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Steel Design Chapter 2 Tension Members (Beams), Slides of Civil Engineering

Technological University of the Philippines (TUP)Civil Engineering

Design of tension members Net area calculations Rupture and yielding checks Step-by-step design procedures

Typology: Slides

2023/2024

Available from 06/05/2025

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imwinter 🇵🇭

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Prepared by;
Gabriel I. Gamana, CE, M.Sc.
Engr. Gabriel Gamana, M.Sc.
Steel Design
Table of Contents
2
1.0 Introduction
2.0 Tension Members
3.0 Compression Members
4.0 Beams
5.0 Beam-Columns
6.0 Connections
2.0 Tension
Members
3
2.1 Introduction
2.2 Tensile Strength
2.3 Effective Area
2.4 Staggered Fasteners
2.5 Block Shear
2.6 Design of Tension Members
2.1 Introduction
4
1 2
3 4
pf3
pf4
pf5
pf8
pf9

Partial preview of the text

Download Steel Design Chapter 2 Tension Members (Beams) and more Slides Civil Engineering in PDF only on Docsity!

Gabriel I. Gamana, CE, M.Sc.

Engr. Gabriel Gamana, M.Sc.

Steel Design

Table of Contents

2 1.0 Introduction 2.0 Tension Members 3.0 Compression Members 4.0 Beams 5.0 Beam-Columns 6.0 Connections

2.0 Tension

Members

3 2.1 Introduction 2.2 Tensile Strength 2.3 Effective Area 2.4 Staggered Fasteners 2.5 Block Shear 2.6 Design of Tension Members

2.1 Introduction

4

Gabriel I. Gamana, CE, M.Sc.

2.1 Introduction

5

  • Tension members are structural elements that are subjected to axial tensile forces. They are used in various types of structures and include truss members, bracing for buildings and bridges, cables in suspended roof systems, and cables in suspension and cable-stayed bridges.
  • Any cross-sectional configuration may be used, because for any given material, the only determinant of the strength of a tension member is the cross-sectional area.

2.1 Introduction

6

  • The stress in an axially loaded tension member is given by 𝜎 = 𝑃/𝐴 where P is the magnitude of the load and A is the cross-sectional area (the area normal to the load).
  • The stress as given by this equation is exact, provided that the cross section under consideration is not adjacent to the point of application of the load, where the distribution of stress is not uniform.

2.2 Tensile Strength

7 A tension member can fail by reaching one of two limit states: excessive deformation or fracture. To prevent excessive deformation, initiated by yielding, the load on the gross section must be small enough that the stress on the gross section is less than the yield stress Fy. To prevent fracture, the stress on the net section must be less than the tensile strength Fu.

  • The nominal strength in yielding is; 𝑃௡ = 𝐹௬𝐴௚
  • The nominal strength in fracture is; 𝑃௡ = 𝐹௨𝐴௘

2.2 Tensile Strength

8

  • The design strength is the resistance factor (LRFD) times the nominal strength is; 𝑃௨ ≤ 𝜙𝑃௡ 𝑃௨ ≤ 0.90𝐹௬𝐴௚, 𝑌𝑖𝑒𝑙𝑑𝑖𝑛𝑔 𝑃௨ ≤ 0.75𝐹௨𝐴௘, 𝐹𝑟𝑎𝑐𝑡𝑢𝑟𝑒
  • In allowable strength design (ASD), the total service load is compared to the allowable strength; 𝑃௨ ≤
2.00 ≤ 0.50𝐹௨𝐴௘,^ 𝐹𝑟𝑎𝑐𝑡𝑢𝑟𝑒

Gabriel I. Gamana, CE, M.Sc.

2.3 Effective Area

13

  • Several factors influencing the performance of a tension member, the manner in which it is connected is the most important. A connection almost always weakens the member, and the measure of its influence is called the joint efficiency.
  • This factor is a function of the ductility of the material, fastener spacing, stress concentrations at holes, fabrication procedure, and a phenomenon known as shear lag. All contribute to reducing the effectiveness of the member, but shear lag is the most important.

2.3 Effective Area

14

For bolted splice plates or gusset plates connecting tension members 𝐴௘ = 𝐴௡ ≤ 0.85𝐴௚

2.3 Effective Area

15 Shear Lag Factors for Connections to Tension Members

2.3 Effective Area

16

Gabriel I. Gamana, CE, M.Sc.

2.3 Effective Area

17

2.3 Effective Area

18 Problem 2- W8x21, A50 steel is to carry a dead load of 30 kips and a live load of 90 kips in tension. The member is 25 ft long. Verify the member strength by both LRFD and ASD with the bolted end connection shown. Verify that the member satisfies the recommended slenderness limit. 𝐴௚ = 6.16 𝑖𝑛ଶ^ , 𝑏௙ = 5.27 𝑖𝑛 , 𝑡௙ = 0.40 𝑖𝑛 , 𝑑 = 8.28 𝑖𝑛 , 𝑟௬ = 1.26 𝑖𝑛 , 𝑦ത = 0.831 (𝑓𝑜𝑟 𝑊𝑇 4 𝑥10.50)

2.3 Effective Area

19 Problem 2- Verify, by both ASD and LRFD, the strength of an L4× 4 ×½, A steel, with one line of (4) ¾ in. diameter bolts in standard holes. The member carries a dead load of 20 kips and a live load of 60 kips in tension. Calculate at what length this tension member would cease to satisfy the recommended slenderness limit. 𝐴௚ = 3.75 𝑖𝑛ଶ, 𝑟௭ = 0.776 𝑖𝑛, 𝑦ത^ = 𝑥̅ = 1.18 𝑖𝑛

2.3 Effective Area

20 Problem 2- A WT6×20, A50 steel, member has a length of 30 ft and carries a dead load of 40 kips and a live load of 120 kips in tension. Assume the end connection is fillet welded and has a length of 16 in. Verify the member strength by both LRFD and ASD. Assume that the gusset plate and the weld have been checked and are satisfactory. 𝐴௚ = 5.84 𝑖𝑛ଶ, 𝑟௭ = 1.57 𝑖𝑛, 𝑦ത = 𝑥̅ = 1.09 𝑖𝑛

Gabriel I. Gamana, CE, M.Sc.

2.4 Staggered Fasteners

25 Problem 2- An angle with staggered fasteners in each leg is shown in figure. A36 steel is used, and holes are for 7Ú8-inch-diameter bolts. Determine the design strength using LRFD and ASD. 𝐴௚ = 6.80 𝑖𝑛ଶ.

2.4 Staggered Fasteners

26 Problem 2- Determine the net area along route ABCDEF for the C15 x 33. (𝐴௚ = 10.00 𝑖𝑛ଶ) shown in figure. Hole are for ¾ in. bolts.

2.5 Block Shear

27

  • For certain connection configurations, a segment or “block” of material at the end of the member can tear out. For example, the connection of the single-angle tension member shown in figure is susceptible to this phenomenon, called block shear.
  • For the case illustrated, the shaded block would tend to fail by shear along the longitudinal section ab and by tension on the transverse section bc.

2.5 Block Shear

28

  • The model used in the AISC Specification assumes that failure occurs by rupture (fracture) on the shear area and rupture on the tension area. Both surfaces contribute to the total strength, and the resistance to block shear will be the sum of the strengths of the two surfaces.
  • The shear rupture stress is taken as 60% of the tensile ultimate stress, so the nominal strength in shear is 0.6𝐹௨𝐴௡௩ and the nominal strength in tension is 𝐹௨𝐴௡௧ 𝑅௡ = 0.6𝐹௨𝐴௡௩ + 𝑈௕௦𝐹௨𝐴௡௧ 𝑅௡ ≤ 0.6𝐹௬𝐴௚௩ + 𝑈௕௦𝐹௨𝐴௡௧ For LRFD 𝑅௨ ≤ 0.75𝑅௡ For ASD 𝑅௨ ≤ 0.5𝑅௡

Gabriel I. Gamana, CE, M.Sc.

2.5 Block Shear

29 Where: 𝐴௡௩ = net area along shear surface 𝐴௚௩ = gross area along shear surface 𝐴௡௧ = net area along tension surface 𝑈௕௦ = 1.0 when tension stress is uniform 0.5 when tension stress is non-uniform

2.5 Block Shear

30 Problem 2- Compute the block shear strength of the tension member shown in figure. The holes are for 7Ú8-inch-diameter bolts, and A36 steel is used.

2.5 Block Shear

31 Problem 2- The single 200 mm x 10 mm steel plate is connected to 12 mm—thick steel plate by four 16 mm diameter rivets as shown in figure. The Steel is ASTM A36 with Fy = 248 MPa and Fu = 400 MPa. Determine the value of P in all possible modes of failure and safe value of P that the connection can resist.

2.6 Design of Tension Members

32

  • The design of a tension member involves finding a member with adequate gross and net areas. If the member has a bolted connection, the selection of a suitable cross section requires an accounting for the area lost because of holes.
  • A secondary consideration in the design of tension members is slenderness. If a structural member has a small cross section in relation to its length, it is said to be slender. A more precise measure is the slenderness ratio, L/r, where L is the member length and r is the minimum radius of gyration of the cross-sectional area.