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Explore the design equations for steel tension members, including gross and net areas, bolt holes, shear lag, and example calculations. Learn about LRFD and ASD approaches for determining tensile strength.
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應力-應變曲線 破壞 應力 Fu 極限強度 Fy 降伏強度 破壞 應變硬化 塑性 頸縮 彈性 應變 10
拉力桿件 What is the maximum P? P P LRFD Equation ASD Equation
拉力破壞模式 P P
P 1.全斷面降伏 Fy P
1.全斷面降伏 Fy Pn=AgFy
2.淨斷面剪壞 Fu
2.淨斷面剪壞 Fu Fy Fu Pn=AeFu
Design of Steel Tension Members Equations for strength of tension members: • For yielding in the gross section: • For fracture in the net section:
1.全斷面降伏 Fy ASD LRFD fPn=0.9AgFy
2.淨斷面剪壞 Ae= An≤ 0.85Ag Fu Fy Fu ASD LRFD fPn=0.75AeFu
Net Area(淨面積) • The net area, An, of a member is the sum of the products of the thickness and the net width of each element computed as follows: In computing net area for tension and shear, the width of a bolt hole shall be taken as 1/ 16 in. (2 mm) greater than the nominal dimensionof the hole. 1/16” 1/16” 孔徑放大 鑽孔損失 D
Determine the net area of the 3/8 × 8-in. plates shown below. The plate is connected at its end with two lines of ¾-in. bolts. 8 in
Net Section for Staggered Bolt Holes Recall definition of Net Area, LRFD p. 16.1-10
s g
Determine the critical net area of the 1/2 -in. thick plates shown below. Using the AISC Spe.(D3.2).The holes are punched for ¾-in. bolts. 21/2 in. 3 in. 3 in. 21/2 in. 3 in.
1/2 -in. thick plates The holes are punched for ¾-in. bolts. CASE 1 21/2 in. 3 in. 3 in. 21/2 in. 3 in.
1/2 -in. thick plates The holes are punched for ¾-in. bolts. CASE 2 21/2 in. 3 in. 3 in. 21/2 in. 3 in.
1/2 -in. thick plates The holes are punched for ¾-in. bolts. CASE 3 21/2 in. 3 in. 3 in. 21/2 in. 3 in.
21/2 in. 3 in. 3 in. 21/2 in. 3 in.
For two lines of bolt holes shown below. Determine the pitch that will give a net area DEFG equal to the one along ABC. The holes are punched for ¾-in. bolts. A D 2in E B 2in F 2in G C s in ABC DEFG
Determine the net area of the W12×16(Ag=4.71 in.2). The holes are punched for 1-in. bolts.
Determine the net area along route ABCDEF for the C15×33.9 (Ag=10.0 in.2). The holes are for ¾-in. bolts.
Design Requirements • Ag– Gross cross-sectional area • Ae– Effective net area If tension load is transmitted directly to each of the cross-sectional elements by fasteners or welds: • Ae = An • An = Net cross-sectional area (gross-section minus bolt holes)
Design Requirements If tension load transmitted through some but not all of the cross-sectional elements: by fasteners, Ae = AnU by welds, Ae = AgU or Ae = AU
Shear Lag 剪力遲滯 • Ae = UAn T T
Shear Lag 剪力遲滯 • Ae = UAn T
Example of tension transmitted by some but not all of cross-section L –shape with bolts in one leg only Reduction coefficient, Where is the connection eccentricity
LRFD • Pu= Pn= 0.9Fy Ag , for yielding in the gross section →Pu = Pn = 0.9Fy Ag = 0.936(95/8) = 182.3 kips • Pu = Pn = 0.75Fu Ae , for fracture in the net section →Pu = Pn = 0.75Fu Ae = 0.75583.98 = 173.1 kips • Ae= UAn = 1.0[9 – 3(3/4+1/8)] = 3.98 in.2
ASD • Pu= Pn/Ω = Fy Ag /1.67 , for yielding in the gross section →Pu = Pn /Ω = Fy Ag /1.67 = 36(95/8)/1.67 =121.3 kips • Pu = Pn /Ω = Fu Ae /2 , for fracture in the net section →Pu = Pn /Ω = Fu Ae /2 = 583.98/2 = 115.42 kips • Ae= UAn = 1.0[9 – 3(3/4+1/8)] = 3.98 in.2
Determine the LRFD tensile design andthe ASD allowable tensile strength of the member (A572 Gr. 50).
6in. 7/8 in. 3/8 in. 6in.
100 80 80 80 100 80 50 50 50 50 20 80 80 80 80 An= Ag=