Structural Steel Design (USING AISC-13Ed. -LRFD )

By

Asst. Prof. Dr. MAY J. HAMOODI

M. J. HAMOODI

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column

beambracing

connection

STRUCTURAL STEEL DESIGN

Introduction

Steel structures may be divided into three general categories:

a) Framed structures

: Most typical building construction is in this category. The multistory building usually consists of beams and columns, either rigidly connected or having simple end connections along with diagonal bracing to provide stability.

b) Shell-type structures

:The vessel used to store liquids, storage bins, tanks are examples for the structures in this category. These structures are participated in carrying loads, where the main stress is tension.

c) Suspension-type structures

:The most common structure of this type is the suspension bridge and cable-stayed bridge where tension cables are the major supporting elements. A roof may be cable- supported.

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L

Steel tests

1- Tensile test,

:

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =𝑃𝑃𝐴𝐴

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝐿𝐿𝑠𝑠−𝐿𝐿𝐿𝐿

L n

= the new length

2- Chemical test.

3- Impact test.

4-Hardness test.

5-X-ray test.

[ see Table2-3,2-4 and 2-5 in AISC-13th

Edition]

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Plastic foam

Roof beam

TileConcrete

Steel deckBeam

Connection

Bolt Shear connector

Column

Column base plateAnchor bolt

Concrete foundation

Chemical contents of steel alloy

Product analysis tolerances

:

Element Max. specified value % Iron Fe 96.08 Carbon C 0.15 Manganese Mn 0.6 Phosphorus P 0.04 Sulfur S 0.05 Silicon Si 0.3 ASTM A6/A 6M

Section in a steel building

:

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Types of structural steel sections

1-Single sections,

:

2-Built-up section,

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Design philosophies

Load and resistance factor design [LRFD] ,is similar to plastic design in that ,at failure, parts of member will be subjected to very large strain ( large enough to put the member into the plastic range). Load factors are applied to the service loads. In addition, the theoretical strength of the member is reduced by the application of a resistance factor. The criterion that must be satisfied in the selection of a member is,

:

Required strength ≤ available strength

Factored load ≤ factored strength

∑ service loads x load factors ≤ resistance x resistance factor

The load factors are greater than unity and the resistance factor is less than unity.

The required strength is determined from the following factored combinations,

Where

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According to the AISC,

𝑅𝑅𝑢𝑢 ≤ 𝜑𝜑𝑅𝑅𝑠𝑠

Where

𝑅𝑅𝑢𝑢 = required strength

𝑅𝑅𝑠𝑠 = nominal strength (resistance)

𝜑𝜑 = resistance factor given by the specification as,

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Beam to Column Connections

Beam to column connection is an eccentric connection, the connection at which the resultant of the applied loads does not pass through the center of gravity of the fasteners or welds. The centroid of the shear area of the fasteners or welds may be used as the reference point, and the perpendicular distance from the line of action of the load to the centroid is called the eccentricity.

There are two types of beam to column connection:

A- Framed beam to column connections,

In a connection such as the one for the tee stub bracket of shown in the Figure, an eccentric load creates a couple that will increase the tension in the upper row of fasteners and decrease it in the lower row.

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If the fasteners are pretensioned high-strength bolts, the contact surface between the column flange and the bracket flange will be uniformly compressed before the external load is applied. The bearing pressure will equal the total bolt tension divided by the area of contact. As the load P is gradually applied, the compression at the top will be relieved and the compression at the bottom will increase, as shown in the Figure.

When the compression at the top has been completely overcome, the components will separate, and the couple Pe will be resisted by tensile bolt forces and compression on the remaining surface of contact, as shown in the Figure. As the ultimate load is approached, the forces in the bolts will approach their ultimate tensile strengths.

A simplified method will be used here. The neutral axis of the con-nection is assumed to pass through the centroid of the bolt areas. Bolts above this axis are subjected to tension, and bolts below the axis are assumed to be subjected to compressive forces, as in the Figure. Since there are two bolts at each level, each force is shown as 2rt . The resultant of the tensile and compressive forces is a couple that equals the resisting moment of the connection. When the resisting moment is equated to the applied moment, the resulting equation can be solved for the unknown bolt tensile force rt .

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Specification,

A beam-to-column connection is made with a structural tee as shown in the Figure. Eight 3⁄4 -inch-diameter, A325, fully tightened bearing-type bolts are used to attach the flange of the tee to the column flange. Investigate the adequacy of this connection (the tee-to-column connection) if it is subjected to a service dead load of 20 kips and a ser-vice live load of 40 kips at an eccentricity of 2.75 inches. Assume that the bolt threads are in the plane of shear. All structural steel is A992.

Example co1

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Solution,

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Compute the tensile force per bolt and then check the tension–shear interaction. Because of symmetry, the centroid of the connection is at middepth. The Figure shows the bolt areas and the distribution of bolt tensile forces.

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B- Moment resisting beam to column connections,

Special measures must be taken to make a connection moment-resisting. Moment-rotation curves for three different connection types are shown in the Figure .

The fixed-end moment, caused by the actual load on a beam with fixed ends , is plotted on the moment axis (rotation =0). If the end were simply supported (pinned end), the moment would be zero. The end rotation corresponding to a simple support and the actual loading is then plotted on the rotation axis (moment =0). The line connecting the two points is the beam line.

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If beam line is superimposed on the three moment-rotation curves, as in the following figure, the rigid connection which has a theoretical moment capacity equal to the fixed-end moment (FEM) of the beam, it will actually have a moment resistance of about 90% of FEM.

Similarly, a connection designed as pinned (simply supported; no moment) would actually be capable of transmitting a moment of about 20% of the fixed-end moment, with a rotation of approximately 80% of the simple support rotation.

The advantage of a partially-restrained connection is that it can equalize the negative and positive moments within a span. Regardless of the support conditions, whether simple, fixed, or something in between, the same static moment of 𝑤𝑤𝑤𝑤2 8⁄ will have to be resisted by a uniformly loaded beam .

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The effect of partially restrained connections is to shift the moment diagram as shown in the Figure. This will increase the positive moment but decrease the negative moment, which is the maximum moment in the beam, thereby potentially resulting in a lighter beam.

The AISC Specification defines three categories of connections,

• Simple (flexible connection)

• Fully Restrained (rigid connection)

• Partially Restrained (semirigid connection)

Types of moment resisting connections,

Examples of commonly used moment resisting connections are illustrated in the following:

most of the moment transfer is through the beam flanges, and most of the moment capacity is developed there. The plate connecting the beam web to the column flange is shop welded to the column and field bolted to the beam. With this arrangement the flanges can be field welded to the column. The plate connection is designed to resist only shear and takes care of the beam reaction. Complete penetration groove welds connect the beam flanges to the column to transfer the moment.

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When a beam is framed into the web of a column rather than its flange the connection shown in Figure (b) can be used. This connection is similar to the one shown in Figure (a) but requires the use of column stiffeners to make the connections to the beam flanges.

The three-plate connection is shown in Figure c. It has the advantage of being completely field bolted. The flange plates and the web plate are all shop welded to the column flange and field bolted to the beam. To provide for variation in the beam depth, the distance between flange plates is made larger than the nominal depth of the beam, usually by about 3/8inch. This gap is filled at the top flange during erection with shims, which are thin strips of steel used for adjusting the fit at joints, Figure (d) .

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Design a three-plate moment connection of the type shown in the Figure for the connection of a W21 ×50 beam to the flange of a W14 ×99 column. Assume a beam setback of 1/2 inch. The connection must transfer the following service load effects: a dead-load moment of 35 ft-kips, a live-load moment of 105 ft-kips, a dead-load shear of 5.5 kips, and a live-load shear of 16.5 kips. All plates are to be shop welded to the column with E70XX electrodes and field bolted to the beam with A325 bearing-type bolts. A36 steel is used for the plates, and A992 steel is used for the beam and column.

Example co2

Solution,

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o.k.

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For the flange connection, first select the bolts. From the Figure, the force at the interface between the beam flange and the plate is,

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The Figure shows the bolt lay-out and two possible block shear failure modes. The shear areas for both cases are,

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The shear areas for both cases are,

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If a transverse tension area between the bolts is considered (Figure a), the width is 3.5 in. If two outer blocks are considered (Figure b), the total tension width is 2(1.5) =3.0 in. This will result in the smallest block shear strength,

Check block shear in the beam flange. The bolt spacing and edge distance are the same as for the plate,

Since 384.3 kips >360.6 kips, block shear in the plate controls. O.k.

Part of the beam flange area will be lost because of the bolt holes.Use the provisions of AISC F13.1 to determine whether we need to account for this loss.

The gross area of one flange is,

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.: Use the connection shown in the Figure (column stiffener requirements should be check).

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Beam-Columns

Most beams and columns are subjected to some degree of both bending and axial load. This is especially true for members in rigid frame. For the rigid frame shown in the figure,

The x-bracing, indicated by dashed lines, prevents sidesway in the lower

story. The members of this frame are beam-columns.

Moment amplification,

The presence of the axial load in the beam-column will produce secondary moments. For the beam–column with an axial load and a transverse uniform load , which is shown in the figure, has a bending moment caused by the uniform load and an additional moment, Py, caused by the axial load acting at an eccentricity from the longitudinal axis of the member. This secondary moment ,Py, is largest where the deflection is largest.

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Ordinary structural analysis methods that do not take the displaced geometry into account are referred to as first-order methods. The methods which is used to find the deflections and secondary moments, usually implemented with a computer program, are called second-order methods. AISC Specification, permit the use of either a second-order analysis or the moment amplification method. This method entails computing the maximum bending moment resulting from flexural loading by the first order analysis, then multiplying by a moment amplification factor to account for the secondary moment.

Braced and unbraced frames,

Two amplification factors are used by AISC Specification: one to account for amplification resulting from the member deflection in braced frame and one to account for the effect of sway when the member is part of an unbraced frame.

a) braced frame b) unbraced frame

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In Figure (a), the member is restrained against sidesway, and the maximum secondary moment is P𝛿𝛿, which is added to the maximum moment within the member. In Figure (b), the frame is actually unbraced, and there is an additional component of the secondary moment, shown in Figure b, that is caused by sidesway. This secondary moment has a maximum value of PΔ, which represents an amplification of the end moment.

To approximate these two effects, two amplification factors, B1 and B2, are used for the two types of moments,(𝑀𝑀𝑛𝑛𝑛𝑛 , no translation moment and 𝑀𝑀𝑙𝑙𝑛𝑛 , lateral translation moment) as follows:

𝑀𝑀𝑟𝑟 = 𝐵𝐵1𝑀𝑀𝑛𝑛𝑛𝑛 + 𝐵𝐵2𝑀𝑀𝑙𝑙𝑛𝑛

Where,

𝑀𝑀𝑟𝑟= required moment strength = 𝑀𝑀𝑢𝑢

𝑀𝑀𝑛𝑛𝑛𝑛 = maximum moment assuming that no sidesway occurs.

𝑀𝑀𝑙𝑙𝑛𝑛 = maximum moment caused by sidesway .This moment can be caused by lateral loads or by unbalanced gravity loads. Gravity load can produce sidesway if the frame is unsymmetrical or if the gravity loads are unsymmetrically placed.

B1

B

= amplification factor for the moments occurring in the member when it is braced against sidesway.

2

= amplification factor for the moments resulting from sidesway.

Evaluation of B1 , B2

𝐵𝐵1 =𝐶𝐶𝑚𝑚

1 − 𝑃𝑃𝑟𝑟 𝑃𝑃𝑒𝑒1⁄≥ 1

and 𝑪𝑪𝒎𝒎 factors,

Where,

𝑃𝑃𝑟𝑟 = 𝑃𝑃𝑛𝑛𝑛𝑛 + 𝐵𝐵2𝑃𝑃𝑙𝑙𝑛𝑛 = required axial load = 𝑃𝑃𝑢𝑢

𝑃𝑃𝑟𝑟 = 𝑃𝑃𝑛𝑛𝑛𝑛 + 𝑃𝑃𝑙𝑙𝑛𝑛 (as an approximation)

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𝑃𝑃𝑛𝑛𝑛𝑛= axial load corresponding to the braced condition.

𝑃𝑃𝑙𝑙𝑛𝑛 = axial load corresponding to the sidesway condition.

𝑃𝑃𝑒𝑒1 = elastic critical buckling resistance = 𝜋𝜋2𝐸𝐸𝐸𝐸

(𝑘𝑘1𝐿𝐿)2

𝑘𝑘1=1 = effective length factor – no lateral translation.

𝐵𝐵2 =1

1 − ∑𝑃𝑃𝑛𝑛𝑛𝑛∑𝑃𝑃𝑒𝑒2

≥ 1

Where,

∑𝑃𝑃𝑒𝑒2=elastic story sidesway buckling resistance = ∑𝜋𝜋2𝐸𝐸𝐸𝐸

(𝑘𝑘2𝐿𝐿)2

𝑘𝑘2= effective length factor – with lateral translation.

∑𝑃𝑃𝑛𝑛𝑛𝑛=total vertical load supported by the story.

𝐶𝐶𝑚𝑚= a coefficient assuming no lateral translation of the frame, whose value shall be taken as follows:

1- for beam-columns not subjected to transverse loading between supports in the plane of bending,

For beam-column AB

𝐶𝐶𝑚𝑚 = 0.6 − 0.4 �𝑀𝑀1𝑀𝑀2

�

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Where,

M1 is the end moment that is smaller in absolute value, M2

is the larger, and the ratio is positive for members bent in reverse curvature and negative for single-curvature bending.

2- for beam-columns subjected to transverse loading between supports in the plane of bending,

For beam-column AB

𝐶𝐶𝑚𝑚 = 1

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Interaction formulas,

The interaction of flexure and compression in doubly symmetric members and singly symmetric members for which 0.1 ≤ (𝐸𝐸𝑦𝑦𝑦𝑦 𝐸𝐸𝑦𝑦⁄ ) ≤ 0.9, are limited by equation H1-1a for large axial load and equation H1-1b for small axial load; where 𝐸𝐸𝑦𝑦𝑦𝑦 is the moment of inertia of the compression flange about the y axis,in4

For 𝑃𝑃𝑢𝑢𝜑𝜑𝑦𝑦𝑃𝑃𝑛𝑛

≥ 0.2

. The equations are :

𝑃𝑃𝑢𝑢𝜑𝜑𝑦𝑦𝑃𝑃𝑛𝑛

+ 89� 𝑀𝑀𝑢𝑢𝑢𝑢𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛𝑢𝑢

+ 𝑀𝑀𝑢𝑢𝑦𝑦𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛𝑦𝑦

� ≤ 1.0 …… H1-1a

For 𝑃𝑃𝑢𝑢𝜑𝜑𝑦𝑦𝑃𝑃𝑛𝑛

< 0.2

𝑃𝑃𝑢𝑢2𝜑𝜑𝑦𝑦𝑃𝑃𝑛𝑛

+ � 𝑀𝑀𝑢𝑢𝑢𝑢𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛𝑢𝑢

+ 𝑀𝑀𝑢𝑢𝑦𝑦𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛𝑦𝑦

� ≤ 1.0 .…. H1-1b

Where,

x and y are bending axes

𝜑𝜑𝑦𝑦 = 𝜑𝜑𝑏𝑏 = 0.9

Design notes,

1- The effective length KL (ft) used in Table 6-1(pag.6-5 – 6-95 of the AISC manual- part 6), is the larger of (𝑘𝑘𝑙𝑙)𝑦𝑦 and (𝑘𝑘𝑙𝑙)𝑦𝑦 𝑒𝑒𝑒𝑒 , where

(𝑘𝑘𝑙𝑙)𝑦𝑦 𝑒𝑒𝑒𝑒 =(𝑘𝑘𝑙𝑙)𝑢𝑢𝑟𝑟𝑢𝑢𝑟𝑟𝑦𝑦

2- the interaction equations can be written as follows:

𝑝𝑝𝑃𝑃𝑢𝑢 + 𝑏𝑏𝑢𝑢𝑀𝑀𝑢𝑢𝑢𝑢 + 𝑏𝑏𝑦𝑦𝑀𝑀𝑢𝑢𝑦𝑦 ≤ 1.0 …. H1-1a

12𝑝𝑝𝑃𝑃𝑢𝑢 +

98

(𝑏𝑏𝑢𝑢𝑀𝑀𝑢𝑢𝑢𝑢 + 𝑏𝑏𝑦𝑦𝑀𝑀𝑢𝑢𝑦𝑦 ) ≤ 1.0 …. H1-1b

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𝑝𝑝, 𝑏𝑏𝑢𝑢 𝑎𝑎𝑛𝑛𝑎𝑎 𝑏𝑏𝑦𝑦 are obtained from Table 6-1 with 𝐹𝐹𝑦𝑦 = 50 ksi only, where

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14 ft

Example BC-1

Select the lightest W14 x w section to carry an axial compression load of 170 kip dead load and 40 kip live load, in combination with a bending moment of 140 ft-kip dead load, 140 ft-kip live load, and 420 ft-kip wind load. The member is part of a braced system with lateral support provided at the top and bottom of a 14 ft length. Assume the moment causes single curvature. Use A992 steel.

Solution

For A992 steel , Fy=50 ksi and Fu=65 ksi

a) * gravity load (case 1),

𝑃𝑃𝑢𝑢=1.2(170)+1.6(40)=268 kip

𝑀𝑀𝑛𝑛𝑛𝑛= 1.2(140)+1.6(140)=392 k-ft

*gravity load + wind load (case 2),

𝑃𝑃𝑢𝑢=1.2(170)+0.5(40)=224 kip

𝑀𝑀𝑛𝑛𝑛𝑛= 1.2(140)+0.5(140)+1.6(420)=910 k-ft

.: use the more severe loading, case 2.

b) choose for a trial section,

𝑘𝑘𝑥𝑥=𝑘𝑘𝑦𝑦=1 (braced system)

(𝑘𝑘𝑘𝑘)𝑦𝑦=14

(𝑘𝑘𝑘𝑘)𝑦𝑦 𝑒𝑒𝑒𝑒 =(𝑘𝑘𝑘𝑘 )𝑥𝑥𝑟𝑟𝑥𝑥𝑟𝑟𝑦𝑦

= 141.9

= 7.3 [from Table 6-1, 𝑟𝑟𝑥𝑥𝑟𝑟𝑦𝑦≈ 1.9 for W14

.: H1-1b applies:

]

.: KL =14

Since the bending moment is dominant, estimate an intermediate 𝑝𝑝 value and solve for the required 𝑏𝑏𝑥𝑥 .

from Table 6-1, 𝑝𝑝 ≈ 0.8,

.: 𝑝𝑝𝑃𝑃𝑢𝑢= 𝑃𝑃𝑢𝑢𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛

= 0.81000

224 = 0.179 < 0.2

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12𝑝𝑝𝑃𝑃𝑢𝑢 +

98

(𝑏𝑏𝑥𝑥𝑀𝑀𝑢𝑢𝑥𝑥 ) ≤ 1.0

12�

0.81000

� 224 +98

𝑏𝑏𝑥𝑥1000

(910) ≤ 1.0

0.0896+1.02375 𝑏𝑏𝑥𝑥=1

.: 𝑏𝑏𝑥𝑥= 0.89

Enter AISC manual Table 6-1, with KL=14 and 𝑏𝑏𝑥𝑥= 0.89,

Try W14 x 159 ( 𝑝𝑝 = 0.541 and 𝑏𝑏𝑥𝑥= 0.826 )

𝑝𝑝𝑃𝑃𝑢𝑢= 𝑃𝑃𝑢𝑢𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛

= 0.5411000

224 = 0.121 < 0.2

.: 12𝑝𝑝𝑃𝑃𝑢𝑢 +98

(𝑏𝑏𝑥𝑥𝑀𝑀𝑢𝑢𝑥𝑥 ) ≤ 1.0

12 �0.5411000

�224 + 98

(0.8261000

)910 ≤ 1.0

0.9< 1.0

Try the next lighter section W14 x 145, ( 𝑝𝑝 = 0.593 and 𝑏𝑏𝑥𝑥= 0.912 )

12 �0.5931000

�224 + 98

(0.9121000

)910 =1.0

0.93< 1.0

.: check for W14 x 145.

c) column effect,

W14 x 145 , A=42.7 𝑟𝑟𝑥𝑥=6.33 𝑟𝑟𝑦𝑦=3.98 𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

= 7.11 ℎ𝑛𝑛𝑤𝑤

= 16.8

𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

≤ 𝜆𝜆𝑟𝑟 ℎ𝑛𝑛𝑤𝑤≤ 𝜆𝜆𝑟𝑟

7.11< 0.56�𝐸𝐸𝐹𝐹𝑦𝑦

16.8 < 1.49�𝐸𝐸𝐹𝐹𝑦𝑦

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7.11< 13.48 o.k 16.8 < 35.88 o.k.

𝑘𝑘𝑥𝑥= 𝑘𝑘𝑦𝑦 = 14

�𝑘𝑘𝑘𝑘𝑟𝑟�𝑥𝑥

= 14∗1∗126.33

= 26.5

�𝑘𝑘𝑘𝑘𝑟𝑟�𝑦𝑦

= 14∗1∗123.98

= 42.2 < 200

�𝑘𝑘𝑘𝑘𝑟𝑟� ≤ 4.71�

𝐸𝐸𝐹𝐹𝑦𝑦

42.2< 113.4

.:𝐹𝐹𝑐𝑐𝑟𝑟 = [0.658𝐹𝐹𝑦𝑦 𝐹𝐹𝑒𝑒⁄ ]𝐹𝐹𝑦𝑦

𝐹𝐹𝑒𝑒 =𝜋𝜋2𝐸𝐸

(𝑘𝑘𝑘𝑘 𝑟𝑟⁄ )2= 𝜋𝜋

229000(42.2)2

= 160 ksi

𝐹𝐹𝑐𝑐𝑟𝑟 = [0.65850 160⁄ ]50 = 43.86 ksi

𝑃𝑃𝑛𝑛=𝐹𝐹𝑐𝑐𝑟𝑟𝐴𝐴 = 43.86 ∗ 42.7 = 1872.82 kip

𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛 = 0.9(1872.82) = 1685.5 kip

Check 𝑃𝑃𝑢𝑢𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛

≥ 0.2

2241685.5

= 0.132 < 0.2

.: use formula H1-1b

d) beam effect,

𝐿𝐿𝑏𝑏 = 14 ft

𝐿𝐿𝑝𝑝 = 1.76 ∗ 3.98 �29000

50 =14 ft

M. J. HAMOODI

11

𝐿𝐿𝑏𝑏 = 𝐿𝐿𝑝𝑝 .: braced member

𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

≤ 𝜆𝜆𝑝𝑝

𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

≤ 0.38�𝐸𝐸𝐹𝐹𝑦𝑦

7.11

M. J. HAMOODI

12

𝐵𝐵1 =1

1−224 17403 .3⁄= 1.01

f) check for the interaction formula ,

𝑀𝑀𝑢𝑢𝑥𝑥 = 𝑀𝑀𝑛𝑛𝑛𝑛 𝐵𝐵1

𝑀𝑀𝑢𝑢𝑥𝑥 = 910(1.01) = 919.1 k-ft

Omitting the bending term for the y-axis,

𝑃𝑃𝑢𝑢2𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛

+ 𝑀𝑀𝑢𝑢𝑥𝑥𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛𝑥𝑥

≤ 1.0 .…. H1-1b

0.1322

+919.1975

≤ 1.0

1=1

.: use W14 x 145

M. J. HAMOODI

13

M. J. HAMOODI

14

M. J. HAMOODI

15

Solution:

F

Example BC-2

Verify if an ASTM A992 W12x45 has sufficient available strength to support the loads applied to the vertical members in the single-story frame shown in the figure. The unbraced frame is subjected to dead load ,roof live load and wind load. The service gravity loads are shown in Fig. a, and the service wind load is shown in Fig. b. bending is about the strong axis, and each column is laterally braced at top and bottom.

y=50 ksi and Fu

=65 ksi

𝑘𝑘𝑦𝑦=1 𝑘𝑘𝑥𝑥 = 𝑘𝑘𝑦𝑦 = 𝑘𝑘𝑏𝑏 = 15 ft

The results of the first order analysis (for the column) are given in the following:

M. J. HAMOODI

16

The load combinations are,

Combination 1:

Combination 2:

Combination 3:

Load combination 2 will obviously govern.

W12 x 45 , A=13.1 𝑟𝑟𝑥𝑥=5.15 𝑟𝑟𝑦𝑦=1.95 𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

= 7 ℎ𝑛𝑛𝑤𝑤

= 29.6

𝑆𝑆𝑥𝑥 = 57.7 𝑍𝑍𝑥𝑥 = 64.2 𝐸𝐸𝑥𝑥 = 348 𝑟𝑟𝑛𝑛𝑡𝑡 = 2.23 ℎ𝑜𝑜 = 11.5 𝑗𝑗 = 1.26

𝐵𝐵1 =𝐶𝐶𝑚𝑚

1 − 𝑃𝑃𝑢𝑢 𝑃𝑃𝑒𝑒1⁄≥ 1

𝐶𝐶𝑚𝑚 = 0.6 − 0.4 �

𝑀𝑀1𝑀𝑀2� = 0.6 − 0.4 � 0

𝑀𝑀2� = 0.6

𝑃𝑃𝑒𝑒1 = 𝜋𝜋2𝐸𝐸𝐸𝐸

(𝑘𝑘1𝐿𝐿)2 = 𝜋𝜋

229000(348)(1∗15∗12)2

= 3074.2 kip

M. J. HAMOODI

17

.: 𝐵𝐵1 =0.6

1−52 3074.2⁄ = 0.61 < 1.0

Use 𝐵𝐵1 = 1

𝐵𝐵2 =1

1 − ∑𝑃𝑃𝑛𝑛𝑛𝑛∑𝑃𝑃𝑒𝑒2

≥ 1

Assume that the column and the beam in the frame have the same section.

𝐺𝐺𝐴𝐴 =𝐸𝐸 15⁄𝐸𝐸 40⁄

= 2.66

𝐺𝐺𝐵𝐵 = 10 , .: 𝑘𝑘𝑥𝑥 = 2.2 (from chart)

𝑃𝑃𝑒𝑒2 =𝜋𝜋2𝐸𝐸𝐸𝐸

(𝑘𝑘2𝐿𝐿)2 = 𝜋𝜋

22900∗348(2.2∗15∗12)2

= 635.16 kip

𝐵𝐵2 =1

1 − 2 ∗ 51.22 ∗ 635.16

= 1.087

.: 𝑀𝑀𝑢𝑢= 184.8(1) + 16(1.087) = 202.2 k-ft

- calculate 𝑃𝑃𝑛𝑛

𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

≤ 𝜆𝜆𝑟𝑟 ℎ𝑛𝑛𝑤𝑤≤ 𝜆𝜆𝑟𝑟

7 < 0.56�𝐸𝐸𝐹𝐹𝑦𝑦

29.6 < 1.49�𝐸𝐸𝐹𝐹𝑦𝑦

7< 13.48 o.k 29.6 < 35.88 o.k.

�𝑘𝑘𝑘𝑘𝑟𝑟�𝑥𝑥

=2.2(15)12

5.15= 76.89

M. J. HAMOODI

18

�𝑘𝑘𝑘𝑘𝑟𝑟�𝑦𝑦

= 1(15)121.95

= 92.3 < 200 o.k.

�𝑘𝑘𝑘𝑘𝑟𝑟� ≤ 4.71�

𝐸𝐸𝐹𝐹𝑦𝑦

92.3 < 113.4

.:𝐹𝐹𝑐𝑐𝑟𝑟 = [0.658𝐹𝐹𝑦𝑦 𝐹𝐹𝑒𝑒⁄ ]𝐹𝐹𝑦𝑦

𝐹𝐹𝑒𝑒 =𝜋𝜋2𝐸𝐸

(𝑘𝑘𝑘𝑘 𝑟𝑟⁄ )2= 𝜋𝜋

229000(92.3)2

= 33.59 ksi

𝐹𝐹𝑐𝑐𝑟𝑟 = [0.65850 33.59⁄ ]50 = 26.8 ksi

𝑃𝑃𝑛𝑛=𝐹𝐹𝑐𝑐𝑟𝑟𝐴𝐴 = 26.8 ∗ 13.1 = 351.28 kip

𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛 = 0.9(351.28) = 316.1 kip

- Calculate 𝑀𝑀𝑛𝑛

𝑏𝑏𝑓𝑓2𝑛𝑛𝑓𝑓

≤ 𝜆𝜆𝑝𝑝= 0.38�𝐸𝐸𝐹𝐹𝑦𝑦

7

M. J. HAMOODI

19

𝐿𝐿𝑟𝑟 = 22.4 ft

.: 𝐿𝐿𝑝𝑝 < 15 < 𝐿𝐿𝑟𝑟

𝑀𝑀𝑛𝑛 = 𝐶𝐶𝑏𝑏 �𝑀𝑀𝑝𝑝 − �𝑀𝑀𝑝𝑝 − 0.7𝐹𝐹𝑦𝑦𝑆𝑆𝑥𝑥� �𝐿𝐿𝑏𝑏−𝐿𝐿𝑝𝑝𝐿𝐿𝑟𝑟−𝐿𝐿𝑝𝑝

�� ≤ 𝑀𝑀𝑝𝑝

𝑀𝑀𝑃𝑃 = Fy 𝑍𝑍𝑥𝑥=50 x 64.2 = 3210 =267.5 k.ft

𝑀𝑀𝑛𝑛= 1.67 �3210 − (3210 − 35 ∗ 57.7) �15−6.88

22.4−6.88�� =360 k.ft > 𝑀𝑀𝑝𝑝

𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛=0.9(267.5)=240.75 k.ft

- Interaction formula,

Check 𝑃𝑃𝑢𝑢𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛

≥ 0.2

52316.1

= 0.16 < 0.2

.: use H1-1b

𝑃𝑃𝑢𝑢2𝜑𝜑𝑐𝑐𝑃𝑃𝑛𝑛

+𝑀𝑀𝑢𝑢𝑥𝑥𝜑𝜑𝑏𝑏𝑀𝑀𝑛𝑛𝑥𝑥

≤ 1.0

12(0.16) +

202.2240.75

≤ 1.0

0.92

M. J. HAMOODI

1

Simple connections

Every structure is an assemblage of individual parts or members that must be fastened together, usually at the ends. When the line of action of the resultant force to be resisted, passes through the center of gravity of the connection, each part of the connection is assumed to resist an equal share of the load, and the connection is called a simple connection. Failure of structural members is rare; most structural failures are the result of poorly designed or detailed connections.

Bolted Connection

:

a) Lap joint b) Butt Joint

Single shear Double shear

High-Strength bolts

A325 and A490 are high-strength bolts according to ASTM. These bolts shall be tightened to a bolt tension not less than that given in[ Table j3.1].

:

M. J. HAMOODI

2

The nominal tensile stress 𝐹𝐹𝑛𝑛𝑛𝑛 and nominal shear stress in bearing-type connections 𝐹𝐹𝑛𝑛𝑛𝑛 for these bolts are given in [Table j3.2].

A connection with high-strength bolts is classified as:

1- Bearing-type connection.

2- Slip-critical connection.

In a bearing-type connection, slip is acceptable, and shear and bearing actually occur. The requirement for the installation of the bolts is that they be tensioned enough so that the surfaces of contact in the connection firmly bear on one another.

In a slip-critical connection no slippage is permitted and the friction force must not be exceeded. In some structures ,like bridges, the load on connections can undergo many cycles of reversal. In such cases, fatigue of the fasteners can become critical if the connection is allowed to slip; therefore a slip-critical connection is recommended.

M. J. HAMOODI

3

F=µN

Size and Use of Holes

The max. sizes of holes are given in [Table j3.3]. There are three types of holes, standard (S.H.), oversized (O.S.H.), short-slotted(S.S.H.) and long-slotted(L.S.H.) .

:

Diameter of S.H. = bolt diameter (d) + 1/16 in

S.H. __ can be used in bearing and slip-critical connections.

O.S.H.__ It is used in slip-critical connection only.

S.S.H. and L.S.H.__ They can be used in slip-critical connection. They can also be used in bearing -type , but the length shall be normal to the direction of the load.

Possible modes of failure of bolted connections

:

M. J. HAMOODI

4

Minimum spacing

The distance between centers of holes (S) ≥ 223 d

:

Where, d is the nominal diameter of the fastener; a distance of [3d ] is preferred.

Minimum edge distance

The distance from the center of a standard hole [S.H.] to an edge of connected part (L

:

e

1- The value from [Table j3.4].

) in any direction shall not be less than :

2- The value required for "Bearing strength at bolt holes".

Le for O.S.H., or S.S.H. or L.S.H = Le for S.H. + C2 (from Table j3.5).

Maximum edge distance

(L

:

e)max.

Where, t = thickness of the connected part under consideration.

= 12xt ≤ 6 in

Shear strength of high-strength bolts

The nominal shear strength is,

:

𝑅𝑅𝑛𝑛 = 𝐹𝐹𝑛𝑛𝑛𝑛 𝐴𝐴𝑏𝑏

Where,

𝐹𝐹𝑛𝑛𝑛𝑛 nominal shear stress [Table j3.2]

𝐴𝐴𝑏𝑏 cross-section area of the unthreaded part of the bolt.

𝑅𝑅𝑛𝑛 = 𝐹𝐹𝑛𝑛𝑛𝑛 2 𝐴𝐴𝑏𝑏 (double shear)

M. J. HAMOODI

5

The Design shear strength, 𝜑𝜑𝑅𝑅𝑛𝑛 = 0.75𝑅𝑅𝑛𝑛

Bearing strength at bolt holes

The nominal bearing strength is,

:

1- when deformation at the bolt hole at service load is a design consideration

𝑅𝑅𝑛𝑛= 1.2 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢 ≤ 2.4 𝑑𝑑𝑛𝑛𝐹𝐹𝑢𝑢

2-when deformation at the bolt hole at service load is not a design consideration

𝑅𝑅𝑛𝑛= 1.5 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢 ≤ 3.0 𝑑𝑑𝑛𝑛𝐹𝐹𝑢𝑢

3- for long-slotted holes with slot perpendicular to the direction of force

𝑅𝑅𝑛𝑛= 1.0 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢 ≤ 2.0 𝑑𝑑𝑛𝑛𝐹𝐹𝑢𝑢

Where,

d = nominal bolt diameter

𝐹𝐹𝑢𝑢 = minimum tensile strength of the connected material

t = thickness of connected material

𝐿𝐿𝑐𝑐 = clear distance, in the direction of the forces, between the edge of the hole and the edge of the adjacent hole or edge of the material.

The design strength, 𝜑𝜑𝑅𝑅𝑛𝑛 = 0.75𝑅𝑅𝑛𝑛

M. J. HAMOODI

6

Bolt strength in slip-critical connections

High-strength bolts in slip-critical connection are designed to prevent slip:

:

1- as serviceability limit state, when the holes are standard or the slots are transverse to the direction of the load. [𝜑𝜑 = 1].

2- at the required strength limit state, when the holes are oversized or the slots are parallel to the direction of the load. [𝜑𝜑 = 0.85]

The nominal slip resistance of a high-strength bolt is,

𝑅𝑅𝑛𝑛 = 𝜇𝜇 𝐷𝐷𝑢𝑢 ℎ𝑠𝑠𝑐𝑐 𝑇𝑇𝑏𝑏𝑁𝑁𝑠𝑠

Where,

𝜇𝜇 = 0.35 , mean slip coefficient for Class A surfaces.

𝜇𝜇 = 0.5, mean slip coefficient for Class B surfaces.

𝐷𝐷𝑢𝑢 = 1.13; a multiplier that reflects the ratio of the mean installed bolt pretension to the specified minimum bolt pretension.

ℎ𝑠𝑠𝑐𝑐 = hole factor determined as follows:

= 1.0 for S.H.

= 0.85 for O.S.H. and S.S.H.

= 0.7 for L.S.H.

𝑁𝑁𝑠𝑠 = number of slip planes (shear planes)

𝑇𝑇𝑏𝑏 = minimum fastener tension given in [Table j3.1].

M. J. HAMOODI

7

g

g

s s

W

a

b

c

de

t

Gross and Net areas

The gross area , 𝐴𝐴𝑔𝑔 , is the total cross-sectional area.

:

𝐴𝐴𝑔𝑔 = thickness ( t ) x width (w)

𝐴𝐴𝑛𝑛 = Net area = thickness ( t ) x net width (wn

Where,

)

net width = width - ∑ ( hole diameter + 116 in ) + ∑ 𝑠𝑠2

4𝑔𝑔

𝐴𝐴𝑛𝑛 ≤ 0.85 𝐴𝐴𝑔𝑔 for splice plates

𝐴𝐴𝑛𝑛 = 𝐴𝐴𝑒𝑒 [effective net area] for splice plates

For the plate shown,

𝐴𝐴𝑔𝑔 = w x t

[𝐴𝐴𝑛𝑛 ]abde

[𝐴𝐴𝑛𝑛 ]

= t[w - 2(hole diameter + 116 in)]

abcde

= t[w - 3(hole diameter + 116 in) + 2 (𝑠𝑠2

4𝑔𝑔)]

M. J. HAMOODI

8

1/2 x 6

3/8 gusset plate

Tensile strength

For yielding nominal strength is,

𝑃𝑃𝑛𝑛 = 𝐹𝐹𝑦𝑦 𝐴𝐴𝑔𝑔

:

𝑃𝑃𝑢𝑢 ≤ 𝜑𝜑𝑛𝑛 𝑃𝑃𝑛𝑛= 0.9𝐹𝐹𝑦𝑦 𝐴𝐴𝑔𝑔

For fracture nominal strength is,

𝑃𝑃𝑛𝑛 = 𝐹𝐹𝑢𝑢 𝐴𝐴𝑒𝑒

𝑃𝑃𝑢𝑢 ≤ 𝜑𝜑𝑛𝑛 𝑃𝑃𝑛𝑛=0.75𝐹𝐹𝑢𝑢 𝐴𝐴𝑒𝑒

The service load D=15kip and L=45kip is an axial tension load carried by a plate 1/2 x 6 in . Design the connection to the gusset plate. Given :

Example 1

3/8 in thickness of gusset plate.

3/4 in- diameter, A325 bolts with threads in the shear plane.

Bearing type connection.

A36 steel.

Solution

The factored load is:

:

Pu

=1.2(15) + 1.6(45)

= 1.2D + 1.6L

Pu

= 90 kip

Min. spacing in any direction,

S=3d

S=3(3/4)=2.25

M. J. HAMOODI

9

Use S=2.5 in

From Table J3.4, (Le) min

Use L

= 1 1/4 in

e

=1.5

Shear strength

𝐴𝐴𝑏𝑏 =𝜋𝜋(3 4)⁄ 2

4 = 0.441 in

:

𝑅𝑅𝑛𝑛 = 𝐹𝐹𝑛𝑛𝑛𝑛 𝐴𝐴𝑏𝑏 = 48 x 0.441 [Table J3.2]

2

= 21.168

kip

Bearing strength

h

:

diameter

The gusset plate is thinner than the plate and will control.

= d + 1/16 =3/4 + 1/16 = 13/16

Use t = 3/8 in

Lc = Le

– h/2 = 1.5 – 13/32 = 1.093

𝑅𝑅𝑛𝑛= 1.2 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢 ≤ 2.4 𝑑𝑑𝑛𝑛𝐹𝐹𝑢𝑢

𝑅𝑅𝑛𝑛= 1.2 x 1.093 x 3/8 x 58 = 28.527 kip

Check upper limit,

2.4 𝑑𝑑𝑛𝑛𝐹𝐹𝑢𝑢=2.4 (3/4)(3/8) x 58

= 39.15

28.527 < 39.15

.: use 𝑅𝑅𝑛𝑛= 28.527

For other bolts,

kip

Lc =S - h =2.5 - (13/16) =1.688 in

M. J. HAMOODI

10

1.5

1.52.5 2.5

1.5

1.5

1.5

𝑅𝑅𝑛𝑛= 1.2 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢

= 1.2(1.688) (3/8) x 58 = 44.05

44.05 > 39.15

.: use 𝑅𝑅𝑛𝑛=

39.15

Shear strength per bolt control ,21.168 kip

The design strength per bolt,

𝜑𝜑 𝑅𝑅𝑛𝑛 = 0.75 x 21.168 =15.876 kip

Number of bolts required = 𝑛𝑛𝑡𝑡𝑛𝑛𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑑𝑑𝑡𝑡𝑡𝑡𝑡𝑡𝑑𝑑 𝑝𝑝𝑒𝑒𝑝𝑝 𝑏𝑏𝑡𝑡𝑡𝑡𝑛𝑛

= 9015.876

= 5.66

Use 6 bolts

Tension on the gross area

𝑃𝑃𝑛𝑛 = 𝐹𝐹𝑦𝑦 𝐴𝐴𝑔𝑔= 36 (6 x 0.5) = 108 kip

:

𝑃𝑃𝑢𝑢≤ 𝜑𝜑𝑛𝑛𝑃𝑃𝑛𝑛 = 0.9 x 108

90 < 97.2 o.k.

M. J. HAMOODI

11

2

2

3

3

1.5

3 3

1.5

Tension on net area

𝐴𝐴𝑛𝑛 = 0.5[ 6 – 2 (13/16 + 1/16)] = 2.125 in

:

0.85 x 𝐴𝐴𝑔𝑔 = 0.85x3=2.55

2

2.125 < 2.55 o.k.

𝑃𝑃𝑛𝑛 = 𝐹𝐹𝑢𝑢 𝐴𝐴𝑒𝑒= 58 x 2.125 = 123.25 kip

𝑃𝑃𝑢𝑢≤ 𝜑𝜑𝑛𝑛𝑃𝑃𝑛𝑛

90 ≤ 0.75 x 123.25

90 ≤ 92.43 o.k.

Example 2

Determine the strength of the connection shown in the fig. ,If no slip is permitted . A500-Gr.C steel is used. A 11/16 in thick tension member is connected to two 3/8 in splice plates. The bolts is A490 with 7/8 in diameter.

:

For A500-Gr.C steel, Fy=46 ksi and Fu

From Table J3.4 , L

=62 ksi . Table 2-3

e

S=3d= 2.625 < 6 and

M. J. HAMOODI

12

Shear strength

𝐴𝐴𝑏𝑏 =𝜋𝜋(7 8)⁄ 2

4 = 0.6 in

:

𝑅𝑅𝑛𝑛 = 𝐹𝐹𝑛𝑛𝑛𝑛 2 𝐴𝐴𝑏𝑏 = 60 x 2 x 0.6 = 72 kip [Table J3.2]

2

For five bolts,

𝑅𝑅𝑛𝑛 =5 x 72 =360 kip

The design strength,

𝜑𝜑 𝑅𝑅𝑛𝑛 = 0.75 x360 =270

kip

Slip-critical strength

Because no slippage is permitted, this connection is classified as slip-critical.

:

From Table J3.1, Tb

Assumed Class A surface and S.H. are used,

= 49 kip

The nominal strength for one bolt is,

𝑅𝑅𝑛𝑛 = 𝜇𝜇 𝐷𝐷𝑢𝑢 ℎ𝑠𝑠𝑐𝑐 𝑇𝑇𝑏𝑏𝑁𝑁𝑠𝑠 = 0.35(1.13) 1x 49 x 2

= 38.759 kip

For 5 bolts,

𝑅𝑅𝑛𝑛= 5 x 38.759 =193.795

The design strength , 𝜑𝜑 𝑅𝑅𝑛𝑛= 1 x 193.795 = 193.795

kip

Bearing strength

The thickness of the two splice plates = 2 x 3/8 = 12/16 in

:

.: use t = 11/16 in [the smaller value]

h=d+1/16= 7/8 + 1/16 = 15/16 in

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13

12

12

for the holes near the edge,

Lc = Le

𝑅𝑅𝑛𝑛= 1.2 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢 = 1.2 (1.031) 11/16 x 62

– h/2 = 1.5 – (15/16)/2 =1.03125

= 52.73 kip

Upper limit = 2.4 𝑑𝑑𝑛𝑛𝐹𝐹𝑢𝑢=2.4 (7/8)(11/16) x 62 = 89.51 kip

52.73 < 89.51 .: use 𝑅𝑅𝑛𝑛=52.73 kip

For the other holes,

𝑅𝑅𝑛𝑛= 1.2 𝐿𝐿𝑐𝑐𝑛𝑛𝐹𝐹𝑢𝑢 = 1.2 (6 - 15/16) 11/16 x 62

=258.9 > 89.51 kip

.: use 𝑅𝑅𝑛𝑛= 89.51 kip

The nominal bearing strength for the connection is,

𝑅𝑅𝑛𝑛 = 2(52.73) + 3(89.51) = 373.99 kip

The design strength , 𝜑𝜑 𝑅𝑅𝑛𝑛= 0.75 x 373.99 = 280.49

kip

Tension on the gross area

𝑃𝑃𝑛𝑛 = 𝐹𝐹𝑦𝑦 𝐴𝐴𝑔𝑔= 46 (10 x 11/16) = 316.25 kip

:

The design strength, 𝜑𝜑𝑛𝑛𝑃𝑃𝑛𝑛 = 0.9 x 316.25 = 284.625

kip

Tension on the net area

𝐴𝐴𝑛𝑛 R 1

:

= 11/16 (10-2(15/16 + 1/16)) = 5.5 in

𝐴𝐴𝑛𝑛 R 2

2

= 11/16 (10- 3 x1+ 2( 32 / (4x2)) = 6.359 in2

Use 𝐴𝐴𝑛𝑛 = 5.5

0.85 𝐴𝐴𝑔𝑔 = 6.875

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5.5 < 6.875 o.k.

𝑃𝑃𝑛𝑛 = 𝐹𝐹𝑢𝑢 𝐴𝐴𝑒𝑒= 62 x 5.5 = 341 kip

The design strength, 𝜑𝜑𝑛𝑛𝑃𝑃𝑛𝑛= 0.75 x 341= 255.75

kip

The strength corresponding to the Slip-critical is the smallest,

.: The design strength of the connection is 193.795 kip

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15

Welded connection

Welding has several advantages over bolting. Connections that are extremely complex with fasteners can become very simple when welds are used.

:

Types of welding

There are three types of welds,

:

a) fillet weld , which can be used for lap joint and in a tee joint as in the figure.

b) Groove weld , this type is used for butt, tee and corner joints. One or both of the connected parts will have "prepared edges'.

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16

c) plug or slot weld , it is used when more weld is needed than length of edge is available. A circular or slotted hole is cut in one of the parts to be connected and is filled with the weld metal.

Shielded metal arc welding (SMAW)

SMAW is one of the most popular welding processes and it is shown schematically in the figure,

:

The current arcs across a gap between the electrode and base metal, heating the connected parts and depositing part of the electrode into the molten base metal.

The strength of the electrode is defined as its ultimate tensile strength, with strengths of 60, 70, 80, 90, 100, 110, and 120 ksi. A typical designation for the electrode would be for example, E70XX, where the

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17

70 ksi is the classification number ( FEXX

For the commonly used grades of steel, only two electrodes need be considered:

) of the electrode and it representing its ultimate tensile strength.

Use E70XX electrodes with steels that have a yield stress less than 60 ksi.

Use E80XX electrodes with steels that have a yield stress of 60 ksi or 65 ksi.

Fillet welds

The design and analysis of fillet welds is based on the assumption that the cross section of the weld is a 45° right triangle, as shown in Figure.

:

w = size of the fillet weld.

L = length of the weld.

A fillet weld is weak in shear and it is always assumed to fail in this mode.

The shear plane = 0.707𝑤𝑤𝑤𝑤

The nominal load capacity of the weld can be written as

𝑅𝑅𝑛𝑛 = 0.707𝑤𝑤𝑤𝑤𝐹𝐹𝑊𝑊

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Where, (𝐹𝐹𝑊𝑊) is the ultimate shearing stress of the weld.

𝐹𝐹𝑊𝑊 = 0.6 FEXX

𝐹𝐹𝑊𝑊 = 0.6 F

( when the load is parallel to the weld )

EXX

Where, (𝜃𝜃) is the angle between the direction of the load and the axis of

(1.0 + 0.5𝑠𝑠𝑠𝑠𝑛𝑛1.5𝜃𝜃)

the weld .

the following Table shows the shearing strength for several values of 𝜃𝜃,

When the load is perpendicular to the weld, the shearing strength is 50% higher.

For simple (concentrically loaded) welded connections with both longitudinal and transverse welds, AISC J2.4c specifies that the larger nominal strength from the following two options will be used:

1. Use the ultimate shearing stress of the weld, FW = 0.6FEXX

R

, for both the longitudinal and the transverse welds:

n = Rwl + Rwt

where

Rwl

R

= strength of the longitudinal weld

wt = strength of the transverse weld

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2. Use the 50% increase for the transverse weld, but reduce the shearing strength of the weld by 15% for the longitudinal welds.

Rn = 0.85Rwl + 1.5Rwt

Minimum size of weld

The minimum size ( 𝑤𝑤 )

:

min.

permitted is a function of the thickness of the thinner connected part and is given in AISC Table J2.4.

Maximum size of weld

Along a part which has a thickness ( t ), the maximum size ( 𝑤𝑤 )

:

max.

If ( t ) < 1/4 in ( 𝑤𝑤 )

may be obtained as follows,

max.

If ( t ) ≥ 1/4 in ( 𝑤𝑤 )

= t

max.

For fillet welds other than those along edges as in the figure, no maximum size specified; but the strength computation would be that limited by the base metal shear strength.

= t – 1/16 in

Minimum length of weld

The minimum permissible length of a fillet weld is four times its size,

:

(4𝑤𝑤) > 1 1/2 in

This limitation is applied to intermittent weld.

For flat bar tension-member connections using longitudinal welds only, as shown in the in the figure,

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The length of the welds in this case cannot be less than the distance between them, thus:

L ≥ W

Maximum length of weld

AISC does not impose a limit on the length of welds; but when the longitudinal end-loaded welds are exceeds 100 times the weld size, a reduced effective length is used in the computation of strength. Thus,

:

reduced effective length = L x β

where;

L = actual length of weld ( > 100 𝑤𝑤)

𝛽𝛽 = 1.2 − 0.002(𝑤𝑤 𝑤𝑤⁄ ) ≤ 1.0

𝑤𝑤 = weld size

If the length is larger than 300 times the weld size, use β = 0.60.

End returns

When a weld extends to the end of a member, it is sometimes continued around the corner, as shown in the Figure. The primary reason for this continuation, called an end return, is to ensure that the weld size is maintained over the full length of the weld.

:

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Shear strength on the base metal

weld shear strength cannot be larger than the base metal shear strength. This requirement can be explained by an examination of the welded connection shown in the Figure.

:

The shear strength of the weld AB cannot exceed the shear strength of the base metal ( gusset plate ) along line AB and corresponding to an area tL.

Weld shear strength: φRn = 0.75(0.707wLFW

Base metal shear strength: φRn = min[1.0(0.6F

)

y tL), 0.75(0.6Fu

tL)]

To work with the strength of one-inch length,

Weld shear strength ≤ Base metal shear strength

0.75(0.707wFW) ≤ min[1.0(0.6Fy t), 0.75(0.6Fu

t)]

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Weld symbols

Welds are specified on design drawings by standard symbols, which provide a convenient method for describing the required weld configuration.

:

Example 3

Design the fillet weld needed for lap joint shown in the figure. The factored axial tension load carried by the flat bar 5/8 x 7 in , is 150 kip. Use E70XX electrode, F

:

y

=42 ksi for A500-Gr.B steel and thickness of gusset plate is 3/8 in.

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23

( 𝑤𝑤 ) min.

( 𝑤𝑤 )

= 3/16 in Table j2.4

max.

Use weld size 𝑤𝑤 = 3/8

= 5/8 – 1/16 = 9/16 in

The nominal strength per inch of weld is,

𝑅𝑅𝑛𝑛 = 0.707𝑤𝑤𝐹𝐹𝑊𝑊= 0.707 x 3/8 (0.6 x 70)

= 11.135 kip/in

The design strength of weld per in, 𝜑𝜑𝑅𝑅𝑛𝑛= 0.75 x 11.135

= 8.35 kip/in

Check the base metal shear per inch,

φRn = φ (0.6Fy

φRn = φ (0.6F

t) = 1 x 0.6 x42 (3/8) = 9.45 kip/in

u

the base metal shear strength is therefore 9.45 kip/in

t) = 0.75 x 0.6 x58 (3/8) = 9.78 kip/in

8.35 < 9.45 o.k., weld shear strength controls.

First option :

Total required length = 150/8.35 = 17.96 in

length of longitudinal welds = (17.96 – 7) / 2 = 5.48 in

Second option :

the strength of longitudinal weld is,

0.85(8.35) = 7.09 kip/in

the strength of transverse weld is,

1.5(8.35) = 12.5 kip/in

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3/8

3/8

7

E70 XX

E70 XX

4.5

the load to be carried by longitudinal weld,

150- 7x 12.5 = 62.5 kip

.: the required length of the longitudinal welds is,

62.5/(7.09x2) = 4.4 in

.: use 7in transverse weld and two 4.5in longitudinal welds.

Example 4

A plate 1/2 x 6 in of A36 steel is used as a tension member to carry a service dead load of 20 kip and a service live load of 42 kip. It is to be attached to a (1/4 in ) gusset plate. Design a welded connection. Use E70XX electrode .

:

𝑃𝑃𝑢𝑢 = 1.2 D + 1.6 L

= 1.2x20 + 1.6x42 = 91.2 kip

( 𝑤𝑤 ) min.

( 𝑤𝑤 )

= 1/8 in

max.

Use weld size 𝑤𝑤 = 1/4 in

= 1/2 – 1/16 = 7/16 in

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25

3/16 11E70 XX

The nominal strength of weld is,

𝑅𝑅𝑛𝑛 = 0.707𝑤𝑤𝐹𝐹𝑊𝑊= 0.707 x 1/4 (0.6 x 70)

= 7.42 kip/in

𝜑𝜑𝑅𝑅𝑛𝑛= 0.75 x 7.42 = 5.56 kip/in The design strength of weld

Check for the base metal shear,

φRn = φ (0.6Fy

φRn = φ (0.6F

t) = 0.6 x36 (1/4) = 5.4 kip/in yield strength

u

the shear strength of the base metal = 5.4 kip/in

t) = 0.45 x58 (1/4) = 6.5 kip/in rupture strength

5.4 < 5.56 not o.k.

.: use 𝑤𝑤 = 3/16

5.4 > 4.17 o.k.

91.2 / 4.17 = 21.8 in required length of weld

.: use two 11 in long side welds for a total length, 2 x 11 = 22 in

Min. length = 4 𝑤𝑤 =4 x 3/16 = 0.75 in < 11 o.k.

the length of side welds ≥ the transverse distance between them

11 > 6 o.k.

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1

Tension members

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.

Circular rods and rolled angle shapes are frequently used. Built-up shapes, either from plates, rolled shapes, or a combination of plates and rolled shapes, are sometimes used when large loads must be resisted. The most common built-up configuration is probably the double-angle section, as shown in Figure.

Tensile strength

In load and resistance factor design [LRFD], the factored load is compared to the design strength. The design strength is the resistance factor times the nominal strength.

:

𝑅𝑅𝑢𝑢 = 𝜑𝜑𝑅𝑅𝑛𝑛

For tension member the above equation can be written as,

𝑃𝑃𝑢𝑢 ≤ 𝜑𝜑𝑡𝑡𝑃𝑃𝑛𝑛

𝑃𝑃𝑢𝑢 is the factored loads. The resistance factor 𝜑𝜑𝑡𝑡 is smaller for fracture than for yielding, reflecting the more serious nature of fracture.

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For yielding, 𝜑𝜑𝑡𝑡 = 0.90

For fracture, 𝜑𝜑𝑡𝑡 = 0.75

Because there are two limit states, both of the following conditions must be satisfied for tension member :

𝑃𝑃𝑢𝑢 ≤ 0.90𝐹𝐹𝑦𝑦𝐴𝐴𝑔𝑔

𝑃𝑃𝑢𝑢 ≤ 0.75𝐹𝐹𝑢𝑢𝐴𝐴𝑒𝑒

The smaller of these is the design strength of the tension member.

Where , 𝐴𝐴𝑔𝑔 is the total cross-sectional area and 𝐴𝐴𝑒𝑒 is the effective area.

Slenderness limitations

There is no maximum slenderness limit for design of members in tension. The slenderness ratio 𝑙𝑙 𝑟𝑟⁄ preferably should not exceed 300.

:

Where, 𝑙𝑙 is the length of the member and 𝑟𝑟 is the minimum radius of gyration of the tension member section.

Effective area

For bolted connections, the effective net area is,

:

𝐴𝐴𝑒𝑒 = 𝐴𝐴𝑛𝑛 𝑈𝑈

For welded connections, effective area is,

𝐴𝐴𝑒𝑒 = 𝐴𝐴𝑔𝑔 𝑈𝑈

Where 𝑈𝑈 is the reduction factor given in [ Table D3.1].

Case 1

𝑈𝑈=1 all tension members where the tension load is transmitted directly to each of cross-sectional elements by fasteners or weld.

,

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Case 2

𝑈𝑈 = 1 − �̅�𝑥 𝑙𝑙� all tension members where the tension load is transmitted

to

,

some

�̅�𝑥=distance from centroid of connected area to the plane of connection.

but not all of the cross-sectional elements by fasteners or longitudinal welds. Where,

𝑙𝑙= length of connection.

Case 3

𝑈𝑈=1 and 𝐴𝐴𝑛𝑛 = area of directly connected elements,

,

all tension member where the tension load is transmitted by transverse welds to some

but not all of the cross-sectional elements.

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Case 4

Plates where the tension load is transmitted by longitudinal welds only,

,

𝑈𝑈=1 𝑙𝑙 ≥ 2𝑤𝑤

𝑈𝑈=0.87 2𝑤𝑤 > 𝑙𝑙 ≥ 1.5𝑤𝑤

𝑈𝑈 =0.75 1.5𝑤𝑤 > 𝑙𝑙 ≥ 𝑤𝑤

𝑤𝑤 > 8 in

Case 5

Round HSS with a single concentric gusset plate,

,

Case 6

Rectangular HSS,

,

a) with a single concentric gusset plate.

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5

b) with two side gusset plates.

Case 7

W, M, S or HP shapes or Tees cut from these shapes,

,

a) with flange connected with 3 or more fasteners per line in direction of loading.

𝑈𝑈=0.9 𝑏𝑏𝑓𝑓 ≥ 23𝑑𝑑

𝑈𝑈= 0.85 𝑏𝑏𝑓𝑓 < 23𝑑𝑑

b) with web connected with 4 or more fasteners in direction of loading.

𝑈𝑈= 0.7

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Case 8

Single angles,

,

a) with 4 or more fasteners in direction of loading.

𝑈𝑈= 0.8

b) with 2 or 3 fasteners in direction of loading.

𝑈𝑈= 0.6

Block shear

For certain connections, a segment or block of material at the end of the member can tear out. The connection of the single-angle tension member shown in the figure is susceptible to block shear.

:

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The shaded block would tend to fail by shear along the longitudinal section ab and by tension on the transverse section bc. AISC Specification assumes that failure occurs by rupture on the shear area and rupture on the tension area.

The resistance to block shear will be the sum of the strengths of the two surfaces, along shear surface and along tension surface :

𝑅𝑅𝑛𝑛 = 0.6𝐹𝐹𝑢𝑢𝐴𝐴𝑛𝑛𝑛𝑛 + 𝑈𝑈𝑏𝑏𝑏𝑏𝐹𝐹𝑢𝑢𝐴𝐴𝑛𝑛𝑡𝑡 ≤ 0.6𝐹𝐹𝑦𝑦𝐴𝐴𝑔𝑔𝑛𝑛 + 𝑈𝑈𝑏𝑏𝑏𝑏𝐹𝐹𝑢𝑢𝐴𝐴𝑛𝑛𝑡𝑡

Where,

𝐴𝐴𝑛𝑛𝑛𝑛 net area along shear surfaces

𝐴𝐴𝑛𝑛𝑡𝑡 net area along tension surface

𝐴𝐴𝑔𝑔𝑛𝑛 gross area along shear surfaces

0.6𝐹𝐹𝑢𝑢 shear rupture stress

0.6𝐹𝐹𝑦𝑦 shear yield stress

𝑈𝑈𝑏𝑏𝑏𝑏 1.0 when the tension stress is uniform.

0.5 when the tension stress is non-uniform .

Build-up members

The longitudinal spacing of connectors between a plate and a shape or two plates are,

:

a) for members not subject to corrosion ,

the spacing ≤ 24t ( t =thickness of thinner plate )

≤ 12 in

b) for members subject to corrosion,

the spacing ≤ 14t ( t =thickness of thinner plate )

≤ 7 in

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The longitudinal spacing of connectors between components should limit the slenderness ratio in any component between the connectors to 300.

Example 5

Design a tension member carried a service dead load of 42 kip and service live load of 98 kip. If ,

:

*W10 x w

*F

section is available.

y

*A490 bolts with 5/8 in diameter, bearing- type. Threads are not excluded from shear plane.

= 36 ksi and L = 23 ft.

*Gusset plate thickness is 1/2 in, the connection is through the flanges only.

𝑃𝑃𝑢𝑢= 1.2D + 1.6L

= 1.2x42 + 1.6x98 = 207.2 kip

Required 𝐴𝐴𝑔𝑔 = 𝑃𝑃𝑢𝑢 / (0.90𝐹𝐹𝑦𝑦 ) = 207.2/(0.9x36) = 6.39 in

Required 𝐴𝐴𝑒𝑒= 𝑃𝑃𝑢𝑢 / (0.75𝐹𝐹𝑢𝑢 ) = 207.2/(0.75x58) = 4.76 in

2

r

2

(min)

≥ L/300 = (23x12)/300 = 0.92

Try W

𝑏𝑏𝑓𝑓=5.75 𝐴𝐴𝑔𝑔 =6.49 𝑟𝑟𝑥𝑥= 4.27

10 x 22

𝑑𝑑= 10.2 𝑡𝑡𝑓𝑓 = 0.36 𝑟𝑟𝑦𝑦=1.33

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6.49 > 6.39 o.k.

1.33 > 0.92 o.k.

𝑏𝑏𝑓𝑓< 2/3 𝑑𝑑

5.75 < 2/3 x 10.2

5.75 < 6.8 , assume 3 bolts or more per line in direction of loading is

used.

.: use U=0.85

𝐴𝐴𝑛𝑛 = 𝐴𝐴𝑔𝑔 – 4 x 𝑡𝑡𝑓𝑓 (bolt diameter + 1/8) = 5.41 in

𝐴𝐴𝑒𝑒 = U 𝐴𝐴𝑛𝑛 = 0.85 x 5.41= 4.598 < 4.76 N.G.

2

Try W

𝐴𝐴𝑔𝑔 = 7.61 𝑑𝑑= 10.3 𝑏𝑏𝑓𝑓=5.77 𝑡𝑡𝑓𝑓 = 0.44

10 x 26

𝐴𝐴𝑛𝑛 = 7.61- 4 x 0.44(3/4) = 6.29

𝑏𝑏𝑓𝑓< 2/3 𝑑𝑑

5.77 < 6.8 .: U=0.85

𝐴𝐴𝑒𝑒 = U 𝐴𝐴𝑛𝑛 = 0.85 x 6.29 = 5.34 > 4.76 o.k.

Bolts

(L

,

e ) min.

Use L

=1 1/8 Table

e

S=3d, S=1.875

=1 1/4 in

Use S=2 in

Shear strength,

𝑅𝑅𝑛𝑛= 𝜋𝜋(5 8)⁄ 2

4 x 60 = 18.4 kip , 𝜑𝜑 𝑅𝑅𝑛𝑛 = 0.75 x 18.4 =13.8 kip

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Bearing at bolt hole,

use t = 0.44

𝑅𝑅𝑛𝑛= 1.2 𝐿𝐿𝑐𝑐𝑡𝑡𝐹𝐹𝑢𝑢 ≤ 2.4 𝑑𝑑𝑡𝑡𝐹𝐹𝑢𝑢

𝑅𝑅𝑛𝑛=1.2[1.25 - 1/2(5/8+1/16)] 0.44x58 = 27.75 kip

2.4 𝑑𝑑𝑡𝑡𝐹𝐹𝑢𝑢=2.4(5/8)0.44x58 = 38.28 kip

27.75 < 38.28 o.k.

𝑅𝑅𝑛𝑛=1.2[2 - (11/16)] 0.44x58 =40.194

40.194>38.28 , 𝑅𝑅𝑛𝑛=38.28

.: 𝜑𝜑 𝑅𝑅𝑛𝑛 = 0.75 x 27.75 =20.8 kip

Use the lower of shear and bearing strength,

Number of bolts =207.2/13.8 =15

Use 16 bolts , 4 bolts in line in the direction of load.

Check for U value, 𝑈𝑈 = 1 − 𝑥𝑥̅𝑙𝑙

�̅�𝑥 = 1.06 from WT

𝑙𝑙 = 3x2 =6 in

5 x 13

𝑈𝑈 = 1 − (1.06/6) = 0.82

0.82 < 0.85 o.k.

.: use W

10 x 26

M. J. HAMOODI

11

3/4

x

y

x

7 @ 2

2 1/4

2 1/2

1 1/4

1 1/2

Check shear block for the gusset plate,

𝑅𝑅𝑛𝑛 = 0.6𝐹𝐹𝑢𝑢𝐴𝐴𝑛𝑛𝑛𝑛 + 𝑈𝑈𝑏𝑏𝑏𝑏𝐹𝐹𝑢𝑢𝐴𝐴𝑛𝑛𝑡𝑡 ≤ 0.6𝐹𝐹𝑦𝑦𝐴𝐴𝑔𝑔𝑛𝑛 + 𝑈𝑈𝑏𝑏𝑏𝑏𝐹𝐹𝑢𝑢𝐴𝐴𝑛𝑛𝑡𝑡

𝐴𝐴𝑔𝑔𝑛𝑛= 2(7.25)0.5 = 7.25

𝐴𝐴𝑛𝑛𝑛𝑛=2(1.25+6-3.5(5/8+1/8)) 0.5= 4.625

𝐴𝐴𝑛𝑛𝑡𝑡=5.77 – 2x1.125 – (5/8+1/8) = 2.77

𝑅𝑅𝑛𝑛= 0.6x58x4.625 + 1x58x2.77= 321.61

0.6x36x7.25+1x58x2.77=317.26

321.61>317.26 not o.k. :. Use 𝑅𝑅𝑛𝑛=317.26

𝜑𝜑 𝑅𝑅𝑛𝑛 = 0.75 x 317.26 =237.9 kip The design block shear strength

237.9 > 103.6 o.k.

Example 6

Determine the load capacity in tension for the 2L6x4x1/2 LLBB separated by 3/4 in, shown in the Fig. Use A529-Gr.50 steel , A325- 7/8 in diameter bolts, bearing type connection, threads are excluded from shear plane. The length of the tension member is 20ft.

:

𝐴𝐴𝑔𝑔 = 9.5 𝑟𝑟𝑥𝑥=1.91 𝑟𝑟𝑦𝑦=1.77

M. J. HAMOODI

12

From Table, 𝐹𝐹𝑦𝑦=50 ksi 𝐹𝐹𝑢𝑢=65 ksi

𝑙𝑙 𝑟𝑟𝑚𝑚𝑚𝑚𝑛𝑛⁄ ≤ 300

(20x12)/1.77 = 135.59 < 300 o.k.

(Le ) min.

= 1 1/8 rolled edge < 1 1/4 o.k.

=1 1/2 sheared edge =1 1/2 o.k.

S= 3d =3(7/8) =2.625 < 4 o.k.

Tensile strength,

𝜑𝜑𝑃𝑃𝑛𝑛=0.90𝐹𝐹𝑦𝑦𝐴𝐴𝑔𝑔 = 0.9 x 50 x 9.5= 427.5

𝜑𝜑𝑃𝑃𝑛𝑛=0.75𝐹𝐹𝑢𝑢𝐴𝐴𝑒𝑒

kip

𝐴𝐴𝑛𝑛 = 𝐴𝐴𝑔𝑔 − 𝑡𝑡�ℎ𝑜𝑜𝑙𝑙𝑒𝑒 𝑑𝑑𝑚𝑚𝑑𝑑𝑚𝑚𝑡𝑡𝑒𝑒𝑟𝑟 +1

16+ 𝑡𝑡�

𝑏𝑏2

4𝑔𝑔

𝐴𝐴𝑛𝑛1= 9.5 -1/2( 2 (7/8+1/8)) =8.5

𝐴𝐴𝑛𝑛2= 9.5 – 1/2(4 x 1) + 1/2 x2( (2)2

4(2.5) ) =7.9

𝐴𝐴𝑒𝑒 =U 𝐴𝐴𝑛𝑛2

𝑈𝑈 = 1 − 𝑥𝑥̅𝑙𝑙 = 1- 0.981/14 = 0.929

𝐴𝐴𝑒𝑒 =0.929 x 7.9 = 7.33 in

𝜑𝜑𝑃𝑃𝑛𝑛=0.75 x 65 x 7.33 =

2

357.33

kip

Bolts strength,

𝑅𝑅𝑛𝑛=2 x 𝜋𝜋(7 8)⁄ 2

4 x 60 = 72.158 kip

Design shear strength = 𝜑𝜑 8 𝑅𝑅𝑛𝑛 = 0.75 x 8 x 72.158 = 432.9 kip

M. J. HAMOODI

13

t = 3/4 and t = 2x1/2 = 1

.: use t = 3/4

Hole diameter = 7/8 + 1/16 =15/16

𝑅𝑅𝑛𝑛= 1.2 (1.5 – 15/32) 3/4 x65 ≤ 2.4 x 7/8 x 3/4 x 65

= 60.328 ≤ 102.375

𝑅𝑅𝑛𝑛= 1.2 (3.5 – 15/32) 3/4 x65 = 177.3 > 102.375

𝑅𝑅𝑛𝑛= 1.2 (4 – 15/16) 3/4 x65 = 179.15 > 102.375

Design bearing strength for 8 bolts = 0.75( 60.328 + 7 x 102.375)

= 582.7

.: the tensile rupture limit state is governed, 357.33 kip

kip

Let the longitudinal spacing of connectors between the two angles is 2ft,

.: 24𝑟𝑟𝑧𝑧(𝑓𝑓𝑜𝑜𝑟𝑟 𝑏𝑏𝑚𝑚𝑛𝑛𝑔𝑔𝑙𝑙𝑒𝑒 𝑑𝑑𝑛𝑛𝑔𝑔𝑙𝑙𝑒𝑒 )

= 24/0.864 = 27.7 < 300 o.k.

Notes

1- when lines of bolts are present in more than one element of the cross section of a rolled shape,

:

then the distance g to be used in the [s2

g=3+2 – 1/2

/4g] term would be,

M. J. HAMOODI

14

2- since 1/10 of the load is transferred by each bolt, the line abceg resists only 9/10 of the load.

If 𝐴𝐴𝑔𝑔=6.75 , hole diameter + 1/16=1

𝐴𝐴𝑛𝑛 R(abceg)=[6.75 – 3x1+(1.5)2

=[5.363] 10/9

/ (4x2.5)] 10/9

=5.959 in2

Example 7

Design a member which is carrying a dead load of 35 kip and a live load of 105 kip in tension. The length of the member is 30 ft. Assume the end is welded to a 1/2 in thick single concentric gusset plate. Use A500 grade B steel, E70XX electrode and a rectangular HSS.

:

For A500-Gr B, 𝐹𝐹𝑦𝑦=46 ksi 𝐹𝐹𝑢𝑢=58 ksi

𝑃𝑃𝑢𝑢 = 1.2(35 )+ 1.6(105) = 210 kip

Required 𝐴𝐴𝑔𝑔 = 𝑃𝑃𝑢𝑢 / (0.90𝐹𝐹𝑦𝑦 ) = 210/(0.9x46) = 5.07 in

Required 𝐴𝐴𝑒𝑒= 𝑃𝑃𝑢𝑢 / (0.75𝐹𝐹𝑢𝑢 ) = 210/(0.75x58) = 4.827 in

2

2

M. J. HAMOODI

15

r(min)

≥ L/300 = (30x12)/300 = 1.2

try HSS6x4x3/8

𝐴𝐴𝑔𝑔 =6.18 t=0.349 𝑟𝑟𝑥𝑥=2.14 𝑟𝑟𝑦𝑦=1.55

6.18>5.07 o.k.

1.55>1.2 o.k.

𝐴𝐴𝑛𝑛 = 𝐴𝐴𝑔𝑔 − 2 �𝑡𝑡𝑝𝑝 +1

16� 𝑡𝑡

= 6.18 – 2(1/2+1/16)0.349 = 5.787

𝐴𝐴𝑒𝑒 =𝑈𝑈𝐴𝐴𝑛𝑛

𝑈𝑈 = 1 −�̅�𝑥𝑙𝑙

�̅�𝑥 = 1.6 in [Table D3.1, B=4 and H=6]

Assume 𝑙𝑙 = 16 in [ 𝑙𝑙 ≥ H ]

𝑈𝑈 = 1 − 1.616

= 0.9

𝐴𝐴𝑒𝑒 = 0.9 x 5.787 = 5.2 > 4.827 o.k.

To find the size of weld,

1(0.6𝐹𝐹𝑦𝑦𝑡𝑡) = 0.6 x 46 x 0.349 =9.63 kip/in

0.75(0.6𝐹𝐹𝑢𝑢𝑡𝑡) =0.75 x0.6 x58 x0.349 =9.1 kip/in

Use 9.1 kip/in base metal shear strength

.: 0.75(0.707 𝑤𝑤 𝐹𝐹𝑊𝑊) ≤ 9.1

M. J. HAMOODI

16

0.75(0.707x 𝑤𝑤 (0.6x70) ) ≤ 9.1

𝑤𝑤 ≤ 0.4 in use 𝑤𝑤 = 5/16

.: the force transmitted by the weld is,

4 x 16 [0.75x0.707(5/16) 42] = 445.41 kip > 210 kip o.k.