Question-2: Determine the design (maximum) tensile capacity N* for the connectio

Question

Question-2: Determine the design (maximum) tensile capacity N* for the connections shown below. All plates and steel profiles (i.e. PFCs and UBs) are Grade 300PLUS. The steel plates comply with AS/NZS3678 standard. Hint: Both connected members and plates must be checked. In case you require data involving steel grades and dimensions, please use this link to look up the data you need: http://infostore.saiglobal.com/store/Portal.aspx?publisher=AS (not clickable). Much appreciated.

20 are a 50 – Bolt M16 150PFC - 50 he —180 10 ( for the LOFFT, Bolt M16 200UB22.3 a 50 ce N* 10 : AF s—180 l 90 | Oo | Oo - 50 or 200UB22.3 - –200 mmH Cross-section Elevation view Top view

Explanation / Answer

• Gross section area = Ag = 5 x ½ = 2.5 in2

• Net section area (An)

- Bolt diameter = db = 7/8 in.

- Nominal hole diameter = dh = 7/8 + 1/16 in. = 15/16 in.

- Hole diameter for calculating net area = 15/16 + 1/16 in. = 1 in.

- Net section area = An = (5 – 2 x (1)) x ½ = 1.5 in2

• Gross yielding design strength = t Pn = t Fy Ag

- Gross yielding design strength = 0.9 x 50 ksi x 2.5 in2 = 112.5 kips

• Fracture design strength = t Pn = t Fu Ae

- Assume Ae = An (only for this problem)

- Fracture design strength = 0.75 x 65 ksi x 1.5 in2 = 73.125 kips

• Design strength of the member in tension = smaller of 73.125 kips and 112.5 kips

• Net section area = An

- Bolt diameter = 5/8 in.

- Nominal hole diameter = 5/8 + 1/16 = 11/16 in.

- Hole diameter for calculating net area = 11/16 + 1/16 = 3/4 in.

- Net section area = Ag – (3/4) x 3/8 = 2.86 – 3/4 x 3/8 = 2.579 in2

• Effective net area = Ae = 0.85 x 2.579 in2 = 2.192 in2

• Gross yielding design strength = t Ag Fy = 0.9 x 2.86 in2 x 36 ksi = 92.664 kips

• Net section fracture = t Ae Fu = 0.75 x 2.192 in2 x 58 ksi = 95.352 kips

8

CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Design strength = 92.664 kips (gross yielding governs)

• Ultimate (design) load acting for the tension member = Pu

- The ultimate (design) load can be calculated using factored load combinations given on

page 2-11 of the AISC manual, or Equations (4.2-1 to 4.2-5) of notes (see pg. 4)

- According to these equations, two loading combinations are important for this problem.

These are: (1) 1.4 D; and (2) 1.2 D + 1.6 L

- The corresponding ultimate (design) loads are:

1.4 x (PD) = 1.4 (35) = 49 kips

1.2 (PD) + 1.6 (PL) = 66 kips (controls)

- The ultimate design load for the member is 66 kips, where the factored dead +

live loading condition controls.

• Compare the design strength with the ultimate design load

- The design strength of the member (92.664 kips) is greater than the ultimate design load

(66 kips).

- t Pn (92.664 kips) > Pu (66 kips)

• The L 4 x 4 x 3/8 in. made from A36 steel is adequate for carrying the factored loads.

• For ASTM A992 material: Fy = 50 ksi; and Fu = 65 ksi

• For the W8 x 24 section:

- Ag = 7.08 in2

- tw = 0.285 in.

- tf = 0.4 in.

• Gross yielding design strength = t Pn = t Ag Fy = 0.90 x 7.08 in2 x 50 ksi = 319 kips

• Net section fracture strength = t Pn = t Ae Fu = 0.75 x Ae x 65 ksi

- Ae = U An - for bolted connection

- An = Ag – (no. of holes) x (diameter of hole) x (thickness of flange)

An = 7.08 – 4 x (diameter of bolt + 1/8 in.) x 0.4 in.

An = 5.68 in2

d = 7.93 in.

bf = 6.5 in.

ry = 1.61 in.

3 in.

5 in.

5 in.

3 in. f d

3 in. 3 in. 3 in. 3 in. 3 in.

• The effective hole diameter is 1 + 1/8 = 1.125 in.

wn = 16.0 – 2 (1.125) = 13.75 in.

wn = 16.0 – 3 (1.125) + 2 x 32/ (4 x 5) = 13.52 in.

• The line a-b-c-d-e governs:

• An = t wn = 0.75 (13.52) = 10.14 in2

• Each fastener resists an equal share of the load

• Therefore different potential failure lines may be subjected to different loads.

• For example, line a-b-c-d-e must resist the full load, whereas i-j-f-h will be subjected to 8/11

of the applied load. The reason is that 3/11 of the load is transferred from the member

before i-j-f-h received any load.

• Step I. Select a section from the Tables

- Go to the TEN section of the AISC manual. See Table 3-1 on pages 3-17 to 3-19.

- From this table, select W8x10 with Ag = 2.96 in2, Ae = 2.22 in2.

- Gross yielding strength = 133 kips, and net section fracture strength=108 kips

- This is the lightest section in the table.

- Assumed U = 0.75. And, net section fracture will govern if Ae < 0.923 Ag

• Step II. Calculate the net section fracture strength for the actual connection

- According to the Figure above, An = Ag - 4 (db + 1/8) x tf

- An = 2.96 - 4 (3/4 + 1/8) x 0.205 = 2.24 in2

- The connection is only through the flanges. Therefore, the shear lag factor U will be the

distance from the top of the flange to the centroid of a WT 4 x 5.

31

CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- See DIM section of the AISC manual. See Table 1-8, on pages 1-50, 1-51

- x = 0.953

- U = 1- x /L = 1 - 0.953 / 4 = 0.76

- Ae = 0.76 An = 0.76 x 2.24 = 1.70 in2

- tPn = 0.75 x Fu x Ae = 0.75 x 65 x 1.70 = 82.9 kips

- Unacceptable because Pu = 100 kips; REDESIGN required

• Step III. Redesign

Many ways to redesign. One way is shown here:

- Assume t Pn > 100 kips

- Therefore, 0.75 x 65 x Ae > 100 kips

- Therefore, Ae > 2.051 in2

- Assume, Ae = 0.76 An (based on previous calculations, step II)

- Therefore An > 2.7 in2

- But, Ag = An + 4 (db + 1/8) x tf (based on previous calculations, step II)

- Therefore Ag > 2.7 + 3.5 x tf

- Go to the section dimension table 1-1 on page 1-22 of the AISC manual. Select next

highest section.

For W 8 x 13, tf = 0.255 in.

Therefore, Ag > 2.7 + 3.5 x 0.255 = 3.59 in2

From Table 1-1, W8 x 13 has Ag = 3.84 in2 > 3.59 in2

Therefore, W8 x 13 is acceptable and is chosen.

32

CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Step IV. Check selected section for net section fracture

- Ag = 3.84 in2

- An = 3.84 - 3.5 x 0.255 = 2.95 in2

- From dimensions of WT4 x 6.5, x = 1.03 in.

- Therefore, U = 1- x /L = 1-1.03/4 = 0.74

- Therefore, Ae = U An = 0.74 x 2.95 = 2.19 in2

- Therefore, net section fracture strength = 0.75 x 65 x 2.19 = 106.7 kips

- Which is greater than 100 kips (design load). Therefore, W 8 x 13 is acceptable.

• Step V. Check the block shear rupture strength

o Identify the block shear path

4

in.

2 in.

1.5 in.

1.5 in.

2 in. 4 in.

- The block shear path is show above. Four blocks will separate from the tension

member (two from each flange) as shown in the figure above.

- Agv = [(4+2) x tf ] x 4 = 6 x 0.255 x 4 = 6.12 in2 - for four tabs

- Anv = {4+2 - 1.5 x (db+1/8)} x tf x 4 = 4.78 in2

- Agt = 1.5 x tf x 4 = 1.53 in2

- Ant = {1.5 - 0.5 x (db+1/8)}x tf x 4 = 1.084 in2

o Identify the governing equation:

33

CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- FuAnt = 65 x 1.084 = 70.4 kips

- 0.6FuAnv = 0.6 x 65 x 4.78 = 186.42 kips , which is > FuAnt

o Calculate block shear strength

- tRn = 0.75 (0.6FuAnv + FyAgt) = 0.75 (186.42 + 50 x 1.53) = 197.2 kips

- Which is greater than Pu = 100 kips. Therefore W8 x 13 is still acceptable

• Summary of solution

Mem. Design Ag An U Ae Yield Fracture Block-shear

load strength strength strength

W8x13 100 kips 3.84 2.95 0.74 2.19 173 kips 106.7 kips 197.2 kips

Design strength = 106.7 kips (net section fracture governs)

W8 x 13 is adequate for Pu = 100 kips and the given connection

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