Chapter 10: Structural Connectionscaed20301.webs.com/CN Chap10 - Structural Connections.pdf ·...

10.1

Chapter 10: Structural Connections

Introduction

Chapter 3 – Analysis of Selected Determinate Structural Systems.

• Free-body diagrams and equations of equilibrium were introduced.

• Forces and support reactions were determined for each structural element.

Chapter 4 - Load Tracing.

• Load paths were traced from the roof to the foundation.

Chapter 7 Bending and Shear in Simple Beams.

• Shear, and moment diagrams were constructed.

• The critical shears and moments used in the design of bending members were

determined.

Chapter 9 – Column Analysis and Design.

• Columns were analyzed to determine the adequacy or critical load capacity.

In previous chapters individual structural members were designed as isolated

elements, assuming that the transfer of loads from one member to the next was

assured.

• All structural elements are interconnected and work in combination with other

members through physical connections.

• Structural members must be joined to permit the safe and effective transfer

of loads from one member to another.

Many structural failures in buildings and bridges occur at the connection of the

members rather than in the members themselves.

• Such failures can occur due to a number of reasons.

- Improperly installed bolts.

- Undersized bolts.

- An improper strength rating of the bolts.

- The bolt holes are too large.

- A bolt is tightened improperly.

- The welds are made with an improper penetration of the weld material.

10.2

To prevent bad joints, or at least limit the number of bad joints, various standard

codes are in effect.

• Timber design - American Institute of Timber Construction (AITC) and the

National Design Specification (NDS) by the American Forest and Paper

Association.

• Steel design - American Institute of Steel Construction (AISC) Manual of Steel

Construction, and the specifications of the America Welding Society (AWS).

10.1 Steel Bolted Connections

Bolts used in building connections are generally subjected to forces that cause

tension, shear, or a combination of the two.

• Typical shear connections - Splices used in trusses, beam-lap splices, and gusset

plates.

• Eccentric shear connections - Brackets and web shear splices.

• Tension connections – Hangar connections.

• Combination tension/shear connections – Beam-to-column moment connections.

Illustrations for typical connections are illustrated in Figures 10.3 through 10.6

(pp. 495 - 496 of the textbook).

Failure of Bolted Connections

Five basic types of failure of bolted connections include:

• Failure of the bolt

- Shear (single shear or double shear) of the bolt.

• Failure of the connected member

- Bearing failure (crushing) of the connected members against the bolt.

- Tension failure of the connected member material.

- End tear-out of the connected member.

- Block shear, a condition that can occur when the top flange of the beam is

coped to intersect with a girder.

Illustrations of each failure mode are depicted in Figures 10.7 through 10.12

(pp. 497 - 500 of the textbook).

Shear failure

• Shear failure is the result of failure of the bolt.

• The shear stresses are not uniform on the bolt cross section.

• The shear stresses are assumed as uniform for design purposes.

10.3

• The calculated resisting shear stress is an average shear stress.

Single shear: (fv)average = P/(n x A) ≤ Fv

where

(fv)average = average shear stress (psi, ksi)

P = load on plate or connection (lbs, kips)

A = cross-sectional area of bolt (in2)

Fv = allowable shear stress of bolt (psi, ksi)

n = number of bolts

Double shear: (fv)average = P/(2A x n) ≤ Fv

where

(fv)average = average shear stress (psi, ksi)

P = load on plate or connection (lbs, kips)

2A = two cross-sectional areas per bolt (in2)

Fv = allowable shear stress of bolt (psi, ksi)

n = number of bolts

Bearing failure

Bearing failure is the result of failure of the connecting members, not the bolt

itself.

The contact surface of the bolt pushing on the plate is taken as follows.

Ap = n x t x d

where

Ap = bearing area of plate (in2)

t = plate thickness (in.)

d = bolt diameter (in.)

n = number of bolts in bearing

The average bearing stress developed between the bolt and plate can be expressed

as follows.

(fp)average = P/Ap ≤ Fp

10.4

where

(fp)average = average bearing stress (psi, ksi)

Ap = bearing area (in2)

P = applied load (lbs, kips)

Fp = allowable bearing stress (psi, ksi)

Fp = 1.2Fu

Fu = ultimate stress (psi, ksi

= 58.0 ksi (for A36 steel) and 65.0 ksi (for A572 Grade 50 steel) (ref.

Table 5.1, p. 273 of the textbook)

Tensile failure

• Tensile failure is the result of failure of the material left to resist the applied

load after deducting the void left by the bolt holes:

Anet = (b – n x D) t

where

Anet = net plate area resisting tension (in2)

b = plate width (in.)

D = diameter of a standard size bolt hole

(D = bolt diameter + 1/16”)

t = plate thickness (in.)

n = number of bolts in a row

The average tension stress developed across the net area of the plate can be

expressed as follows.

(ft)net = P/Anet ≤ Ft

where

(ft)net = average tension stress across the net area of the member (psi, ksi)

P = applied load (lbs, kips)

Ft = allowable tension stress across the net area (psi, ksi)

Ft = 0.5Fu

Fu = ultimate tensile stress (psi, ksi)

10.5

End tear-out and block shear are possible failure modes:

• When high, allowable fastener values are used in conjunction with relatively thin

material (i.e. strong bolts and thin member).

• However, failure is less likely to occur because of the design specifications

requiring ample edge and end distances and minimum bolt spacing.

Figure 10.13 (p. 500 of the textbook) illustrates, for preliminary design purposes,

the minimum pitch (measured perpendicular to the axis of the member), gauge

(measured parallel to the axis of the member), and edge distances specified by

AISC to avoid these failure modes.

• Pitch and gauge (center-to-center) spacing

- Absolute minimum bolt spacing: 2-2/3 times the bolt diameter

- Preferred bolt spacing: 3 times the bolt diameter

• Edge distance: 1.5 times the bolt diameter

• In some cases, a dimension of 3 inches is used for all sizes of bolts up to 1-inch

diameter.

Shear and bearing failures, the result of excessive stresses on the bolts, are

avoided by providing a sufficient number of fasteners to keep the stresses within

the allowable limits.

Design Stresses for Bolts

Bolting has replaced riveting as a means of structurally connecting members.

• Bolting is quieter and simpler.

• Bolting can be performed more quickly with smaller crews, resulting in lower

labor costs.

The most common construction method is to use shop welding in combination with

field bolting.

There are three primary types of bolts used in steel construction today.

1. Common bolts (also known as rough or unfinished bolts)

• Designated as ASTM A307 (American Society of Testing Materials)

• Common bolts are suitable for use in light steel frame structures where

vibration and impact do not have to be considered.

- Where these conditions exist, nuts may become loose, impairing the

strength of the connection.

10.6

• Unfinished bolts are not permitted under some building codes in the

construction of buildings exceeding certain height limits.

2. High strength bolts: ASTM A325

3. High strength bolts: ASTM A490

High strength bolts are the most widely used fasteners for structural steel

connections made in the field.

Bolts are available in a variety of sizes from 5/8” to 1-1/2” diameter.

• The most typically used sizes in building construction are 3/4” and 7/8”.

• Larger diameter bolts generally require special equipment and increased spacing

and edge distances for proper bolt placement.

Mechanically fastened connections are categorized as follows.

• Slip-critical (SC).

- Slip-critical connections, also formerly referred to as friction-type connections, depend upon sufficiently high clamping force to prevent slip of

the connected parts under anticipated service conditions.

• Bearing type (N or X).

- Bearing-type connections are based on the contact (bearing) between the

bolt(s) and the sides of the holes to transfer the load from one connected

element to another.

Tightening of high-strength bolts in slip-critical connections is accomplished by

special torque or impact wrenches.

• The bolts are tightened to a tension equal to 70 percent of the specified

minimum tensile strength of the bolt.

• The Research Council on Structural Connections (RCSC) provides four methods

of controlling bolt tension.

1. The calibrated wrench.

2. The twist-off-type tension-control bolt.

3. The direct tension indicator.

4. The turn of the nut method (which requires specified additional nut rotation

after the bolts are “snug tight”).

“Snug tight” is defined as the full effort of an ironworker with an ordinary

spud wrench that brings the connected members into firm contact.

10.7

The allowable shear stresses for A325 and A490 (high strength) bolts used in

standard round holes are as follows.

A325-SC Fv = 17 ksi (slip-critical connection, standard round hole).

A325-N Fv = 21 ksi (bearing-type connection with threads included in the shear

plane).

A325-X Fv = 30 ksi (bearing-type connection with threads excluded from the

shear plane).

A490-SC Fv = 21 ksi (slip-critical connection, standard round hole).

A490-N Fv = 28 ksi (bearing-type connection with threads included in the

shear plane).

A490-X Fv = 40 ksi (bearing-type connection with threads excluded from the

shear plane).

Table 10.1 (p. 504 of the textbook) lists the allowable AISC shear capacities of

bolts.

Table 10.2 (p. 505 of the textbook) lists the allowable AISC bearing capacities of

bolts.

Several types of holes are typically used for bolted connections.

• Standard round holes

- Standard round holes are 1/16” larger than the diameter of the bolt.

• Oversized holes

- Bolts with diameters of 7/8” and less - Oversized holes have nominal

diameters up to 3/16” larger than bolt diameter.

- Bolts with a diameter of 1” – Oversize holes have a diameter of 1-1/4”.

- Oversize holes are permitted in slip-critical connections only.

• Short-slotted holes

- Short-slotted holes are 1/16” wider than the bolt diameter.

- Length of short-slotted holes are specified as follows.

◦ Bolts with diameters of 7/8” and less – The length of the hole (or slot)

can be no more than 1/4" longer than the bolt diameter.

◦ Bolts with a diameter of 1” – The length of the hole (or slot) can be no

more than 1-5/16”.

10.8

- Short-slotted holes can be used in either bearing- or slip-critical-type

connections.

- If used in bearing, the slots have to be perpendicular to the direction of the

load.

• Long-slotted holes

- Long-slotted holes are 1/16” wider than the bolt diameter.

- The length of long-slotted holes cannot exceed 2-1/2 times the bolt

diameter.

- This type of connection may be used in slip-critical connections without

regard for the load direction.

- In bearing-type connections, the load must be perpendicular to the slot

direction.

- Long-slotted holes are particularly useful where some amount of field

adjustment is necessary.

- Long-slotted holes can only be used in one of the connected members at a

joint.

◦ The other member must use a standard round hole.

Oversize and slotted holes are typically used in the field to reduce the fit-up and

assembly time during erection.

Standard Framed Beam Connections

Standard AISC tables are available to cover the design of a large number of

typical structural connections where beams frame into girders or girders frame

into columns.

• This type of standard shear connection

consists of two clip angles placed back to

back on either side of the beam or girder

web.

Framed beam connections are generally

designed for shear, bearing and web tear-

out or block shear.

A sample AISC table is shown as Table 10.3 (pp. 515 - 516) of the textbook.

• The AISC table is used in designing slip-critical and bearing-type connections

based on the shear capacity of bolts using standard-size holes.

10.9

• The AISC table provides bolt type, bolt size, hole type, number of bolts, clip

angle thickness, and clip angle length.

• High-strength bolts, in either slip-critical or bearing-type connections, provide

- Double shear condition through the beam web

- Single shear attachment to the column flange or girder web.

• Clip angle thickness and length are dependent on the fastener size, the

magnitude of the applied load, and space limitation within the beam’s flanges.

- Clip angles must be able to fit between the clearance of top and bottom

flange fillets.

- Clip angle lengths are generally at least half of the beam’s depth to provide

some resistance to end rotation at the end of the beam.

10.10

Example Problems - Steel Bolted Connections

Problem 10.4 (p. 512)

Given: Truss joint with members

loaded as shown.

Bolts: 3/4”dia A325-X (NSL)

(Bearing threads excluded; slotted holes

normal to the load.)

Members: A36 steel.

Find: Number of bolts required for

each connection based on shear

and bearing only.

(Tension and tear-out not critical.)

Solution

Member a

Load carried by the member: P = 105 k – 63 k = 42 k

Determine the number of bolts required for double shear (2 angles).

BVs = 26.5 kips/bolt (ref. Table 10.1, p. 504)

n = P/BVs = 42/26.5 = 1.58 2 bolts required

Determine the number of bolts required for bearing.

BVp = 19.6 kips/bolt for thickness = 3/8” (ref. Table 10.2, p. 505)

Fu = 58 ksi (A36 steel, Table 5.1, p. 273)

n = P/BVp = 42/19.6 = 2.14 3 bolts required Governs

Use 3 bolts

Member b

Load carried by the member: P = 26 k

Determine the number of bolts required for single shear (1 angle).



10.11





Use 3 bolts

Member c









Use 2 bolts

Member d









Use 4 bolts

10.12

Problem 10.5

Given: Butt splice shown.

Bolts: 4 – ¾” dia. A325-SC

Holes: Standard round

Plates: A36 steel

Find: Load capacity (P) of the connection

based on shear, bearing, and net tension.

Solution

Determine the capacity based on double shear: Pv = n BVs


Pv = n BVs = 4 (15.0) = 60 k Governs

Determine the capacity based on bearing: Pp = n Fp



Pp = n BVp = 4 (26.1) = 104.4 k

Determine the capacity based on net tension: Pt = Ft Anet

Anet = [8 – 2 (3/4 + 1/16)] (1/2) = 3.1875 in2

Ft = 0.5 Fu = 0.5 (58) = 29 ksi

Pt = Ft Anet = 29 (3.1875) = 92.4 k

Determine the capacity based on tensile yield strength: Pa = Fa Agross

Agross = 8.0 (0.5) = 4.0 in2

Fa = 0.6 Fy = 0.6 (36) = 21.6 ksi

Pa = Fa Agross = 21.6 (4.0) = 86.4 kips

10.13

10.2 Welded Connections

Welding is a method of joining metals by fusion (i.e. the melting of metals) without

the application of pressure.

• The metal at the joint and the additional metal supplied in the form of filler

metal is melted forming a small pool or crater.

• Upon cooling, the weld and base metal form a continuous and almost

homogeneous joint.

Many welding processes are recognized by the American Welding Society (AWS).

• Arc welding is the method generally used in welding structural steel used in

building construction.

• Arc welding was first made possible by the discovery of the electric arc by Sir

Humphrey Davy in the 19th century.

In recent years, great progress has been made in automatic and semiautomatic

welding processes.

• Manual welding today is generally limited to short welds and field welding

(i.e. welding done at the project site).

• Shop welding is a common practice.

- Connecting devices (e.g. clip angles and bearing plates) are welded to one

member, then delivered to the project site, put into place and fastened to a

connecting member in the field using bolts.

There are cases (e.g. in moment-resisting connections) where fully welded joints

are a good option.

• Welded members are connected directly without the use of connecting plates

or angles.

• The welded connection is usually simpler, more compact, and requires a smaller

crew.

• Holes for bolts are eliminated.

- The full cross-sectional area (rather than the net cross-section) is available

to resist the tensile force in the member.

Occasionally, problems may arise in the application of welding to structural

connections.

• Factors that may contribute to improper welds include the following.

- The selection of the wrong electrode.

- Using an improper amperage/voltage setting on the welding machine.

10.14

- The cooling rate of the weld is too rapid.

◦ The result may be the development of internal stresses from differential

cooling.

Seriously flawed work has been substantially eliminated by

• Qualified workers.

- Each welder is required to pass rigid qualification tests.

• Inspection of completed work.

- Each welder must submit his or her work to the careful scrutiny of a trained

inspector.

• Testing the safety of welded joints.

- Ultrasonic testing (UT) and magnetic particle inspection (MT) may be used

to test the integrity of the completed weld.

Types of Welded Joints

There are various types of welded joints in common use.

• Fillet welds and groove (butt) welds are the two most commonly used weld types

in building construction.

• Plug and slot welds are used for special circumstances.

The selection of the type of weld depends on the following factors:

• The magnitude of the load at the joint.

• The direction of the applied load.

- Examples: Torsion, compression, or shear.

• The configuration of the joint.

• The difficulty of the joint preparation.

• The cost of the installation.

Examples of the various welds are illustrated in Figures 10.35 through 10.38

(pp. 520 – 521 of the textbook).

Common symbols used for designating the type of weld are shown in Table 10.4

(p. 522 of the textbook).

• The table is reproduced from the American Institute of Steel Construction

(AISC).

10.15

Fillet welds and groove (butt) welds differ primarily from the manner in which the

stress transfer takes place.

• Groove welds are normally in direct tension or compression.

• Fillet welds are generally subjected to shear as well as tension or compression.

Groove (butt welds)

• Groove welds are generally used in structural assemblies in which full-

strength welds are required.

- The groove weld is actually stronger than the base material in shear,

tension, and compression.

- The strength of full-penetration groove welds is conservatively assumed

as equal to the base material.

• Groove welds are more expensive to produce than fillet welds.

- Groove welds require relatively large amounts of weld metal and can

sometimes experience problems during the welding process.

- Groove welds also require the cutting of structural members to more

exact lengths for the ends to be in the proper position.

- Groove welds require extensive edge preparation.

Fillet welds

• The fillet weld is one of the most commonly used welds.

• Most structural connections are joined by fillet welding.

- The fillet weld is used by steel fabricators to join plate materials

together in the production of built-up beams and girders.

- The fillet weld is used to join beams to columns or to join beams to

girders.

• Fillet welds allow for greater fit-up tolerances and generally require no edge

preparation before welding.

The most commonly used electrodes for welding A36 steel are E60XX and E70XX.

• The prefix E denotes electrode.

• The first two digits indicate the ultimate tensile strength in thousands of

pounds per square inch.

- For example: An E70XX electrode has an ultimate tensile capacity of 70 ksi.

10.16

• The next-to-last digit indicates the weld position in which the electrode is

capable of making satisfactory welds.

- For example: E701X All positions.

E702X Flat position and horizontal fillets.

Fillet weld strength is based on the allowable shear stress for the weld metal

across the effective throat area as well as the size and length of the weld.

• The effective throat area is illustrated in Figure 10.40 (page 552 of the

textbook).

• Table 10.5 (p 524 of the textbook) lists the allowable strength of fillet welds

per inch of weld for various size welds.

Welding code provisions are addressed in the AISC Manual of Steel Construction (Allowable Stress Design) and in the structural code of the American Welding

Society (AWS), and include the following.

• The maximum size of a fillet weld applied to a square edge of a plate or section

is determined as follows.

Plate thickness Size of the fillet weld

¼” or greater 1/16” less than the plate thickness

Less than ¼” Maximum weld size may equal plate thickness

• The minimum size of a fillet weld is dependent on the thicker of the two

members being welded but cannot exceed the thickness of the thinner member

and is determined as follows.

1/8” for material 1/4” or less

3/16” for material thickness over 1/4” and up to 1/2”

1/4” for material thickness over 1/2” and up to 3/4”

5/16” for material thickness over 3/4”

In the 13th edition of the AISC Manual, the minimum fillet weld size is based on the

thickness of the thinner plate.

• The minimum effective length of fillet welds should be four times the nominal

size.

- For example: The minimum effective length of a ¼” weld is 1”.

• If two or more welds are parallel to each other, the length for each weld must

be at least equal to the perpendicular distance between the welds.

10.17

• The minimum length of intermittent fillet welds should not be less than four

times the weld size (with a minimum length of 1-1/2”).

• Side or end fillet welds terminating at the ends or sides should be “returned”

(i.e. wrapped), if practical, around the corners for a distance not less than two

times the nominal weld size.

- Added strength is given to fillet welds with end returns.

Economic considerations may influence the size of the fillet weld that is used.

• Weld metal volume has a direct correlation to the labor costs involved in

depositing the weld.

• The most economical weld minimizes weld metal volume and reduces the heat

input and associated shrinkage and distortion of the joint.

• Minimizing the weld metal also minimizes the potential for weld defects.

• Fillet weld sizes should be 5/16” or less.

- The 5/16” weld is the largest weld that can be deposited in one pass.

- Larger fillet welds generally require two or more passes.

In practice, it is advisable to maintain the same size fillet weld throughout the

connection.

• A change of fillet size requires a change of welding rods.

• A change of welding rods slows the work and may cause errors.

10.18

Example Problems - Welded Connections

Problem 10.10 (p. 530)

Given: Welded plate shown.

Base metal is A36 steel.

Electrodes used are E70XX.

Find: Shear capacity of the fillet weld.

Solution

Determine the tensile yield capacity of the plate.

P = 22.0 ksi (5”)(5/16”) = 34.4 k

Determine the length and size of the fillet welds.

Minimum weld size: 3/16” (S = 2.78 k/in Table 10.5, p. 524)

Maximum weld size: 5/16 – 1/16” = 1/4" (S = 3.71 k/in Table 10.5, p. 524)

(Note: Only two weld sizes to consider.)

Try Lmin = 5” (distance between longitudinal welds)

Total minimum weld length = 15”

Required weld strength = P/L = 34.4/15 = 2.29 k/in

Therefore, use 3/16” weld with L = 5”. (Note: The use of a ¼” weld would be a waste.)

Determine the shear capacity of the 3/16” fillet weld.

Shear capacity = 15” (2.78 k/in) = 41.7 kips > 34.4 kips

10.19

Problem 10.12 (p. 530)

Given: Field welded connection shown.

Base metal is A36 steel.

Electrodes used are E70XX.

Find: a) Capacity of the 1/4" fillet weld

connection.

b) Capacity of a full-penetration

groove weld.

Solution

a) Determine the capacity of the 1/4” fillet weld connection.

For 1/4” fillet weld, S = 3.71 k/in (ref. Table 10.5, p. 524)

Total weld length = Perimeter = 4 x 4” = 16”

Weld capacity = S x L = 3.71 (16) = 59.4 k

b) Determine the capacity of a full-penetration groove weld.

• The strength of a full-penetration groove weld is equal to the tensile

capacity of the tube (ref. p. 521 of the textbook).

4” x 4” x ¼” square tube, A36 steel

Ft = 22.0 ksi (ref. Table 5.2, p. 278 of the textbook)

A = 3.59 in2 (ref. 9th edition of the AISC Manual, p. 1-96)

Tensile capacity of the structural tube and capacity of full-penetration weld:

Pt = A x Ft = 3.59 in2 (22.0 ksi) = 79.0 k

10.20

Eccentricity in Welded Joints

The most common example of

eccentrically loaded welded joints is

that of a structural angle welded to a

gusset plate.

• Example: Truss member.

The load P in the angle is assumed to

act along the centroidal axis of the

angle. Angle welded to gusset plate

Since an angle is an asymmetrical cross section, the welds marked L1 and L2 are

made unequal in length to eliminate torsion at the connection.

• The forces in the welds are proportional to the distributed area of the angle.

The procedure to determine the length of each weld follows.

• Write the equations of equilibrium for the applied force and weld resistances,

R1 and R2.

∑Fx = 0 = P - R1 - R2 P = R1 + R2

where

R1 = S1 L1

R2 = S2 L2

Equation 1: P = S1 L1 + S2 L2 (where P is known)

Define the strength of the weld as S, and write an equation summing moments

about point O.

∑Mo = 0 = R1 a - R2 b R1 a = R2 b

so S1 L1 a = S2 L2 b

If the weld size is constant, then let S1 = S2 = S

Equation 2: L1 = L2 (b/a)

• Finally, combine equations 1 and 2 to determine the required weld lengths.

P = S1 L2 (b/a) + S2 L2 = S L2 (b/a + 1)

where

P, S, b, and a are known.

Solve for L2. Then using Equation 2, find L1.

10.21

Example Problem - Eccentricity in Welded Joints

Problem 10.13 (p. 530)

Given: Angle welded to plate as shown.

Find: Length and size of the welds to

develop the full tensile strength of

the angle.

Solution

Obtain the section properties for the 5” x 3-1/2” x ½” angle.

A = 4.00 in2, y = 1.66” (ref. Table A6, p. 575 of the textbook)

Determine the tensile strength of the angle.

P = A x Ft = 4.00 (22.0) = 88.0 k

Determine the size of the fillet welds.

Minimum weld size: 3/16” (S = 2.78 k/in Table 10.5)

Maximum weld size: 1/2" – 1/16” = 7/16" (S = 6.49 k/in Table 10.5)

Write the equations of equilibrium for the applied force and weld resistances, R1

R2, and Re.

∑Fx = 0 = P - R1 - R2 - Re P = R1 + R2 + Re

∑Mo = 0 = R1 a - R2 b + Re c R1 (3.34) - R2 (1.66) + Re (2.5 – 1.66) = 0

Try a 3/16” weld (S = 2.78 k/in).

Re = S x Le = 2.78 k/in (5”) = 13.90 k

From ∑Fx = 0: P = R1 + R2 + Re

88.0 = R1 + R2 + 13.90

R1 + R2 = 88.0 – 13.90

R1 + R2 = 74.10 k Equation 1

From ∑Mo = 0: R1 (3.34) - R2 (1.66) + Re (2.5 – 1.66) = 0

R1 (3.34) - R2 (1.66) + (13.90) (2.5 – 1.66) = 0

R1 (3.34) - R2 (1.66) + 11.68 = 0

2.01 R1 - R2 + 7.04 = 0 Equation 2

10.22

Solve the equations simultaneously to determine the forces in the welds.

R1 + R2 = 74.10 Equation 1

2.01 R1 - R2 = - 7.04 Equation 2

3.01 R1 = 67.06

R1 = 22.28 k

R2 = 74.10 – R1 = 74.10 – 22.28 R2 = 51.82 k

Re = 13.90 k (previously determined)

Determine the weld lengths.

R1 = S x L1: L1 = R1/S = 22.28/2.78 = 8.01”

R2 = S x L2: L2 = R2/S = 51.82/2.78 = 18.64”

Minimum length of weld must be 5”. Thus, the 3/16” weld is OK.

Try 1/4” weld (S = 3.71 k/in).

Re = S x Le = 3.71 k/in (5”) = 18.55 k

From ∑Fx = 0: P = R1 + R2 + Re

88.0 = R1 + R2 + 18.55

R1 + R2 = 88.0 – 18.55

R1 + R2 = 69.45 k Equation 1

From ∑Mo = 0: R1 (3.34) - R2 (1.66) + Re (2.5 – 1.66) = 0

R1 (3.34) - R2 (1.66) + (18.55) (2.5 – 1.66) = 0

R1 (3.34) - R2 (1.66) + 15.58 = 0

2.01 R1 - R2 + 9.39 = 0 Equation 2

Solve the equations simultaneously to determine the weld forces.

R1 + R2 = 69.45 Equation 1

2.01 R1 - R2 = - 9.39 Equation 2

3.01 R1 = 60.06

R1 = 19.95 k

R2 = 69.45 – R1 = 69.45 – 19.95 R2 = 49.50 k

Re = 18.55 k (previously determined)

Determine the weld lengths.

R1 = S x L1 : L1 = R1/S = 19.95/3.71 = 5.38”

R2 = S x L2 : L2 = R2/S = 49.50/3.71 = 13.34”

Minimum length of weld must be 5”. Thus, the 1/4” weld is OK as well.

10.23

10.3 Common Framing Details in Steel

Structural analysis theory assumes that connections between beams and columns

or between columns and foundations are either

• Rigid (fixed) - with no relative rotation between connected elements.

• Pinned (hinged) - with no moment resistance.

In reality, structural design deals with connections that fit neither of these

assumptions fully. Most connections exhibit

• Some degree of moment resistance.

• Varying degrees of relative joint rotation.

The strength and stability of the structural framework depend on

• The connections, and

• Lateral resisting strategy.

The three basic lateral resisting systems for steel framed structures include the

following.

1. The rigid frame.

• Rigid frame systems require the use of rigid, moment resisting connections.

2. The braced frame.

• Braced frames are generally designed as pin connections.

• Material and labor costs are generally much lower in pin connections as

compared to rigid, moment connections.

3. The shearwall system.

Examples of Connection Details

Figures 10.54 (p. 534 of the textbook) shows standard details that are frequently

used in detailed drawings of steel buildings.

The example structure shown in Figure 10.51 (p. 532 of the textbook) illustrates

the following lateral stability strategies.

• A strategy that utilizes a rigid frame system.

• A centrally located concrete shear core.

• Lateral stability is achieved using a braced frame and shearwall system in the

perpendicular direction.

10.24

Rigid Frames

Rigid frames are constructed with beam and columns rigidly attached with moment-

resisting connections.

• The strength of the rigid frame to resist gravity and lateral loads is from the

moment interaction between beams and columns.

• Beam-column connections maintain their relative 90° orientation to each other

under load, even if the connection assembly rotates as a unit.

• Bending moments are shared by both beams and columns.

• Member sizes are generally larger than would be found in braced frame

systems.

The following compares rigid frames with braced frames or shearwall structures.

• Rigid frames have the advantage of providing unobstructed space.

• Under high seismic loading, the rigid frame may undergo large deformations

that, in turn, may cause distress to the architectural finishes.

• Rigid frame systems are effective for low- to medium-rise buildings.

Braced Frames

Braced frames found in use today are illustrated in Figure 10.56 (p. 567 of the

textbook).

• Braced frames are more cost effective when compared to rigid frames.

• Two basic categories of braced frames are the concentrically and eccentrically

braced frame.

Concentrically braced frame

Lateral bracing includes the use of diagonal bracing, X-bracing, and K- or Vee-

bracing.

- Members act as a vertical truss system.

- Diagonal tension X-bracing is typically analyzed as having only tension forces.

- The inverted Vee (K-bracing) allows clearance for doorways, corridors, and

rooms.

Eccentrically braced frame

Members are connected to the beam to form a short “link beam” between the

brace and the column or between two opposing braces.

- Link beams act as a “fuse” to prevent other elements in the frame from

being over-stressed.

10.25

- In major earthquakes, the link beam is designed to yield, thereby absorbing

larger quantities of seismic energy, and preventing buckling of the other

bracing members.

Shearwalls

A shearwall acts as a vertical cantilever beam to provide lateral support.

• Shearwalls for steel framed buildings are made of reinforced concrete or

concrete block.

• Concrete walls used as fire-resisting compartment walls, stairway and elevator

cores, and vertical service shafts, may also stiffen the building against lateral

loads.

A common structural strategy uses a combination of braced or rigid frames in

combination with concrete walls.

• The beams of the steel frame must be adequately connected to the concrete

cores or shearwalls to provide stability against vertical and horizontal forces.

• Careful detailing of the framing elements and connections to the shearwall is

essential.

Chapter 10: Structural Connectionscaed20301.webs.com/CN Chap10 - Structural Connections.pdf ·...

Documents

Transcript of Chapter 10: Structural Connectionscaed20301.webs.com/CN Chap10 - Structural Connections.pdf ·...