﻨﮕﺮ ﺧﻤﻴﺮﻱ = ‪= Mp = Plastic moment‬‬ ‫‪= Fy Zx‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ = ‪Zx = plastic section modulus‬‬ ‫‪13‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻨﺸﻬﺎﻱ ﺧﻤﺸﻲ‬ ‫‪Bending Stresses‬‬ ‫ﺑﺮﺭﺳﻲ ﺭﻓﺘﺎﺭ ﺗﻴﺮﻱ ﺑﺎ ﻣﻘﻄﻊ ﻣﺴﺘﻄﻴﻠﻲ‪ ،‬ﻛﻪ ﺑﺨﺶ ﻓﺸﺎﺭﻱ ﺁﻥ ﺩﺍﺭﺍﻱ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻛﺎﻣﻞ ﺍﺳﺖ‪،‬‬ ‫ﺗﺤﺖ ﺍﺛﺮ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻭﺍﺭﺩﻩ‪:‬‬ ‫ ﺗﻨﺶ ﺩﺭ ﺗﻴﺮ ﺩﺭ ﺣﻮﺯﻩ ﺍﺭﺗﺠﺎﻋﻲ )‪ (elastic limit‬ﺑﻪ ﺻﻮﺭﺕ ﺧﻄﻲ ﺗﻐﻴﻴﺮ ﻛﺮﺩﻩ ﻭ ﻣﻘﺪﺍﺭ ﺁﻥ‬ ‫ﺩﺭ ﺗﺎﺭ ﺧﺎﺭﺟﻲ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‪:‬‬ ‫‪fb = M c / I = M / S‬‬ ‫ﺗﻨﺶ ﺧﻤﺸﻲ = ‪fb = Bending Stress‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺍﺭﺗﺠﺎﻋﻲ = ‪S = I /c = Section Modulus‬‬ ‫‪14‬‬ ‫ﺗﻨﺸﻬﺎﻱ ﺧﻤﺸﻲ‬ ‫‪Bending Stresses‬‬ ‫ ﺗﻐﻴﻴﺮﺍﺕ ﺧﻄﻲ ﺗﻨﺶ ﺗﺎ ﻫﻨﮕﺎﻣﻲ ﻛﻪ ﺗﺎﺭ ﺧﺎﺭﺟﻲ ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺑﺮﺳﺪ ﺑﺮﻗﺮﺍﺭ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﻟﻨﮕﺮﻱ ﻛﻪ ﺩﺭ ﺁﻥ ﺗﺎﺭ ﺧﺎﺭﺟﻲ ﻣﻘﻄﻊ ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺑﺮﺳﺪ ﻟﻨﮕﺮ ﺗﺴﻠﻴﻢ ‪،Yield Moment‬‬ ‫‪ ،My‬ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ‪:‬‬ ‫‪My = S Fy‬‬ ‫ ﺩﺭ ﺻﻮﺭﺕ ﺍﻓﺰﺍﻳﺶ ﻟﻨﮕﺮ ﻭﺍﺭﺩﻩ ﺍﺯ ﻟﻨﮕﺮ ﺗﺴﻠﻴﻢ‪ ،‬ﺩﺭ ﺗﻴﺮﻱ ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺍﺯ ﻓﻮﻻﺩ ﻣﻌﻤﻮﻟﻲ‪:‬‬ ‫ ﺗﺎﺭﻫﺎﻱ ﺧﺎﺭﺟﻲ ﺗﻴﺮ ﻛﻪ ﻗﺒﻼ ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺭﺳﻴﺪﻩﺍﻧﺪ ﺩﺭ ﻫﻤﺎﻥ ﺗﻨﺶ ﺑﺎﻗﻲ ﻣﺎﻧﺪﻩ ﻭﻟﻲ‬ ‫ﺍﺯﺩﻳﺎﺩ ﻃﻮﻝ ﭘﻴﺪﺍ ﺧﻮﺍﻫﻨﺪ ﻛﺮﺩ‪،‬‬ ‫‬ ‫‬ ‫ﺗﺎﺭﻫﺎﻱ ﺩﻳﮕﺮ ﻣﻘﻄﻊ ﻭﻇﻴﻔﻪ ﺗﺎﻣﻴﻦ ﻟﻨﮕﺮ ﺍﺿﺎﻓﻲ ﻣﻘﺎﻭﻡ ﺭﺍ ﻋﻬﺪﻩﺩﺍﺭ ﺧﻮﺍﻫﻨﺪ ﺷﺪ‪،‬‬ ‫ﺗﺎﺭﻫﺎﻱ ﺑﻴﺸﺘﺮﻱ ﺍﺯ ﻣﻘﻄﻊ ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﻟﻨﮕﺮ ﺟﺎﺭﻱ ﺧﻮﺍﻫﻨﺪ ﺷﺪ‪.‬‬ ‫ ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﻟﻨﮕﺮ ﻭﺍﺭﺩﻩ ﺩﺭ ﻧﻬﺎﻳﺖ ﻛﻞ ﻣﻘﻄﻊ ﺟﺎﺭﻱ ﺷﺪﻩ‪ ،‬ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ‪ Plastic Hinge‬ﺩﺭ‬ ‫ﺗﻴﺮ ﺍﻳﺠﺎﺩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺑﺎ ﺗﺸﻜﻴﻞ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺗﻴﺮ ﻗﺎﺩﺭ ﺑﻪ ﺗﺤﻤﻞ ﻟﻨﮕﺮ ﺍﺿﺎﻓﻲ ﻧﺒﻮﺩﻩ ﻭ ﺗﺤﺖ ﻛﻮﭼﻜﺘﺮﻳﻦ ﻟﻨﮕﺮ‬ ‫ﺍﺿﺎﻓﻲ ﺩﺭ ﻣﻘﻄﻊ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺩﻭﺭﺍﻥ ﺧﻮﺍﻫﺪ ﻛﺮﺩ‪.‬‬ ‫ﺗﻨﺸﻬﺎﻱ ﺧﻤﺸﻲ‬ ‫‪Bending Stresses‬‬ ‫ ﻟﻨﮕﺮﻱ ﻛﻪ ﺩﺭ ﺍﺛﺮ ﺁﻥ ﻛﻞ ﻣﻘﻄﻊ ﺟﺎﺭﻱ ﺷﺪﻩ ﻭ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺗﺸﻜﻴﻞ ﻣﻲﺷﻮﺩ ﻟﻨﮕﺮ ﺧﻤﻴﺮﻱ‬ ‫‪ ، Mp ، Plastic Moment‬ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ‪:‬‬ ‫‪Mp = Z Fy‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ = ‪Z = Plastic Section Modulus‬‬ ‫ ﻧﺴﺒﺖ ﻟﻨﮕﺮ ﺧﻤﻴﺮﻱ ﺑﻪ ﻟﻨﮕﺮ ﺗﺴﻠﻴﻢ‪ ،‬ﺿﺮﻳﺐ ﺷﻜﻞ ‪ Shape Factor‬ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ‪:‬‬ ‫‪Mp / My = Z Fy / S Fy = Z/S‬‬ ‫ ﺿﺮﻳﺐ ﺷﻜﻞ ﺑﺮﺍﻱ ﻣﻘﻄﻊ ﻣﺴﺘﻄﻴﻠﻲ ﺑﺮﺍﺑﺮ ﺑﺎ ‪ 1.5‬ﺍﺳﺖ‪.‬‬ ‫‪16‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ‬ ‫‪Plastic Hinge‬‬ ‫ﺑﺮﺭﺳﻲ ﺗﺸﻜﻴﻞ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺩﺭ ﺗﻴﺮ ﺳﺎﺩﻩ ﺗﺤﺖ ﺑﺎﺭ ﻣﺘﻤﺮﻛﺰ ﺩﺭ ﻭﺳﻂ ﺗﻴﺮ‪:‬‬ ‫ ﺩﺭ ﺍﺛﺮ ﺍﻓﺰﺍﻳﺶ ﺗﺪﺭﻳﺠﻲ ﺑﺎﺭ‪ ،‬ﺑﺎ ﺭﺳﻴﺪﻥ ﺗﻨﺶ ﺩﺭ ﺗﺎﺭ ﺧﺎﺭﺟﻲ ﻣﻘﻄﻊ ﺗﻴﺮ ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ‪،‬‬ ‫ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺑﻪ ﻟﻨﮕﺮ ﺗﺴﻠﻴﻢ‪ ،My ،‬ﻣﻲ ﺭﺳﺪ‪.‬‬ ‫ ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﻣﺠﺪﺩ ﺑﺎﺭ‪ ،‬ﺗﺎﺭﻫﺎﻱ ﺧﺎﺭﺟﻲ ﺩﻳﮕﺮﻱ ﺍﺯ ﺗﻴﺮ ﺷﺮﻭﻉ ﺑﻪ ﺗﺴﻠﻴﻢ ﻧﻤﻮﺩﻩ ﻭ ﺧﻤﻴﺮﻱ‬ ‫ﺷﺪﻥ ﺗﻴﺮ ﺑﻪ ﻣﻘﺎﻃﻊ ﺩﻳﮕﺮﻱ ﻏﻴﺮ ﺍﺯ ﻣﻘﻄﻊ ﻟﻨﮕﺮ ﺣﺪﺍﻛﺜﺮ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﮔﺴﺘﺮﺵ ﻣﻲﻳﺎﺑﺪ‪.‬‬ ‫ ﺩﺭ ﻃﻲ ﺍﻓﺰﺍﻳﺶ ﺑﺎﺭ‪ ،‬ﺗﺎﺭﻫﺎﻱ ﺩﺍﺧﻠﻲ ﻧﻴﺰ ﺩﺭ ﻣﻘﻄﻊ ﻟﻨﮕﺮ ﺣﺪﺍﻛﺜﺮ‪ ،‬ﺑﺘﺪﺭﻳﺞ ﺟﺎﺭﻱ‬ ‫ﻣﻲﺷﻮﻧﺪ ﺗﺎ ﺍﻳﻨﻜﻪ ﻛﻞ ﺍﺭﺗﻔﺎﻉ ﻧﻴﻤﺮﺥ ﺩﺭ ﺁﻥ ﻣﻘﻄﻊ ﺟﺎﺭﻱ ﺷﺪﻩ ﻭ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﻃﺒﻖ‬ ‫ﺷﻜﻞ ﺍﻳﺠﺎﺩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪17‬‬ ‫ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ‬ ‫‪Plastic‬‬ ‫‪Hinge‬‬ ‫ ﻃﻮﻝ ﻧﺎﺣﻴﻪ ﺟﺎﺭﻱ ﺷﺪﻩ ﺗﻴﺮ ﺑﺴﺘﮕﻲ ﺑﻪ ﻭﺿﻌﻴﺖ ﺑﺎﺭﮔﺬﺍﺭﻱ ﻭ ﺷﻜﻞ ﻧﻴﻤﺮﺥ ﺩﺍﺭﺩ‪:‬‬ ‫‬ ‫‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻭﺍﺭﺩﻩ ﻣﺘﻤﺮﻛﺰ ﺑﻮﺩﻩ ﻭ ﺷﻜﻞ ﻣﻘﻄﻊ ﻣﺴﺘﻄﻴﻠﻲ ﺑﺎﺷﺪ ﻃﻮﻝ ﻗﺴﻤﺖ ﺧﻤﻴﺮﻱ ﺷﺪﻩ ﺗﻴﺮ‬ ‫ﺑﻪ ‪ 1 ⁄ 3‬ﻃﻮﻝ ﺩﻫﺎﻧﻪ ﺗﻴﺮ ﻣﻲﺭﺳﺪ‪،‬‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻭﺍﺭﺩﻩ ﻣﺘﻤﺮﻛﺰ ﺑﻮﺩﻩ ﻭ ﺷﻜﻞ ﻣﻘﻄﻊ ‪ IPE‬ﻭ ‪ W‬ﺑﺎﺷﺪ ﻃﻮﻝ ﻗﺴﻤﺖ ﺧﻤﻴﺮﻱ ﺷﺪﻩ‬ ‫ﺗﻴﺮ ﺑﻪ ‪ 1 ⁄ 8‬ﻃﻮﻝ ﺩﻫﺎﻧﻪ ﺗﻴﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ ﺍﮔﺮ ﭼﻪ ﺍﺛﺮ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺩﺭ ﻃﻮﻟﻲ ﺍﺯ ﺩﻫﺎﻧﻪ ﺗﻴﺮ ﮔﺴﺘﺮﺵ ﻣﻲﻳﺎﺑﺪ ﻭﻟﻲ ﻓﺮﺽ ﻣﻲﺷﻮﺩ‬ ‫ﻛﻪ ﺍﻳﻦ ﻣﻔﺼﻞ ﻓﻘﻂ ﺩﺭ ﻳﻚ ﻣﻘﻄﻊ ﺍﺯ ﺗﻴﺮ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ‪.‬‬ ‫ ﻃﻮﻝ ﻧﺎﺣﻴﻪ ﺧﻤﻴﺮﻱ‪ ،‬ﺩﺭ ﻣﺤﺎﺳﺒﻪ ﺗﻐﻴﻴﺮﻣﻜﺎﻧﻬﺎ ﻭ ﻃﺮﺍﺣﻲ ﻣﻬﺎﺭﻫﺎ ﺩﺍﺭﺍﻱ ﺍﻫﻤﻴﺖ ﺑﺴﻴﺎﺭﻱ‬ ‫ﺍﺳﺖ‪.‬‬ ‫‪18‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ‬ ‫‪The Plastic Modulus‬‬ ‫ﻣﺤﺎﺳﺒﻪ ﻟﻨﮕﺮ ﺗﺴﻠﻴﻢ‪ ، My ،‬ﺑﻪ ﻳﻜﻲ ﺍﺯ ﺩﻭ ﺭﻭﺵ ﺯﻳﺮ ﺍﻣﻜﺎﻥﭘﺬﻳﺮ ﺍﺳﺖ‪:‬‬ ‫ ﺍﺯ ﺣﺎﺻﻠﻀﺮﺏ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺩﺭ ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺍﺭﺗﺠﺎﻋﻲ‬ ‫‪My = Fy S = Fy I/c = Fy bd2/6‬‬ ‫ ﺑﺎ ﻣﺤﺎﺳﺒﻪ ﻟﻨﮕﺮ ﻣﻘﺎﻭﻡ ﺩﺍﺧﻠﻲ ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ )ﺣﺎﺻﻠﻀﺮﺏ ‪ C‬ﻳﺎ ‪ T‬ﺩﺭ ﺑﺎﺯﻭﻱ ﻟﻨﮕﺮ ﺑﻴﻦ ﺁﻥ ﺩﻭ(‬ ‫‪My = (Fy bd/4) (2d/3) = Fy bd2/6‬‬ ‫‪19‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ‬ ‫‪The Plastic Modulus‬‬ ‫ ﻣﺤﺎﺳﺒﻪ ﻟﻨﮕﺮ ﺧﻤﻴﺮﻱ‪ ) ، Mp ،‬ﻛﻪ ﺩﺭ ﺻﻮﺭﺕ ﺗﺎﻣﻴﻦ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﺑﺮﺍﺑﺮ ﺑﺎ ﻟﻨﮕﺮ ﺍﺳﻤﻲ‬ ‫‪ Mn‬ﻧﻴﺰ ﻫﺴﺖ( ﺑﻪ ﻃﺮﻳﻘﻲ ﻣﺸﺎﺑﻪ ﺭﻭﺵ ﺩﻭﻡ ﺍﻧﺠﺎﻡ ﻣﻲﮔﻴﺮﺩ‪:‬‬ ‫‪Mp = Mn = T d/2 = C d/2 = (Fy bd/2) (d/2) = Fy bd2/4‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ × ﺗﻨﺶ ﺗﺴﻠﻴﻢ = ‪Mp = Fy Z‬‬ ‫ ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ ﺑﺮﺍﻱ ﻣﻘﻄﻊ ﻣﺴﺘﻄﻴﻠﻲ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‪:‬‬ ‫‪Z = bd2/4‬‬ ‫‪20‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪The Modulus Plastic‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ‪ ، Z ،‬ﺑﺮﺍﺑﺮ ﺑﺎ ﺟﻤﻊ ﻟﻨﮕﺮ ﺳﻄﺢ‪ ،‬ﺳﻄﻮﺡ ﺗﺤﺖ ﻛﺸﺶ ﻭ ﻓﺸﺎﺭ ﺣﻮﻝ ﺗﺎﺭ‬ ‫ﺧﻨﺜﻲ ﺍﺳﺖ‪.‬‬ ‫ﺗﻨﻬﺎ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﻣﻘﻄﻊ ﺷﻜﻞ ﻣﺘﻘﺎﺭﻧﻲ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ﻣﺤﻞ ﺗﺎﺭ ﺧﻨﺜﻲ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﺧﻤﻴﺮﻱ ﺑﺮ‬ ‫ﻣﺤﻞ ﺗﺎﺭ ﺧﻨﺜﻲ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﺍﺭﺗﺠﺎﻋﻲ ﻣﻨﻄﺒﻖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﺩﺭ ﺗﻌﻴﻴﻦ ﻣﺤﻞ ﺗﺎﺭ ﺧﻨﺜﻲ ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﺧﻤﻴﺮﻱ ﺑﺎﻳﺪ ﺑﻪ ﺍﻳﻦ ﻧﻜﺘﻪ ﺗﻮﺟﻪ ﺷﻮﺩ ﻛﻪ ﺳﻄﺢ ﺗﺤﺖ‬ ‫ﻓﺸﺎﺭ ﺑﺎﻳﺪ ﺑﺎ ﺳﻄﺢ ﺗﺤﺖ ﻛﺸﺶ ﺑﺮﺍﺑﺮ ﺑﺎﺷﺪ‪.‬‬ ‫ﺿﺮﻳﺐ ﺷﻜﻞ)‪ (shape factor‬ﺑﺮﺍﻱ ﻣﻘﻄﻊ ﻣﺴﺘﻄﻴﻠﻲ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ‪:‬‬ ‫‪1.5‬‬ ‫‪bd 2 / 4‬‬ ‫‪bd 2 / 6‬‬ ‫‪Z‬‬ ‫‪S‬‬ ‫‪Fy Z‬‬ ‫‪Mp‬‬ ‫‪Fy S‬‬ ‫‪My‬‬ ‫ ﺿﺮﻳﺐ ﺷﻜﻞ ﺑﺮﺍﻱ ﺩﻳﮕﺮ ﺍﻧﻮﺍﻉ ﻣﻘﺎﻃﻊ ﺭﺍ ﻧﻴﺰ ﻣﻴﺘﻮﺍﻥ ﺑﺮﺍﺳﺎﺱ ﺭﻭﺵ ﻓﻮﻕ ﺗﻌﻴﻴﻦ ﻛﺮﺩ‪ .‬ﺑﺮﺍﻱ ﻣﻘﺎﻃﻊ‬ ‫ﺩﺍﻳﺮﻩ ﺍﻱ ﺷﻜﻞ ﻣﻘﺪﺍﺭ ﺁﻥ ‪ 1.7‬ﻭ ﺑﺮﺍﻱ ﻣﻘﺎﻃﻊ ‪ I‬ﺷﻜﻞ ﺑﻴﻦ ‪ 1.1‬ﺗﺎ ‪ 1.2‬ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫‪21‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺯﻳﺮ‬ Urmia University, M. Sheidaii 22 Urmia University, M. Sheidaii 23 Urmia University, M. Sheidaii 24 ‫ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ)ﺷﻜﺴﺖ(‪ -‬ﺩﺭ ﺳﺎﺯﻩ ﻫﺎﻱ ﻣﻌﻴﻦ‬ ‫‪The Collapse Mechanism‬‬ ‫ ﺩﺭ ﺗﻴﺮ ﺳﺎﺩﻩ ﺍﻱ ﻛﻪ ﺗﺤﺖ ﺑﺎﺭ ﻣﺘﻤﺮﻛﺰ ﺩﺭ ﻭﺳﻂ ﺩﻫﺎﻧﻪ ﻗﺮﺍﺭ ﺩﺍﺭﺩ ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﺑﺎﺭ ﻭ ﺭﺳﻴﺪﻥ ﺁﻥ ﺑﻪ‬ ‫ﺗﺮﺍﺯ ﺑﺎﺭ ‪ ،Pn‬ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺩﺭ ﻣﻘﻄﻊ ﻟﻨﮕﺮ ﺣﺪﺍﻛﺜﺮ)ﺯﻳﺮ ﺑﺎﺭ( ﺍﻳﺠﺎﺩ ﺷﺪﻩ ﻭ ﺳﺎﺯﻩ ﻧﺎﭘﺎﻳﺪﺍﺭ‬ ‫ﻣﻲﮔﺮﺩﺩ‪.‬‬ ‫ ﺍﻓﺰﺍﻳﺶ ﻣﺠﺪﺩ ﺑﺎﺭ ﺑﺎﻋﺚ ﺧﺮﺍﺑﻲ ﺳﺎﺯﻩ ﺷﺪﻩ ﻭ ﺑﺎﺭ ﺣﺪﺍﻛﺜﺮ‪ ،‬ﺑﺎﺭ ﺍﺳﻤﻲ ‪ Pn‬ﻗﺎﺑﻞ ﺗﺤﻤﻞ ﺗﻮﺳﻂ ﺳﺎﺯﻩ‬ ‫ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﺑﻨﺎﺑﺮﺍﻳﻦ‪:‬‬ ‫ﺳﺎﺯﻩﺍﻱ ﻛﻪ ﺍﺯ ﻧﻈﺮ ﺍﺳﺘﺎﺗﻴﻜﻲ ﻣﻌﻴﻦ ﺑﺎﺷﺪ ﺑﺎ ﺗﺸﻜﻴﻞ ﻳﻚ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺩﭼﺎﺭ ﺧﺮﺍﺑﻲ ﺧﻮﺍﻫﺪ ﺷﺪ‪.‬‬ ‫ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ‪ -‬ﺩﺭ ﺳﺎﺯﻩ ﻫﺎﻱ ﻧﺎﻣﻌﻴﻦ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪26‬‬ ‫ﺩﺭ ﻳﻚ ﺳﺎﺯﻩ ﻧﺎﻣﻌﻴﻦ ﭘﺲ ﺍﺯ ﺗﺸﻜﻴﻞ ﻳﻚ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ‪ ،‬ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﭘﻴﻜﺮﺑﻨﺪﻱ ﻫﻨﺪﺳﻲ‬ ‫ﺳﺎﺯﻩ ﺍﺟﺎﺯﻩ ﺩﻫﺪ‪ ،‬ﺍﻣﻜﺎﻥ ﺍﻓﺰﺍﻳﺶ ﺑﺎﺭ ﺑﺪﻭﻥ ﺧﺮﺍﺑﻲ ﺳﺎﺯﻩ ﻭﺟﻮﺩ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ‪.‬‬ ‫ﺩﺭ ﺯﻣﺎﻥ ﺍﻓﺰﺍﻳﺶ ﺑﺎﺭ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﻋﻴﻨﺎ ﻣﺎﻧﻨﺪ ﻳﻚ ﻣﻔﺼﻞ ﻭﺍﻗﻌﻲ ﻋﻤﻞ ﻛﺮﺩﻩ‪ ،‬ﻗﺎﺩﺭ ﺑﻪ ﺗﺤﻤﻞ‬ ‫ﻟﻨﮕﺮ ﺍﺿﺎﻓﻲ ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ ﻟﺬﺍ ﺑﺎﺯﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺩﺭ ﺳﺎﺯﻩ ﺍﺗﻔﺎﻕ ﻣﻲﺍﻓﺘﺪ‪.‬‬ ‫ﺑﺎ ﺗﺸﻜﻴﻞ ﻣﻔﺼﻞﻫﺎﻱ ﺧﻤﻴﺮﻱ ﺑﻴﺸﺘﺮ ﻭ ﺭﺳﻴﺪﻥ ﺗﻌﺪﺍﺩ ﺁﻧﻬﺎ ﺑﻪ ﺣﺪ ﻻﺯﻡ‪ ،‬ﺳﺎﺯﻩ ﺩﭼﺎﺭ ﺧﺮﺍﺑﻲ‬ ‫ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﺁﺭﺍﻳﺸﻲ ﺍﺯ ﻣﻔﺼﻞﻫﺎﻱ ﺧﻤﻴﺮﻱ ﻭ ﻭﺍﻗﻌﻲ ﻛﻪ ﺳﺒﺐ ﺧﺮﺍﺑﻲ ﺳﺎﺯﻩ ﻣﻲﺷﻮﺩ ﻣﻜﺎﻧﻴﺰﻡ‬ ‫)‪ (mechanism‬ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﺩﺭ ﺳﺎﺯﻩﻫﺎﻱ ﻧﺎﻣﻌﻴﻦ ﺑﺮﺍﻱ ﺧﺮﺍﺑﻲ ﺳﺎﺯﻩ )ﺗﺸﻜﻴﻞ ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ( ﻣﻌﻤﻮﻻ ﺑﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ﻳﻚ‬ ‫ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺍﻳﺠﺎﺩ ﺷﻮﺩ‪.‬‬ ‫ﺩﺭ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ ﺑﺎﻳﺪ ﻫﻤﻪ ﻣﻜﺎﻧﻴﺰﻣﻬﺎﻱ ﻣﻤﻜﻦ ﺧﺮﺍﺑﻲ ﻣﻮﺭﺩ ﺑﺮﺳﻲ ﻗﺮﺍﺭ ﮔﻴﺮﺩ ﺗﺎ ﻣﻜﺎﻧﻴﺰﻡ‬ ‫ﺧﺮﺍﺑﻲ ﺻﺤﻴﺢ ﺑﻪ ﺍﺯﺍﻱ ﻛﻤﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺑﺎﺭ ﺗﻌﻴﻴﻦ ﺷﻮﺩ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ‪ -‬ﺩﺭ ﺳﺎﺯﻩ ﻫﺎﻱ ﻧﺎﻣﻌﻴﻦ‬ ‫ ﻳﻚ ﺗﻴﺮ ﺩﻭﺳﺮ ﮔﻴﺮﺩﺍﺭ ﺑﺎ ﺗﺸﻜﻴﻞ ﺳﻪ‬ ‫ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺑﻪ ﻳﻚ ﻣﻜﺎﻧﻴﺰﻡ ﺗﺒﺪﻳﻞ‬ ‫ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺩﺭ ﻳﻚ ﺗﻴﺮ ﻳﻚ ﺳﺮ ﮔﻴﺮﺩﺍﺭ‪ -‬ﻳﻚ ﺳﺮ‬ ‫ﻣﻔﺼﻞ ﺑﺎ ﺗﺸﻜﻴﻞ ﺩﻭ ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ‪،‬‬ ‫ﻳﻚ ﻣﻜﺎﻧﻴﺰﻡ ﺍﻳﺠﺎﺩ ﻣﻲﺷﻮﺩ)ﻭﺟﻮﺩ ﺳﻪ‬ ‫ﻣﻔﺼﻞ ﺩﺭ ﻳﻚ ﺍﻣﺘﺪﺍﺩ(‪.‬‬ ‫‪27‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻓﺮﺿﻴﺎﺕ ﺗﺌﻮﺭﻱ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ‬ ‫ ﻧﻤﻮﺩﺍﺭ ﺗﻨﺶ‪-‬ﻛﺮﻧﺶ ﺑﻪ ﺻﻮﺭﺕ ﺍﻳﺪﻩ ﺁﻝ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ‬ ‫ ﺍﻧﻄﺒﺎﻕ ﻧﻘﻄﻪ ﺗﺴﻠﻴﻢ ﻭ ﻧﻘﻄﻪ ﺣﺪ ﺧﻄﻲ‬ ‫ ﻭﺟﻮﺩ ﻧﺎﺣﻴﻪ ﺳﺨﺖ ﺷﺪﮔﻲ ﻛﺮﻧﺶ ) ﻛﻪ ﺩﺭ ﺁﻥ ﻓﻮﻻﺩ ﻗﺎﺩﺭ‬ ‫ﺑﻪ ﺗﺤﻤﻞ ﺗﻨﺶ ﺑﺎﻻﺗﺮﻱ ﺍﺳﺖ(‪ .‬ﺍﻟﺒﺘﻪ ﺩﺭ ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﻻﺯﻡ‬ ‫ﺑﻪ ﺗﺬﻛﺮ ﺍﺳﺖ ﻛﻪ‪:‬‬ ‫ ﻋﻤﻼ ﺩﺭ ﺍﻳﻦ ﺣﻮﺯﻩ‪ ،‬ﻛﺮﻧﺶ ﺁﻧﭽﻨﺎﻥ ﺑﺎﻻﺳﺖ ﻛﻪ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻋﻤﻠﻲ ﻧﺪﺍﺭﺩ‪،‬‬ ‫ ﻫﺮﭼﻨﺪ ﻛﻪ ﻣﻘﺪﺍﺭ ﺳﺨﺖﺷﺪﮔﻲ ﻛﺮﻧﺸﻲ ﺩﺭ ﻧﻤﻮﺩﺍﺭ‬ ‫ﻗﺎﺑﻞ ﺗﻮﺟﻪ ﺍﺳﺖ ﻟﻴﻜﻦ ﺑﻪ ﺩﻟﻴﻞ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﺭﺗﺠﺎﻋﻲ‬ ‫ﻗﺪﺭﺕ ﺑﺎﺭﺑﺮﻱ ﻣﻘﻄﻊ ﺑﻴﺶ ﺍﺯ ‪ Mn‬ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫‪28‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺵ ﻛﺎﺭ ﻣﺠﺎﺯﻱ ﺑﺮﺍﻱ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ ﺳﺎﺯﻩ ﻫﺎ‬ ‫‪The Virtual –work Method‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﺭﻭﺵ‪ ،‬ﺍﺑﺘﺪﺍ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﺳﺎﺯﻩ ﺗﺤﺖ ﺑﺎﺭﻱ ﺑﻪ ﻣﻴﺰﺍﻥ ﻇﺮﻓﻴﺖ ﺧﻤﺸﻲ ﺧﻮﺩ ﻗﺮﺍﺭ ﺩﺍﺷﺘﻪ‬ ‫ﺑﺎﺷﺪ‪،‬‬ ‫ ﺳﭙﺲ ﻓﺮﺽ ﻣﻲﮔﺮﺩﺩ ﺳﺎﺯﻩ ﺗﺤﺖ ﺑﺎﺭ ﻧﻬﺎﻳﻲ ﻭﺍﺭﺩﻩ‪ ،‬ﺗﻐﻴﺮﻣﻜﺎﻥ ﺍﺿﺎﻓﻲ ﻛﻮﭼﻜﻲ ﻳﺎﺑﺪ‪،‬‬ ‫ ﻛﺎﺭ ﺑﺎﺭﻫﺎﻱ ﺧﺎﺭﺟﻲ ﺩﺭ ﻃﻲ ﺍﻳﻦ ﺗﻐﻴﻴﺮﻣﻜﺎﻥ ﺍﺿﺎﻓﻲ‪ ،‬ﺑﺮﺍﺑﺮ ﺑﺎ ﻛﺎﺭ ﺩﺍﺧﻠﻲ ﺟﺬﺏ ﺷﺪﻩ ﺗﻮﺳﻂ‬ ‫ﻣﻔﺼﻞﻫﺎ ﻗﺮﺍﺭ ﺩﺍﺩﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﺭﻭﺵ ﺍﺯ ﻧﻈﺮﻳﻪ ﺯﻭﺍﻳﺎﻱ ﻛﻮﭼﻚ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲﺷﻮﺩ )‪.( sin T≈T , tan T≈T‬‬ ‫‪29‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example 1: Uniformly loaded fixed-ended beam Urmia University, M. Sheidaii 30 Example 2: Propped cantilever beam Urmia University, M. Sheidaii 31 Example 3: Uniformly loaded propped cantilever beam Urmia University, M. Sheidaii 32 Urmia University, M. Sheidaii 33 ‫ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ)ﻳﻜﺴﺮﻩ(‬ ‫‪Continuous Beams‬‬ ‫ ﺗﺤﻠﻴﻞ ﺍﺭﺗﺠﺎﻋﻲ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻧﺎﻣﻌﻴﻨﻲ ﺍﺳﺘﺎﺗﻴﻜﻲ ﺁﻧﻬﺎ ﭘﻴﭽﻴﺪﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺍﻣﺎ ﺍﻋﻤﺎﻝ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ ﺑﺮ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ ﻣﺎﻧﻨﺪ ﺍﻋﻤﺎﻝ ﺁﻥ ﺑﺮ ﺗﻴﺮﻫﺎﻱ ﻳﻚ ﺩﻫﺎﻧﻪ ﺑﻮﺩﻩ ﻭ‬ ‫ﺍﻓﺰﺍﻳﺶ ﺗﻌﺪﺍﺩ ﺩﻫﺎﻧﻪﻫﺎ ﺗﺎﺛﻴﺮ ﭼﻨﺪﺍﻧﻲ ﺑﺮ ﺣﺠﻢ ﻣﺤﺎﺳﺒﺎﺕ ﻧﺪﺍﺭﺩ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ ﻛﺎﻓﻲ ﺍﺳﺖ ﺭﻭﺵ ﻛﺎﺭ ﻣﺠﺎﺯﻱ ﺑﻪ ﻃﻮﺭ ﺟﺪﺍﮔﺎﻧﻪ ﺑﻪ ﻫﺮ‬ ‫ﻳﻚ ﺍﺯ ﺩﻫﺎﻧﻪﻫﺎﻱ ﺗﻴﺮ ﺍﻋﻤﺎﻝ ﺷﻮﺩ‪.‬‬ ‫ ﻣﻜﺎﻧﻴﺰﻡ ﺻﺤﻴﺢ ﺑﺮﺍﻱ ﺳﺎﺯﻩ‪ ،‬ﻣﻜﺎﻧﻴﺰﻣﻲ ﺍﺳﺖ ﻛﻪ ﻣﻨﺠﺮ ﺑﻪ ﻛﻮﭼﻜﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺑﺎﺭ)ﻣﺘﻨﺎﻇﺮ ﺑﺎ‬ ‫ﺑﺰﺭﮔﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ‪ ( Mp‬ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪34‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 35 36 ‫ﻗﺎﺑﻬﺎﻱ ﺳﺎﺧﺘﻤﺎﻧﻲ‬ ‫‪Building Frames‬‬ ‫ ﺭﻭﺵ ﻛﺎﺭ ﻣﺠﺎﺯﻱ ﺑﺮﺍﻱ ﻗﺎﺑﻬﺎ ﻧﻴﺰ ﻫﻤﭽﻮﻥ ﺗﻴﺮﻫﺎ ﻗﺎﺑﻞ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺩﺭﻗﺎﺑﻬﺎ ﻣﻜﺎﻧﻴﺰﻣﻬﺎﻱ ﺩﻳﮕﺮﻱ ﻏﻴﺮ ﺍﺯ ﻣﻜﺎﻧﻴﺰﻣﻬﺎﻱ ﺗﻴﺮ ﻧﻈﻴﺮ ﻣﻮﺍﺭﺩ ﺯﻳﺮ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪:‬‬ ‫ ﻣﻜﺎﻧﻴﺰﻡ ﺍﻧﺘﻘﺎﻝ ﺟﺎﻧﺒﻲ‬ ‫ ﻣﻜﺎﻧﻴﺰﻡ ﺗﺮﻛﻴﺒﻲ ﺗﻴﺮ ﻭ ﺍﻧﺘﻘﺎﻝ ﺟﺎﻧﺒﻲ‬ ‫ ﺑﺮ ﺧﻼﻑ ﺗﺤﻠﻴﻞ ﺍﺭﺗﺠﺎﻋﻲ ﺩﺭ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ ﺍﺻﻞ ﺟﻤﻊ ﺁﺛﺎﺭ ﻗﺎﺑﻞ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻴﺴﺖ‪.‬‬ ‫ ﺩﺭ ﻣﺤﻞ ﺍﺗﺼﺎﻝ ﺩﻭ ﻋﻀﻮ ﺑﺎ ﺳﺎﻳﺰﻫﺎﻱ ﻣﺨﺘﻠﻒ‪ ،‬ﻣﻔﺼﻞ ﺧﻤﻴﺮﻱ ﺑﺮ ﺭﻭﻱ ﻋﻀﻮ ﺿﻌﻴﻒﺗﺮ ﺗﺸﻜﻴﻞ‬ ‫ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪37‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example : Possible mechanisms for a portal frame 38 ‫ﺑﺎﺯ ﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ‬ ‫‪Moment Redistribution in Design of Continuous Beams‬‬ ‫ ﺩﺭ ﺗﻴﺮﻫﺎﻱ ﺑﺎ ﻣﻘﻄﻊ ﻓﺸﺮﺩﻩ ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺍﺯ ﻓﻮﻻﺩ ﺑﺎ ﺗﻨﺶ ﺗﺴﻠﻴﻢ‬ ‫)‪ Fy d 4500kgf/cm2(≡65ksi‬ﻛﻪ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺁﻧﻬﺎ ﺍﺯ ﺗﻜﻴﻪ ﮔﺎﻫﻬﺎﻱ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ‬ ‫ﺑﺮﺧﻮﺭﺩﺍﺭ ﺑﺎﺷﺪ ﺍﻣﻜﺎﻥ ﺑﺎﺯﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺣﺎﺻﻞ ﺍﺯ ﺑﺎﺭﻫﺎﻱ ﺍﺿﺎﻓﻪ ﻭﺍﺭﺩﻩ ﻭﺟﻮﺩ ﺧﻮﺍﻫﺪ‬ ‫ﺩﺍﺷﺖ‪.‬‬ ‫ ﺩﺭ ﻃﺮﺡ ﺧﻤﻴﺮﻱ ﺗﻴﺮﻫﺎ ﻋﻤﻼ ﭼﻨﻴﻦ ﻣﺰﻳﺘﻲ ﺩﺭ ﺗﺤﻠﻴﻞ ﻟﺤﺎﻅ ﻣﻲ ﮔﺮﺩﺩ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﺍﻣﻜﺎﻥ ﺑﻬﺮﻩﻣﻨﺪﻱ ﺍﺯ ﻣﺰﺍﻳﺎﻱ ﻧﺎﺷﻲ ﺍﺯ ﺑﺎﺯﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺩﺭ ﻃﺮﺡ ﺍﺭﺗﺠﺎﻋﻲ ﺗﻴﺮ‬ ‫ﻧﻴﺰ ﻓﺮﺍﻫﻢ ﺁﻳﺪ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﻗﺎﻋﺪﻩ ﻣﻮﺳﻮﻡ ﺑﻪ ‪ 0.9‬ﻟﻨﮕﺮ ﻣﻨﻔﻲ ﻛﻪ ﺩﺭ‬ ‫‪ AISC05 App.1.3 p.16.1-151‬ﻣﻄﺮﺡ ﮔﺮﺩﻳﺪﻩ ﺍﺳﺖ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ‪.‬‬ ‫‪39‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺯ ﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ‬ ‫‪Moment Redistribution in Design of Continuous Beams‬‬ ‫ ﻗﺎﻋﺪﻩ ‪ 0.9‬ﻟﻨﮕﺮ ﺩﺭ ﻃﺮﺡ ﺍﺭﺗﺠﺎﻋﻲ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ‪:‬‬ ‫ ﻟﻨﮕﺮﻫﺎﻱ ﻣﻨﻔﻲ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﻧﺎﺷﻲ ﺍﺯ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺛﻘﻠﻲ ﻛﻪ ﺑﺎ ﺗﺤﻠﻴﻞ ﺍﻻﺳﺘﻴﻚ‬ ‫ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﻧﺪ ﺩﺭ ﺿﺮﻳﺐ ‪ 0.9‬ﺿﺮﺏ ﻣﻲ ﺷﻮﻧﺪ‪،‬‬ ‫ ﻟﻨﮕﺮﻫﺎﻱ ﻣﺜﺒﺖ ﻭﺳﻂ ﺩﻫﺎﻧﻪ ﻧﻴﺰ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ‪ 0.1‬ﻣﺘﻮﺳﻂ ﻟﻨﮕﺮﻫﺎﻱ ﻣﻨﻔﻲ‬ ‫ﺗﻜﻴﻪ ﮔﺎﻫﻬﺎﻱ ﻃﺮﻓﻴﻦ ﺍﻓﺰﺍﻳﺶ ﻣﻲ ﻳﺎﺑﺪ‪.‬‬ ‫‪40‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺯ ﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ‬ ‫‪Moment Redistribution in Design of Continuous Beams‬‬ ‫ ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﻗﺎﻋﺪﻩ ‪ 0.9‬ﻟﻨﮕﺮ‪:‬‬ ‫‪41‬‬ ‫‬ ‫ﺍﻳﻦ ﻗﺎﻋﺪﻩ ﺗﻨﻬﺎ ﺩﺭ ﺣﺎﻟﺖ ﺍﻋﻤﺎﻝ ﺑﺎﺭﻫﺎﻱ ﺛﻘﻠﻲ ﺑﻜﺎﺭ ﺭﻓﺘﻪ ﻭ ﺩﺭ ﻣﻮﺭﺩ ﺑﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻧﻈﻴﺮ ﺑﺎﺭ‬ ‫ﺑﺎﺩ ﻭ ﺯﻟﺰﻟﻪ ﺑﻜﺎﺭ ﻧﻤﻲ ﺭﻭﺩ‪،‬‬ ‫‬ ‫ﺍﻳﻦ ﻛﺎﻫﺶ ﺩﺭ ﻣﻮﺭﺩ ﻟﻨﮕﺮﻫﺎﻱ ﻧﺎﺷﻲ ﺍﺯ ﺑﺎﺭﻫﺎ ﺭﻭﻱ ﻛﻨﺴﻮﻟﻬﺎ ﻣﺠﺎﺯ ﻧﻴﺴﺖ‪،‬‬ ‫‬ ‫ﺍﻳﻦ ﻛﺎﻫﺶ ﺩﺭ ﻃﺮﺍﺣﻲ ﺑﺮ ﺍﺳﺎﺱ ﺑﺨﺶ ﻫﺎﻱ ‪ 1-8‬ﺗﺎ ‪ AISC05 App.1-4‬ﻣﺠﺎﺯ‬ ‫ﻧﻴﺴﺖ)ﺑﺨﺶ ﻫﺎﻱ ﻃﺮﺍﺣﻲ ﻏﻴﺮﺍﻻﺳﺘﻴﻚ ﺑﺮﺍﻱ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ‪ ،‬ﭘﺎﻳﺪﺍﺭﻱ ﻭ ﺗﺤﻠﻴﻞ ﻣﺮﺗﺒﻪ‬ ‫ﺩﻭﻡ‪ ،‬ﺳﺘﻮﻧﻬﺎ ﻭﺳﺎﻳﺮ ﺍﻋﻀﺎﻱ ﻓﺸﺎﺭﻱ‪ ،‬ﺗﻴﺮﻫﺎ ﻭ ﺳﺎﻳﺮ ﺍﻋﻀﺎﻱ ﺧﻤﺸﻲ‪ ،‬ﻭ ﺍﻋﻀﺎﻱ ﺗﺤﺖ ﻧﻴﺮﻭﻫﺎﻱ‬ ‫ﺗﺮﻛﻴﺒﻲ(‪،‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺑﺎﺯ ﺗﻮﺯﻳﻊ ﻟﻨﮕﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎﻱ ﺳﺮﺍﺳﺮﻱ‬ ‫‪Moment Redistribution in Design of Continuous Beams‬‬ ‫‬ ‫‪42‬‬ ‫ﺍﮔﺮ ﻟﻨﮕﺮ ﻣﻨﻔﻲ ﺑﻮﺳﻴﻠﻪ ﮔﻴﺮﺩﺍﺭ ﻛﺮﺩﻥ ﺳﺘﻮﻧﻬﺎ ﺑﻪ ﺗﻴﺮﻫﺎ ﻭ ﺷﺎﻫﺘﻴﺮﻫﺎ)ﻗﺎﺏ ﺧﻤﺸﻲ( ﻣﻬﺎﺭ‬ ‫ﺷﻮﺩ‪ ،‬ﻛﺎﻫﺶ ‪ 0.1‬ﻟﻨﮕﺮ ﺑﺮﺍﻱ ﭼﻨﻴﻦ ﺳﺘﻮﻧﻲ ﻛﻪ ﺗﺤﺖ ﺗﺮﻛﻴﺐ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﻭ ﺧﻤﺶ‬ ‫ﺍﺳﺖ ﻗﺎﺑﻞ ﺍﻋﻤﺎﻝ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪ .‬ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺩﺭ ﻃﺮﺍﺣﻲ ﺑﺮ ﺍﺳﺎﺱ ‪ LRFD‬ﻣﻘﺪﺍﺭ ﻧﻴﺮﻭﻱ‬ ‫ﻣﺤﻮﺭﻱ ﺳﺘﻮﻥ ﺍﺯ ‪ 0.15IcFyAg‬ﻭ ﺩﺭﻃﺮﺍﺣﻲ ﺑﺮ ﺍﺳﺎﺱ ‪ ،ASD‬ﻣﻘﺪﺍﺭ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ‬ ‫ﺳﺘﻮﻥ ﺍﺯ ‪ 0.15FyAg/:c‬ﻧﺒﺎﻳﺪ ﺑﻴﺸﺘﺮ ﺷﻮﺩ‪.‬‬ ‫‪ = Ag‬ﺳﻄﺢ ﻣﻘﻄﻊ ﻧﺎﺧﺎﻟﺺ)ﻛﻞ( ﻋﻀﻮ‬ ‫‪ = Fy‬ﺣﺪﺍﻗﻞ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺑﺎﻝ‬ ‫‪ = Ic‬ﺿﺮﻳﺐ ﻛﺎﻫﺶ ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ = ‪0.9‬‬ ‫‪ = :c‬ﺿﺮﻳﺐ ﺍﻃﻤﻴﻨﺎﻥ ﻓﺸﺎﺭﻱ = ‪1.67‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 43 Urmia University, M. Sheidaii 44 Urmia University, M. Sheidaii 45 Design of B E A M S for Flexure ‫ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎ ﺑﺮﺍﻱ ﺧﻤﺶ‬:‫ ﻗﺴﻤﺖ ﺩﻭﻡ‬-‫ﻓﺼﻞ ﭘﻨﺠﻢ‬ Urmia University, M. Sheidaii 1 ‫ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﺗﻴﺮﻫﺎ‬ ‫‪Lateral-Torsional Buckling (LTB) of Beams‬‬ ‫ ﺗﺤﺖ ﺑﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻭﺍﺭﺩﻩ ﻗﺴﻤﺖ ﻓﺸﺎﺭﻱ ﺗﻴﺮ ﻣﺎﻧﻨﺪ ﻳﻚ ﺳﺘﻮﻥ ﻋﻤﻞ ﻣﻲﻛﻨﺪ ﻭ ﺩﺭ ﺻﻮﺭﺕ ﻋﺪﻡ‬ ‫ﺗﺎﻣﻴﻦ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﻛﻤﺎﻧﻪ ﻛﻨﺪ‪ .‬ﻛﻤﺎﻧﺶ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﻗﺎﺋﻢ ﺍﺗﻔﺎﻕ ﻧﻤﻲﺍﻓﺘﺪ ﺯﻳﺮﺍ‪:‬‬ ‫¾‬ ‫¾‬ ‫ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺩﺭ ﺧﻤﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ﺍﻓﻘﻲ ‪ x‬ﺑﺴﻴﺎﺭ ﺑﻴﺸﺘﺮ ﺍﺯ ﻣﺤﻮﺭ ﻗﺎﺋﻢ ‪ y‬ﺑﻮﺩﻩ‪،‬‬ ‫ﺟﺎﻥ ﻭ ﺑﺎﻝ ﻛﺸﺸﻲ ﺗﻴﺮ ﺑﻪ ﻣﺜﺎﺑﻪ ﺗﻜﻴﻪﮔﺎﻩ ﺗﻴﺮ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﻗﺎﺋﻢ ﻋﻤﻞ ﻣﻲﻛﻨﻨﺪ‪.‬‬ ‫ ﺍﻣﺎ ﻛﻤﺎﻧﺶ ﺗﻴﺮ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺍﻓﻘﻲ ﺍﻣﻜﺎﻥﭘﺬﻳﺮ ﺑﻮﺩﻩ ﻭ ﺍﻳﻦ‬ ‫ﭘﺪﻳﺪﻩ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺍﻳﻦ ﭘﺪﻳﺪﻩ ﻣﺘﻘﺎﺭﻥ ﻧﻤﻲﺑﺎﺷﺪ ﺯﻳﺮﺍ ﺗﻴـﺮ ﺍﺯ ﻳـﻚ ﻗﺴـﻤﺖ‬ ‫ﻛﺸﺸﻲ ﻭ ﻳﻚ ﻗﺴﻤﺖ ﻓﺸﺎﺭﻱ ﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻛـﻪ‬ ‫ﻗﺴﻤﺖ ﻛﺸﺸﻲ ﻛﻤﺎﻧﻪ ﻧﻤﻲﻛﻨﺪ ﺍﻣﺎ ﻗﺴﻤﺖ ﻓﺸـﺎﺭﻱ ﺩﺭ‬ ‫ﺭﺍﺳــﺘﺎﻱ ﺍﻓﻘــﻲ ﻛﻤــﺎﻧﺶ ﻣــﻲﻛﻨــﺪ ﺩﺭ ﻧﺘﻴﺠــﻪ ﻗﺴــﻤﺖ‬ ‫ﻓﺸﺎﺭﻱ ﺑﻪ ﺻﻮﺭﺕ ﺟﺎﻧﺒﻲ ﺣﺮﻛـﺖ ﻛـﺮﺩﻩ ﻭ ﺗﻴـﺮ ﺷـﻜﻞ‬ ‫ﺗﺎﺑﻴﺪﻩ ﺷﺪﻩ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ‪.‬‬ ‫ ﺑﺪﻳﻬﻲ ﺍﺳﺖ ﻛﻪ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﺑﺎﻋﺚ ﻛـﺎﻫﺶ‬ ‫ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻋﻀﻮ ﺗﻴﺮ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪2‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺷﻬﺎﻱ ﭘﻴﺸﮕﻴﺮﻱ ﺍﺯ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﺗﻴﺮﻫﺎ‬ ‫ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻘﺎﻃﻊ ﺑﺎ ﺑﺎﻟﻬﺎﻱ ﭘﻬﻦﺗﺮ ﻭ ﻳﺎ ﻣﻘﺎﻃﻊ ﺑﺴﺘﻪ‬ ‫‪TUBE‬‬ ‫‪INP‬‬ ‫ ﺗﺎﻣﻴﻦ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﭘﻴﻮﺳﺘﻪ ﺑﺎ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺗﻴﺮ ﺩﺭ ﺗﻮﺩﻩ ﺗﻮﭘﺮ )ﻣﺜﻼ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺗﻴﺮﭼﻪ ﻭ ﺑﻠﻮﻙ ﺑﻴﻦ‬ ‫ﺩﻭ ﺗﻴﺮ ﻭ ﺍﻧﺠﺎﻡ ﺑﺘﻦﺭﻳﺰﻱ( ﻳﺎ ﺍﺗﺼﺎﻝ ﭘﻴﻮﺳﺘﻪ ﺑﺎ ﺩﺍﻝ ﺑﺘﻨﻲ ﻛﻒ ﻳﺎ ﻋﺮﺷﻪ ﻓﻠﺰﻱ‬ ‫ ﺗﻌﺒﻴﻪ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺑﺮﺍﻱ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺩﺭ ﻓﻮﺍﺻﻞ ﻣﺨﺘﻠﻒ ﺗﻮﺳﻂ ﺗﻴﺮﻫﺎﻱ ﻋﺮﺿﻲ‪ ،‬ﻗﺎﺑﻬﺎﻱ‬ ‫ﻋﺮﺿﻲ ﻳﺎ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‪ .‬ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ‪ ،‬ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ)‪(unbraced length‬‬ ‫ﻧﺎﻣﻴﺪﻩ ﺷﺪﻩ ﻭ ﺑﺎ ‪ Lb‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪3‬‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺑﺎﻝ ﻓﺸﺎﺭﻱ‬ ‫‪Local Buckling of Compression Flange‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﻻﻏﺮ ﺑﻮﺩﻥ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺗﻴﺮ‪ ،‬ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺗﻴﺮ ﻗﺒﻞ ﺍﺯ ﺭﺳﻴﺪﻥ ﺗﻨﺶﻫﺎ ﺑﻪ ﺗﻨﺶ‬ ‫ﺗﺴﻠﻴﻢ ‪ Fy‬ﺍﺗﻔﺎﻕ ﻣﻲﺍﻓﺘﺪ‪.‬‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺗﻴﺮ ﺷﺪﻩ ﻭ ﺍﺯ ﺭﺳﻴﺪﻥ ﺁﻥ ﺑﻪ ﻇﺮﻓﻴﺖ ﺧﻤﻴﺮﻱ ﻛﺎﻣﻞ‬ ‫ﺟﻠﻮﮔﻴﺮﻱ ﻣﻲﻛﻨﺪ‪.‬‬ ‫ﺑﺮﺍﻱ ﭘﻴﺸﮕﻴﺮﻱ ﺍﺯ ﺣﺎﻟﺖ ﺣﺪﻱ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﻧﺴﺒﺖ ﻋﺮﺽ ﺑﻪ ﺿﺨﺎﻣﺖ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ )‪(λ =b/t‬‬ ‫ﺑﺎﻳﺪ ﺑﺮ ﺍﺳﺎﺱ ﻣﻘﺎﺩﻳﺮ ﺫﻛﺮ ﺷﺪﻩ ﺩﺭ ‪ AISC05 Table B4.1 p.16.1-14‬ﻣﺤﺪﻭﺩ ﮔﺮﺩﺩ‪.‬‬ ‫ﺩﺭ ﻣﻘﺎﻃﻊ ﻓﺸﺮﺩﻩ ‪ ) compact‬ﺑﺎ ‪ ( λ≤ λp‬ﻗﺒﻞ ﺍﺯ ﻭﻗﻮﻉ ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺩﺭ ﺍﺟﺰﺍ‪ ،‬ﻛﻞ ﻣﻘﻄﻊ ﺑﻪ ﺣﺎﻟﺖ‬ ‫ﺗﺴﻠﻴﻢ ﻣﻲﺭﺳﺪ‪.‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪4‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﻣﻮﺿﻌﻲ ﺑﺎﻝ ﻓﺸﺎﺭﻱ‬ ‫‪Local Buckling of Compression Flange‬‬ ‫ ﺷﺮﺍﻳﻂ ﻓﺸﺮﺩﻩ ﺑﻮﺩﻥ ﻧﻴﻤﺮﺧﻬﺎﻱ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ‪:‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺑﺎﻝ‪:‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﺎﻥ‪:‬‬ ‫§‬ ‫¨‬ ‫¨‬ ‫©‬ ‫‪E‬‬ ‫‪0.38‬‬ ‫‪Fy‬‬ ‫‪bf‬‬ ‫‪d O pf‬‬ ‫‪2t f‬‬ ‫·‬ ‫¸‬ ‫¸‬ ‫‪¹‬‬ ‫*‬ ‫‪§ 5365‬‬ ‫¸· ‪640‬‬ ‫¨‬ ‫‪or‬‬ ‫¨‬ ‫‪Fy‬‬ ‫‪Fy ¸¹‬‬ ‫©‬ ‫‪E‬‬ ‫‪3.76‬‬ ‫‪Fy‬‬ ‫‪h‬‬ ‫‪d O pw‬‬ ‫‪tw‬‬ ‫*‬ ‫‪545‬‬ ‫‪65‬‬ ‫‪or‬‬ ‫‪Fy‬‬ ‫‪Fy‬‬ ‫‪Of‬‬ ‫‪Ow‬‬ ‫* ﻳﻌﻨﻲ ﺩﺭ ﺳﻴﺴﺘﻢ ﺁﺣﺎﺩ ﺍﻣﺮﻳﻜﺎﻳﻲ‬ ‫ ﺩﺭ ﺍﺩﺍﻣﻪ ﺍﻳﻦ ﻓﺼﻞ ﺑﻪ ﺑﺮﺭﺳﻲ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﺗﻴﺮﻫﺎﻱ ﺑﺎ ﻣﻘﻄﻊ ﻓﺸﺮﺩﻩ ﺑﺎ ﻓﻮﺍﺻﻞ‬ ‫ﻣﺨﺘﻠﻒ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﭘﺮﺩﺍﺧﺘﻪ ﺧﻮﺍﻫﺪ ﺷﺪ‪.‬‬ ‫‪6‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻓﺘﺎﺭ ﻛﻤﺎﻧﺸﻲ ﺗﻴﺮﻫﺎ‬ ‫ ﺭﻓﺘﺎﺭ ﻛﻤﺎﻧﺸﻲ ﺗﻴﺮﻫﺎ ﺑﺮ ﺍﺳﺎﺱ ﻓﻮﺍﺻﻞ ﺑﻴﻦ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺑﻪ ﺳﻪ ﺻﻮﺭﺕ ﻣﺨﺘﻠﻒ‬ ‫ﺯﻳﺮ ﻣﻤﻜﻦ ﺍﺳﺖ ﻇﺎﻫﺮ ﺷﻮﺩ‪:‬‬ ‫ ﺭﻓﺘﺎﺭ ﭘﻼﺳﺘﻴﻚ )ﻧﺎﺣﻴﻪ ‪(1‬‬ ‫ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﻻﺳﺘﻴﻚ )ﻧﺎﺣﻴﻪ ‪(2‬‬ ‫ ﻛﻤﺎﻧﺶ ﺍﻻﺳﺘﻴﻚ )ﻧﺎﺣﻴﻪ ‪(3‬‬ ‫‪7‬‬ ‫ﺭﻓﺘﺎﺭ ﭘﻼﺳﺘﻴﻚ )ﻧﺎﺣﻴﻪ ‪(1‬‬ ‫ ﺍﮔﺮ ﻧﻴﻤﺮﺧﻲ ﻓﺸﺮﺩﻩ ﺩﺍﺭﺍﻱ ﺗﻜﻴﻪﮔﺎﻫﻬﺎﻱ ﻣﻤﺘﺪ ﺟﺎﻧﺒﻲ ﻭ ﻳﺎ ﻓﻮﺍﺻﻞ ﻛﻮﺗﺎﻩ ﻣﻬﺎﺭﻱ ﺑﺎﺷﺪ‬ ‫ﻣﻲﺗﻮﺍﻥ ﺁﻥ ﺭﺍ ﺗﺎ ﺗﺸﻜﻴﻞ ﻟﻨﮕﺮ ﭘﻼﺳﺘﻴﻚ ﻛﺎﻣﻞ ﺑﺎﺭﮔﺬﺍﺭﻱ ﻛﺮﺩ‪.‬‬ ‫ ﺍﻋﻤﺎﻝ ﺑﺎﺭ ﺑﻴﺸﺘﺮ ﭘﺲ ﺍﺯ ﺗﺸﻜﻴﻞ ﻣﻔﺼﻞ ﭘﻼﺳﺘﻴﻚ ﺑﺎﻋﺚ ﺩﻭﺭﺍﻥ ﻣﻘﻄﻊ ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ‬ ‫ﺗﻮﺯﻳﻊ ﻣﺠﺪﺩ ﻟﻨﮕﺮ ﻣﻲﺷﻮﺩ ﻟﺬﺍ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﺗﺤﻠﻴﻞ ﭘﻼﺳﺘﻴﻚ)ﺧﻤﻴﺮﻱ( ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﺑﺘﻮﺍﻥ ﺗﻴﺮ ﺭﺍ ﺗﺎ ﺭﺳﻴﺪﻥ ﺑﻪ ﻟﻨﮕﺮ ﭘﻼﺳﺘﻴﻚ ﻛﺎﻣﻞ ‪ Mp‬ﺑﺎﺭﮔﺬﺍﺭﻱ ﻛﺮﺩ‬ ‫ﻓﺎﺻﻠﻪ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ )‪ (Lb‬ﻧﺒﺎﻳﺪ ﺍﺯ ﻣﻘﺪﺍﺭ ﻣﻌﻴﻨﻲ ﻛﻪ ‪ Lp‬ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ ﺗﺠﺎﻭﺯ‬ ‫ﻛﻨﺪ‪.‬‬ ‫ ﻣﻘﺪﺍﺭ ‪ Lp‬ﺑﻪ ﻣﺸﺨﺼﺎﺕ ﻫﻨﺪﺳﻲ ﻣﻘﻄﻊ ﺗﻴﺮ ﻭ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﺁﻥ ﺑﺴﺘﮕﻲ ﺩﺍﺭﺩ‪.‬‬ ‫‪8‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﻻﺳﺘﻴﻚ )ﻧﺎﺣﻴﻪ ‪(2‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﻧﺎﺣﻴﻪ ﺍﻣﻜﺎﻥ ﺍﻳﻨﻜﻪ ﺑﺮﺧﻲ ﺍﺯ ﺗﺎﺭﻫﺎﻱ ﻓﺸﺎﺭﻱ ﻣﻘﻄﻊ ﻗﺒﻞ ﺍﺯ ﻛﻤﺎﻧﺶ ﺑﻪ ﺗﺴﻠﻴﻢ ﺑﺮﺳﻨﺪ‬ ‫ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻭﻟﻲ ﻧﻤﻲﺗﻮﺍﻥ ﻛﻠﻴﻪ ﺗﺎﺭﻫﺎﻱ ﻣﻘﻄﻊ ﺭﺍ ﺑﻪ ﺣﺪ ﺗﺴﻠﻴﻢ)‪ (Fy‬ﺭﺳﺎﻧﻴﺪ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﻇﺮﻓﻴﺖ ﻣﻘﻄﻊ ﺩﺭ ﺑﺮﺍﺑﺮ ﺩﻭﺭﺍﻥ ﺁﻧﻘﺪﺭ ﺑﺎﻻ ﻧﻴﺴﺖ ﻛﻪ ﺍﻣﻜﺎﻥ ﺗﻮﺯﻳﻊ ﻣﺠﺪﺩ ﻛﺎﻣﻞ‬ ‫ﻟﻨﮕﺮ ﺩﺭ ﺗﻴﺮ ﻓﺮﺍﻫﻢ ﺁﻳﺪ ﻭ ﻟﺬﺍ ﺍﺯ ﺗﺤﻠﻴﻞ ﭘﻼﺳﺘﻴﻚ)ﺧﻤﻴﺮﻱ( ﻧﻴﺰ ﻧﻤﻲﺗﻮﺍﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﻧﺎﺣﻴﻪ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮ ﺍﺳﺎﺱ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﻏﻴﺮﺍﻻﺳﺘﻴﻚ ﺗﻌﻴﻴﻦ‬ ‫ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺣﺪﺍﻛﺜﺮ ﻓﺎﺻﻠﻪ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺩﺭ ﺍﻳﻦ ﻧﺎﺣﻴﻪ ﺑﺎ ‪ Lr‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﻲﺷﻮﺩ ﻛﻪ ﺩﺭ ﺁﻥ ﺑﻪ ﻣﺤﺾ‬ ‫ﺭﺳﻴﺪﻥ ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﺑﻪ ﺣﺪﻱ ﻛﻪ ﻣﻮﺟﺐ ﺗﺴﻠﻴﻢ ﺷﺪﻥ ﺗﺎﺭﻱ ﺍﺯ ﻣﻘﻄﻊ ﮔﺮﺩﺩ )ﻛﻪ ﻋﻤﻼ‬ ‫ﺑﻪ ﺩﻟﻴﻞ ﻭﺟﻮﺩ ﺗﻨﺶ ﭘﺴﻤﺎﻧﺪ ﻛﻤﺘﺮ ﺍﺯ ‪ Fy‬ﺍﺳﺖ( ﻣﻘﻄﻊ ﻛﻤﺎﻧﺶ ﻣﻲﻛﻨﺪ‪.‬‬ ‫ ﻣﻘﺪﺍﺭ ‪ Lr‬ﺑﻪ ﻣﺸﺨﺼﺎﺕ ﻫﻨﺪﺳﻲ ﻣﻘﻄﻊ ﺗﻴﺮ‪ ،‬ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﻓﻮﻻﺩ ﻭ ﻧﺤﻮﻩ ﺗﻮﺯﻳﻊ ﺗﻨﺶﻫﺎﻱ‬ ‫ﭘﺴﻤﺎﻧﺪ ﺩﺭ ﻣﻘﻄﻊ ﺗﻴﺮ ﺑﺴﺘﮕﻲ ﺩﺍﺭﺩ‪.‬‬ ‫‪9‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻛﻤﺎﻧﺶ ﺍﻻﺳﺘﻴﻚ )ﻧﺎﺣﻴﻪ ‪(3‬‬ ‫ ﺍﮔﺮ ﻓﻮﺍﺻﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺑﺰﺭﮔﺘﺮ ﺍﺯ ‪ Lr‬ﺑﺎﺷﺪ‪ ،‬ﻗﺒﻞ ﺍﺯ ﺭﺳﻴﺪﻥ ﺗﻨﺶ ﺩﺭ‬ ‫ﻣﻘﻄﻊ ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ‪ ،‬ﺗﻴﺮ ﺑﻪ ﺻﻮﺭﺕ ﺍﺭﺗﺠﺎﻋﻲ ﻛﻤﺎﻧﻪ ﺧﻮﺍﻫﺪ ﻛﺮﺩ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﻧﺎﺣﻴﻪ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮ ﺍﺳﺎﺱ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ‬ ‫ﺍﻻﺳﺘﻴﻚ ﺗﻌﻴﻴﻦ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﻟﻨﮕﺮ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﻪ ﻣﻘﺎﻭﻣﺖ ﭘﻴﭽﺸﻲ ﻭ ﻣﻘﺎﻭﻣﺖ ﺗﺎﺑﻴﺪﮔﻲ )‪ (warping‬ﺗﻴﺮ‬ ‫ﺑﺴﺘﮕﻲ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ‪.‬‬ ‫‪10‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ Beams Design Flexural Strength ‫ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﺤﺎﺳﺒﻪ‬AISC05 ‫ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﻃﺒﻖ ﺿﻮﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ‬ :‫ﻣﻲﺷﻮﺩ‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﻣﻮﺟﻮﺩ‬Available Design Flexural Strength = Ib Mn (LRFD method) ‫ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﻣﺠﺎﺯ‬Allowable Design Flexural Strength = Mn / :b (ASD method) ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ‬ Mn = nominal flexural moment Ib = 0.9 (LRFD) :b = 1.67 (ASD) Urmia University, M. Sheidaii 11 ‫ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ‪ -‬ﻟﻨﮕﺮ ﭘﻼﺳﺘﻴﻚ ﻛﺎﻣﻞ)ﻧﺎﺣﻴﻪ ‪(1‬‬ ‫‪Fully Braced Compact Beams‬‬ ‫ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺗﻴﺮﻱ ﺑﺎ ﻣﻘﻄﻊ ﻓﺸﺮﺩﻩ ﺩﺍﺭﺍﻱ ﻣﻬﺎﺭ ﺟﺎﻧﺒﻲ ﻛﺎﻓﻲ ﺑﺎﺷﺪ ﺑﺎ ﺗﺴﻠﻴﻢ ﻛﻞ ﻣﻘﻄﻊ ﺗﻴﺮ‬ ‫ﺧﺮﺍﺏ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﺍﻳﻦ ﺣﺎﻟﺖ‪ ،‬ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ ﻭ ﺳﺎﺩﻩﺗﺮﻳﻦ ﺣﺎﻟﺖ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﻣﻲﺑﺎﺷﺪ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮ ﺍﺳﺎﺱ ﻟﻨﮕﺮ ﭘﻼﺳﺘﻴﻚ ﻣﻘﻄﻊ ‪ Mp‬ﺗﻌﻴﻴﻦ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﻃﺒﻖ ﺿﻮﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC05‬ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﺤﺎﺳﺒﻪ‬ ‫ﻣﻲﺷﻮﺩ‪:‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ = ‪Mn = nominal flexural moment‬‬ ‫ﻟﻨﮕﺮ ﺧﻤﻴﺮﻱ = ‪= Mp = Plastic moment‬‬ ‫‪= FyZ < 1.6 FyS‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ = ‪Z = plastic section modulus‬‬ ‫ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺍﺭﺗﺠﺎﻋﻲ = ‪S = elastic section modulus‬‬ ‫‪12‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻌﻴﻴﻦ ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺧﻤﻴﺮﻱ ‪Z‬‬ ‫ ﺑﺮﺍﻱ ﻧﻴﻤﺮﺧﻬﺎﻱ ﺍﻣﺮﻳﻜﺎﻳﻲ ‪ Zx‬ﻭ ‪ Zy‬ﻣﺴﺘﻘﻴﻤﺎ ﺍﺯ ﺟﺪﺍﻭﻝ ﻣﺮﺑﻮﻃﻪ ﻗﺎﺑﻞ ﺍﺳﺘﺨﺮﺍﺝ ﺍﺳﺖ‬ ‫)ﻣﻄﺎﺑﻖ ﺍﺳﻼﻳﺪ ﺑﻌﺪﻱ(‪.‬‬ ‫ ﺑﺮﺍﻱ ﻧﻴﻤﺮﺧﻬﺎﻱ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ ﻣﺘﺪﺍﻭﻝ ﺩﺭ ﺍﻳﺮﺍﻥ‪:‬‬ ‫‪ 9‬ﻣﺤﺎﺳﺒﻪ ‪ : Zx‬ﺑﺎ ﺩﻭﺑﺮﺍﺑﺮ ﻛﺮﺩﻥ ))ﻟﻨﮕﺮ ﺍﺳﺘﺎﺗﻴﻜﻲ ﻧﻴﻢ ﺳﻄﺢ ﻣﻘﻄﻊ ﺣﻮﻝ ﻣﺤﻮﺭ ‪(( x‬‬ ‫‪ 9‬ﻣﺤﺎﺳﺒﻪ ‪ : Zy‬ﺑﻮﺍﺳﻄﻪ ﺻﺮﻓﻨﻈﺮ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﺧﻤﻴﺮﻱ ﺟﺎﻥ ﻭ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺍﺑﻌﺎﺩ ﺑﺎﻝ ﺑﻪ‬ ‫ﺻﻮﺭﺕ ﻣﺴﺘﻄﻴﻠﻲ )ﺗﻮﺳﻂ ﺭﺍﺑﻄﻪ ‪( 2tfbf2/4‬‬ ‫‪13‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 14 ‫ﻣﺤﺪﻭﺩﻳﺖ ﻓﻮﺍﺻﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ‪ ،‬ﻧﺎﺣﻴﻪ ‪1‬‬ ‫ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺩﺭ ﻧﺎﺣﻴﻪ ‪ 1‬ﻣﻴﺘﻮﺍﻥ ﺍﺯ ﺗﺤﻠﻴﻞ ﺍﺭﺗﺠﺎﻋﻲ)ﺍﻻﺳﺘﻴﻚ( ﻳﺎ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ‬ ‫)ﭘﻼﺳﺘﻴﻚ( ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫ ﺩﺭ ﻃﺮﺍﺣﻲ ﺑﺮ ﺍﺳﺎﺱ ﺗﺤﻠﻴﻞ ﺍﺭﺗﺠﺎﻋﻲ ﻓﺎﺻﻠﻪ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ)‪ (Lb‬ﻧﺒﺎﻳﺪ ﺍﺯ ‪ Lp‬ﺗﺠﺎﻭﺯ ﻛﻨﺪ‪:‬‬ ‫*‬ ‫¸· ‪300 ry‬‬ ‫‪E §¨ 2510ry‬‬ ‫‪or‬‬ ‫¨‬ ‫‪Fy‬‬ ‫‪Fy‬‬ ‫‪Fy ¸¹‬‬ ‫©‬ ‫‪1.76 ry‬‬ ‫‪Lp‬‬ ‫ﺑﺮﺍﻱ ﻧﻴﻤﺮﺧﻬﺎﻱ ‪ I‬ﻭ ‪ U‬ﺩﺭ ﺧﻤﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ﻗﻮﻱ‬ ‫ ﺩﺭ ﻃﺮﺍﺣﻲ ﺑﺮ ﺍﺳﺎﺱ ﺗﺤﻠﻴﻞ ﺧﻤﻴﺮﻱ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ )‪ (Lb‬ﻧﺒﺎﻳﺪ ﺍﺯ ‪ Lpd‬ﺗﺠﺎﻭﺯ ﻛﻨﺪ‪:‬‬ ‫‪ª‬‬ ‫· ‪§ M1 · º § E‬‬ ‫¨¨‬ ‫‪¸¸» ¨ ¸ ry‬‬ ‫‬ ‫‪0‬‬ ‫‪.‬‬ ‫‪12‬‬ ‫‪0‬‬ ‫‪.‬‬ ‫‪076‬‬ ‫«‬ ‫¸ ¨‬ ‫‪© M 2 ¹¼ © Fy ¹‬‬ ‫¬‬ ‫‪L pd‬‬ ‫·‪§E‬‬ ‫‪ª‬‬ ‫· ‪§ M1 · º § E‬‬ ‫¨‬ ‫¸‬ ‫¨‬ ‫¸‬ ‫¨ ‪«0.17 0.10‬‬ ‫‪» ¨ ¸ ry t 0.1¨¨ ¸¸ ry‬‬ ‫¸‬ ‫‪© M 2 ¹¼ © Fy ¹‬‬ ‫¬‬ ‫‪© Fy ¹‬‬ ‫‪L pd‬‬ ‫‪15‬‬ ‫ﺑﺮﺍﻱ ﻧﻴﻤﺮﺧﻬﺎﻱ ‪ I‬ﺩﺭ ﺧﻤﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ﻗﻮﻱ‬ ‫ﺑﺮﺍﻱ ﻣﻴﻠﻪ ﻫﺎﻱ ﻣﺴﺘﻄﻴﻠﻲ ﺗﻮﭘﺮ ﻭ ﺗﻴﺮﻫﺎﻱ ﻗﻮﻃﻲ‬ ‫ﺷﻜﻞ ﻣﺘﻘﺎﺭﻥ ﺩﺭ ﺧﻤﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ﻗﻮﻱ‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺤﺪﻭﺩﻳﺖ ﻓﻮﺍﺻﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ‪ ،‬ﻧﺎﺣﻴﻪ ‪) 1‬ﺍﺩﺍﻣﻪ(‬ ‫ﻛﻪ‪:‬‬ ‫‪ = M1‬ﻟﻨﮕﺮ ﻛﻮﭼﻜﺘﺮ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ﺗﻴﺮ‬ ‫‪ = M2‬ﻟﻨﮕﺮ ﺑﺰﺭﮔﺘﺮ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ﺗﻴﺮ‬ ‫‪ = ry‬ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﺣﻮﻝ ﻣﺤﻮﺭ ﺿﻌﻴﻒ‬ ‫ﺗﻮﺟﻪ‪ (M1/M2) :‬ﻭﻗﺘﻲ ﻣﺜﺒﺖ ﺍﺳﺖ ﻛﻪ ﻟﻨﮕﺮﻫﺎ ﺑﺎﻋﺚ ﺍﻳﺠﺎﺩ ﺍﻧﺤﻨﺎﻱ ﻣﻀﺎﻋﻒ ~‬ ‫ﻭ ﻭﻗﺘﻲ ﻣﻨﻔﻲ ﺍﺳﺖ ﻛﻪ ﺑﺎﻋﺚ ﺍﻳﺠﺎﺩ ﺍﻧﺤﻨﺎﻱ ﻣﻨﻔﺮﺩ ﺩﺭ ﺗﻴﺮ ﺷﻮﻧﺪ‪.‬‬ ‫ﺩﺭ ﺗﻴﺮ ﺷﻮﻧﺪ‬ ‫ﺑﺮ ﺍﺳﺎﺱ ﺗﻮﺿﻴﺤﺎﺕ ‪ AISC05 Appendix 1-2‬ﺗﺤﻠﻴﻞ ﻭ ﻃﺮﺍﺣﻲ ﻏﻴﺮ ﺍﻻﺳﺘﻴﻚ ﺗﻨﻬﺎ ﺑﺮﺍﻱ‬ ‫ﺍﻋﻀﺎﻳﻲ ﻗﺎﺑﻞ ﭘﺬﻳﺮﺵ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻧﻬﺎ ‪ Fy ≤ 4500 kgf/cm2‬ﺑﺎﺷﺪ ﺑﺪﻳﻦ ﻣﻔﻬـﻮﻡ ﻛـﻪ ﻓـﻮﻻﺩ‬ ‫ﻣﺼﺮﻓﻲ ﺑﺎﻳﺪ ﺍﺯ ﻧﻮﻉ ﺷﻜﻞ ﭘﺬﻳﺮ ﺑﻮﺩﻩ ﻭ ﺩﺍﺭﺍﻱ ﭘﻠﻪ ﺧﻤﻴﺮﻱ ﺑﺎﺷﺪ‪.‬‬ ‫ﻫﻴﭻ ﺣﺪﻱ ﺑﺮﺍﻱ ﻓﻮﺍﺻﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ )‪ (Lb‬ﺩﺭ ﺗﻴﺮﻫﺎﻱ ﺑﺎ ﻣﻘﻄﻊ ﻣﺮﺑﻊ‪ ،‬ﺩﺍﻳﺮﻩ ﻭ ﻳﺎ ﺩﺭﺗﻴـﺮ‬ ‫ﺗﺤﺖ ﺧﻤﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ﺿﻌﻴﻒ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‪.‬‬ ‫‪16‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻮﺭﺍﺥ ﺩﺭ ﺗﻴﺮﻫﺎ‬ ‫ ﺍﻏﻠﺐ ﻻﺯﻡ ﺍﺳﺖ ﺩﺭ ﺗﻴﺮﻫﺎﻱ ﻓﻮﻻﺩﻱ ﺳﻮﺭﺍﺧﻬﺎﻳﻲ ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﭘﻴﭽﻲ ﻳﺎ ﻋﺒﻮﺭ ﻟﻮﻟﻪﻫﺎ‪ ،‬ﻛﺎﻧﺎﻟﻬﺎ ﻭ‬ ‫‪ ...‬ﺍﻳﺠﺎﺩ ﮔﺮﺩﺩ‪.‬‬ ‫ ﺍﻳﺠﺎﺩ ﺳﻮﺭﺍﺥ ﺩﺭ ﺟﺎﻥ ﺗﻴﺮ‪ ،‬ﻛﺎﻫﺶ ﺷﺪﻳﺪﻱ ﺩﺭ ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﻭ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻣﻘﻄﻊ‬ ‫ﺍﻳﺠﺎﺩ ﻧﻤﻲﻛﻨﺪ ﺍﻣﺎ ﻭﺟﻮﺩ ﺳﻮﺭﺍﺥ ﺑﺰﺭگ ﺩﺭ ﺟﺎﻥ‪ ،‬ﻣﻘﺎﻭﻣﺖ ﺑﺮﺷﻲ ﻣﻘﻄﻊ ﻓﻮﻻﺩﻱ ﺭﺍ ﺑﺸﺪﺕ ﻛﺎﻫﺶ‬ ‫ﻣﻲﺩﻫﺪ‪.‬‬ ‫ ﺍﮔﺮ ﻧﻴﺮﻭﻱ ﺑﺮﺷﻲ ﺩﺭ ﺗﻴﺮ ﻛﻮﭼﻚ ﺑﺎﺷﺪ ﺑﻬﺘﺮ ﺍﺳﺖ‬ ‫ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺟﺎﻥ ﺗﻴﺮ ﺍﻳﺠﺎﺩ ﺷﻮﻧﺪ‪.‬‬ ‫ ﺍﮔﺮ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺩﺭ ﺗﻴﺮ ﻛﻮﭼﻚ ﺑﺎﺷﺪ ﺑﻬﺘﺮ ﺍﺳﺖ‬ ‫ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺑﺎﻝ ﺗﻴﺮ ﺍﻳﺠﺎﺩ ﺷﻮﻧﺪ‪.‬‬ ‫‪17‬‬ ‫ﺳﻮﺭﺍﺥ ﺩﺭ ﺗﻴﺮﻫﺎ )ﺍﺩﺍﻣﻪ(‬ ‫‬ ‫‬ ‫ﺍﺛﺮﻛﺎﻫﺸﻲ ﻭﺟﻮﺩ ﺳﻮﺭﺍﺥ ﺩﺭ ﺑﺎﻝ ﻛﺸﺸﻲ ﺗﻴﺮﻫﺎ ﺑﺮ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ ‪ Mn‬ﺑﺮ ﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ‬ ‫‪ AISC05 F13.1 p 16.1-61‬ﺑﺸﺮﺡ ﺯﻳﺮ ﺗﻌﻴﻴﻦ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﺩﺭ ﺻﻮﺭﺕ ﻭﺟﻮﺩ ﺳﻮﺭﺍﺥ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ ‪ Mn‬ﺑﺎﻳﺪ ﺑﺮ ﻃﺒﻖ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺑﺎﻝ‬ ‫ﺗﺤﺖ ﻛﺸﺶ ﻧﻴﺰ ﻣﺤﺪﻭﺩ ﺷﻮﺩ‪:‬‬ ‫‪(a‬‬ ‫‪(b‬‬ ‫‪18‬‬ ‫ﺍﮔﺮ ‪ Fu A f n ≥ Yt Fy Afg‬ﺑﺎﺷﺪ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻛﺸﺶ ﺍﻋﻤﺎﻝ ﻧﻤﻲﺷﻮﺩ)ﺗﺴﻠﻴﻢ‬ ‫ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ ﺑﺎﻝ ﭘﻴﺶ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺑﺎﻝ(‪.‬‬ ‫ﺍﮔﺮ ‪ Fu A f n < Yt Fy Afg‬ﺑﺎﺷﺪ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ ‪ Mn‬ﺩﺭ ﻣﺤﻞ ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺑﺎﻝ‬ ‫ﻛﺸﺸﻲ ﻧﺒﺎﻳﺪ ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻣﻘﺪﺍﺭ ﺯﻳﺮ ﺩﺭ ﻧﻈﺮﮔﺮﻓﺘﻪ ﺷﻮﺩ‪:‬‬ ‫ﻛﻪ ﺩﺭ ﺁﻥ‪:‬‬ ‫‪ = Afg‬ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ)ﻧﺎﺧﺎﻟﺺ( ﺑﺎﻝ ﻛﺸﺸﻲ‬ ‫‪ = Afn‬ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺑﺎﻝ ﻛﺸﺸﻲ‬ ‫‪ = Sx‬ﺍﺳﺎﺱ ﻣﻘﻄﻊ ﺍﺭﺗﺠﺎﻋﻲ ﺣﻮﻝ ﻣﺤﻮﺭ ‪x‬‬ ‫‪ 1.0 = Yt‬ﺑﺮﺍﻱ ‪Fy / Fu ≤ 0.8‬‬ ‫= ‪ 1.1‬ﺑﺮﺍﻱ ﺳﺎﻳﺮ ﻧﺴﺒﺘﻬﺎ‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻮﺭﺍﺥ ﺩﺭ ﺗﻴﺮﻫﺎ )ﺍﺩﺍﻣﻪ(‬ ‫ﭘﺲ ﺑﺮﺍﻱ ﻓﻮﻻﺩ ﻣﻌﻤﻮﻟﻲ )‪ ( Fu=3700kgf/cm2 , Fy=2333kgf/cm2‬ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ‪:‬‬ ‫‪Fy/Fu=0.63 < 0.8 → Yt=1.0‬‬ ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ ‪ Yt Fy / Fu=0.63‬ﺍﺳﺖ‪:‬‬ ‫ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻛﺸﺶ ﺍﻋﻤﺎﻝ ﻧﻤﻴﺸﻮﺩ → ‪If A f n ≥0.63 Afg‬‬ ‫‪Fu A fn‬‬ ‫‪Sx‬‬ ‫‪A fg‬‬ ‫‪19‬‬ ‫‪Mn d‬‬ ‫→‬ ‫‪If A f n < 0.63 Afg‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 20 Urmia University, M. Sheidaii 21 ‫ﺗﻜﻴﻪﮔﺎﻩ ﺟﺎﻧﺒﻲ ﺗﻴﺮﻫﺎ‬ ‫‪Lateral Support of Beams‬‬ ‫‬ ‫ﺗﻌﺒﻴﻪ ﺗﻜﻴﻪﮔﺎﻫﻬﺎﻱ ﺟﺎﻧﺒﻲ ﺑﺮﺍﻱ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺑﺎﻋﺚ ﺍﻓﺰﺍﻳﺶ ﻣﻘﺎﻭﻣﺖ ﺗﻴﺮ ﺩﺭ ﺑﺮﺍﺑﺮ ﻛﻤﺎﻧﺶ‬ ‫ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‬ ‫ﺍﻧﻮﺍﻉ ﻣﻬﺎﺭﺑﻨﺪﻱ ﺟﺎﻧﺒﻲ‬ ‫‪(a‬‬ ‫‪22‬‬ ‫ﻣﻬﺎﺭﺑﻨﺪﻱ ﺟﺎﻧﺒﻲ ﭘﻴﻮﺳﺘﻪ ﺑﺎﻝ ﻓﺸﺎﺭﻱ‪ :‬ﻛﻪ ﺩﺭ ﺁﻥ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺩﺭ ﻓﻮﺍﺻﻞ ﺑﺴﻴﺎﺭ‬ ‫ﻛﻮﺗﺎﻩ ﺗﻮﺳﻂ ﺩﺍﻟﻬﺎﻱ ﺑﺘﻦﺁﺭﻣﻪ ﻳﺎ ﻋﺮﺷﻪﻫﺎﻱ ﻓﻠﺰﻱ ﻭ ‪ ...‬ﺗﺎﻣﻴﻦ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻜﻴﻪﮔﺎﻩ ﺟﺎﻧﺒﻲ ﺗﻴﺮﻫﺎ‬ ‫‪Lateral Support of Beams‬‬ ‫‪ (b‬ﻣﻬﺎﺭﺑﻨﺪﻱ ﺟﺎﻧﺒﻲ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺩﺭ ﻓﻮﺍﺻﻞ‪ :‬ﻛﻪ ﺩﺭ ﺁﻥ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺩﺭ‬ ‫ﻓﻮﺍﺻﻞ ﻣﺠﺰﺍﻳﻲ ﺗﻮﺳﻂ ﺩﻳﮕﺮ ﺍﻋﻀﺎﻱ ﺳﻴﺴﺘﻢ ﻣﻬﺎﺭﺑﻨﺪﻱ ﻓﺮﺍﻫﻢ ﻣﻲﺷﻮﻧﺪ‪.‬‬ ‫‪23‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻧﻜﺎﺕ ﻭﻳﮋﻩ ﺩﺭ ﻃﺮﺍﺣﻲ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ‬ ‫ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺑﺎﻳﺪ ﺩﺍﺭﺍﻱ ﺳﺨﺘﻲ ﻭ ﻣﻘﺎﻭﻣﺖ ﻛﺎﻓﻲ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺍﻧﺘﻘﺎﻝ ﺟﺎﻧﺒﻲ ﻣﻘﻄﻊ‬ ‫ﻋﺮﺿﻲ ﺑﺎﺷﻨﺪ‪.‬‬ ‫ ﺧﻮﺷﺒﺨﺘﺎﻧﻪ ﻧﻴﺮﻭﻱ ﻋﻀﻮ ﻣﻬﺎﺭﻱ ﻣﻘﺪﺍﺭ ﺑﺰﺭﮔﻲ ﻧﺒﻮﺩﻩ ﻭ ﻣﻘﺪﺍﺭ ﻣﺘﺪﺍﻭﻝ ﺁﻥ ‪2.0-2.5%‬‬ ‫ﻧﻴﺮﻭﻱ ﻓﺸﺎﺭﻱ ﺩﺭ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﻣﻴﺒﺎﺷﺪ‪.‬‬ ‫ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺑﺎﻳﺪ ﺑﻪ ﻧﺤﻮﻱ ﻃﺮﺍﺣﻲ ﺷﻮﻧﺪ ﻛﻪ ﺩﺭ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ‬ ‫ﻣﻮﺛﺮ ﻭﺍﻗﻊ ﺷﻮﻧﺪ‪.‬‬ ‫‪24‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻧﻜﺎﺕ ﻭﻳﮋﻩ ﺩﺭ ﻃﺮﺍﺣﻲ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ‬ ‫‪25‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻴﺮﻫﺎﻱ ﻓﺸﺮﺩﻩ ﻣﻬﺎﺭﻧﺸﺪﻩ )ﻧﻮﺍﺣﻲ ‪2‬ﻭ‪(3‬‬ ‫‪Unbraced Compact Beams‬‬ ‫ ﺍﮔﺮ ﻓﺎﺻﻠﻪ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ‪ Lb‬ﺑﺰﺭﮔﺘﺮ ﺍﺯ ‪ Lp‬ﺑﺎﺷﺪ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﭘﻴﭽﺸﻲ ﺗﻴﺮ ﺑﻪ ﺻﻮﺭﺕ‬ ‫ﻏﻴﺮﺍﻻﺳﺘﻴﻚ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﭘﻴﭽﺸﻲ ﻏﻴﺮﺍﻻﺳﺘﻴﻚ‬ ‫‪Lp< Lb ≤ Lr o‬‬ ‫ ﺍﮔﺮ ﻓﺎﺻﻠﻪ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ‪ Lb‬ﺑﺰﺭﮔﺘﺮ ﺍﺯ ‪ Lr‬ﺑﺎﺷﺪ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﭘﻴﭽﺸﻲ ﺗﻴﺮ ﺑﻪ ﺻﻮﺭﺕ‬ ‫ﺍﻻﺳﺘﻴﻚ ﺑﻮﺩﻩ ﺩﺭ ﻫﻨﮕﺎﻡ ﻛﻤﺎﻧﺶ‪ ،‬ﺗﻨﺶ ﻫﻴﭻ ﺗﺎﺭﻱ ﺍﺯ ﻣﻘﻄﻊ ﺗﻴﺮ ﺑﻪ ﺗﻨﺶ ﺗﺴﻠﻴﻢ ﻧﺨﻮﺍﻫﺪ ﺭﺳﻴﺪ‪.‬‬ ‫‪Lr < L b o‬‬ ‫ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﭘﻴﭽﺸﻲ ﺍﻻﺳﺘﻴﻚ‬ ‫ ﺑﻪ ﺩﻟﻴﻞ ﻭﺟﻮﺩ ﺗﻨﺸﻬﺎﻱ ﭘﺴﻤﺎﻧﺪ ﺩﺭ ﺗﻴﺮ‪ ،‬ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﺟﺎﺭﻱ‬ ‫ﺷﺪﻥ ﻣﻘﻄﻊ ﺩﺭ ﺗﻨﺸﻲ ﺑﺮﺍﺑﺮ ﺑﺎ ‪ Fy-Fr=0.7Fy‬ﺍﻳﺠﺎﺩ ﺷﻮﺩ‪.‬‬ ‫‪ Fr‬ﺗﻨﺶ ﭘﺴﻤﺎﻧﺪ ﻓﺸﺎﺭﻱ ﻣﻘﻄﻊ ﺍﺳﺖ ﻛﻪ ﺑﺮﺍﻱ ﺳﺎﺩﮔﻲ ﺑﺮﺍﺑﺮ ﺑﺎ‬ ‫‪ 0.3Fy‬ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪26‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﺮﻳﺐ ﺍﺻﻼﺣﻲ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ‪Cb‬‬ ‫‪Modification Factor for Non uniform Moment‬‬ ‫ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺑﺴﻴﺎﺭ ﻣﺤﺘﻤﻞ ﺑﺮﺍﻱ ﺍﻳﺠﺎﺩ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﭘﻴﭽﺸﻲ ﺑﺎﺭﮔﺬﺍﺭﻳﻲ ﺍﺳﺖ ﻛﻪ ﺍﻳﺠﺎﺩ ﻟﻨﮕﺮ‬ ‫ﺧﻤﺸﻲ ﻳﻜﻨﻮﺍﺧﺖ ﺩﺭ ﺗﻴﺮ ﻣﻲ ﻧﻤﺎﻳﺪ ﺯﻳﺮﺍ ﺩﺭ ﭼﻨﻴﻦ ﺣﺎﻟﺘﻲ ﺍﺳﺖ ﻛﻪ ﻛﻞ ﺑﺎﻝ ﻓﻮﻗﺎﻧﻲ ﺗﺤﺖ‬ ‫ﻓﺸﺎﺭ ﻳﻜﻨﻮﺍﺧﺖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ)ﺗﻴﺮ ﺑﺎ ﻳﻚ ﺍﻧﺤﻨﺎ ﺧﻤﻴﺪﻩ ﺧﻮﺍﻫﺪ ﺷﺪ(‪.‬‬ ‫ ﺩﺭ ﺣﺎﻻﺗﻲ ﻛﻪ ﻟﻨﮕﺮ ﺩﺭ ﻃﻮﻝ ﺗﻴﺮ ﺛﺎﺑﺖ ﻧﺒﺎﺷﺪ‪ ،‬ﺑﺎﻟﻬﺎﻱ ﺗﻴﺮ ﺗﺤﺖ ﻓﺸﺎﺭ ﻣﺘﻐﻴﺮ ﺑﻮﺩﻩ ﻭ ﻣﻤﻜﻦ‬ ‫ﺍﺳﺖ ﺍﺯ ﻛﺸﺶ ﺑﻪ ﻓﺸﺎﺭ ﺗﻐﻴﻴﺮ ﻳﺎﺑﻨﺪ ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﻓﻘﻂ ﺑﺨﺸﻲ ﺍﺯ ﺑﺎﻝ ﺗﻴﺮ ﺗﺤﺖ ﻓﺸﺎﺭ ﺑﺎﺷﺪ)ﺗﻴﺮ‬ ‫ﺑﺎ ﻳﻚ ﺍﻧﺤﻨﺎ ﺧﻤﻴﺪﻩ ﻧﺸﻮﺩ(‪.‬‬ ‫‪27‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﺮﻳﺐ ﺍﺻﻼﺣﻲ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ‪Cb‬‬ ‫‪Modification Factor for Non uniform Moment‬‬ ‫ ﺍﻏﻠﺐ ﺗﻴﺮﻫﺎ ﺑﺎ ﻳﻚ ﺍﻧﺤﻨﺎ ﺧﻤﻴﺪﻩ ﻧﻤﻲ ﺷﻮﻧﺪ )ﺑﻪ ﻋﻠﺖ ﺷﺮﺍﻳﻂ ﺗﻜﻴﻪ ﮔﺎﻫﻲ ﻭ ﺑﺎﺭﮔﺬﺍﺭﻱ ﻣﺨﺘﻠﻒ(‬ ‫ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﻋﻤﻼ ﺗﻴﺮﻫﺎ ﺩﺍﺭﺍﻱ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺑﻴﺸﺘﺮﻱ ﺧﻮﺍﻫﻨﺪ ﺑﻮﺩ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺍﺛﺮ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺭﻭﻱ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ ﭘﻴﭽﺸﻲ‪ ،‬ﺩﺭ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﺍﺯ‬ ‫ﺿﺮﻳﺐ ﺍﺻﻼﺣﻲ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ‪ Cb‬ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﺿﺮﻳﺐ ‪ Cb‬ﺑﺰﺭﮔﺘﺮ ﻳﺎ ﻣﺴﺎﻭﻱ ﻭﺍﺣﺪ ﺑﻮﺩﻩ‪ ،‬ﺩﺭ ﻟﻨﮕﺮ ﻣﻘﺎﻭﻡ ﺍﺳﻤﻲ ‪ Mn‬ﺿﺮﺏ ﻣﻲ ﺷﻮﺩ ﻭ ﺩﺭ‬ ‫ﻧﺘﻴﺠﻪ ﺑﻪ ﻇﺮﻓﻴﺖ ﺧﻤﺸﻲ ﺑﺎﻻﻱ ﺗﻴﺮ ﻣﻨﺠﺮ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﻣﻲ ﺗﻮﺍﻥ ﺑﻄﻮﺭ ﻣﺤﺎﻓﻈﻪ ﻛﺎﺭﺍﻧﻪ ﺑﺮﺍﻱ ﻫﻤﻪ ﺣﺎﻻﺕ ‪ Cb =1‬ﻓﺮﺽ ﻧﻤﻮﺩ ﻭﻟﻲ ﺑﺎ ﭼﻨﻴﻦ ﻓﺮﺿﻲ‪،‬‬ ‫ﻋﻤﻼ ﺍﺯ ﺻﺮﻓﻪ ﺟﻮﻳﻲ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪ ﺍﻱ ﺩﺭ ﻭﺯﻥ ﻓﻮﻻﺩ ﻣﺼﺮﻓﻲ ﺻﺮﻓﻨﻈﺮ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪28‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﺮﻳﺐ ﺍﺻﻼﺣﻲ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ‪Cb‬‬ ‫‪ = Mmax‬ﻗﺪﺭ ﻣﻄﻠﻖ ﺣﺪﺍﻛﺜﺮ ﻟﻨﮕﺮ ﺩﺭ ﻗﺴﻤﺖ ﻣﻬﺎﺭ ﻧﺸﺪﻩ‪،‬‬ ‫‪ = MA‬ﻗﺪﺭ ﻣﻄﻠﻖ ﻟﻨﮕﺮ ﺩﺭ ﻧﻘﻄﻪ ﻳﻚ ﭼﻬﺎﺭﻡ ﻗﺴﻤﺖ ﻣﻬﺎﺭ ﻧﺸﺪﻩ‪،‬‬ ‫‪ = MB‬ﻗﺪﺭ ﻣﻄﻠﻖ ﻟﻨﮕﺮ ﺩﺭ ﻣﺮﻛﺰ ﻗﺴﻤﺖ ﻣﻬﺎﺭ ﻧﺸﺪﻩ‪،‬‬ ‫‪ = MC‬ﻗﺪﺭ ﻣﻄﻠﻖ ﻟﻨﮕﺮ ﺩﺭ ﻧﻘﻄﻪ ﺳﻪ ﭼﻬﺎﺭﻡ ﻗﺴﻤﺖ ﻣﻬﺎﺭ ﻧﺸﺪﻩ‪،‬‬ ‫‪ = Rm‬ﭘﺎﺭﺍﻣﺘﺮ ﺗﻚ ﺗﻘﺎﺭﻧﻲ ‪ monosymmetry‬ﻣﻘﻄﻊ‪،‬‬ ‫= ‪ 1.0‬ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﺩﺍﺭﺍﻱ ﺗﻘﺎﺭﻥ ﺩﻭﮔﺎﻧﻪ ‪،doubly symmetric members‬‬ ‫= ‪ 1.0‬ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﺩﺍﺭﺍﻱ ﺗﻘﺎﺭﻥ ﺳﺎﺩﻩ ‪singly symmetric‬ﺗﺤﺖ ﺧﻤﺶ ﺑﺎ ﺍﻧﺤﻨﺎﻱ ﻣﻨﻔﺮﺩ ‪l‬‬ ‫= ‪ 0.5+(Iyc/Iy) 2‬ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﺩﺍﺭﺍﻱ ﺗﻘﺎﺭﻥ ﺳﺎﺩﻩ ﺗﺤﺖ ﺧﻤﺶ ﺑﺎ ﺍﻧﺤﻨﺎﻱ ﻣﻀﺎﻋﻒ~‬ ‫‪ = Iy‬ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺣﻮﻝ ﻣﺤﻮﺭ ﺍﺻﻠﻲ‪،y‬‬ ‫‪ = Iyc‬ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺑﺎﻝ ﺗﺤﺖ ﻓﺸﺎﺭ ﺣﻮﻝ ﻣﺤﻮﺭ ﺍﺻﻠﻲ ‪ .y‬ﺑﺮﺍﻱ ﺧﻤﺶ ﺑﺎ ﺍﻧﺤﻨﺎﻱ ﻣﻀﺎﻋﻒ‪ ،‬ﺍﺯ ﺑﺎﻝ ﻓﺸﺎﺭﻱ‬ ‫ﻛﻮﭼﻜﺘﺮ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪.‬‬ ‫‪29‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺜﺎﻝ‪ :‬ﻣﻄﻠﻮﺑﺴﺖ ﺗﻌﻴﻴﻦ ‪ Cb‬ﺑﺮﺍﻱ ﺗﻴﺮﻱ ﺑﺎ ﺗﻜﻴﻪ ﮔﺎﻫﻬﺎﻱ ﺳﺎﺩﻩ ﺗﺤﺖ ﺑﺎﺭﮔﺬﺍﺭﻱ ﮔﺴﺘﺮﺩﻩ ﻳﻜﻨﻮﺍﺧﺖ ﺑﺎ‬ ‫ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻓﻘﻂ ﺩﺭ ﺩﻭ ﺍﻧﺘﻬﺎ‪.‬‬ ‫ﺣﻞ‪ :‬ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺗﻘﺎﺭﻥ‪ ،‬ﻟﻨﮕﺮ ﺣﺪﺍﻛﺜﺮ ﺩﺭ ﻭﺳﻂ ﺩﻫﺎﻧﻪ ﺍﺳﺖ‪:‬‬ ‫ﻫﻤﭽﻨﻴﻦ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺗﻘﺎﺭﻥ‪ ،‬ﻟﻨﮕﺮ ﺩﺭ ﻧﻘﻄﻪ ﻳﻚ ﭼﻬﺎﺭﻡ ﺑﺎ ﻟﻨﮕﺮ ﻧﻘﻄﻪ ﺳﻪ ﭼﻬﺎﺭﻡ ﺑﺮﺍﺑﺮ ﺍﺳﺖ‬ ‫‪30‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺿﺮﻳﺐ ﺍﺻﻼﺣﻲ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ‪ – Cb‬ﻣﻘﺎﺩﻳﺮ ﻧﻤﻮﻧﻪ‬ ‫‪31‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺗﻴﺮﻫﺎﻱ ﻣﻬﺎﺭﻧﺸﺪﻩ‬ ‫‪Strength of Unbraced Beams‬‬ ‫ ﻳﻚ ﺗﻴﺮ ﻗﻮﻱ ﺗﺮ ﺧﻮﺍﻫﺪ ﺷﺪ ﺍﮔﺮ‪:‬‬ ‫ﺍﻟﻒ( ﻓﻮﺍﺻﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ )‪ (Lb‬ﻛﻮﭼﻜﺘﺮ ﺷﻮﺩ‪،‬‬ ‫ﺏ( ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ﺑﺰﺭﮔﺘﺮ ﺷﻮﺩ‪.‬‬ ‫ﺩﺭ ﺍﺩﺍﻣﻪ ﺑﻪ ﺑﺮﺭﺳﻲ ﺍﺛﺮ ﺍﻳﻦ ﺩﻭ ﻋﺎﻣﻞ ﺩﺭ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺗﻴﺮ ﭘﺮﺩﺍﺧﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪32‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Lb ‫ ﺍﺛﺮ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ‬-‫ﺍﻟﻒ‬ Effect of Unbraced Length Urmia University, M. Sheidaii 33 ‫ﺍﻟﻒ‪ -‬ﺍﺛﺮ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ‪Lb‬‬ ‫‪Effect of Unbraced Length‬‬ ‫ ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﺭﺍﺑﻄﻪ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﺍﺳﻤﻲ ﻭ ﻃﻮﻝ ﻣﻬﺎﺭ ﻧﺸﺪﻩ ‪ Lb‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬ ‫‪34‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻟﻒ‪ -‬ﺍﺛﺮ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ‪Lb‬‬ ‫‪Effect of Unbraced Length‬‬ ‫ ﻣﻌﺎﺩﻟﻪ ﺗﺌﻮﺭﻳﻚ ﺑﺮﺍﻱ ﻣﻘﺎﻭﻣﺖ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﺍﻻﺳﺘﻴﻚ ﺑﺮﺍﻱ ﻣﻘﺎﻃﻊ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ ﺗﺤﺖ ﺑﺎﺭ‬ ‫ﺩﺭ ﺻﻔﺤﻪ ﺟﺎﻥ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‪:‬‬ ‫ﻛﻪ‪:‬‬ ‫‪S‬‬ ‫‪S2 EC w‬‬ ‫‪EI y GJ 1 2‬‬ ‫‪Lb‬‬ ‫‪L b GJ‬‬ ‫‪M cr‬‬ ‫‪ Lb‬ﻃﻮﻝ ﻣﻬﺎﺭ ﻧﺸﺪﻩ ﺑﺨﺸﻲ ﺍﺯ ﺗﻴﺮ ﺗﺤﺖ ﻟﻨﮕﺮﻫﺎﻱ ﺍﻧﺘﻬﺎﻳﻲ ﻣﺴﺎﻭﻱ‪،‬‬ ‫‪ Iy‬ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﺣﻮﻝ ﻣﺤﻮﺭ ﺿﻌﻴﻒ ﺗﺮ‪،‬‬ ‫‪ G‬ﻣﺪﻭﻝ ﺑﺮﺷﻲ ﻣﺼﺎﺍﻟﺢ‪،‬‬ ‫ﻣﺪﻭﻝ ﺍﺭﺗﺠﺎﻋﻲ ﻣﺼﺎﺍﻟﺢ‪،‬‬ ‫‪E‬‬ ‫ﻣﻤﺎﻥ ﺍﻳﻨﺮﺳﻲ ﻗﻄﺒﻲ ﻣﻘﻄﻊ)‪) (cm4‬ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ ﺛﺎﺑﺖ ﭘﻴﭽﺶ ﺳﻦ ﻭﻧﺎﻥ ﺑﺮﺍﺑﺮ ﺑﺎ‬ ‫‪J‬‬ ‫‪،(∑bt3/3‬‬ ‫‪ Cw‬ﺛﺎﺑﺖ ﭘﻴﭽﺶ ﺗﺎﺑﻴﺪﮔﻲ ﻣﻘﻄﻊ)‪.(cm6‬‬ ‫‪35‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻟﻒ‪ -‬ﺍﺛﺮ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ‪Lb‬‬ ‫‪Effect of Unbraced Length‬‬ ‫ ﻣﻌﺎﺩﻟﻪ ﺗﺌﻮﺭﻳﻚ ﻣﺰﺑﻮﺭ ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺶ ﺗﻴﺮﻫﺎﻳﻲ ﻛﻪ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ﺁﻧﻬﺎ‬ ‫ﺑﺰﺭﮔﺘﺮ ﺍﺯ ‪ Lr‬ﺑﺎﺷﺪ ﻗﺎﺑﻞ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺳﺖ)ﻧﺎﺣﻴﻪ ﻛﻤﺎﻧﺶ ﺍﻻﺳﺘﻴﻚ(‪.‬‬ ‫ ‪ Lr‬ﻓﺎﺻﻠﻪ ﺍﻱ ﺍﺯ ﻣﻬﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﺑﻪ ﻣﺤﺾ ﺭﺳﻴﺪﻥ ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﺗﻴﺮ ﺑﻪ‬ ‫ﺗﻨﺶ ﺗﺴﻠﻴﻢ‪ ،‬ﻛﻤﺎﻧﺶ ﺍﺗﻔﺎﻕ ﻣﻲ ﺍﻓﺘﺪ‪.‬‬ ‫ ﺑﺎ ﻓﺮﺽ ﻭﺟﻮﺩ ﺗﻨﺶ ﭘﺴﻤﺎﻧﺪ ﺣﺪﺍﻛﺜﺮﻱ ﺑﻪ ﻣﻴﺰﺍﻥ ‪ ، Fr=0.3Fy‬ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺍﺳﻤﻲ ﻣﻘﻄﻊ‬ ‫ﺑﻪ ﺍﺯﺍﻱ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ‪ Lr‬ﺑﺮﺍﺑﺮ ﺑﺎ ‪ Mr=(Fy-Fr)Sx=0.7FySx‬ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ‪ Lr‬ﻛﺎﻓﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺗﺌﻮﺭﻳﻚ ﻓﻮﻕ ﺍﻟﺬﻛﺮ‪ ،‬ﻣﻘﺪﺍﺭ ‪ Lb‬ﺑﻪ ﺍﺯﺍﻱ‬ ‫‪ Mn=Mr=0.7FySx‬ﺗﻌﻴﻴﻦ ﺷﻮﺩ‪،‬‬ ‫ ﻇﺮﻓﻴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻣﻘﻄﻊ ﺩﺭ ﻧﺎﺣﻴﻪ ‪) II‬ﻧﺎﺣﻴﻪ ﻛﻤﺎﻧﺶ ﻏﻴﺮﺍﻻﺳﺘﻴﻚ( ﻳﻌﻨﻲ ﻭﻗﺘﻲ‬ ‫‪ Lp d LbdLr‬ﺍﺳﺖ ﺭﻭﻱ ﺧﻂ ﻣﺴﺘﻘﻴﻤﻲ ﻛﻪ ﺑﻴﻦ ‪ Mp‬ﺩﺭ ‪ ، Lp‬ﻭ ‪ Mr‬ﺩﺭ ‪ Lr‬ﻭﺍﻗﻊ ﺷﺪﻩ‬ ‫ﺍﺳﺖ ﻗﺮﺍﺭ ﻣﻲ ﮔﻴﺮﺩ‪.‬‬ ‫‪36‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺏ‪ -‬ﺍﺛﺮ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ‪Cb‬‬ ‫‪Effect of Moment Gradient‬‬ ‫ ﺍﺛﺮ ﺗﻐﻴﻴﺮﺍﺕ ﻟﻨﮕﺮ ﺑﻪ ﺳﺎﺩﮔﻲ ﺑﺎ ﺿﺮﺏ ﻛﺮﺩﻥ ﺿﺮﻳﺐ ﺍﺻﻼﺣﻲ ﻟﻨﮕﺮ ‪ Cb‬ﺑﻪ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ‬ ‫ﺍﺳﻤﻲ ‪ ، Mn‬ﺍﻋﻤﺎﻝ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﺣﺎﺻﻞ ﻫﻴﭽﮕﺎﻩ ﻧﺒﺎﻳﺪ ﺑﻴﺸﺘﺮ ﺍﺯ ‪ Mp‬ﺷﻮﺩ‪.‬‬ ‫‪.‬‬ ‫‪37‬‬ ‫}‬ ‫‪ Mn ) , Mp‬ﺣﺎﺻﻞ ﺍﺯ ﺗﺤﻠﻴﻞ ‪×( LTB‬‬ ‫‪{C‬‬ ‫‪b‬‬ ‫‪Mn = Min‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC‬ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﻣﻘﺎﻃﻊ ﻓﺸﺮﺩﻩ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ‬ ‫ ﺿﻮﺍﺑﻂ ﻣﺮﺑﻮﻁ ﺑﻪ ﻃﺮﺍﺣﻲ ﻣﻘﺎﻃﻊ ﻓﺸﺮﺩﻩ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ ﺩﺭ ‪ AISC05 F2 p.16.1-47‬ﻗﻴﺪ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬ ‫‪When Lb d Lp‬‬ ‫)‪a‬‬ ‫) ‪When Lp Lb d Lr , Inelastic LTB ( Zone II‬‬ ‫)‪b‬‬ ‫‪When Lr Lb‬‬ ‫)‪c‬‬ ‫) ‪, No LTB ( Zone I‬‬ ‫) ‪, Elastic LTB ( Zone III‬‬ ‫ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﻋﻴﻨﺎ ﻣﺘﻨﺎﻇﺮ ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺗﺌﻮﺭﻳﻚ ﻛﻤﺎﻧﺶ ﺟﺎﻧﺒﻲ‪-‬ﭘﻴﭽﺸﻲ ﺍﻻﺳﺘﻴﻚ ﺫﻛﺮ ﺷﺪﻩ ﺩﺭ ﻗﺴﻤﺖ ﻗﺒﻞ ﺍﺳﺖ‪.‬‬ ‫ﺭﻭﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC‬ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﻣﻘﺎﻃﻊ ﻓﺸﺮﺩﻩ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ‬ ‫ ﻃﻮﻟﻬﺎﻱ ﻣﺤﺪﻭﺩ ﻛﻨﻨﺪﻩ ‪ Lp‬ﻭ ‪ Lr‬ﺍﺯ ﺭﻭﺍﺑﻂ ﺯﻳﺮ ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﺩ‪:‬‬ ‫ ﻗﺴﻤﺖ ﺭﺍﺩﻳﻜﺎﻟﻲ ﻣﻌﺎﺩﻟﻪ )‪ (F2-4‬ﻣﻲ ﺗﻮﺍﻧﺪ ﺑﻄﻮﺭ ﺧﻴﻠﻲ ﻣﺤﺎﻓﻈﻪ ﻛﺎﺭﺍﻧﻪ ﻣﻌﺎﺩﻝ ﺑﺎ ‪1.0‬‬ ‫ﺩﺭﻧﻈﺮﮔﺮﻓﺘﻪ ﺷﻮﺩ‪ .‬ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﻣﻌﺎﺩﻟﻪ )‪ (F2-6‬ﺗﺒﺪﻳﻞ ﻣﻲ ﺷﻮﺩ ﺑﻪ‪:‬‬ ‫ ‪ rts‬ﺩﺭ ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‪:‬‬ ‫‪39‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ‪ AISC‬ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﻣﻘﺎﻃﻊ ﻓﺸﺮﺩﻩ ‪ I‬ﻭ ‪ U‬ﺷﻜﻞ‬ ‫ ﺑﺮﺍﻱ ﻣﻘﺎﻃﻊ ﺷﻜﻞ ‪ I‬ﺩﺍﺭﺍﻱ ﺗﻘﺎﺭﻥ ﺩﻭﮔﺎﻧﻪ ﺑﺎ ﺑﺎﻟﻬﺎﻱ ﻣﺴﺘﻄﻴﻞ ﺷﻜﻞ‪ ،‬ﺩﺍﺭﻳﻢ‪:‬‬ ‫ﻟﺬﺍ ﻣﻌﺎﺩﻟﻪ )‪ (F2-7‬ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﺯﻳﺮ ﺗﺒﺪﻳﻞ ﻣﻲ ﺷﻮﺩ‪:‬‬ ‫‪ = h0‬ﻓﺎﺻﻠﻪ ﻣﺮﺍﻛﺰ ﺑﺎﻟﻬﺎ = ‪d-tf‬‬ ‫ ‪ rts‬ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺪﻗﺖ ﻭ ﺑﻄﻮﺭ ﻣﺤﺎﻓﻈﻪ ﻛﺎﺭﺍﻧﻪ ﺑﻪ ﻋﻨﻮﺍﻥ ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﺑﺎﻝ ﻓﺸﺎﺭﻱ ﺑﻪ ﻋﻼﻭﻩ‬ ‫ﻳﻚ ﺷﺸﻢ ﺟﺎﻥ ﺗﻘﺮﻳﺐ ﺯﺩ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ‪:‬‬ ‫ ﺿﺮﻳﺐ ‪ c‬ﺩﺭ ﺍﻳﻦ ﺭﻭﺍﺑﻂ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‪:‬‬ ‫‪40‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﺑـﺮﺍﻱ ﺍﻧـﻮﺍﻉ ﻣﻘـﺎﻃﻊ ﺍﻣﺮﻳﻜـﺎﻳﻲ ﺑـﻪ ﺍﺯﺍﻱ‬ ‫ﻃﻮﻟﻬﺎﻱ ﻣﻬﺎﺭﻧﺸﺪﻩ ﻣﺨﺘﻠﻒ‪ ،‬ﺗﻌﻴﻴﻦ ﺷﺪﻩ ﻭ ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻣﺮﺑﻮﻃـﻪ‬ ‫ﻛـــﻪ ﻧﻤﻮﺩﺍﺭﻫـــﺎﻱ ﻃﺮﺍﺣـــﻲ ﺗﻴـــﺮ ﻧﺎﻣﻴـــﺪﻩ ﻣـــﻲ ﺷـــﻮﺩ ﺩﺭ‬ ‫‪ Manual of Steel Construction‬ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺩﺭ ﺍﻳﻦ ﻧﻤﻮﺩﺍﺭﻫﺎ ﻧﻘﻄﻪ ﻣﺘﻨﺎﻇﺮ ﺑﺎ ﻃﻮﻝ ﻣﻬﺎﺭﻧﺸﺪﻩ ‪ Lb‬ﺑﻪ ﺻﻮﺭﺕ‬ ‫ﺩﺍﻳﺮﻩ ﺗﻮﭘﺮ ‪ x‬ﻭ ﻧﻘﻄﻪ ﻣﺘﻨﺎﻇﺮ ﺑﺎ ‪ Lr‬ﺑـﺎ ﻧﻤـﺎﺩ ﺩﺍﻳـﺮﻩ ﺗﻮﺧـﺎﻟﻲ ‪R‬‬ ‫ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺍﻳﻦ ﻧﻤﻮﺩﺍﺭﻫﺎ ﻓﻘﻂ ﺑﺮﺍﻱ ﻃﺮﺍﺣﻲ ﺗﺤـﺖ ﺧﻤـﺶ ﺑـﻮﺩﻩ ﻭ ﺑـﺪﻭﻥ‬ ‫ﺗﻮﺟﻪ ﺑﻪ ﻣﻮﺍﺭﺩﻱ ﻫﻤﭽﻮﻥ ﺧﻴﺰ ﻭ ﺑﺮﺵ ﻭ‪ ...‬ﻛﻪ ﺩﺭ ﺑﻌﻀﻲ ﻣـﻮﺍﺭﺩ‬ ‫ﻣﻲ ﺗﻮﺍﻧﻨﺪ ﺣﺎﻛﻢ ﺑﺮ ﻃﺮﺍﺣﻲ ﺑﺎﺷﻨﺪ ﺍﺳﺘﺨﺮﺍﺝ ﺷﺪﻩ ﺍﻧﺪ‪.‬‬ ‫‪41‬‬ ‫ ﺍﻳــﻦ ﻧﻤﻮﺩﺍﺭﻫــﺎ ﺑــﺮ ﻣﺒﻨــﺎﻱ ‪ Cb=1.0‬ﻃﺮﺍﺣــﻲ ﺷــﺪﻩ ﺍﻧــﺪ ﺍﮔــﺮ‬ ‫‪ Cb>1.0‬ﺑﺎﺷﺪ ﻣﻘﺪﺍﺭ ﺑﺪﺳﺖ ﺁﻣﺪﻩ ﺍﺯ ﻧﻤﻮﺩﺍﺭ ﺑﺮﺍﻱ ﻟﻨﮕﺮ ﺧﻤﺸﻲ‬ ‫ﻃﺮﺍﺣﻲ ﻋﻀﻮ‪ ،‬ﺑﺎﻳﺪ ﺑﺎ ﺿﺮﺏ ﻧﻤﻮﺩﻥ ﺩﺭ ‪ Cb‬ﺗﺼﺤﻴﺢ ﺷﻮﺩ‪ ،‬ﺍﻟﺒﺘﻪ‬ ‫ﻣﻘﺪﺍﺭ ﺣﺎﺻﻞ ﻧﺒﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ‪ Ib FyZ‬ﻣﻨﻈﻮﺭ ﺷﻮﺩ‪.‬‬ ‫‪Design Charts‬‬ ‫ﻧﻤﻮﺩﺍﺭﻫﺎﻱ ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫ ﺑﺮﺍﻱ ﺍﻧﺘﺨﺎﺏ ﻳﻚ ﻣﻘﻄﻊ ﻣﻨﺎﺳﺐ ﺗﻮﺳـﻂ ﻧﻤﻮﺩﺍﺭﻫـﺎﻱ ﻃﺮﺍﺣـﻲ‬ ‫ﻛﺎﻓﻴﺴﺖ ﺑﺎ ﺩﺍﺷﺘﻦ ‪ Lb‬ﻭ ‪) Mu‬ﺍﻟﺒﺘﻪ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ‪Cb≠1.0‬‬ ‫ﺑﺎﺷﺪ ﺑـﺎ ‪ ( Mu /Cb‬ﻧﻘﻄـﻪ ﻣﺘﻨـﺎﻇﺮ ﺑـﺮ ﺭﻭﻱ ﻧﻤـﻮﺩﺍﺭ ﻣﻨﺎﺳـﺐ‬ ‫ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﺩ‪ .‬ﻫﺮ ﻣﻘﻄﻌﻲ ﻛﻪ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﻭ ﺑﺎﻻﻱ ﻧﻘﻄﻪ‬ ‫ﻣﺰﺑﻮﺭ )‪ (Ü‬ﻗﺮﺍﺭ ﮔﻴﺮﺩ ﻗﺎﺑﻞ ﻗﺒﻮﻝ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ ﺧﻄﻮﻁ ﻣﻘﻄﻊ ) ‪ ( - - -‬ﺩﺭ ﻣﺤﺪﻭﺩﻩ ﻏﻴﺮﺍﻗﺘﺼـﺎﺩﻱ ﺑـﻮﺩﻩ ﻭ ﺑـﺎ‬ ‫ﺣﺮﻛﺖ ﺑﻴﺸﺘﺮ ﺑﻪ ﺑﺎﻻ ﻭ ﺭﺍﺳﺖ‪ ،‬ﺍﻭﻟﻴﻦ ﺧﻂ ﭘﺮ )—— ( ﻣﻌـﺮﻑ‬ ‫ﺳﺒﻜﺘﺮﻳﻦ ﻣﻘﻄﻊ ﻗﺎﺑﻞ ﻗﺒﻮﻝ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫‪42‬‬ ‫‪Design Charts‬‬ ‫ﺟﺪﺍﻭﻝ ﻃﺮﺍﺣﻲ ﺗﻴﺮ‬ ‫‪43‬‬ ‫ﺟﺪﺍﻭﻝ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮﺍﻱ ﻣﻘﺎﻃﻊ ‪ IPE‬ﻣﺘﺪﺍﻭﻝ ﺩﺭ ﺍﻳﺮﺍﻥ‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪44‬‬ ‫ﺟﺪﺍﻭﻝ ﻃﺮﺍﺣﻲ ﺗﻴﺮ ﺑﺮﺍﻱ ﻣﻘﺎﻃﻊ ‪ IPB‬ﻣﺘﺪﺍﻭﻝ ﺩﺭ ﺍﻳﺮﺍﻥ‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪45‬‬ Urmia University, M. Sheidaii 46 Urmia University, M. Sheidaii 47 Urmia University, M. Sheidaii 48 Urmia University, M. Sheidaii 49 Urmia University, M. Sheidaii 50 Urmia University, M. Sheidaii 51 Urmia University, M. Sheidaii 52 Urmia University, M. Sheidaii 53 Urmia University, M. Sheidaii 54 17.5ft Urmia University, M. Sheidaii 55 Urmia University, M. Sheidaii 56 Urmia University, M. Sheidaii 57 Urmia University, M. Sheidaii 58 Urmia University, M. Sheidaii 59 Tension Members ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬: ‫ﻓﺼﻞ ﺳﻮﻡ‬ Urmia University, M. Sheidaii 1 ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Tension Members‬‬ ‫ ﺍﻋﻀﺎﻱ ﺳﺎﺯﻩ ﺍﻱ ﻛﻪ ﺗﺤﺖ ﻧﻴﺮﻭﻱ ﻛﺸﺸﻲ ﻣﺤﻮﺭﻱ ﻫﺴﺘﻨﺪ )ﺍﻋﻀﺎﻱ‬ ‫ﺧﺮﭘﺎ ‪ ،‬ﻛﺎﺑﻠﻬﺎ ﺩﺭ ﭘﻠﻬﺎﻱ ﻣﻌﻠﻖ ‪ ،‬ﻣﻬﺎﺭﺑﻨﺪﻱ ﺳﺎﺧﺘﻤﺎﻧﻬﺎ ‪( ... ،‬‬ ‫‪2‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻧﻮﺍﻉ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Types of Tension Members‬‬ ‫ ﻫﺮ ﺷﻜﻞ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﮔﻴﺮﺩ‪ ،‬ﭼﻮﻥ‬ ‫ﺗﻨﻬﺎ ﻋﺎﻣﻞ ﺗﺎﺛﻴﺮﮔﺬﺍﺭ ﺩﺭ ﻣﻘﺎﻭﻣﺖ ﻋﻀﻮ‪ ،‬ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﺍﺳﺖ‪.‬‬ ‫ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻴﻤﺮﺥ ﻫﺎﻱ ﻧﻮﺭﺩﺷﺪﻩ ﻧﺒﺸﻲ ﺷﻜﻞ ﻭ ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺩﺍﻳﺮﻩﺍﻱ‬ ‫ﺑﺴﻴﺎﺭ ﻣﺘﺪﺍﻭﻝ ﺍﺳﺖ‪.‬‬ ‫‪3‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Holes in Tension Members‬‬ ‫ ﺳﻮﺭﺍﺧﻬﺎ ﺩﺭ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﻭ ﺍﺛﺮ ﻣﻬﻢ ﺩﺍﺭﻧﺪ ‪:‬‬ ‫‪ (1‬ﺗﻤﺮﻛﺰ ﺗﻨﺶ )‪(Stress Concentration‬‬ ‫‪ (2‬ﻛﺎﻫﺶ ﻣﻘﻄﻊ ﻋﺮﺿﻲ )‪(area reduction‬‬ ‫ ﺭﻭﺵ ﻣﺘﺪﺍﻭﻝ ﺳﻮﺭﺍﺧﻜﺎﺭﻱ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﻣﺘﻪ ﺯﺩﻥ ﻳﺎ ﭘﺎﻧﭻ ﻛﺮﺩﻥ ﺳﻮﺭﺍﺥ ﻫﺎﻱ ﺍﺳﺘﺎﻧﺪﺍﺭﺩﻱ ﺑﺎ‬ ‫ﻗﻄﺮﻱ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ‪1.5‬ﻣﻴﻠﻴﻤﺘﺮ) ‪ 1/16‬ﺍﻳﻨﭻ( ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻗﻄﺮ ﭘﻴﭻ‪.‬‬ ‫ ﺑﺮﺍﻱ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻥ ﻧﺎﻫﻤﻮﺍﺭﻱ ﻫﺎ ﻭ ﺯﺑﺮﻱ ﺩﻭﺭ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ‪ AISC ،‬ﻣﻠﺰﻡ ﻣﻲ ﻧﻤﺎﻳﺪ ‪ 1.5‬ﻣﻴﻠﻴﻤﺘﺮ‬ ‫)‪ 1/16‬ﺍﻳﻨﭻ( ﺩﻳﮕﺮ ﻋﻤﻼ ﺑﺮﺍﻱ ﻗﻄﺮ ﺳﻮﺭﺍﺥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﻮﺩ‪.‬‬ ‫ ﺑﻨﺎﺑﺮﺍﻳﻦ ﻗﻄﺮ ﻣﻮﺛﺮ ﺳﻮﺭﺍﺥ‪ 3 ،‬ﻣﻴﻠﻴﻤﺘﺮ )‪ 1/8‬ﺍﻳﻨﭻ( ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻗﻄﺮ ﭘﻴﭻ ﻣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫‪4‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ﻭ ﺳﻄﺢ ﻣﻘﻄﻊ‬ ‫ﺧﺎﻟﺺ )‪(Ag, An‬‬ ‫‪Gross and Net Areas‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ )‪ (Ag‬ﻳﻚ ﻋﻀﻮ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻛﻞ‬ ‫ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺁﻥ‪.‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ = ‪Ag = Gross Area‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ = ‪An = Net Area‬‬ ‫‪An = Ag - Aholes‬‬ ‫‪5‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﺜﺎﻝ ‪:‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ﻭ ﺧﺎﻟﺺ ﻭﺭﻕ ‪ 1.5×20 cm‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ‪.‬‬ ‫ﺍﺗﺼﺎﻝ ﻭﺭﻕ ﺩﺭ ﺍﻧﺘﻬﺎ ﺑﺎ ﺩﻭ ﺭﺩﻳﻒ ﭘﻴﭻ ‪ 1.9‬ﺳﺎﻧﺘﻴﻤﺘﺮﻱ ﺍﻧﺠﺎﻡ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪ Ag = 20 ×1.5 = 30 mm2‬‬ ‫‪ An = 30 - 2(1.9 + 0.3)(1.5) = 23.4 mm2‬‬ ‫‪6‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Design Strength of Tension Members‬‬ ‫ ﺩﻭ ﻣﻜﺎﻧﻴﺰﻡ ﺧﺮﺍﺑﻲ ﻣﻬﻢ ﺑﺮﺍﻱ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪ .‬ﺍﻳﻦ ﺣﺎﻟﺖ ﻫﺎﻱ ﺣﺪﻱ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ ‪:‬‬ ‫‪ (1‬ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ)‪(yielding on the gross area‬‬ ‫‪ (2‬ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪(fracture on the net section‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ﻭ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﻫﺎﻱ ﺍﺿﺎﻓﻲ ﻧﺎﺷﻲ ﺍﺯ ﺁﻥ‪ ،‬ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫)‪ (Ag‬ﺑﺎﻳﺪ ﻛﻮﭼﻜﺘﺮ ﺍﺯ ﺗﻨﺶ ﺗﺴﻠﻴﻢ )‪ (Fy‬ﺑﺎﺷﺪ‪.‬‬ ‫‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪ ،‬ﺗﻨﺶ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪ (An‬ﺑﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﺗﻨﺶ‬ ‫ﻧﻬﺎﻳﻲ)‪ (Fu‬ﺑﺎﺷﺪ‪.‬‬ ‫‪7‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ‬ ‫‪Nominal Strength‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺩﺭ ﺗﺴﻠﻴﻢ ‪،‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺩﺭ ﮔﺴﻴﺨﺘﮕﻲ ‪،‬‬ ‫‪P n = F y Ag‬‬ ‫‪Pn = Fu Ae‬‬ ‫‪ Ae‬ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺼﻲ ﺍﺳﺖ ﻛﻪ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﺩﺭ ﻣﻘﻄﻊ ﮔﺬﺭﻧﺪﻩ ﺍﺯ‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎ ﺩﺭ ﺑﺮﺍﺑﺮ ﻛﺸﺶ ﻣﻘﺎﻭﻣﺖ ﻣﻲ ﻛﻨﺪ‪.‬‬ ‫‪ Ae‬ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﺨﺎﻃﺮ ﻋﻮﺍﻣﻠﻲ ﻫﻤﭽﻮﻥ ﺗﻤﺮﻛﺰ ﺗﻨﺶ ﻭ ﺑﺮﺧﻲ ﻋﻮﺍﻣﻞ‬ ‫ﺩﻳﮕﺮ ‪ ،‬ﻗﺪﺭﻱ ﻛﻤﺘﺮ ﺍﺯ ‪ An‬ﺑﺎﺷﺪ‪.‬‬ ‫‪8‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫‪Yielding on Gross Area‬‬ ‫‪ = It Pn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ‪ LRFD‬ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﺩﺭ ﻛﺸﺶ‬ ‫‪ =Pn / Ω‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﺠﺎﺯ ‪ ASD‬ﺑﺮﺍﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ﺩﺭ ﻛﺸﺶ‬ ‫ﻛﻪ ‪:‬‬ ‫)‪It = 0.90 (LRFD‬‬ ‫)‪Ω = 1.67 (ASD‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ = ‪Pn‬‬ ‫‪= FyAg‬‬ ‫‪9‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ‬ ‫‪Fracture on Net Section‬‬ ‫‪ = It Pn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ‪ LRFD‬ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﻛﺸﺶ‬ ‫‪ =Pn / Ω‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﺠﺎﺯ ‪ ASD‬ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﻛﺸﺶ‬ ‫ﻛﻪ ‪:‬‬ ‫)‪It = 0.75 (LRFD‬‬ ‫)‪Ω = 2.00(ASD‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻋﻀﻮ = ‪Pn‬‬ ‫‪= FuAe‬‬ ‫ﺿﺮﻳﺐ ﻛﺎﻫﺶ ﻣﻘﺎﻭﻣﺖ ﺑﺮﺍﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻛﻤﺘﺮ ﺍﺯ ﺗﺴﻠﻴﻢ ﺍﺳﺖ‪ .‬ﭼﻮﻥ ﺭﺳﻴﺪﻥ ﺑﻪ ﺣﺎﻟﺖ‬ ‫ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺑﺴﻴﺎﺭ ﺧﻄﺮﻧﺎﻛﺘﺮ ﺍﺳﺖ‪.‬‬ ‫‪10‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ )‪(LRFD‬‬ ‫‪Design Strength‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﺩﺭ ﻣﻘﻄﻊ ﻛﻞ ﻛﻪ ﻫﺪﻑ ﺍﺯ ﺍﻋﻤﺎﻝ ﺁﻥ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺍﻓﺰﺍﻳﺶ‬ ‫ﻃﻮﻝ ﺑﻴﺶ ﺍﺯ ﺣﺪ ﻋﻀﻮ ﺍﺳﺖ ‪:‬‬ ‫‪ It = 0.90‬؛ ‪Pu ≤ It Fy Ag‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺩﺭ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺳﻮﺭﺍﺧﻬﺎﻱ ﭘﻴﭻ ﻳﺎ‬ ‫ﭘﺮچ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ‪:‬‬ ‫‪ It = 0.75‬؛ ‪Pu ≤ It Fu Ae‬‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ﻛﻤﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺍﺯ ﺩﻭ ﻣﻘﺪﺍﺭ ﻓﻮﻕ ﺍﺳﺖ‪.‬‬ ‫‪11‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ )‪(Ae‬‬ ‫‪Effective Net Area‬‬ ‫ ﻫﻨﮕﺎﻣﻲ ﻛﻪ ﻫﻤﻪ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﺍﺗﺼﺎﻝ ﻧﻴﺎﻓﺘﻪ ﺑﺎﺷﻨﺪ )ﻣﺜﻼ ‪ :‬ﻓﻘﻂ ﻳﻚ ﭘﺎﻱ‬ ‫ﻧﺒﺸﻲ ﺑﻪ ﻭﺭﻕ ﻟﭽﻜﻲ ﭘﻴﭽﻜﺎﺭﻱ ﺷﻮﺩ( ‪ ،‬ﭘﺪﻳﺪﻩ ﺗﺎﺧﻴﺮ ﺑﺮﺷﻲ)‪ (shear lag‬ﭘﻴﺶ‬ ‫ﻣﻲﺁﻳﺪ‪.‬‬ ‫ ﺍﺟﺰﺍﻱ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ ﺗﺤﺖ ﺑﺎﺭ ﺑﻴﺸﺘﺮﻱ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﻧﺪ ﻭ ﻗﺴﻤﺖ ﻫﺎﻱ ﺍﺗﺼﺎﻝ ﻧﻴﺎﻓﺘﻪ‬ ‫ﺗﺤﺖ ﺗﻨﺶ ﻛﺎﻣﻞ ﻗﺮﺍﺭ ﻧﻤﻲﮔﻴﺮﻧﺪ)‪.(not fully stressed‬‬ ‫ ﺑﺮﺍﻱ ﺑﻪ ﺣﺴﺎﺏ ﺁﻭﺭﺩﻥ ﺍﻳﻦ ﭘﺪﻳﺪﻩ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﻳﻚ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﻛﺎﻫﺶﻳﺎﻓﺘﻪ ‪ ،‬ﻳﺎ ﻣﻮﺛﺮ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫‪12‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ )‪(Ae‬‬ ‫‪Effective Net Area‬‬ ‫‪Ae = U An‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ ‪Ae = effective net area‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ )‪An = net area (see AISC p. 16.1-14‬‬ ‫ﺿﺮﻳﺐ ﺗﺎﺧﻴﺮ ﺑﺮﺵ ”‪U = reduction factor considering “shear lag‬‬ ‫)ﻳﺎ ﺟﺪﻭﻝ ‪ 1-3-2-10‬ﺹ ‪ 163‬ﻣﺒﺤﺚ ﺩﻫﻢ( ﻳﺎ ‪= See AISC Table D3.1 p. 16.1-29‬‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻛﺸﺸﻲ ﻣﺴﺘﻘﻴﻤﺎ ﺑﻪ ﻫﺮ ﺟﺰء ﺗﻮﺳﻂ ﭘﻴﭻ ﻫﺎ ﻳﺎ ﺟﻮﺵﻫﺎ ﺍﻧﺘﻘﺎﻝ ﻳﺎﺑﺪ ‪= 1.0‬‬ ‫‪x‬‬ ‫ﺍﮔﺮ ﺑﺎﺭ ﻛﺸﺸﻲ ﺑﻪ ﺗﻌﺪﺍﺩﻱ ﺍﺯ ﺍﺟﺰﺍ )ﻭ ﻧﻪ ﻫﻤﻪ ﺁﻧﻬﺎ( ﺗﻮﺳﻂ ﭘﻴﭽﻬﺎ ﻳﺎ ﺟﻮﺵﻫﺎ ﺍﻧﺘﻘﺎﻝ ﻳﺎﺑﺪ ‪= 1‬‬ ‫"‬ ‫ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺍﺗﺼﺎﻝ ‪x = connection eccentricity‬‬ ‫ﻃﻮﻝ ﺍﺗﺼﺎﻝ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ = "‬ ‫‪13‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 14 Urmia University, M. Sheidaii 15 Urmia University, M. Sheidaii 16 Urmia University, M. Sheidaii 17 ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ‬ ‫‪ x‬ﺑﺮﻭﻥ ﻣﺤﻮﺭﻱ ﺍﺗﺼﺎﻝ ﺍﺳﺖ )ﻓﺎﺻﻠﻪ ﻣﺮﻛﺰ ﺳﻄﺢ ﻗﻄﻌﻪ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ ﺗﺎ ﺻﻔﺤﻪ ﺍﺗﺼﺎﻝ(‬ ‫ﺍﮔﺮ ﻋﻀﻮﻱ ﺩﺍﺭﺍﻱ ﺩﻭ ﺻﻔﺤﻪ ﺍﺗﺼﺎﻝ‪ ،‬ﻛﻪ ﺑﻄﻮﺭ ﻣﺘﻘﺎﺭﻥ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪﺍﻧﺪ‪ ،‬ﺑﺎﺷﺪ ‪ x‬ﺍﺯ ﻣﺮﻛﺰ ﻧﺰﺩﻳﻜﺘﺮﻳﻦ‬ ‫ﻧﻴﻤﻪ ﻣﻘﻄﻊ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‬ ‫‬ ‫‪18‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ‬ ‫ﻃﻮﻝ ﺍﺗﺼﺎﻝ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ = "‬ ‫‪19‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪ Ae‬ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﭘﻴﭽﻲ ﻭ ﭘﺮﭼﻲ‬ ‫‪Ae for Bolted and Riveted Connections‬‬ ‫‪Ae = U An‬‬ ‫ ﺍﮔﺮ ﻫﻤﻪ ﺍﺟﺰﺍﻱ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﺘﺼﻞ ﺷﺪﻩ ﺑﺎﺷﻨﺪ ‪U=1 :‬‬ ‫ ﺩﺭ ﻏﻴﺮ ﺍﻳﻦ ﺻﻮﺭﺕ ﺍﺯ ﻣﻘﺎﺩﻳﺮ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺿﺮﻳﺐ ﻛﺎﻫﺸﻲ ‪ U‬ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ ‪:‬‬ ‫)‪(AISC05 Table D3.1 case 2 or cases 7,8‬‬ ‫ ﺑﺮﺍﻱ ﺻﻔﺤﺎﺕ ﻭﺻﻠﻪ ﭘﻴﭽﻲ ﺩﺭ ﺍﺗﺼﺎﻻﺕ ‪:‬‬ ‫‪(AISC05 J4.1 p16.1-112): U=1, Ae=And0.85Ag‬‬ ‫‪20‬‬ ‫‪Urmia University, M. Sheidaii‬‬ U ‫ﻣﻘﺎﺩﻳﺮ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺑﺮﺍﻱ‬ Recommended Values for U Urmia University, M. Sheidaii 21 ‫‪ Ae‬ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﺟﻮﺷﻲ‬ ‫‪Ae for welded connections‬‬ ‫‪transverse‬‬ ‫‪ Ae = U Ag‬‬ ‫‪longitudinal‬‬ ‫‪ (1‬ﺑﺮﺍﻱ ﻫﺮ ﻧﻴﻤﺮﺥ ‪ I‬ﻳﺎ ‪ T‬ﺷﻜﻞ ﻛﻪ ﻓﻘﻂ ﺑﺎ ﺟﻮﺵ ﺟﺎﻧﺒﻲ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ‬ ‫ﺑﺎﺷﺪ ‪:‬‬ ‫‪U=1‬‬ ‫ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﺍﺟﺰﺍﻳﻲ ﻛﻪ ﻣﺴﺘﻘﻴﻤﺎ ﺍﺗﺼﺎﻝ ﻳﺎﻓﺘﻪ ﺍﻧﺪ = ‪Ae‬‬ ‫‪22‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪ Ae‬ﺑﺮﺍﻱ ﺍﺗﺼﺎﻻﺕ ﺟﻮﺷﻲ‪ -‬ﺍﺩﺍﻣﻪ‬ ‫‪Ae for welded connections‬‬ ‫‪ (2‬ﺑﺮﺍﻱ ﻭﺭﻗﻬﺎ ﻳﺎ ﻣﻴﻠﻪ ﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﺍﻧﺘﻬﺎﻳﺸﺎﻥ ﺑﺎ ﺟﻮﺷﻬﺎﻱ ﻃﻮﻟﻲ ﺍﺗﺼﺎﻝ‬ ‫ﻳﺎﻓﺘﻪ ﺍﻧﺪ ‪:‬‬ ‫‪L t 2w‬‬ ‫‪2w ! L t 1.5w‬‬ ‫‪1.5w ! L t w‬‬ ‫‪U=1‬‬ ‫‪U=0.87‬‬ ‫‪U=0.75‬‬ ‫¾‬ ‫¾‬ ‫¾‬ ‫‪ t w‬ﻃﻮﻝ ﺯﻭﺝ ﺟﻮﺷﻬﺎ = ‪L‬‬ ‫ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺟﻮﺷﻬﺎ = ‪w‬‬ ‫‪23‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example 1: Determine the tensile design strength of a IPB 20 with two lines of ¾ inch(1.9cm) diameter in each flange using ST37 steel with Fy=2333 kgf/cm2 and Fu=3700 kgf/cm2.There are assumed at least 3 bolts in each line 4 in on center. PL 210×300×10 Pu/2 IPB 20 Solution. Pu Pu/2 Using an IPB 20 (Ag=78.1cm2, d=20cm, bf=20cm, tf=15mm) (a) Gross section yield Pu ≤ Øt Fy Ag = (0.9) (2333) (78.1) = 164 ton (b) Net section fracture An = 78.1 - (4) (1.9 + 0.3) (1.5) = 64.9 cm2 bf > 2d /3 → U= 0.9 Ae = U An = 0.9 (64.9) = 58.41 cm2 Pu ≤ Øt Fu Ae = (0.75) (3700) (58.41) = 162 ton ← USE 24 Example 2: The tension member of Example 1 is assumed to be connected at its ends with two 300×10mm plates, as shown in Figure. Determine the design tensile force which the plates PL 210×300×10 can transfer. Pu/2 IPB 20 Pu Pu/2 Solution. (a) Gross section yield Pu ≤ Øt Fy Ag = (0.9) (2333) (2×1×30) = 125982 kgf ← USE (b) Net section fracture An of 2 plates= [(1×30) - (2.2×2×1)](2) = 51.2 cm2 An≤ 0.85 Ag= 0.85 (2×1×30) = 51.0 cm2 → An = 51.0 cm2 Pu ≤ Øt Fu An = (0.75) (3700) (51.0) = 141525 kgf Pu ≤ 125982 kgf ≈ 126 ton 25 26 Urmia University, M. Sheidaii 27 ‫ﺳﻮﺭﺍﺧﻬﺎﻱ ﻧﺎﻣﻨﻈﻢ‬ ‫‪Staggered Holes‬‬ ‫ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺑﻴﺸﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺭﺍ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ ﺍﮔﺮ ﭘﻴﭽﻬﺎ ﺩﺭ ﻳﻚ ﺧﻂ ﺑﺎﺷﻨﺪ‪.‬‬ ‫ ﺍﮔﺮ ﺑﻴﺶ ﺍﺯ ﻳﻚ ﺧﻂ ﭘﻴﭻ ﺿﺮﻭﺭﺕ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ )ﻃﻮﻝ ﺍﺗﺼﺎﻝ ﻣﺤﺪﻭﺩ( ‪ ،‬ﺍﻳﺠﺎﺩ‬ ‫ﺳﻮﺭﺍﺥ ﻫﺎﻱ ﻧﺎﻣﻨﻈﻢ )ﻗﻄﺮﻱ ﻳﺎ ﺯﻳﮕﺰﺍگ( ﺑﺎﻋﺚ ﺑﻪ ﺣﺪﺍﻗﻞ ﺭﺳﻴﺪﻥ ﻛﺎﻫﺶ ﺩﺭ ﺳﻄﺢ‬ ‫ﻣﻘﻄﻊ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ ﻣﺴﻴﺮﻫﺎﻱ ﺧﺮﺍﺑﻲ ﻣﻤﻜﻦ ‪ABC or ABDE :‬‬ ‫ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺣﺪﺍﻗﻞ ﺍﻧﺘﺨﺎﺏ ﺷﻮﺩ‪.‬‬ ‫‪28‬‬ ‫ﺭﻭﺵ ﺗﺠﺮﺑﻲ ﻣﺤﺎﺳﺒﻪ ﻣﻘﻄﻊ ﺧﺎﻟﺺ‬ ‫‪net width = gross width – 6 d + 6 s2 / 4g‬‬ ‫ﻋﺮﺽ ﻛﻞ‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫ﻋﺮﺽ ﺧﺎﻟﺺ‬ ‫‪ d‬ﻗﻄﺮ ﺳﻮﺭﺍﺥ ) " ‪ dh + 1/16‬ﻳﺎ ‪(db + 1/8 " ,‬‬ ‫‪ s2/4g‬ﺑﻪ ﺍﺯﺍﻱ ﻫﺮ ﮔﺎﻡ ﻋﺮﺿﻲ ﺩﺭ ﺯﻧﺠﻴﺮﻩ ﻣﻮﺭﺩ ﻧﻈﺮ ‪ ،‬ﺍﺿﺎﻓﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪) s‬ﮔﺎﻡ ﻃﻮﻟﻲ( ‪ :‬ﻓﺎﺻﻠﻪ ﻃﻮﻟﻲ ﻣﺮﻛﺰ ﺑﻪ ﻣﺮﻛﺰ ﻫﺮ ﺩﻭ ﺳﻮﺭﺍﺥ ﻣﺠﺎﻭﺭ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ‬ ‫‪) g‬ﮔﺎﻡ ﻋﺮﺿﻲ( ‪ :‬ﻓﺎﺻﻠﻪ ﻣﺮﻛﺰ ﺑﻪ ﻣﺮﻛﺰ ﺳﻮﺭﺍﺧﻬﺎ ﻋﻤﻮﺩ ﺑﺮ ﺭﺍﺳﺘﺎﻱ ﺑﺎﺭﮔﺬﺍﺭﻱ‬ ‫ﺿﺨﺎﻣﺖ ﻭﺭﻕ × ﻋﺮﺽ ﺧﺎﻟﺺ = )‪ (An‬ﻣﻘﻄﻊ ﺧﺎﻟﺺ‬ ‫‪ (Ae) = U An‬ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﻣﻮﺛﺮ‬ ‫)‪ = Øt Ae Fu (Øt = 0.75‬ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ‬ ‫‪29‬‬ Example : Compute the smallest net area for the plate shown below: Plate thickness = 12 mm Bolt diameter = 19 mm Solution. Urmia University, M. Sheidaii 30 Example : Compute the Minimum pitch (smin) in the other words determine the pitch will give a net area DEFG equal to the one along ABC. Solution. cm Urmia University, M. Sheidaii cm 31 Example : Compute the smallest net area for the angle shown below: Bolt diameter = 19 mm Solution. The unfolding is done at the middle surface to obtain an equivalent plate (with gross width equal to the sum of the leg lengths minus the angle thickness). mm mm cm2 Urmia University, M. Sheidaii 32 Example : Determine the net area along route ABCDEF for the C380×100×54.5 (Ag =69.39 cm2). Holes are for 19 mm Bolts. Solution. The unfolding is done at the middle surface to obtain an equivalent plate. /2 – tf /2 /2 – 16 /2 136.8 136.8 Urmia University, M. Sheidaii 33 ‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪BLOCK SHEAR‬‬ ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻫﻤﭽﻨﻴﻦ ﻣﻤﻜﻦ ﺍﺳﺖ ﺑﻪ ﻋﻠﺖ ﭘﺎﺭﮔﻲ )‪ (tear-out‬ﻣﺼﺎﻟﺢ ﺩﺭ ﻧﻮﺍﺣﻲ‬ ‫ﻣﺠﺎﻭﺭ ﺍﺗﺼﺎﻝ ﺩﭼﺎﺭ ﺧﺮﺍﺑﻲ ﺷﻮﻧﺪ‪ .‬ﺑﻪ ﺍﻳﻦ ﭘﺪﻳﺪﻩ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻣﻲ ﮔﻮﻳﻨﺪ‪.‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻋﻀﻮ ﻛﺸﺸﻲ‬ ‫ﺗﻚ ﻧﺒﺸﻲ‬ ‫‪34‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪BLOCK SHEAR‬‬ ‫ ﻣﺜﺎﻝﻫﺎﻱ ﻣﺘﺪﺍﻭﻝ ﺩﺭ ﺷﻜﻞﻫﺎﻱ ﻣﻘﺎﺑﻞ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫ ﺧﺮﺍﺑﻲ ﺩﺭ ﻃﻮﻝ ﻣﺴﻴﺮﻫﺎﻱ ﺯﻳﺮ ﺍﺗﻔﺎﻕ ﻣﻲﺍﻓﺘﺪ‪:‬‬ ‫‪ .i‬ﻛﺸﺶ ﺩﺭ ﺻﻔﺤﻪ ﺍﻱ ﻋﻤﻮﺩ ﺑﺮ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭ‬ ‫‪ .ii‬ﺑﺮﺵ ﺩﺭ ﺻﻔﺤﻪ ﺍﻱ ﻣﻮﺍﺯﻱ ﺑﺎ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭ‬ ‫ ﻣﻘﺎﻭﻣﺖ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻣﺠﻤﻮﻉ ‪:‬‬ ‫‪ .i‬ﻣﻘﺎﻭﻣﺖ ﺑﺮﺷﻲ ﺩﺭ ﻳﻚ ﻣﺴﻴﺮ ﺧﺮﺍﺑﻲ‬ ‫‪ .ii‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﺭﻭﻱ ﻗﻄﻌﻪﺍﻱ ﻋﻤﻮﺩ ﺑﺮ‬ ‫ﻣﺴﻴﺮ ﺧﺮﺍﺑﻲ ﻓﻮﻕ ﺍﻟﺬﻛﺮ‬ ‫‪35‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬-‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‬ BLOCK SHEAR Urmia University, M. Sheidaii 36 ‫ﻣﻘﺎﻭﻣﺖ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﻣﻘﺎﻭﻣﺖ ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪Block Shear Rupture Strength‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ )‪ (Rn‬ﺩﺭ ﻃﻮﻝ ﺳﻄﺢ ﻳﺎ ﺳﻄﻮﺡ ﺑﺮﺷﻲ ﻭ ﻛﺸﺸﻲ‬ ‫ﻋﻤﻮﺩ ﺑﺮ ﺁﻥ ﺍﺯ ﺭﺍﺑﻄﻪ ﺫﻳﻞ ﺗﻌﻴﻴﻦ ﻣﻲ ﺷﻮﺩ )‪ AISC p. 16.1-112‬ﻭ ﻳﺎ ﺻﻔﺤﻪ ‪ 318‬ﻣﺒﺤﺚ ﺩﻫﻢ(‬ ‫‪Rn = 0.6FuAnv + UbsFuAnt ≤ 0.6FyAgv + UbsFuAnt‬‬ ‫ﻭ ﺑﺮ ﺍﻳﻦ ﺍﺳﺎﺱ ‪:‬‬ ‫ﻛﻪ ‪:‬‬ ‫‪37‬‬ ‫‪LRFD available block shear rupture strength = Ø Rn‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻣﻮﺟﻮﺩ ‪LRFD‬‬ ‫‪ASD allowable block shear rupture strength = Rn / Ω‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﻣﺠﺎﺯ ‪ASD‬‬ ‫)‪Ø = 0.75 (LRFD‬‬ ‫)‪Ω = 2.00 (ASD‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺗﺤﺖ ﺑﺮﺵ ‪Anv = net area subjected to shear‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﺗﺤﺖ ﻛﺸﺶ ‪Ant = net area subjected to tension‬‬ ‫ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻠﻲ ﺗﺤﺖ ﺑﺮﺵ ‪Agv = gross area subjected to shear‬‬ ‫ﺿﺮﻳﺐ ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﻛﺸﺸﻲ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ‪Ubs = 1.0‬‬ ‫ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﻏﻴﺮﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﻛﺸﺸﻲ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ‪= 0.5‬‬ ‫ﺣﻮﺯﻩ ﺑﺮﺷﻲ‪ -‬ﺑﺮﺵ ﻗﺎﻟﺒﻲ‬ ‫‪BLOCK SHEAR‬‬ ‫ﺿﺮﻳﺐ ﻛﺎﻫﺶ ‪ Ubs‬ﺩﺭ ﻣﻌﺎﺩﻟﻪ‪ ،‬ﺑﺮﺍﻱ ﺗﻘﺮﻳﺐ ﺳﺎﺯﻱ ﺗﻮﺯﻳﻊ ﻏﻴﺮ ﻳﻜﻨﻮﺍﺧﺖ ﺗﻨﺶ ﺩﺭ ﺳﻄﺢ ﻛﺸﺸﻲ ﺁﻭﺭﺩﻩ‬ ‫ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫‪38‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Example : A steel angle L6x6x½ using A36 steel is subjected to tensile load. Determine the block shear rupture strength. Urmia University, M. Sheidaii 39 Solution: Anv = net area subjected to shear (in2) = (Matl. Thickness)[Lv – (# bolts)(Bolt dia. + 1/8”)] = ½”[10” – (2.5 holes)(¾” + 1/8”)] = 3.91 in2 Ant = net area subjected to tension (in2) = (Matl. Thickness)[Lt – (# bolts)(Bolt dia. + 1/8”)] = ½”[2½” – (0.5 holes)(¾” + 1/8”)] = 1.03 in2 Agv = gross area subjected to shear (in2) = (Matl. Thickness)(Lv) = (½”)(10”) = 5.0 in2 Rn = Available block shear = 0.6FuAnv + UbsFuAnt = 0.6(58 KSI)(3.91 in2) + (1.00)(58 KSI)(1.03 in2) = 195.8 Kips Check if Rn ≤ 0.6FyAgv + UbsFuAnt ≤ 0.6(36 KSI)(5 in2) + (1.00)(58 KSI)(1.03 in2) ≤ 167.7 Kips 40 Pbs = Øt Rn = 0.75 (167.7) = 125.8 Kips ‫ﺳﺨﻦ ﺁﺧﺮ‬ ‫‪Bottom line:‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪41‬‬ ‫ﻫﺮ ﻛﺪﺍﻡ ﺍﺯ ﺳﻪ ﺣﺎﻟﺖ ﺣﺪﻱ )ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪ ،‬ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ‬ ‫ﺧﺎﻟﺺ ‪ ،‬ﻳﺎ ﺧﺮﺍﺑﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ( ﻣﻲ ﺗﻮﺍﻧﺪ ﺣﺎﻛﻢ ﺑﺮ ﻃﺮﺍﺣﻲ ﺑﺎﺷﺪ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺑﺮﺍﻱ ﻫﺮ ﺳﻪ ﺣﺎﻟﺖ ﺣﺪﻱ ﺑﺎﻳﺪ ﻣﺤﺎﺳﺒﻪ ﺷﻮﺩ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ) ‪ ØPn‬ﻳﺎ ‪ (Pn /Ω‬ﻛﻤﺘﺮﻳﻦ ﺳﻪ ﻣﻘﺪﺍﺭ ﻣﺤﺎﺳﺒﻪ‬ ‫ﺷﺪﻩ ﻓﻮﻕ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻋﻀﻮ ﺑﺎﻳﺪ ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ ﻋﻀﻮ ﻛﺸﺸﻲ‬ ‫ﺑﺎﺷﺪ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 42 Urmia University, M. Sheidaii 43 Urmia University, M. Sheidaii 44 Urmia University, M. Sheidaii 45 ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‬ ‫‪Design of Tension Members‬‬ ‫ﻃﺮﺡ ﻋﻀﻮ ﻛﺸﺸﻲ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﭘﻴﺪﺍ ﻛﺮﺩﻥ ﺳﺒﻚﺗﺮﻳﻦ ﻣﻘﻄﻊ ﻓﻮﻻﺩﻱ )ﻧﺒﺸﻲ‪،I ،‬‬ ‫ﻧﺎﻭﺩﺍﻧﻲ ﻭ ‪ ( ...‬ﺑﺎ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﺑﺰﺭﮔﺘﺮ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ‪.‬‬ ‫ﺑﻪ ﺭﻭﺵ ‪Ø Pn ≥ Pu LRFD‬‬ ‫ﺑﻪ ﺭﻭﺵ ‪Pn / Ω ≥ Pa ASD‬‬ ‫ ‪ Pn‬ﻣﻘﺎﻭﻣﺖ ﺍﺳﻤﻲ ﻃﺮﺍﺣﻲ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪،‬‬ ‫ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ﻭ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ﺍﺳﺖ‪.‬‬ ‫ ‪ Pu‬ﻣﻘﺎﻭﻣﺖ ﻧﻬﺎﻳﻲ ﻻﺯﻡ )ﺑﺎﺭ ﻧﻬﺎﻳﻲ( ﻛﻪ ﺍﺯ ﺗﺤﻠﻴﻞ ﺳﺎﺯﻩ ﺗﺤﺖ ﺗﺮﻛﻴﺒﺎﺕ‬ ‫ﺑﺎﺭﮔﺬﺍﺭﻱ ﺿﺮﻳﺒﺪﺍﺭ )ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭ ‪ (LRFD‬ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ‪.‬‬ ‫ ‪ Pa‬ﻣﻘﺎﻭﻣﺖ ﻻﺯﻡ )ﺑﺎﺭ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ( ﻛﻪ ﺍﺯ ﺗﺤﻠﻴﻞ ﺳﺎﺯﻩ ﺗﺤﺖ ﺗﺮﻛﻴﺒﺎﺕ‬ ‫ﺑﺎﺭﮔﺬﺍﺭﻱ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ )ﺗﺮﻛﻴﺒﺎﺕ ﺑﺎﺭ ‪ (ASD‬ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ‪.‬‬ ‫‪46‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﻪ ﺭﻭﺵ ‪LRFD‬‬ ‫‪Design of Tension Members - LRFD method‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪ØPn = 0.9 Ag Fy ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.9 Ag Fy ≥ Pu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ‪Ag ≥ Pu / 0.9 Fy‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ‪ØPn = 0.75 Ae Fu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.75 Ae Fu ≥ Pu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪ Ae ≥ Pu / 0.75 Fu ،‬ﻳﺎ‬ ‫‪An ≥ Pu / 0.75 Fu U‬‬ ‫ ﻛﻨﺘﺮﻝ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ‪،‬‬ ‫) ‪ØRn = 0.75( 0.6FuAnv + UbsFuAnt)≤ 0.75(0.6FyAgv + UbsFuAnt‬‬ ‫¾‬ ‫‪47‬‬ ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ‪ØRn ≥ Pu ،‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﻪ ﺭﻭﺵ ‪ASD‬‬ ‫‪Design of Tension Members - ASD method‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﻣﻘﻄﻊ ﻛﻞ ‪Pn/Ω = AgFy /1.67 = 0.6AgFy ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.6Ag Fy ≥ Pa ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﺴﻠﻴﻢ ‪Ag ≥ Pa / 0.6 Fy‬‬ ‫ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﻣﻘﻄﻊ ﺧﺎﻟﺺ ‪Pn/Ω = AeFu /2.00 = 0.5AeFu ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪0.5Ae Fu ≥ Pa ،‬‬ ‫¾ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﮔﺴﻴﺨﺘﮕﻲ ‪Ae ≥ Pa / 0.5Fu ،‬‬ ‫ ﻛﻨﺘﺮﻝ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺣﺪﻱ ﮔﺴﻴﺨﺘﮕﻲ ﺣﻮﺯﻩ ﺑﺮﺷﻲ ‪،‬‬ ‫) ‪Rn/Ω = Rn/2.00 = 0.5( 0.6FuAnv + UbsFuAnt)≤ 0.5(0.6FyAgv + UbsFuAnt‬‬ ‫¾‬ ‫‪48‬‬ ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ‪Rn/Ω ≥ Pa ،‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ‪ -‬ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﻻﻏﺮﻱ‬ ‫‪Design of Tension Members - Slenderness Limitations‬‬ ‫ ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪ AISC‬ﺣﺪﻱ ﺑﺮﺍﻱ ﻻﻏﺮﻱ ﺣﺪﺍﻛﺜﺮ ﺩﺭ ﻃﺮﺍﺣﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‪.‬‬ ‫ ﺍﻣﺎ ﺍﮔﺮ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﺩﺭ ﻳﻚ ﻋﻀﻮ ﻛﺸﺸﻲ ﻻﻏﺮ ﺑﺮﺩﺍﺷﺘﻪ ﺷﺪﻩ ﻭ ﺑﺎﺭﻫﺎﻱ ﺟﺎﻧﺒﻲ ﻛﻮﭼﻜﻲ ﺍﻋﻤﺎﻝ‬ ‫ﺷﻮﺩ ‪ ،‬ﺍﺭﺗﻌﺎﺷﺎﺕ ﻭ ﺧﻴﺰﻫﺎﻱ ﻧﺎﻣﻄﻠﻮﺑﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﭘﺪﻳﺪ ﺁﻳﺪ‪ .‬ﻟﺬﺍ ‪ AISC‬ﭘﻴﺸﻨﻬﺎﺩ ﻣﻲ ﻧﻤﺎﻳﺪ ‪:‬‬ ‫)‪L/r t 300 ( not for cables or rods‬‬ ‫ﻛﻪ ‪ r‬ﺷﻌﺎﻉ ژﻳﺮﺍﺳﻴﻮﻥ ﺣﺪﺍﻗﻞ ﻭ ‪ L‬ﻃﻮﻝ ﻋﻀﻮ ﺍﺳﺖ‪.‬‬ ‫ ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻓﻮﻻﺩ ﺍﻳﺮﺍﻥ )ﺻﻔﺤﻪ ‪ 160‬ﻣﺒﺤﺚ ﺩﻫﻢ( ‪ :‬ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﺣﺪﺍﻛﺜﺮ‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻧﺒﺎﻳﺪ ﺍﺯ ‪ 300‬ﺗﺠﺎﻭﺯ ﻛﻨﺪ‪.‬‬ ‫ ﺩﺭ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎﻱ ﻛﺸﺸﻲ ﻛﻪ ﺩﺍﺭﺍﻱ ﭘﻴﺶ ﺗﻨﻴﺪﮔﻲ ﺍﻭﻟﻴﻪ ﺑﻪ ﻣﻘﺪﺍﺭ ﻛﺎﻓﻲ ﺑﺎﺷﻨﺪ ‪ ،‬ﺭﻋﺎﻳﺖ‬ ‫ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﻻﻏﺮﻱ ﻻﺯﻡ ﻧﻴﺴﺖ‪ ،‬ﻟﻴﻜﻦ ﻧﺴﺒﺖ ﻃﻮﻝ ﺑﻪ ﻗﻄﺮ ﺍﻳﻦ ﺍﻋﻀﺎ ﻧﺒﺎﻳﺪ ﺍﺯ ‪ 300‬ﺗﺠﺎﻭﺯ‬ ‫ﻛﻨﺪ‪.‬‬ ‫‪49‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺭﺯﻭﻩ ﺷﺪﻩ ﻭ ﻛﺎﺑﻠﻬﺎ‬ ‫‪Threaded Rods and Cables‬‬ ‫‬ ‫‬ ‫‬ ‫ﺭﺷﺘﻪ‬ ‫‪50‬‬ ‫ﻛﺎﺑﻞ ﺳﻴﻤﻲ‬ ‫‬ ‫ﺍﮔﺮ ﻻﻏﺮﻱ ﻋﻀﻮ ﻣﻮﺭﺩ ﺗﻮﺟﻪ ﻗﺮﺍﺭ ﻧﮕﻴﺮﺩ ‪ ،‬ﺍﻏﻠﺐ ﺍﺯ‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎ ﻳﺎ ﻛﺎﺑﻞﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲﺷﻮﺩ‪) .‬ﺁﻭﻳﺰﻫﺎ ‪،‬‬ ‫ﭘﻠﻬﺎﻱ ﻣﻌﻠﻖ(‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎ ﻋﻨﺎﺻﺮﻱ ﺑﺎ ﻣﻘﻄﻊ ﺗﻮﭘﺮ ﺑﻮﺩﻩ ﻭ ﻛﺎﺑﻞﻫﺎ‬ ‫ﺍﺯ ﺭﺷﺘﻪ ﻫﺎﻱ )ﺳﻴﻢ ﺑﺎﻓﺘﻪ( ﺟﺪﺍﮔﺎﻧﻪ ﭘﻴﭽﻴﺪﻩ ﺑﻪ‬ ‫ﻫﻢ ﺳﺎﺧﺘﻪ ﻣﻲﺷﻮﻧﺪ‪.‬‬ ‫ﻫﺮ ﺭﺷﺘﻪ ﺍﺯ ﺳﻴﻢﻫﺎﻱ ﺟﺪﺍﮔﺎﻧﻪﺍﻱ ﻛﻪ ﺩﻭﺭ ﻳﻚ‬ ‫ﻫﺴﺘﻪ ﻣﺮﻛﺰﻱ ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﺭﭘﻴﭽﻲ ﭘﻴﭽﻴﺪﻩﺍﻧﺪ‬ ‫ﺗﺸﻜﻴﻞ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﻳﻚ ﻛﺎﺑﻞ ﺳﻴﻤﻲ ﺧﻮﺩ ﺍﺯ ﭼﻨﺪﻳﻦ ﺭﺷﺘﻪ ﻛﻪ ﺩﻭﺭ‬ ‫ﻳﻚ ﻫﺴﺘﻪ ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﺭﭘﻴﭽﻲ ﭘﻴﭽﺎﻧﺪﻩ ﺷﺪﻩﺍﻧﺪ‬ ‫ﺗﺸﻜﻴﻞ ﻣﻲﺷﻮﺩ‪.‬‬ ‫‪Threaded Rod‬‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺭﺯﻭﻩ ﺷﺪﻩ ﻭ ﻛﺎﺑﻠﻬﺎ‬ ‫‪Threaded Rods and Cables‬‬ ‫ﻣﻌﻤﻮﻻ ﺍﻳﻦ ﺍﻋﻀﺎ ﺑﻪ ﺳﺎﺯﻩ ﺑﻪ ﻃﺮﻕ ﻣﺨﺘﻠﻔﻲ ﻫﻤﭽﻮﻥ ﺟﻮﺷﻜﺎﺭﻱ ﻳﺎ ﺍﺗﺼﺎﻻﺕ ﭘﻨﺠﻪ‬ ‫ﻣﻔﺼﻠﻲ )‪ (bolted clevises‬ﻭ ﭘﻴﭻﻫﺎ‪ ،‬ﻭﺻﻞ ﻣﻲﺷﻮﻧﺪ‪.‬‬ ‫‪51‬‬ ‫ﻣﻴﻠﮕﺮﺩﻫﺎﻱ ﺭﺯﻭﻩ ﺷﺪﻩ ﻭ ﻛﺎﺑﻠﻬﺎ‬ ‫‪Threaded Rods and Cables‬‬ ‫ ﺭﺯﻭﻩ ﻛﺮﺩﻥ ﺍﻧﺘﻬﺎﻱ ﻋﻀﻮ ﺑﺎﻋﺚ ﻛﺎﻫﺶ ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻲ ﺷﻮﺩ )ﻗﻄﻮﺭ ﻛﺮﺩﻥ ﺍﻧﺘﻬﺎﻱ‬ ‫ﻋﻀﻮ ﺍﺯ ﭼﻨﻴﻦ ﻛﺎﺭﻱ ﺟﻠﻮﮔﻴﺮﻱ ﻣﻲ ﻛﻨﺪ ‪ ،‬ﻭﻟﻲ ﭘﺮﻫﺰﻳﻨﻪ ﺍﺳﺖ(‬ ‫ ﻣﻘﺎﻭﻣﺖ ﻃﺮﺍﺣﻲ ﻛﺸﺸﻲ ﺑﺮﺍﻱ ﻳﻚ ﻣﻴﻠﮕﺮﺩ ﺭﺯﻭﻩ ﺷﺪﻩ ‪:‬‬ ‫)‪I Pn = 0.75 (0.75 AD Fu‬‬ ‫ ﺳﻄﺢ ﻣﻘﻄﻊ ﺭﺯﻭﻩ ﻧﺸﺪﻩ )ﻛﻞ( ﻣﻴﻠﮕﺮﺩ = ‪AD‬‬ ‫ ﺑﻨﺎﺑﺮﺍﻳﻦ ‪AD ≥ Pu / Ø 0.75 Fu ; Ø=0.75‬‬ ‫ ﺑﻪ ‪ Example 3.14 p.70 Segui‬ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ‪.‬‬ ‫‪52‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﺭ ﺳﻘﻒ ﺧﺮﭘﺎﻳﻲ‬ ‫‪Tension Members in Roof Truss‬‬ ‫ ﺧﺮﭘﺎﻫﺎ ﺩﺭ ﻣﻮﺍﺭﺩﻱ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﻧﺪ ﻛﻪ ﻗﻴﻤﺖ ﻭ ﻭﺯﻥ‬ ‫ﺗﻴﺮ ﮔﺮﺍﻥ ﻭ ﺑﺎﺯﺩﺍﺭﻧﺪﻩ ﺑﺎﺷﺪ‪) .‬ﺩﻫﺎﻧﻪ ﻫﺎﻱ ﻃﻮﻳﻞ(‬ ‫ ﻣﻲ ﺗﻮﺍﻥ ﺧﺮﭘﺎ ﺭﺍ ﺗﻴﺮ ﻋﻤﻴﻘﻲ ﺗﺼﻮﺭ ﻛﺮﺩ ﻛﻪ ﺑﺨﺶ ﻋﻤﺪﻩ ﺍﻱ ﺍﺯ ﺟﺎﻥ‬ ‫ﺁﻥ ﺑﺮﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺩﺭ ﺳﻘﻒﻫﺎﻱ ﺧﺮﭘﺎﻳﻲ ﺷﺎﻣﻞ ﺑﻌﻀﻲ ﺍﺯ ﺍﻋﻀﺎﻱ ﺧﺮﭘﺎ ﻭ‬ ‫ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫‪53‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ Sag Rods Urmia University, M. Sheidaii 54 ‫ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ ‫‪Sag Rods‬‬ ‫ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﺑﺮﺍﻱ ﺗﺎﻣﻴﻦ ﺗﻜﻴﻪﮔﺎﻩ ﺟﺎﻧﺒﻲ ﺑﺮﺍﻱ ﻻﭘﻪﻫﺎ ﺑﻜﺎﺭ ﻣﻲﺭﻭﻧﺪ )ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ‬ ‫ﺷﻜﻢ ﺩﺍﺩﻥ ﻻﭘﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﻮﺍﺯﻱ ﺑﺎ ﺷﻴﺐ ﺑﺎﻡ ﺗﺤﺖ ﺑﺎﺭﻫﺎﻱ ﻗﺎﺋﻢ ﻭﺍﺭﺩﻩ(‬ ‫‪55‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ ‫‪Design of Sag Rods‬‬ ‫ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﺑﺮﺍﻱ ﺗﺤﻤﻞ ﺑﺨﺸﻲ ﺍﺯ ﺑﺎﺭﻫﺎﻱ ﺑﺎﻡ ﻛﻪ ﺑﻪ ﻣﻮﺍﺯﺍﺕ ﺑﺎﻡ ﻋﻤﻞ ﻣﻲ ﻛﻨﻨﺪ ﻃﺮﺍﺣﻲ‬ ‫ﻣﻲ ﺷﻮﻧﺪ‪.‬‬ ‫ ﺑﻨﺎ ﺑﻪ ﻓﺮﺽ ﻫﺮ ﻗﻄﻌﻪﺍﻱ ﺍﺯ ﻣﻴﻞ ﻣﻬﺎﺭ ﺑﻴﻦ ﻻﭘﻪﻫﺎ ‪ ،‬ﺑﺎﺭ ﺗﻤﺎﻡ ﻗﺴﻤﺖﻫﺎﻱ ﺯﻳﺮﻳﻦ ﺭﺍ ﺗﺤﻤﻞ‬ ‫ﻣﻲﻛﻨﺪ ؛ ﻟﺬﺍ ﻣﻴﻞ ﻣﻬﺎﺭ ﻓﻮﻗﺎﻧﻲ ﺑﺎﻳﺪ ﺑﺮﺍﻱ ﺳﻄﺢ ﺑﺎﺭﮔﻴﺮ ﻣﻴﻠﻪ ﻛﻪ ﺍﺯ ﺭﺍﺱ ﺗﺎ ﭘﺎﻱ ﺧﺮﭘﺎ‬ ‫ﺍﺳﺖ ‪ ،‬ﻃﺮﺍﺣﻲ ﺷﻮﺩ‪.‬‬ ‫‪56‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ‬ ‫‪Design of Sag Rods‬‬ ‫ ﻣﻴﻞ ﻣﻬﺎﺭ ﻭﺍﻗﻊ ﺩﺭ ﺭﺍﺱ ﺧﺮﭘﺎ ﺑﺎﻳﺪ ﺑﺎﺭ ﻫﻤﻪ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎﻱ ﺩﻳﮕﺮ ﺩﺭ ﻫﺮ ﺩﻭ ﻃﺮﻑ ﺧﺮﭘﺎ ﺭﺍ‬ ‫ﺗﺤﻤﻞ ﻛﻨﺪ‪.‬‬ ‫‪57‬‬ ‫‪T2 = T1 / Cosα‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Urmia University, M. Sheidaii 58 ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﻟﻮﻻﻳﻲ‬ ‫‪PIN-CONNECTED MEMBERS‬‬ ‫ ﺑﺮﺍﻱ ﺍﻳﺠﺎﺩ ﻳﻚ ﺍﺗﺼﺎﻝ ﻋﺎﺭﻱ ﺍﺯ ﺧﻤﺶ ‪ ،‬ﺳﻮﺭﺍﺧﻲ ﺩﺭ ﻋﻀﻮ ﻭ ﻗﻄﻌﺎﺕ ﻣﺘﺼﻞ ﺑﻪ‬ ‫ﺁﻥ ﺍﻳﺠﺎﺩ ﻛﺮﺩﻩ ﻭ ﻳﻚ ﭘﻴﻦ ﺍﺯ ﺳﻮﺭﺍﺥ ﮔﺬﺭﺍﻧﺪﻩ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ‪ ØPn‬ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ‪ ، LRFD‬ﻭ ﻧﻴﺰ ‪ Pn /Ω‬ﻣﻘﺎﻭﻣﺖ‬ ‫ﻛﺸﺸﻲ ﻣﺠﺎﺯ ﺑﻪ ﺭﻭﺵ ‪ ASD‬ﺑﺎﻳﺪ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻻﺕ ﺣﺪﻱ ﺯﻳﺮ ﺗﻌﻴﻴﻦ ﺷﻮﺩ ‪:‬‬ ‫)‪ AISC05 D5.1 p.16.1-28‬ﻭ ﺻﻔﺤﻪ ‪ 167‬ﻣﺒﺤﺚ ﺩﻫﻢ(‬ ‫ﺍﻟﻒ( ﮔﺴﻴﺨﺘﮕﻲ ﻛﺸﺸﻲ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ﺧﺎﻟﺺ ‪:‬‬ ‫‪Øt = 0.75 , Ωt= 2.00 , Pn = 2tbeffFu‬‬ ‫ﺏ( ﮔﺴﻴﺨﺘﮕﻲ ﺑﺮﺷﻲ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻮﺛﺮ ‪:‬‬ ‫‪Øsf = 0.75, Ωsf= 2.00 , Pn = 0.6Fu Asf‬‬ ‫‪59‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﻟﻮﻻﻳﻲ‬ ‫‪PIN-CONNECTED MEMBERS‬‬ ‫پ( ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﺩﺭ ﺳﻄﺢ ﺗﺼﻮﻳﺮ ﺷﺪﻩ ﻗﻠﻢ ﻟﻮﻻ )ﭘﻴﻦ( ‪:‬‬ ‫‪Ø = 0.75 , Ω = 2.00 , Pn = 1.8Fy Apb‬‬ ‫ﺕ( ﺗﺴﻠﻴﻢ ﺩﺭ ﺳﻄﺢ ﻣﻘﻄﻊ ﻛﻞ ‪:‬‬ ‫‪Øt = 0.75, Ωt= 1.67 , Pn = Fy Ag‬‬ ‫ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺭﻭﺍﺑﻂ ‪:‬‬ ‫ﺿﺨﺎﻣﺖ ﻭﺭﻕ = ‪t‬‬ ‫‪beff = 2t + 0.63, in. (= 2t + 16, mm) d b‬‬ ‫ﻓﺎﺻﻠﻪ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ﺗﺎ ﻟﺒﻪ ﻋﻀﻮ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﻋﻤﻮﺩ ﺑﺮ ﻧﻴﺮﻭﻱ ﻭﺍﺭﺩﻩ = ‪b‬‬ ‫‪= 2t(a + d/2),‬‬ ‫ﻛﻮﺗﺎﻩ ﺗﺮﻳﻦ ﻓﺎﺻﻠﻪ ﻟﺒﻪ ﺳﻮﺭﺍﺥ ﺗﺎ ﻟﺒﻪ ﻋﻀﻮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻧﻴﺮﻭ =‬ ‫ﻗﻄﺮ ﭘﻴﻦ =‬ ‫‪ = dt‬ﺳﻄﺢ ﺗﺼﻮﻳﺮ ﺷﺪﻩ ﭘﻴﻦ =‬ ‫‪60‬‬ ‫‪Asf‬‬ ‫‪a‬‬ ‫‪d‬‬ ‫‪Apb‬‬ ‫ﺍﻋﻀﺎﻱ ﺑﺎ ﺍﺗﺼﺎﻻﺕ ﻟﻮﻻﻳﻲ – ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﺍﺑﻌﺎﺩﻱ ﺗﺴﻤﻪ ﻫﺎﻱ ﻟﻮﻻ ﺷﺪﻩ‬ ‫‪PIN-CONNECTED MEMBERS - DIMENSIONAL REQUIREMENTS‬‬ ‫ ﻣﺤﺪﻭﺩﻳﺖ ﻫـﺎﻱ ﺍﺑﻌـﺎﺩﻱ ﺑـﺮﺍﻱ ﺍﻋﻀـﺎﻱ ﺑـﺎ‬ ‫ﺍﺗﺼــــــــﺎﻻﺕ ﻣﻔﺼــــــــﻠﻲ ﻃﺒــــــــﻖ‬ ‫‪ AISC05 D5.2 p.16.1-30‬ﻭ ﻧﻴﺰ ﻣﺒﺤـﺚ‬ ‫ﺩﻫﻢ ﺻﻔﺤﻪ ‪ 166‬ﺩﺭ ﺷﻜﻞ ﺯﻳـﺮ ﻧﺸـﺎﻥ ﺩﺍﺩﻩ‬ ‫ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫‪61‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﺴﻤﻪ ﺳﺮﭘﻬﻦ‬ ‫‪EYEBARS‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪62‬‬ ‫ﺗﺴﻤﻪ ﺳﺮﭘﻬﻦ ﻧﻮﻉ ﺧﺎﺻﻲ ﺍﺯ ﻋﻀﻮ ﺑﺎ ﺍﺗﺼﺎﻝ ﻣﻔﺼﻠﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﻗﻄﺮ ﺍﻧﺘﻬﺎﻱ ﺳﻮﺭﺍﺧﺪﺍﺭ ﺗﻮﺳﻌﻪ‬ ‫ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﻛﺸﺸﻲ ﻣﻮﺟﻮﺩ ﺗﺴﻤﻪ ﻫﺎﻱ ﺳﺮﭘﻬﻦ ﺑﻪ ﻃﺮﻳﻘﻲ ﻣﺸﺎﺑﻪ ﺣﺎﻟﺖ ﻛﻠﻲ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﺗﻌﻴﻴﻦ‬ ‫ﻣﻲ ﺷﻮﺩ‪) .‬ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪،( AISC D2‬‬ ‫ﺑﺎ ﺍﻳﻦ ﺗﻔﺎﻭﺕ ﻛﻪ‪ Ag ،‬ﺑﺮﺍﺑﺮ ﺑﺎ ﺳﻄﺢ ﻣﻘﻄﻊ ﺑﺪﻧﻪ ﺗﺴﻤﻪ ﺩﺭﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ‪.‬‬ ‫ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﺍﺑﻌﺎﺩﻱ ﺑﺮﺍﻱ ﺗﺴﻤﻪ ﻫﺎﻱ ﺳﺮﭘﻬﻦ ﻃﺒﻖ ﺿﻮﺍﺑﻂ ‪ AISC05 D6.2 p.16.1-30‬ﻭ ﻧﻴﺰ‬ ‫ﻣﺒﺤﺚ ﺩﻫﻢ ﺻﻔﺤﻪ ‪ 171‬ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ‪:‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ‪-‬‬ ‫ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪AISC‬‬ ‫‪Built-Up Tension Members-According to AISC D.4 p16.1-28‬‬ ‫ ﻫﺮﭼﻨﺪ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ﭼﻨﺪﺍﻥ ﻣﺘﺪﺍﻭﻝ ﻧﻴﺴﺘﻨﺪ ﻟﻴﻜﻦ ﺿﻮﺍﺑﻂ ‪AISC‬‬ ‫ﺳﺎﺧﺖ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻗﻴﺪﻫﺎﻱ )ﺑﺴﺖ ﻫﺎ‪ ،‬ﺻﻔﺤﺎﺕ(‬ ‫ﻣﻮﺍﺯﻱ‪ ،‬ﻣﻮﺭﺏ ﻭ ﺻﻔﺤﺎﺕ ﭘﻮﺷﺸﻲ ﺳﻮﺭﺍﺧﺪﺍﺭ ﺭﺍ ﻣﺠﺎﺯ ﻣﻲﺩﺍﻧﺪ‪.‬‬ ‫ ﺩﺭ ﺭﺍﺑﻄﻪ ﺑﺎ ﻣﺤﺪﻭﺩﻳﺖﻫﺎﻱ ﻭﺿﻊ ﺷﺪﻩ ﺩﺭ ﺯﻣﻴﻨﻪ ﻓﺎﺻﻠﻪ ﻃﻮﻟﻲ ﺍﺑﺰﺍﺭ ﺍﺗﺼﺎﻝ‬ ‫)‪ (connectors‬ﺍﺟﺰﺍﻱ ﺑﺎ ﺍﺗﺼﺎﻝ ﻣﻤﺘﺪ ﻣﺘﺸﻜﻞ ﺍﺯ ﻳﻚ ﻭﺭﻕ ﻭ ﻳﻚ ﻧﻴﻤﺮﺥ‪ ،‬ﻳﺎ ﺩﻭ‬ ‫ﻭﺭﻕ‪ ،‬ﺑﻪ ﺿﻮﺍﺑﻂ ‪ AISC J3.5‬ﻣﺮﺍﺟﻌﻪ ﺷﻮﺩ‪.‬‬ ‫ ﺩﺭ ﻭﺟﻪ ﺑﺎﺯ ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻭﺭﻕ ﻫﺎﻱ ﭘﻮﺷﺸﻲ ﺳﻮﺭﺍﺧﺪﺍﺭ ﻭ ﻳﺎ‬ ‫ﺑﺴﺖﻫﺎﻱ ﻣﻮﺍﺯﻱ ﺑﺪﻭﻥ ﭼﭗ ﻭ ﺭﺍﺳﺖ ﻣﺠﺎﺯ ﺍﺳﺖ‪.‬‬ ‫‪63‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺍﻋﻀﺎﻱ ﻛﺸﺸﻲ ﻣﺮﻛﺐ ‪-‬‬ ‫ﺑﺮﺍﺳﺎﺱ ﺿﻮﺍﺑﻂ ‪AISC‬‬ ‫‪Built-Up Tension Members-According to AISC D.4 p16.1-28‬‬ ‫ ﻃﻮﻝ ﺻﻔﺤﺎﺕ ﺑﺴﺖ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﺩﻭ ﺳﻮﻡ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺧﻄﻮﻁ ﺟﻮﺵ ﻳﺎ ﭘﻴﭻ ﻫﺎﻱ ﺍﺗﺼﺎﻝ ﺁﻧﻬﺎ ﺑﻪ ﺍﺟﺰﺍﻱ ﻋﻀﻮ ﻣﺮﻛﺐ ﺑﺎﺷﺪ‪.‬‬ ‫ ﺿﺨﺎﻣﺖ ﺍﻳﻦ ﺻﻔﺤﺎﺕ ﺑﺴﺖ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﻳﻚ ﭘﻨﺠﺎﻫﻢ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺧﻄﻮﻁ ﺟﻮﺵ ﻳﺎ ﭘﻴﭽﻬﺎﻱ ﺍﺗﺼﺎﻝ ﺁﻧﻬﺎ ﺑﻪ ﺍﺟﺰﺍﻱ ﻋﻀﻮ‬ ‫ﻣﺮﻛﺐ ﺑﺎﺷﺪ‪.‬‬ ‫ ﻓﺎﺻﻠﻪ ﻃﻮﻟﻲ ﺟﻮﺵ ﻫﺎ ﻳﺎ ﭘﻴﭽﻬﺎﻱ ﻣﺘﻨﺎﻭﺏ ﺩﺭ ﺻﻔﺤﺎﺕ ﺑﺴﺖ ﻧﺒﺎﻳﺪ ﺑﻴﺶ ﺍﺯ ‪ 150‬ﻣﻴﻠﻴﻤﺘﺮ )‪ 6‬ﺍﻳﻨﭻ( ﺑﺎﺷﺪ‪.‬‬ ‫ ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺍﺑﺰﺍﺭ ﺍﺗﺼﺎﻝ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺑﺎﻳﺪ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﺍﻱ ﺑﺎﺷﺪ ﻛﻪ ﺿﺮﻳﺐ ﻻﻏﺮﻱ ﻫﺮﻳﻚ ﺍﺯ ﺍﺟﺰﺍﻱ ﻛﺸﺸﻲ ﻣﺘﺼﻞ ﺷﺪﻩ ﺑﻴﻦ ﺍﻳﻦ‬ ‫ﺍﺗﺼﺎﻻﺕ ﺗﺮﺟﻴﻬﺎ ﺍﺯ ‪ 300‬ﺑﻴﺸﺘﺮ ﻧﺸﻮﺩ‪.‬‬ ‫ ﺁﻳﻴﻦ ﻧﺎﻣﻪ ﻓﻮﻻﺩ ﺍﻳﺮﺍﻥ ﻧﻴﺰ ﺿﻮﺍﺑﻂ ﺗﻘﺮﻳﺒﺎ ﻣﺸﺎﺑﻬﻲ ﺭﺍ ﺩﺭ ﺻﻔﺤﻪ ‪ 166‬ﻣﺒﺤﺚ ﺩﻫﻢ ﻣﻄﺮﺡ ﻧﻤﻮﺩﻩ ﺍﺳﺖ‪.‬‬ ‫‪t 2h/3‬‬ ‫‪t h/50‬‬ ‫‪h‬‬ ‫‪150 mm‬‬ ‫‪64‬‬ ‫‪Urmia University, M. Sheidaii‬‬ Base Plates for Concentrically Loaded Columns ‫ ﺻﻔﺤﺎﺕ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ‬: ‫ﻓﺼﻞ ﺷﺸﻢ‬ ‫ﺗﺤﺖ ﺑﺎﺭ ﻣﺤﻮﺭﻱ‬ Urmia University, M. Sheidaii 1 Urmia University, M. Sheidaii 3 ‫ﻣﻘﺪﻣﻪ‬ ‫ ﺑﺎﺭ ﺳﺘﻮﻥ ﺑﺎﻳﺪ ﺍﺯ ﻃﺮﻳﻖ ﺷﺎﻟﻮﺩﻩ ﺑﺘﻨﻲ ﺑﻪ ﭘﻲ ﺳﺎﺧﺘﻤﺎﻥ ﻣﻨﺘﻘﻞ ﮔﺮﺩﺩ‪.‬‬ ‫ ﺻﻔﺤﻪ ﺯﻳﺮ ﺳﺘﻮﻥ ﺑﻴﻦ ﺳﺘﻮﻥ ﻭ ﺷﺎﻟﻮﺩﻩ ﺑﺘﻨﻲ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﻭ ﺑﺎ ﺗﻮﺯﻳﻊ ﺑﺎﺭ ﺳﺘﻮﻥ ﺩﺭ‬ ‫ﺳﻄﺤﻲ ﻭﺳﻴﻊﺗﺮ ﺑﺎﻋﺚ ﻣﻲﺷﻮﺩ ﺗﻨﺸﻬﺎ ﺩﺭ ﺷﺎﻟﻮﺩﻩ ﺍﺯ ﺣﺪ ﻗﺎﺑﻞ ﺗﺤﻤﻞ ﺗﻮﺳﻂ ﻣﺼﺎﻟﺢ‬ ‫ﺷﺎﻟﻮﺩﻩ)ﺣﺪﻭﺩ ‪ 50‬ﺗﺎ ‪ (kgf/cm2 150‬ﺗﺠﺎﻭﺯ ﻧﻨﻤﺎﻳﺪ‪.‬‬ ‫ ﺷﺎﻟﻮﺩﻩ ﺑﺘﻨﻲ ﻧﻴﺰ ﺑﻪ ﻃﺮﻳﻘﻲ ﻣﺸﺎﺑﻪ‪ ،‬ﺑﺎﺭ ﺭﺍ ﺭﻭﻱ ﭘﻲ ﺑﻨﺤﻮﻱ ﺗﻮﺯﻳﻊ ﻣﻲﻛﻨﺪ ﻛﻪ ﺗﻨﺶ ﺩﺭ‬ ‫ﺧﺎﻙ ﺯﻳﺮ ﭘﻲ ﺍﺯ ﺣﺪ ﻻﺯﻡ ﺗﺠﺎﻭﺯ )ﺣﺪﻭﺩ ‪ 0.5‬ﺗﺎ ‪ (kgf/cm2 5‬ﻧﻜﻨﺪ‪.‬‬ ‫‪4‬‬ ‫ﻣﻘﺪﻣﻪ‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪5‬‬ ‫ﺍﮔﺮ ﻣﺤﻞ ﺍﺗﺼﺎﻝ ﺳﺘﻮﻥ ﻛﺎﻣﻼ ﺻﺎﻑ ﻭ ﮔﻮﻧﻴﺎ ﺑﺎﺷﺪ ﺳﺘﻮﻥ ﺭﺍ ﻣﻲﺗﻮﺍﻥ ﻣﺴﺘﻘﻴﻤﺎ ﺑﻪ‬ ‫ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺟﻮﺵ ﻛﺮﺩ ﺩﺭ ﻏﻴﺮﺍﻳﻨﺼﻮﺭﺕ ﺑﺎﻳﺪ ﺍﺯ ﻧﺒﺸﻲﻫﺎﻱ ﺍﺗﺼﺎﻝ ﻳﺎ ﻭﺭﻗﻬﺎﻱ‬ ‫ﺫﻭﺯﻧﻘﻪﺍﻱ ﺷﻜﻞ ﺑﺮﺍﻱ ﺍﻧﺘﻘﺎﻝ ﺑﺎﺭ ﺳﺘﻮﻥ ﺑﻪ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫ﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺷﺎﻟﻮﺩﻩ ﻭ ﺻﻔﺤﻪ ﺯﻳﺮ ﺳﺘﻮﻥ ﺗﻮﺳﻂ ﻣﻼﺕ ﻣﺎﺳﻪ ﺳﻴﻤﺎﻥ ﺑﺴﻴﺎﺭ ﻧﺮﻡ ﺑﺮﺍﻱ‬ ‫ﺻﺎﻑ ﻛﺮﺩﻥ ﺳﻄﺢ ﺗﻤﺎﺱ ﻭ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺗﻤﺮﻛﺰ ﺗﻨﺶ ﭘﺮ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ﺻﻔﺤﺎﺕ ﺯﻳﺮ ﺳﺘﻮﻥ ﺑﺎﻳﺪ ﺗﻮﺳﻂ ﺣﺪﺍﻗﻞ ‪ 4‬ﻣﻴﻞ ﻣﻬﺎﺭ)‪ (anchor‬ﻛﻪ ﺩﺭ ﺑﺘﻦ ﭘﻲ ﻛﺎﺭ‬ ‫ﮔﺬﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﺩﺭ ﺟﺎﻱ ﺧﻮﺩ ﺛﺎﺑﺖ ﻭ ﺗﻨﻈﻴﻢ ﮔﺮﺩﻧﺪ‪.‬‬ ‫ﺩﺭ ﺳﺘﻮﻧﻬﺎﻳﻲ ﻛﻪ ﻓﻘﻂ ﺗﺤﺖ ﺍﺛﺮ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ ﺧﺎﻟﺺ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ‪ ،‬ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﭘﺲ ﺍﺯ‬ ‫ﺍﺟﺮﺍﻱ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻧﻘﺶ ﺑﺎﺭﺑﺮﻱ ﻧﺪﺍﺷﺘﻪ‪ ،‬ﻭﻟﻲ ﺩﺭ ﻫﻨﮕﺎﻡ ﺍﺟﺮﺍ ﻭ ﻧﺼﺐ ﺳﺘﻮﻥ‪،‬‬ ‫ﺑﺮﺍﻱ ﺟﻠﻮﮔﻴﺮﻱ ﺍﺯ ﺍﻓﺘﺎﺩﻥ ﺳﺘﻮﻥ ﺩﺍﺭﺍﻱ ﻧﻘﺶ ﺍﺳﺎﺳﻲ ﻫﺴﺘﻨﺪ‪.‬‬ ‫ﺩﺭ ﺳﺘﻮﻧﻬﺎﻳﻲ ﻛﻪ ﻋﻼﻭﻩ ﺑﺮ ﻧﻴﺮﻭﻱ ﻣﺤﻮﺭﻱ‪ ،‬ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻧﻴﺰ ﺍﺯ ﻃﺮﻳﻖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﺑﻪ ﭘﻲ ﻣﻨﺘﻘﻞ ﻣﻲﮔﺮﺩﺩ ﻋﻤﻠﻜﺮﺩ ﻣﻴﻞ ﻣﻬﺎﺭﻫﺎ ﺩﺭ ﺗﺤﻤﻞ ﻧﻴﺮﻭ ﺩﺍﺭﺍﻱ ﺍﻫﻤﻴﺖ ﺍﺳﺖ‪.‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﻃﺮﺍﺣﻲ ﺻﻔﺤﺎﺕ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ ﺿﻮﺍﺑﻂ ﻣﺮﺑﻮﻁ ﺑﻪ ﻃﺮﺍﺣﻲ ﺻﻔﺤﺎﺕ ﺯﻳﺮ ﺳﺘﻮﻥ ﺩﺭ ‪ AISC J8 p.16.1-70‬ﺁﻣﺪﻩ ﺍﺳﺖ‪.‬‬ ‫ ﺩﻭ ﺣﺎﻟﺖ ﺣﺪﻱ ﻣﻬﻢ ﺩﺭ ﻃﺮﺍﺣﻲ ﺻﻔﺤﺎﺕ ﺯﻳﺮ ﺳﺘﻮﻥ ﻣﺪﻧﻈﺮ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﺩ‪:‬‬ ‫‪ 9‬ﺣﺎﻟﺖ ﺣﺪﻱ ﺷﻜﺴﺖ ﺑﺘﻦ)‪(crushing‬‬ ‫‪ 9‬ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﺧﻤﺸﻲ ﺻﻔﺤﻪ ﺯﻳﺮ ﺳﺘﻮﻥ‬ ‫ ﺑﺎﺭ ﻭﺍﺭﺩ ﺑﺮﺳﺘﻮﻥ ‪ ) Pu‬ﻳﺎ ‪ ( Pa‬ﺳﺘﻮﻥ ﺍﺯ ﻃﺮﻳﻖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﻪ ﺷﺎﻟﻮﺩﻩ ﻓﺸﺎﺭﻱ ﺑﺮﺍﺑﺮ ‪Pu/A‬‬ ‫)ﻳﺎ ‪ ( Pa/A‬ﻭﺍﺭﺩ ﻣﻲﻛﻨﺪ )‪ =BN=A‬ﻣﺴﺎﺣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ(‪ .‬ﻋﻜﺲﺍﻟﻌﻤﻞ ﺷﺎﻟﻮﺩﻩ ﺭﻭﻱ‬ ‫ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻓﺸﺎﺭ ﺭﻭ ﺑﻪ ﺑﺎﻻﻱ ‪) Pu/A‬ﻳﺎ ‪ (Pa/A‬ﺍﺳﺖ ﻛﻪ ﺗﺎﺛﻴﺮ ﺁﻥ ﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ‪:‬‬ ‫ﺧﻤﻴﺪﻩ ﺷﺪﻥ ﻟﺒﻪﻫﺎﻱ ﺑﻴﺮﻭﻧﻲ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﺑﺎﻻ ﺭﺍﻧﺪﻥ ﺳﻄﺢ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺩﺭ ﻧﺎﺣﻴﻪ ﺑﻴﻦ ﺑﺎﻟﻬﺎﻱ ﺳﺘﻮﻥ‬ ‫‪6‬‬ ‫‪ Pa‬ﻳﺎ‬ ‫‪u‬‬ ‫‪ Pa/A‬ﻳﺎ ‪Pu/A‬‬ ‫ﻃﺮﺍﺣﻲ ﺻﻔﺤﺎﺕ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ ﺑﻴﺸﺘﺮﻳﻦ ﺗﻨﺶ ﺩﺭ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺮﺍﻱ ﺳﺘﻮﻧﻬﺎﻱ ‪ I‬ﺷﻜﻞ‪ ،‬ﺩﺭ‬ ‫ﻣﺤﺪﻭﺩﻩ ﺗﻘﺮﻳﺒﻲ ‪ 0.95d , 0.8bf‬ﺭﺥ ﻣﻲﺩﻫﺪ‪.‬‬ ‫ ﻟﻨﮕﺮ ﺣﺪﺍﻛﺜﺮ ﺩﺭ ﺍﻳﻦ ﻣﺤﺪﻭﺩﻩ ﺩﺭ ﻓﻮﺍﺻﻞ ‪ m‬ﻭ ‪ n‬ﺍﺯ ﻟﺒﻪﻫﺎﻱ ﺁﺯﺍﺩ‬ ‫ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺗﻌﻴﻴﻦ ﺷﺪﻩ‪ ،‬ﻭ ﺑﺮﺍﺳﺎﺱ ﺁﻥ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ‬ ‫ﻣﺤﺎﺳﺒﻪ ﻣﻲﺷﻮﺩ‪.‬‬ ‫ ﻓﻮﺍﺻﻞ ‪ m‬ﻭ ‪ n‬ﭘﻴﺸﻨﻬﺎﺩﻱ ﺑﺮﺍﻱ ﺳﺎﻳﺮ ﺳﺘﻮﻧﻬﺎ ﺩﺭ ﺷﻜﻠﻬﺎﻱ ﺯﻳﺮ‬ ‫ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬ ‫‪7‬‬ ‫ﺗﻌﻴﻴﻦ ﺳﻄﺢ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﺳﻄﺢ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺷﻜﺴﺖ ﺑﺘﻦ ﺗﻌﻴﻴﻦ ﻣﻲﮔﺮﺩﺩ ﺑﺪﻳﻦ ﻣﻌﻨﻲ ﻛﻪ‬ ‫ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﻃﺮﺍﺣﻲ )‪ (design bearing strength‬ﺑﺘﻦ ﺯﻳﺮ ﻛﻒ ﺳﺘﻮﻥ ﺑﺎﻳﺪ ﺣﺪﺍﻗﻞ ﺑﺮﺍﺑﺮ ﺑﺎ ﺑﺎﺭ‬ ‫ﻭﺍﺭﺩﻩ)ﺑﺎﺭ ﻧﻬﺎﻳﻲ ﺿﺮﻳﺒﺪﺍﺭ ‪ Pu‬ﺑﺮﺍﻱ ‪ ، LRFD‬ﻳﺎ ﺑﺎﺭ ﺑﻬﺮﻩ ﺑﺮﺩﺍﺭﻱ ‪ Pa‬ﺑﺮﺍﻱ ‪ ( ASD‬ﺑﺎﺷﺪ‪.‬‬ ‫‪=ØcPp ; Øc=0.6‬ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ‪LRFD‬‬ ‫‪=Pp/:c ; :c =2.5‬ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﻃﺮﺍﺣﻲ ﺑﻪ ﺭﻭﺵ ‪ASD‬‬ ‫‪ ،Pp‬ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﺍﺳﻤﻲ ﺍﺳﺖ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺗﻌﻴﻴﻦ ﻣﻲﺷﻮﺩ‪:‬‬ ‫ﺍﻟﻒ‪ -‬ﺍﮔﺮ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻛﻞ ﺳﻄﺢ ﻓﻮﻗﺎﻧﻲ ﺷﺎﻟﻮﺩﻩ ﺭﺍ ﺑﭙﻮﺷﺎﻧﺪ‪:‬‬ ‫ﺏ‪ -‬ﺍﮔﺮ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻛﻞ ﺳﻄﺢ ﻓﻮﻗﺎﻧﻲ ﺷﺎﻟﻮﺩﻩ ﺭﺍ ﻧﭙﻮﺷﺎﻧﺪ‪:‬‬ ‫ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺭﻭﺍﺑﻂ‪:‬‬ ‫‪8‬‬ ‫= ﻣﻘﺎﻭﻣﺖ ﻓﺸﺎﺭﻱ ﻣﺸﺨﺼﻪ ﺑﺘﻦ‪،‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻌﻴﻴﻦ ﺳﻄﺢ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫‪ =A1‬ﻣﺴﺎﺣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻣﺘﻜﻲ ﺑﺮﺷﺎﻟﻮﺩﻩ ﺑﺘﻨﻲ = ‪B × N‬‬ ‫‪ =A2‬ﺣﺪﺍﻛﺜﺮ ﻣﺴﺎﺣﺖ ﻧﺎﺣﻴﻪﺍﻱ ﺍﺯ ﺳﻄﺢ ﺷﺎﻟﻮﺩﻩ ﻛﻪ ﺍﺯ ﻧﻈﺮ ﻫﻨﺪﺳﻲ ﻣﺘﺸﺎﺑﻪ ﻭ ﻫﻢﻣﺮﻛﺰ ﺑﺎ ﺻﻔﺤﻪ‬ ‫ﺯﻳﺮﺳﺘﻮﻥ ﺍﺳﺖ‪.‬‬ ‫‪1‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻌﻴﻴﻦ ﺳﻄﺢ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ A1 A2 A1 A1 A2 Urmia University, M. Sheidaii A2 10 ‫ﺭﻭﺍﺑﻂ ﺗﻌﻴﻴﻦ ﺳﻄﺢ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ ﻣﻘﺎﻭﻣﺖ ﺍﺗﻜﺎﻳﻲ ﻃﺮﺍﺣﻲ ﺑﺘﻦ ﺯﻳﺮ ﻛﻒ ﺳﺘﻮﻥ ﺑﺎﻳﺪ ﺣﺪﺍﻗﻞ ﺑﺮﺍﺑﺮ ﺑﺎ ﺑﺎﺭ‬ ‫ﺿﺮﻳﺒﺪﺍﺭ ﻭﺍﺭﺩﻩ ﺑﺎﺷﺪ‪:‬‬ ‫‪Pu‬‬ ‫) ‪Ic (0.85 f cc‬‬ ‫‪P :‬‬ ‫‪A1 t a c‬‬ ‫‪0.85 f cc‬‬ ‫‪A1 t‬‬ ‫ﻭ ﻳﺎ‬ ‫‪A2‬‬ ‫‪d2‬‬ ‫‪A1‬‬ ‫‪A2‬‬ ‫‪d2‬‬ ‫‪A1‬‬ ‫;‬ ‫‪Pu‬‬ ‫‪A2‬‬ ‫‪A1‬‬ ‫;‬ ‫‪A1 t‬‬ ‫‪o‬‬ ‫‪o‬‬ ‫‪o‬‬ ‫) ‪Ic (0.85 f cc‬‬ ‫‪Pa : c‬‬ ‫‪A2‬‬ ‫‪A1‬‬ ‫) ‪(0.85 f cc‬‬ ‫‪A1 t‬‬ ‫‬ ‫) ‪°°LRFD : Pu d Ic Pp Ic (0.85 f ccA1‬‬ ‫®‬ ‫‪Pp 0.85 f ccA1‬‬ ‫‪°‬‬ ‫‪ASD : Pa d‬‬ ‫¯‪°‬‬ ‫‪:c‬‬ ‫‪:c‬‬ ‫‪o‬‬ ‫‪A2‬‬ ‫‪A1‬‬ ‫‪A2‬‬ ‫‪A1‬‬ ‫‬ ‫) ‪°LRFD : Pu d Ic Pp Ic (0.85 f ccA1‬‬ ‫‪°‬‬ ‫‪°‬‬ ‫‪°‬‬ ‫®‬ ‫) ‪(0.85 f ccA1‬‬ ‫‪°‬‬ ‫‪Pp‬‬ ‫‪° ASD : Pa d‬‬ ‫‪:c‬‬ ‫‪:c‬‬ ‫‪°‬‬ ‫‪°‬‬ ‫¯‬ ‫ ﻫﻤﭽﻨﻴﻦ ﻣﺴﺎﺣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻧﺒﺎﻳﺪ ﻛﻤﺘﺮ ﺍﺯ ﺣﺎﺻﻠﻀﺮﺏ ﻋﻤﻖ ﺳﺘﻮﻥ ﺩﺭ ﻋﺮﺽ‬ ‫ﺁﻥ ﺑﺎﺷﺪ‪:‬‬ ‫‪11‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫‪A1 t bf d‬‬ ‫ﺑﻬﻴﻨﻪ ﺳﺎﺯﻱ ﺍﺑﻌﺎﺩ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﺑﺮﺍﻱ ﻃﺮﺡ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻧﻲ ﺑﺎ ﺣﺪﺍﻗﻞ ﻭﺯﻥ )ﺣﺪﺍﻗﻞ ﺿﺨﺎﻣﺖ( ﻛﺎﻓﻴﺴﺖ ﻟﻨﮕﺮ ﺧﻤﺸﻲ ﻃﺮﻩﻫﺎﻱ‬ ‫ﻓﺮﺿﻲ ﺩﺭ ﻫﺮ ﺩﻭ ﺍﻣﺘﺪﺍﺩ ﺗﻘﺮﻳﺒﺎ ﺑﺮﺍﺑﺮ ﺑﺎﺷﺪ ﺑﺪﻳﻦ ﻣﻨﻈﻮﺭ ﺑﺎﻳﺪ ‪ m=n‬ﺑﺎﺷﺪ ﺗﺤﺖ ﭼﻨﻴﻦ ﺷﺮﺍﻳﻄﻲ‪:‬‬ ‫' ‪N | A1‬‬ ‫) ‪' 0.5 (0.95d 0.8b f‬‬ ‫‪A1‬‬ ‫‪N‬‬ ‫‪12‬‬ ‫|‪B‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺗﻌﻴﻴﻦ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺮﺍﺳﺎﺱ ﺣﺎﻟﺖ ﺣﺪﻱ ﺗﺴﻠﻴﻢ ﺧﻤﺸﻲ ﺩﺭ ﺍﻧﺘﻬﺎﻱ ﻃﺮﻩﻫﺎﻱ ﭘﻴﺶ ﺁﻣﺪﻩ‬ ‫ﺑﻪ ﻃﻮﻟﻬﺎﻱ ‪ m‬ﻭ ‪ n‬ﺩﺭ ﻫﺮ ﺩﻭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﺎﺳﺒﻪ ﻣﻲﺷﻮﺩ‪ .‬ﺍﻳﻦ ﻣﻘﺎﺩﻳﺮ ﻟﻨﮕﺮ ﺑﺮﺍﻱ ﻋﺮﺽ ﻭﺍﺣﺪ ﺻﻔﺤﻪ‬ ‫ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ) ‪ P=Pu‬ﺩﺭ ﺭﻭﺵ ‪ ، LRFD‬ﻭ ‪ P=Pa‬ﺩﺭ ﺭﻭﺵ ‪:( ASD‬‬ ‫‪P‬‬ ‫‪n P n2‬‬ ‫) ()‪(n‬‬ ‫‪BN‬‬ ‫‪2 2BN‬‬ ‫‪,‬‬ ‫‪m P m2‬‬ ‫‪P‬‬ ‫) ()‪(m‬‬ ‫‪BN‬‬ ‫‪2 2BN‬‬ ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﻣﻮﺟﻮﺩ ﺑﺮﺍﻱ ﻋﺮﺽ ﻭﺍﺣﺪ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ‪:‬‬ ‫‪Ib Fy Z x 0.9 Fy t 2 / 4‬‬ ‫‪Fy Z x‬‬ ‫‪0.6 Fy t 2 / 4‬‬ ‫‪1.67‬‬ ‫‪LRFD : I b M n‬‬ ‫‪°‬‬ ‫‪Mn‬‬ ‫®‬ ‫‪° ASD : :‬‬ ‫‪b‬‬ ‫¯‬ ‫ﻣﻘﺎﻭﻣﺖ ﺧﻤﺸﻲ ﻃﺮﺍﺣﻲ ﻣﻮﺟﻮﺩ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺎﻳﺪ ﺣﺪﺍﻗﻞ ﺑﺮﺍﺑﺮ ﺑﺎ ﺑﺰﺭﮔﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺍﺯ ﻣﻘﺎﺩﻳﺮ‬ ‫ﻟﻨﮕﺮ ﻭﺍﺭﺩﻩ ﻣﺬﻛﻮﺭ ﺑﺎﺷﺪ‪:‬‬ ‫‪13‬‬ ‫‪2Pu‬‬ ‫‪0.9Fy BN‬‬ ‫‪tt n‬‬ ‫‪o‬‬ ‫‪2Pu‬‬ ‫‪Pu n 2‬‬ ‫‪2‬‬ ‫‪tt m‬‬ ‫‪; 0.9 Fy t / 4 t‬‬ ‫‪0.9Fy BN‬‬ ‫‪2BN‬‬ ‫‪o‬‬ ‫‪2Pa‬‬ ‫‪0.6Fy BN‬‬ ‫‪ttn‬‬ ‫‪o‬‬ ‫‪2Pa‬‬ ‫‪Pa n 2‬‬ ‫‪2‬‬ ‫‪ttm‬‬ ‫‪; 0.6 Fy t / 4 t‬‬ ‫‪0.6Fy BN‬‬ ‫‪2BN‬‬ ‫‪o‬‬ ‫‬ ‫‪Pu m 2‬‬ ‫‪2‬‬ ‫‪°LRFD : 0.9 Fy t / 4 t‬‬ ‫‪2BN‬‬ ‫‪°‬‬ ‫®‬ ‫‪2‬‬ ‫‪° ASD : 0.6 F t 2 / 4 t Pa m‬‬ ‫‪y‬‬ ‫‪°‬‬ ‫‪2BN‬‬ ‫¯‬ ‫ﺗﻌﻴﻴﻦ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ :‫ﻭ ﺑﺼﻮﺭﺕ ﺧﻼﺻﻪ ﻣﻲﺗﻮﺍﻥ ﻧﻮﺷﺖ‬ °LRFD : ° ® °ASD : ° ¯ t min t L 2Pu 0.9Fy BN , L Max (m, n ) t min t L 2Pa 0.6Fy BN , L Max (m, n ) 14 Urmia University, M. Sheidaii 15 Urmia University, M. Sheidaii 16 Urmia University, M. Sheidaii 17 18 Urmia University, M. Sheidaii 19 ‫ﺭﻭﺵ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪Thornton‬‬ ‫ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ ﺍﮔﺮ ﺳﺘﻮﻥ ﺩﺍﺭﻱ ﺑﺎﺭ ﻛﻤﻲ ﺑﺎﺷﺪ ﻣﺴﺎﺣﺖ ﻣﺤﺎﺳﺒﻪ ﺷﺪﻩ ﺑـﻪ ﺭﻭﺵ ﻗﺒـﻞ ﺑـﺮﺍﻱ ﺻـﻔﺤﻪ‬ ‫ﺯﻳﺮﺳﺘﻮﻥ ﻧﺴﺒﺘﺎ ﻛﻮﭼﻚ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‪ .‬ﻃﻮﻝ ﻗﺴﻤﺖ ﭘﻴﺶ ﺁﻣﺪﻩ ﺍﺯ ﻟﺒﻪﻫﺎﻱ ﺳﺘﻮﻥ ﺧﻴﻠﻲ‬ ‫ﻛﻮﭼﻚ ﻭ ﻟﻨﮕﺮﻫﺎﻱ ﻣﺤﺎﺳﺒﻪ ﺷﺪﻩ ﻭ ﺿﺨﺎﻣﺖ ﻭﺭﻕ ﻻﺯﻡ ﺑﺴﻴﺎﺭ ﻛﻮﭼﻚ ﺧﻮﺍﻫﺪ ﮔﺮﺩﻳﺪ‪.‬‬ ‫ ﺩﺭ ﭼﻨﻴﻦ ﻣﻮﺍﺭﺩﻱ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﺭﻭﺵ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪ Thornton‬ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ‪.‬‬ ‫ ‪ Thornton‬ﺑﺎ ﺗﺮﻛﺐ ﺳﻪ ﺭﻭﺵ‪ ،‬ﻃﺮﻳﻖ ﻭﺍﺣﺪﻱ ﺭﺍ ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ﺻـﻔﺤﺎﺕ ﺯﻳﺮﺳـﺘﻮﻥ‬ ‫ﺗﺤﺖ ﺑﺎﺭﻫﺎﻱ ﺳﺒﻚ ﻭﺳﻨﮕﻴﻦ ﺍﺭﺍﺋﻪ ﻧﻤﻮﺩﻩ ﺍﺳﺖ‪.‬‬ ‫‪20‬‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺵ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪Thornton‬‬ ‫ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫ﺩﺭ ﺭﻭﺵ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪ Thornton‬ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺑﺮﺍﺳﺎﺱ ﺑﺰﺭﮔﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺍﺯ‬ ‫ﻣﻘﺎﺩﻳﺮ ‪ m , n , Onc‬ﻛﻪ ‪ ℓ‬ﻧﺎﻣﻴﺪﻩ ﻣﻲﺷﻮﺩ ﺗﻌﻴﻴﻦ ﻣﻲﮔﺮﺩﺩ‪:‬‬ ‫‪ Onc‬ﺑﺮ ﺍﺳﺎﺱ ﺭﻭﺍﺑﻂ ﺫﻳﻞ ﺗﻌﻴﻴﻦ ﻣﻲﺷﻮﺩ‪:‬‬ ‫)‪ℓ = Max (m, n, Onc‬‬ ‫‪O d bf‬‬ ‫‪4‬‬ ‫‪2 X‬‬ ‫‪d1‬‬ ‫‪1 1 X‬‬ ‫ﺍﻟﺒﺘﻪ ﻣﻲﺗﻮﺍﻥ ﻣﻘﺪﺍﺭ ‪ O‬ﺭﺍ ﺑﻄﻮﺭ ﻣﺤﺎﻓﻈﻪﻛﺎﺭﺍﻧﻪ ﺑﺮﺍﻱ ﻫﻤﻪ ﺣﺎﻟﺖﻫﺎ ﺑﺮﺍﺑﺮ ﺑﺎ ‪ 1‬ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺖ‪.‬‬ ‫)‪0.6‬‬ ‫‪( LRFD Method with Ic‬‬ ‫‪or‬‬ ‫)‪2.5‬‬ ‫‪21‬‬ ‫‪( ASD Method with : c‬‬ ‫‪O nc‬‬ ‫‪O‬‬ ‫‪ª 4 d b f º Pu‬‬ ‫«‪°‬‬ ‫»‪2‬‬ ‫‪°¬ (d b f ) ¼ Ic Pp‬‬ ‫® ‪X‬‬ ‫‪° ª 4 d b f º : c Pa‬‬ ‫‪° «¬ (d b f ) 2 »¼ Pp‬‬ ‫¯‬ ‫‪Urmia University, M. Sheidaii‬‬ ‫ﺭﻭﺵ ﭘﻴﺸﻨﻬﺎﺩﻱ ‪Thornton‬‬ ‫ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ‬ ‫)ﺍﮔﺮ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻛﻞ ﺳﻄﺢ ﻓﻮﻗﺎﻧﻲ ﺷﺎﻟﻮﺩﻩ ﺭﺍ ﺑﭙﻮﺷﺎﻧﺪ(‬ ‫)ﺍﮔﺮ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﻛﻞ ﺳﻄﺢ ﻓﻮﻗﺎﻧﻲ ﺷﺎﻟﻮﺩﻩ ﺭﺍ ﻧﭙﻮﺷﺎﻧﺪ(‬ ‫‪A2‬‬ ‫‪d2‬‬ ‫‪A1‬‬ ‫‪ 0.85 f ccA1‬‬ ‫‪°‬‬ ‫; ‪®(0.85 f cA ) A 2‬‬ ‫‪c 1‬‬ ‫‪°‬‬ ‫‪A1‬‬ ‫¯‬ ‫‪Pp‬‬ ‫ﺩﺭ ﻧﻬﺎﻳﺖ ﭘﺲ ﺍﺯ ﻣﺤﺎﺳﺒﻪ ‪ ، ℓ‬ﻣﻲﺗﻮﺍﻥ ﺿﺨﺎﻣﺖ ﺻﻔﺤﻪ ﺯﻳﺮﺳﺘﻮﻥ ﺭﺍ ﺍﺯ ﻫﺮ ﻳﻚ ﺍﺯ ﺭﻭﺍﺑﻂ ﺯﻳﺮ ﺗﻌﻴﻴﻦ‬ ‫ﻧﻤﻮﺩ‪:‬‬ ‫‪22‬‬ ‫; )‪(LRFD Method‬‬ ‫‪2 Pu‬‬ ‫‪0.9 Fy B N‬‬ ‫"‬ ‫‪t‬‬ ‫)‪(ASD Method‬‬ ‫‪2 Pa‬‬ ‫‪0.6 Fy B N‬‬ ‫"‬ ‫‪t‬‬ 23 24 25 26 27 28

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