第二章 拉力桿件

拉力桿件

常見受拉桿件型式

10 應力 - 應變曲線 應力 - 應變曲線 降伏強度 極限強度 應變 應力 破壞 彈性 塑性 應變硬化 頸縮 FuFu FyFy

What is the maximum P? P P 拉力桿件 LRFD Equation ASD Equation

拉力破壞模式 P P

1. 全斷面降伏 FyFy P P

FyFy P n =A g F y

2. 淨斷面剪壞 FuFu

FyFy P n =A e F u FuFu FuFu

Design of Steel Tension Members Equations for strength of tension members: a)For yielding in the gross section: b)For fracture in the net section:

1. 全斷面降伏 FyFy LRFD  P n =0.9A g F y ASD

2. 淨斷面剪壞 FyFy FuFu FuFu A e = A n ≤ 0.85A g LRFD  P n =0.75A e F u ASD

Net Area( 淨面積 ) The net area, A n, 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 dimension of the hole. D 孔徑放大 鑽孔損失 1/16”

螺栓孔徑使用之大小

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

g s

3 in. 2 1 / 2 in. 3 in. 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.

3 in. 2 1 / 2 in. 3 in. CASE 1 1/2 -in. thick plates The holes are punched for ¾-in. bolts.

3 in. 2 1 / 2 in. 3 in. CASE 2 1/2 -in. thick plates The holes are punched for ¾-in. bolts.

3 in. 2 1 / 2 in. 3 in. CASE 3 1/2 -in. thick plates The holes are punched for ¾-in. bolts.

3 in. 2 1 / 2 in. 3 in.

A B C D E F G 2in s 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. ABC DEFG

Determine the net area of the W12×16(A g =4.71 in. 2 ). The holes are punched for 1-in. bolts.

Determine the net area along route ABCDEF for the C15×33.9 (A g =10.0 in. 2 ). The holes are for ¾-in. bolts.

Design Requirements A g – Gross cross-sectional area A e – Effective net area If tension load is transmitted directly to each of the cross-sectional elements by fasteners or welds: A e = A n A n = 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, A e = A n U by welds, A e = A g U orA e = AU

Shear Lag 剪力遲滯 A e = UA n T T

Shear Lag 剪力遲滯 A e = UA n 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

P max = ?

LRFD P u =  P n = 0.9F y A g, for yielding in the gross section →P u =  P n = 0.9F y A g = 0.9  36  (9  5/8) = kips P u =  P n = 0.75F u A e, for fracture in the net section →P u =  P n = 0.75F u A e = 0.75  58  3.98 = kips A e = UA n = 1.0  [9 – 3  (3/4+1/8)] = 3.98 in. 2

ASD P u = P n /Ω = F y A g /1.67, for yielding in the gross section →P u = P n /Ω = F y A g /1.67 = 36  (9  5/8)/1.67 =121.3 kips P u = P n /Ω = F u A e /2, for fracture in the net section →P u = P n /Ω = F u A e /2 = 58  3.98/2 = kips A e = UA n = 1.0  [9 – 3  (3/4+1/8)] = 3.98 in. 2

Determine the LRFD tensile design and the ASD allowable tensile strength of the member (A572 Gr. 50).

6in. 3 / 8 in. 7 / 8 in.

Ag=Ag= An=An=

3. 塊剪

(a) When (b) When 3. 塊剪

AISC 塊剪 R n = 0.6F u A nv + U bs F u A nt ≤ 0.6F y A gv + U bs F u A nt A gv = gross area subject to shear, in.`(mm2) A nt = net area subject to tension, in.2 (mm2) A nv = net area subject to shear, in.2 (mm2) Where the tension stress is uniform, U bs = 1; where the tension stress is nonuniform,U bs = 0.5.

T T cm 9cm L155×100×13 mm F y =2.5 t/cm 2 F u =4.1 t/cm 2

20 t =1.3 mm 10 F y =2.5 t/cm 2 F u =4.1 t/cm 2 T T

Tension Analysis Example Determine the factored strength of a 12 ” x 1.5 ”, A36 steel plate connected with one row of 4 – ¾” diameter bolts positioned transversely in a single line. 12 in. A36, 1½” PL

Design Example ( top p. 9 notes) Design a 1000 mm long splice plate to carry a tensile live load of 130 kN and dead load due to a mass of 4500 kg. The bolts will be ¾” in diameter and there will be at least three of them in a row parallel to the direction of force at each end. Space constraints require you to keep the width of the plate ≤ 100 mm. Use A36 steel in conformance with the rest of the building.

Design Example ( bottom p. 9 notes) Design a tension member for a live load of 67.4 kips and a dead load of 22.0 kips. It is part of a web system of a truss and will be 14.8 ft long between connections. The end connections will require two rows of ¾” diameter bolts, with three bolts per row. As a truss web member, an angle section seems most appropriate. Use A36 steel to conform to the rest of the truss. :

LRFD p

Design Example ( p. 10 notes) Same as previous example but with double angles back to back. Assume that they will be bolted to 3/8 in. thick gusset plates, straddling them at each end. 3/8 in. gusset plate

Staggered Bolt Hole Example (p. 12 notes) Consider the 7 x 4 x ½ angle shown. The holes are 7/8” diameter, on normal gage lines. The holes are the U.S. standard 3” c.c. in each row, but the holes in the interior row in the 7” leg are offset by 1 ½ “ from the other holes, which line up with each other. Find A n for both a two hole and a three hold tear line.