Slide 1

TUTORIAL 2

Axially Loaded Members Tension members End Connections

21 Feb 2012Problem 1: Determine the tension capacity of the angle section shown. It is connected on the longer leg75 x 50 x 8 angle ; 16 mm bolts are used.Solution: For single angle connected on one leg, the tension capacity is given ( clause 3.10.3; EN 3 part 1-8; Eq. 3.11 )

Assuming S275 grade steel; fu = 430N/mm2 (Table 3.1, EN 3 part 1-1, )

M2= 1.25, M1 = M0= 1.0 ( Section 6.1, EN 3 part 1-1)

Assuming e2 = 33mm, Nu,Rd = 2.0(33-0.5x18) 8x430/1.25 = 132096 N= 132.10kNNet area of angle connected by one leg allowing for holes Anet = (h-d0)t + (b-t)t Anet = (75-18)8 + (50-8)8 = 57x8+42x8 = 792 mm2Gross area= 941mm2 ( section tables)Design plastic resistance (clause 6.2.3; EN 3 part 1-1: Eq. 6.6) Npl.RD= A*fy/ M0 = 941x275/1.0 = 258775 N = 258.78kN.

ANSWER : The smallest value is 132.1kN.Problem 2:A FLAT BAR 200 mm wide and 25 mm thick is used a tie. Erection conditions require that the bar be constructed from two lengths connected together with a lap splice using M20 bolts. Determine the tension capacity of bar assuming grade S275 steel. P =100 mm200 mmS= 90 mm90 mm90 mm90 mmSolution:For nominal material thickness ( t = 25 mm) of less than or equal to 40 mm: fy= 275N/mm2 and fu= 430N/mm2 (EN 3 part 1-1, table 3.1)Partial safety factors : M0 = 1.00 and M2 = 1.25Gross area: Ag = 200 x 25 = 5000 mm2For calculating net areas, All possible failure patterns must be considered to determine the critical caseHole diameter = bolt diameter + 2 = 22 mmThe different failure patterns are shown:90 mm100 mm200 mmS=90 mm90 mm90 mmAABBAreas to be deducted:

Path AA : Area deducted = t x d = 25 x 22 = 550 mm2

Path BB :

Path BB is critical since the deduction is maximum for itTherefore :An = Gross area deduction = 5000 594 = 4406mm2Tension capacity is determined using:clause. 6.2.3 ; EN 3 part 1-1 :and clause. 3.4.2 ; EN 3 part 1-8Design plastic resistanceNpl,Rd = Afy/M0 = 5000x275/1 = 1375kNDesign ultimate tensile resistance of the net cross section (clause. 6.2.3 ; EN 3 part 1-1; Eq. 6.7):Nu,Rd =0.9Anet fu/M2 = 0.9x4406x430/1.25 = 1364kN

ANSWER : Plate capacity ( lowest) = 1364 kNE.g. 3: Determine the tension capacity of the angle section shown. It is connected on the longer leg by welding75 x 50 x 8 angle ; welds are used.Weld size is not givenif weld connection is to be checked, its size has to be givenSolution:For single angle connected on the longer leg, the effective area = gross area (clause 4.13; EN 3 part 1-8)

Gross area = 9.41 cm2 (given)

Net area = gross area = 9.41cm2weldTension capacity is determined using:clause. 6.2.3 ; EN 3 part 1-1 and clause 4.13; EN 3 part 1-8

Assuming : S275 grade steel; fy= 275 N/mm2, fu=430 N/mm2 M0 =1.0 and M2 =1.25

Design plastic resistance : Npl,Rd = Afy/M0 = 941x275/1 =258775N = 258.775kN

Design ultimate resistance of the net cross section at holes for fasteners :Nu,Rd = 0.9Anet fu/M2 = 0.9x941x430/1.25 = 291333.6N = 291.333 kN

ANSWER : The Angle Capacity = 258.775kN E.g. 4: Determine the tension capacity of the double angle section (2 / 75 x 50 x 8 ) connected to each side of a gusset plate by 12 mm diameter bolts.75 x 50 x 8 double angleSolution:When angle is connected by the smaller leg, the net are allowing for holes.

Note: For an unequal-leg angle connected by a smaller leg, the net area should be taken as equal to the net section area of an equivalent equal-leg angle section with leg size equal to the smaller leg.( clause 3.10.3, EN 3 part 1-8)

Gross area = 2 x 9.41cm2 = 1882 mm23 D VIEW OF JOINT75 mm75 mm50 mmSection AA Tension capacity is determined using: clause 6.2.3 of EN 1993-1-1 and clause 3.10.3 of EN 1993-1-8 Assuming S275 grade steel; fy= 275 N/mm2 , fu = 430N/mm2 and M0 =1.0 and M2 =1.25

Design plastic resistance:Npl,Rd=Afy/M0 = 1882 x 275/1= 517.55kN

Design ultimate resistance of the net cross section at holes for fasteners:single rows with 2 bolts (clause 3.10.3 of EN 1993-1-8 , Eq 3.12)Nu,Rd=2Anet fu/M2 = 0.4 x1248x430/1.25 = 171.72kN

ANSWER : Therefore angle capacity = 171.72 kNE.g. 5: Determine the tension capacity of the double angle section (2 / 75 x 50 x 8 ) connected with the smaller leg to each side of a gusset plate by welds75 x 50 x 8 double angleSolution:For double angle connected on each side of gusset plate, the effective are is taken as the gross area of an equivalent equal angle section with leg size equal to the smaller leg

Tension capacity is determined using:clause. 6.2.3 ; EN 3 part 1-1 and clause 4.13; EN 3 part 1-8

Assuming S275 grade steel; fy=275 N/mm2 , fu = 430N/mm2 And M0 =1.0 and M2 =1.25

Design plastic resistance :

Npl,Rd = Afy/M0 = 1472x275/1= 404.8kN

Design ultimate resistance of the net cross section at holes for fasteners =

Nu,Rd = 0.9Anet fu/M2 = 0.9x1472 x 430/1.25 = 455.73 kN

ANSWER : Therefore angle capacity = 455.73 kN E.g. 6: An internal member of a truss system is subject to a tensile force of 260 kN from truss analysis, Py= 275 N/mm2. Propose a suitable cross section for it if:(a) the end connections are welded(b) the end connections are bolted (18 mm diameter bolts)260kNSTEP 1: Preliminary size of member is obtained by:Ft = 260 kN ; py = 275 N/mm2Approximate area needed = Ft / py = 260 x1000/275 = 945.45mm2Since some area is going to be lost due to holes, designer has to provide extra area say 10-15% more 945.45 x 1.15 = 1087mm2Looking at tables choose a suitable angle with area more than 1087mm2Choosing or trying L 100 x 65 x 7 whose Ag=11.2 cm2

STEP 2: Welded connectionAssume that the longer leg of the section is welded to the gusset plate. The centroid axis of section is eccentric.

For angle connected to gusset plate by longer leg, the effective area is taken as the gross area of angle sectionTherefore Ae= Ag=11.2 cm2

Tension capacity is determined using clause 6.2.3 of BS EN 1993-1-1:2005 and clause 3.4.2 of BS EN 1993-1-8:2005

Assuming S275 grade steel; fy=275 N/mm2 , fu=430N/mm2 and M0 =1.0 and M2 =1.25

Design plastic resistance, Npl,Rd=Afy/M0 = 1120x275/1=308.00kN

Design ultimate resistance of the net cross section at holes for fasteners Nu,Rd=0.9Anet fu/M2 =0.9x1120x430/1.25 = 346.75kN

The angle capacity = 308.00 kNSince Nt,Rd = 308kN > 260 kN (Nt,Ed) the design is safebut can be optimised.STEP 3: Bolted connection Assume that the longer leg of the section is welded to the gusset plate. The centroid axis of section is eccentric.

For single angle connected on one leg, the tension capacity is given by the equation

Assuming:16 mm diameter bolts; S275 grade steel; fy=275 N/mm2 , fu=430N/mm2 and e2 = 33mm, M2= 1.25

Design ultimate resistance: Nu,Rd = 2.0(33-0.5x18)7x430/1.25=115584N= 115.58kN

Design plastic resistance: Npl,Rd = Afy/M0 = 1120x275/1 = 308.00kN

Considering the lower value:

Nt,Rd = 115.58 kN < 260 kN (Nt,Ed), So select a higher section and check.

E.g. 7:(Net area of a bolted universal column section member)Both flanges of a universal column section member have 22 mm diameters holes arranged as shown below. If the gross area of the section is 201x102 mm2 and the flange thickness is 25 mm, determine the net area Anet of the member which is effective in tension. Solution:

Minimum stagger = (No. of bolts taken x S x bolt dia)0.5

72.7 mm > 30 mm = use, s= 30 mm

2)The failure path through each flange is therefore staggered, and by inspection, it includes four holes and two staggers. The net area is calculated by clause 6.2.2.2(4) EC3 as:

Anet = Ag [No. of stagger x No. of bolt (bolt dia x t)] + No. of stagger x S2 x t)/ 4 p= [201*102]- [2*4*(22*25)]+[2*2*302*25)/(4*60)]

= 161*102 mm2

E.g. 8: ( Checking a bolted universal column section member).Determine the tension resistance of the tension member of above example assuming it is of S355 steel

Solution:Tf= 25 mm, fy=345 N/mm2 , fu=490 N/mm2 (EN 10025-2)Assume M0 =1.0 Npl,Rd = Afy/M0 = 201*102 * 345/1.0 = 6935 kN ( 6.2.3 (2)a

Anet = 161 *102 mm2 ( example 7)

Nu,Rd = 0.9 Anet fu / M2 = 0.9 * 161*102*490/1.1 = 6445 kN (6.2.3(2) b

Tension capacity 6445 kN

E.g. 9: ( Checking a bolted universal beam section member)A 610 x 229 UB 125 tension member of S355 steel is connected through both flanges by 20 mm bolts (in 22 mm diameter bolt holes) in four lines, two in each flange as shown in figure. Check the member for a design tension force of Nt,Ed= 4000 kN

Solution:

Solution:tf= 19.6 mm , fy = 345 N/mm2 , fu = 490 N/mm2 ( EN 2)

A= 15 900 mm2

Assume M0 =1.0, M2 =1.1

Npl,Rd = Afy/M0 = 15900 x 345 /1.0 = 5486 kN Cl 6.2.3 (2)a

Anet = 15900 (4 x 22 x 19.6 ) = 14175 mm2 Cl 6.2.2.2 (3)

Nu,Rd = 0.9 Anet fu / M2= 0.9 x 14175 x 490/ 1.1 = 5683 kN Cl. 6.2.3 (2)b

Nt,Rd = 5486 kN > 4000 kN (Nt,Ed) Design Safe

E.g. 10: Checking an eccentrically connected single ( unequal angle)A tension member consists of a 150 x 75 x 10 single unequal angle whose ends are connected to gusset plates through the larger leg by a single row of four 22 mm bolts in 24 mm holes at 60 mm centres. Check the member for a design tension force of Nt,Ed= 340 kN, if the angle is of S355 steel and has a gross area of 21.7 cm2 use method of EC 3 -1-8 3.10.3(2)

Solution: T = 10 mm, fy= 355 N/mm2, fu= 490 N/mm2 (EN 2) Ag= 2170 mm2 Assume M0 =1.0, M2 =1.1

Npl,Rd = Afy/M0 = 2170 x 355 / 1.0 = 770.4 kN Cl. 6.2.3 (2)a

Anet= 2170-(24x10) = 1930 mm2 Using table 3.8 of EN 1:8 page 33: 3= 0.5 ( since pitch p1 = 60 mm = 2.5 d0) EC 3 -1-8 Cl. 3.10.3(2)

Nu,Rd = 3 Anet fu / M2 6.2.3(2)b= 0.5 x 1930 x 490 /1.1 = 429.9 kN

Nt,RD = 429.9 kN > 340kN (Nt,Ed) 6.2.3 (2)Design is Safe.

E.g. 11: An angle in tension connected by one leg. A 100x65x6 mm single angle tie is connected through the smaller leg by two 20 mm diameter bolts in line with a pitch of 2.5 d0. Determine the design ultimate resistance of the angle assuming S275 steel and material factors of M2 = 1.25 and M0= 1

Solution:1-Net area of angle connected by the smaller leg and allowing for holes Anet = ( b- d0) t + (b-t)t = (65-22) x 6 + (65-6) x 6 = 612 mm2

Compare this value with gross area = 1120 mm2 ( section tables)

2) Design ultimate tensile resistance of the net cross-section for a two bolt angle connection eq. 6.7 EN 1-1:

Using table 3.8 of EN 1:8 page 33: 2 = 0.4 ( since pitch p1 = 60 mm = 2.5 d0) EC 3 -1-8 Cl. 3.10.3(2)

Nu,Rd = 2 Anet fu / M2 = 0.4 x 612 x 430 / 1.25 = 84.24kN

Npl,Rd = Afy/ M0 = [(65 x 6)+(65-6)x6] x275 / 1.0) = 204.6 kN

Note: a check is must for strength of bolts