LRFD

Ultimate Limit State Design of Steel Structures

Lecture 3STR 552

Graduate Course

Dr. Maha ModdatherStructural Engineering Department

Faculty of Engineering Cairo [email protected]

Spring 2013

Graduate Course

Design Philosophy.

Limit State Design.

LAST LECTURE

Advantages of Limit State Design Method.

General Design Requirements.

Dr. Maha Moddather

Stability Requirements and Calculations.

Second-Order Effects.

Outline

Local Buckling & Classification of Sections.

Design of Tension Members.

Dr. Maha Moddather

Buckling Length of members with Well-Defined End Conditions

Dr. Maha Moddather

Effective Buckling Length of Compression Members

Dr. Maha Moddather

Truss with a Compression Member Laterally Unbraced For a simply supported truss, with laterally unsupported compression

chords and with no cross-frames but with each end of the truss

adequately restrained, the effective buckling length (KL), shall be takenequal to 0.75 of the truss span.

Dr. Maha Moddather

Truss with a Compression Member Laterally Unbraced For a bridge truss where the compression chord is laterally restrained

by U-frames composed of the cross girders and verticals of the trusses,

the effective buckling length of the compression chord (b) is:

Where:E : The Youngs modulus (t/cm2).

Dr. Maha Moddather

E : The Youngs modulus (t/cm2).Iy : The moment of inertia of the chord member about the Y-Yaxis (cm4).a : The distance between the U-frames (cm). : The flexibility of the U-frame: the lateral deflection near the mid-span at the level of the considered chords centroid due to a unit load acting laterally at each chord connected to the U-frame. The unit load is applied only at the point at which is being calculated. The direction ofeach unit load shall produce a maximum value for (cm).

Truss with a Compression Member Laterally Unbraced

Dr. Maha Moddather

The U-frame is considered to be free and unconnected at all points

except at each point of intersection between cross girder and vertical of

the truss where this joint is considered to be rigidly connected.

Truss with a Compression Member Laterally UnbracedIn case of symmetrical U-frame with constant moment of inertia for each of the cross girder and the verticals through their own length, may be taken from:

Where:d1 : The distance from the centroid of the compression chord to thenearest face of the cross girder of the U-frame.

Dr. Maha Moddather

nearest face of the cross girder of the U-frame.d2 : The distance from the centroid of the compression chord to thecentroidal axis of the cross girder of the U-frame.I1 : The second moment of area of the vertical member forming thearm of the U-frame about the axis of bending.I2 : The second moment of area of the cross girder about the axis ofbending.B :The distance between centres of consecutive main girdersconnected by the U-frame.

Truss with a Compression Member Laterally Unbraced

Dr. Maha Moddather

Buckling Length of Compression Flange of Beams Generally, a beam resists transverse loads by bending action.

In a typical building frame, main beams are employed to span

between adjacent columns; secondary beams when used transmit the

Dr. Maha Moddather

floor loading on to the main beams.

In general, it is necessary to consider only the bending effects in

such cases, any torsional loading effects being relatively insignificant.

Buckling Length of Compression Flange of Beams If the laterally unrestrained length of the compression flange of the

beam is relatively long, then a phenomenon, known as lateral

buckling or lateral torsional buckling of the beam may take place. The

beam would fail well before it could attain its full moment capacity.

This phenomenon has a close similarity to the Euler buckling of

Dr. Maha Moddather

This phenomenon has a close similarity to the Euler buckling of

columns, triggering collapse before attaining its squash load (fullcompressive yield load).Lateral buckling of beams has to be accounted for at all stages of

construction, to eliminate the possibility of premature collapse of the

structure or component

Buckling Length of Compression Flange of Beams For example, in the construction of steel-concrete composite

buildings, steel beams are designed to attain their full moment

capacity based on the assumption that the flooring would provide the

necessary lateral restraint to the beams.

Dr. Maha Moddather

However, during the erection stage of the structure, beams may not

receive as much lateral support from the floors as they get after the

concrete hardens. Hence, at this stage, they are prone to lateral

buckling, which has to be consciously prevented.

Main Failure Modes of Hot Rolled Beams1. Excessive bending triggering collapse

This is the basic failure mode provided:

(a) the beam is prevented from buckling laterally,(b) the component elements are at least compact, so that they donot buckle locally.

Dr. Maha Moddather

not buckle locally.

Such stocky beams will collapse by plastic hinge formation.

y

y

Main Failure Modes of Hot Rolled Beams2. Lateral torsional buckling of long beams which are not

suitably braced in the lateral direction.(i.e. unrestrained beams)

Failure occurs by a combination of lateral deflection and twist. The

proportions of the beam, support conditions and the way the load is

Dr. Maha Moddather

proportions of the beam, support conditions and the way the load is

applied are all factors, which affect failure by lateral torsional

buckling.

Main Failure Modes of Hot Rolled Beams3. Failure by local buckling of a flange in compression or

web due to shear or web under compression due to

concentrated loadsUnlikely for hot rolled sections, which are generally stocky.Fabricated box sections may require flange stiffening to preventpremature collapse. Web stiffening may be required for plate

Dr. Maha Moddather

premature collapse. Web stiffening may be required for plategirders to prevent shear buckling. Load bearing stiffeners aresometimes needed under point loads to resist web buckling.

Main Failure Modes of Hot Rolled Beams4. Local failure by (1) shear yield of web (2) local crushing

of web (3) buckling of thin flangesShear yield can only occur in very short spans and suitable web

stiffeners will have to be designed.

Local crushing is possible when concentrated loads act on

Dr. Maha Moddather

Local crushing is possible when concentrated loads act on

unstiffened thin webs. Suitable stiffeners can be designed.

Main Failure Modes of Hot Rolled Beams4. Local failure by (1) shear yield of web (2) local crushing

of web (3) buckling of thin flangesBuckling of Thin Flanges: This is a problem only when very wide

flanges are employed. Welding of additional flange plates will

reduce the plate b / t ratio and thus flange buckling failure can be

Dr. Maha Moddather

reduce the plate b / t ratio and thus flange buckling failure can be

avoided.

Similarity of Column Buckling and Lateral Buckling of Beams It is well known that slender members under compression are

prone to instability. When slender structural elements are loaded

in their strong planes, they have a tendency to fail by buckling in

their weaker planes. Both axially loaded columns and transversely

Dr. Maha Moddather

their weaker planes. Both axially loaded columns and transversely

loaded beams exhibit closely similar failure characteristics due to

buckling.

Similarity of Column Buckling and Lateral Buckling of Beams Consider a simply supported and laterally unsupported (exceptat ends) beam of short-span subjected to incremental transverseload at its mid section. The beam will deflect downwards i.e. in the

direction of the load.

Dr. Maha Moddather

Similarity of Column Buckling and Lateral Buckling of Beams The direction of the load and the direction of movement of the

beam are the same. This is similar to a short column under axial

compression.

Dr. Maha Moddather

Similarity of Column Buckling and Lateral Buckling of Beams On the other hand, a long-span beam, when incrementally

loaded will first deflect downwards, and when the load exceeds a

particular value, it will tilt sideways due to instability of the

compression flange and rotate about the longitudinal axis.

Dr. Maha Moddather

compression flange and rotate about the longitudinal axis.

Similarity of Column Buckling and Lateral Buckling of Beams Displacement and rotation that take place as the midsection of

the beam undergoes lateral torsional buckling.

The characteristic feature of lateral buckling is that the entire

Dr. Maha Moddather

The characteristic feature of lateral buckling is that the entire

cross section rotates as a rigid disc without any cross sectional

distortion. This behaviour is very similar to an axially compressed

long column, which after initial shortening in the axial direction,

deflects laterally when it buckles.

Similarity of Column Buckling and Lateral Buckling of Beams In the case of axially loaded columns, the deflection takes place

sideways and the column buckles in a pure flexural mode.

A beam, under transverse loads, has a part of its cross section in

compression and the other in tension. The part under compression

Dr. Maha Moddather

compression and the other in tension. The part under compression

becomes unstable while the tensile stresses elsewhere tend to

stabilize the beam and keep it straight.

Thus, beams when loaded exactly in the plane of the web, at a

particular load, will fail suddenly by deflecting sideways and then

twisting about its longitudinal axis.

Similarity of Column Buckling and Lateral Buckling of Beams This form of instability is more complex (compared to columninstability) since the lateral buckling problem is 3-dimensional innature. It involves coupled lateral deflection and twist.

Dr. Maha Moddather

When the beam deflects laterally, the applied moment exerts a

torque about the deflected longitudinal axis, which causes the

beam to twist. The bending moment at which a beam fails by

lateral buckling when subjected to a uniform end moment is calledits elastic critical moment (Mcr).

Similarity of Column Buckling and Lateral Buckling of Beams

Dr. Maha Moddather

Column Buckling Beam Buckling

Factors Affecting Lateral Stability The elastic critical moment (Mcr) is applicable only to a beam ofI section which is simply supported and subjected to end moments. This case is considered as the basic case. In practical situations,

support conditions, beam cross section, loading etc. vary from the

basic case.

Dr. Maha Moddather

Factors affecting on the lateral stability include:

Support Conditions.

Effective Length.

Level of Application of Transverse Loads.

Influence of Type of Loading.

Effect of Cross-Sectional Shape.

Factors Affecting Lateral StabilitySupport Conditions: Lateral buckling involves three kinds of deformations, namely

lateral bending, twisting and warping, it is feasible to think of

various types of end conditions.

The supports should either completely prevent or offer no

Dr. Maha Moddather

The supports should either completely prevent or offer no

resistance to each type of deformation.

Solutions for partial restraint conditions are complicated. The

effect of various support conditions is taken into account by way of

a parameter called effective length.

Factors Affecting Lateral StabilityEffective Length: The concept of effective length incorporates the various types of

support conditions.

For the beam with simply supported end conditions and no

intermediate lateral restraint, the effective length is equal to the

Dr. Maha Moddather

intermediate lateral restraint, the effective length is equal to the

actual length between the supports.

When a greater amount of lateral and torsional restraints is

provided at supports, the effective length is less than the actual

length and alternatively, the length becomes more when there is

less restraint.

Factors Affecting Lateral StabilityEffective Length:The effective length factor would indirectly account for the

increased lateral and torsional rigidities provided by the restraints.

Torsional restraint prevents rotation about the longitudinal axis.

Dr. Maha Moddather

Warping restraint prevents rotation of the flange in its plane.

Buckling Length of Compression Flange of BeamSimply Supported Beams:The effective buckling length of compression flange of simplysupported beams shall be considered as follows :

1. Compression Flange With No Intermediate Lateral Support:

Dr. Maha Moddather

Free to rotate in plan at bearings

Not free to rotate in plan at bearings

Not free to rotate in plan at bearings

Restraint against Torsion Web or Flange cleats.

Bearing stiffeners acting in conjunction with the bearing of thebeam.

Lateral end frames or other external supports to the ends of the

compression flanges.

Dr. Maha Moddather

Beams being built into walls.

The end restraint element shall be able to safely resist in addition to wind

and other external applied loads, a horizontal force a compression force

acting at the bearing in a direction normal to the compression flange of

the beam at the level of centroid and having a value equal to 2.5% of the

maximum force occurring in the flange.

Buckling Length of Compression Flange of BeamSimply Supported Beams: 2. Compression Flange WithIntermediate Lateral Support:

Dr. Maha Moddather

Buckling Length of Compression Flange of BeamCantilever Beams with Intermediate Lateral Supports:The effective buckling length of compression flange of cantilever beams with intermediate lateral supports shall be similar to that of simply supported beams having lateral supports.

Dr. Maha Moddather

Buckling Length of Compression Flange of BeamCantilever Beams without Intermediate Lateral Supports:The effective buckling length of compression flange of cantilever

beams without intermediate lateral supports shall be according to

Table 4.10. The loading condition (normal or destabilizing) is

Dr. Maha Moddather

defined by the point of application of the load.

Destabilizing load conditions exist when a load is applied to the

top flange of a beam or cantilever and both the load and the flange

are free to deflect laterally (and possibly rotationally also) relativeto the centroid of the beam.

Factors Affecting Lateral StabilityLevel of Load Application: The lateral stability of a transversely loaded beam is dependent

on the arrangement of the loads as well as the level of application

of the loads with respect to the centroid of the cross section.

A load applied above the centroid of the cross section causes an

Dr. Maha Moddather

additional overturning moment and becomes more critical than the

case when the load is applied at the centroid.

On the other hand, if the load is applied below the centroid, it

produces a stabilising effect. Thus, a load applied below or above

the centroid can change the buckling load by 40%.

Factors Affecting Lateral StabilityLevel of Load Application: The figure shows a centrally loaded beam experiencing either

destabilising or restoring effect when the cross section is twisted.

Dr. Maha Moddather

Buckling Length of Compression Flange of BeamCantilever Beams without Intermediate Lateral Supports:

Dr. Maha Moddather

Buckling Length of Compression Flange of BeamAt Support:

Dr. Maha Moddather

Buckling Length of Compression Flange of BeamAt Tip of Cantilever:

Dr. Maha Moddather

Local Buckling Cross-sections subject to compression due to axial load or bendingmoment should be classified into Class 1 compact, Class 2 non-

compact, Class 3 slender, depending on their width to thickness ratios

of section elements and hence, their susceptibility against local

buckling.

Dr. Maha Moddather

Cross-sections should be classified to determine whether local

buckling influences their section capacity, without calculating their

local buckling resistance.

Local Buckling A distinction should be made between the following two types of

element,

(a) Outstand elements are attached to adjacent elements at one edge only while the other edge being free.

(b) Internal elements are attached to other elements on both

Dr. Maha Moddather

(b) Internal elements are attached to other elements on both longitudinal edges and including:

Webs comprising internal elements perpendicular to

the axis of bending.

Flanges comprising internal elements parallel to the

axis of bending.

Local Buckling Class 1: Compact Sections

Cross-sections with plastic moment capacity. The plastic moment

capacity can be developed, without local buckling of any of their

compression elements.

For a section to qualify as a compact section:

y

y

Dr. Maha Moddather

Its flanges must be continuously connected to the web or

webs.

The limiting width-to-thickness ratios (p) of compressionmembers must be smaller than a limiting value.

The unbraced length should not exceed a certain value.

Local Buckling Class 2: Non-Compact Sections

Cross-sections that can achieve yield moment capacity without

local buckling of any of their compression elements.

For a section to qualify as a non-compact section:

The limiting width-to-thickness ratios (r) of compression

Dr. Maha Moddather

r

members must be smaller than a limiting value.

y

y

Local Buckling Class 3: Slender Sections

Those Cross-sections that can not achieve yield moment capacity

without local buckling of any of their compression elements.

When any of the compression elements of a cross-section is

classified as class 3, the whole cross section shall be designed as

Dr. Maha Moddather

class 3 cross section.

Slender sections shall be designed same as non-compact sections

except that the section properties used in design shall (be) based onthe effective width be of compression elements.

be = b


be = b

Dr. Maha Moddather


Dr. Maha Moddather

Local Buckling

Dr. Maha Moddather

Local BucklingEffective Width and BucklingFactor for Stiffened CompressionElements

Dr. Maha Moddather

Local BucklingEffective Width and BucklingFactor for UnstiffenedCompression Elements

Dr. Maha Moddather

Design of Tension Members Tension member are structural elements subjected to axialtensile forces.

Generally they are used in:

Truss members

Bracing for building and bridges

Dr. Maha Moddather

Bracing for building and bridges

Cables such as:

Suspended roof systems.

Suspension.

Bridges.

Design of Tension Members Any cross sectional configuration may be used, circular rod

and rolled angle shapes are frequently used.

Other shapes may be used when large load must be resisted.

Dr. Maha Moddather

The stress in an axially loaded tension member is given by:

f = P/A

Where:

P: the magnitude of load.

A: the cross sectional area normal to the load.

Design of Tension Members The stress in a tension member is uniform throughout the

cross-section except: Near the point of application of load, andAt the cross-section with holes for bolts.

The cross sectional area will be reduced by amount equal to the area removed by holes.

Dr. Maha Moddather

the area removed by holes.

Tension members are frequently connected at their ends with bolts.

The typical design problem is to select a member withsufficient cross sectional area:

Factored load < factored strength

Design of Tension MembersConsider an 8 x 0.5 cm bar connected to a gusset plate andloaded in tension.

M16 Bolts

Anet

Dr. Maha Moddather

Area of Bar at section (a-a) = 8 x 0.5 = 4 cm2Area of Bar at section (b-b) = [8 2x(1.6+0.2)]x0.5 = 2.2 cm2

Agross

Design of Tension MembersA tension member can fail by reaching one of two limit states:

Excessive deformation:Excessive deformation can occur due to the yielding of the gross section along the length of the member.

To prevent excessive deformation, the stress at the gross sectional area must be smaller than yielding stress (f < Fy)

Dr. Maha Moddather

y

Fracture:Fracture of the net section can occur if the stress at the net section reaches the ultimate stress Fu.

To prevent Fracture, the stress at the net sectional area must be smaller than ultimate stress (f < Fu).

Design of Tension Members The nominal strength in yielding is:

Pn = Fy * Ag

And the nominal strength in fracture is:

Pn = Fu* AeWhere A is the effective net area: A A

Dr. Maha Moddather

Where Ae is the effective net area: Ae An

The resistance factor = t is smaller for fracture than for

yielding reflecting the more serious nature of reaching the limit

state of fracture:For yielding t =0.85For fracture t =0.70

Design of Tension Members Why is fracture (& not yielding) the relevant limit state at thenet section?

Yielding will occur first in the net section. However, the

deformations induced by yielding will be localized around the

net section. These localized deformations will not cause

Dr. Maha Moddather

net section. These localized deformations will not cause

excessive deformations in the complete tension member. Hence,

yielding at the net section will not be a failure limit state.

Design of Tension MembersExample:

FD.L. = 5 ton

FL.L. = 10 ton

FW.L. = 3 ton

Using St.37, choose a suitable section.

2.0 m

2.0 m

Dr. Maha Moddather

Fult = 1.4 x FD.L. = 1.4 x 5 = 7 ton

Fult = 1.2 x FD.L. + 1.6 x FL.L = 1.2 x 5 + 1.6 x 10 = 22 ton

Fult = 1.2 x FD.L. + 1.6 x FL.L + 0.8 x FW.L = 1.2 x 5 + 1.6 x 10 + 0.8 x 3 =

24.4 ton

Fult = 1.2 x FD.L. + 0.5 x FL.L + 1.3 x FW.L = 1.2 x 5 + 0.5 x 10 + 1.3 x 3 =

14.9 ton


FD.L. = 5 ton

FL.L. = 10 ton

FW.L. = 3 ton

Try 2 < 60x6 Welded

2.0 m

2.0 m

Dr. Maha Moddather

Try 2 < 60x6 Welded

Agross = 2x6.91 = 13.82 cm2

Kl/r = 1x283 / 1.8 = 159 < 300

Pu = 0.85 x 2.4x 13.82 = 28.2 ton

Or = 0.7 x 3.7 x 13.82 = 35.8 ton

Pu = 28.2 ton > 24.4 ton O.K.


FD.L. = 5 ton

FL.L. = 10 ton

FW.L. = 3 ton

Try 2 < 60x6 Bolted M16

2.0 m

2.0 m

Dr. Maha Moddather


Agross = 2x6.91 = 13.82 cm2

Anet = 13.82 2x1.8x0.6

= 11.66 cm2

Kl/r = 1x283 / 1.8 = 159 < 300

Pu = 0.85 x 2.4x 13.82 = 28.2 ton

Or = 0.7 x 3.7 x 11.66 = 30.2 tonPu = 28.2 ton > 24.4 ton O.K.


FD.L. = 5 ton

FL.L. = 10 ton

FW.L. = 3 ton


2.0 m

2.0 m

Dr. Maha Moddather


Agross = 19.2 cm2

Ae = U x Anet = 0.75 (19.2 -2.2x1)= 12.75 cm2

Kl/r = 1x283 / 2 = 141.5 < 300

Pu = 0.85 x 2.4x 19.2 = 39.2 ton

Or = 0.7 x 3.7 x 12.75 = 33.02 ton

Pu = 33.02 ton > 24.4 ton O.K.

Design of Tension Members Effective Net Area

The connection has a significant influence on the performance of

a tension member. A connection almost always weakens the

member, and a measure of its influence is called joint efficiency.

Joint efficiency is a function of:

Dr. Maha Moddather

Joint efficiency is a function of: Material ductility.

Fastener spacing.

Stress concentration at holes.

Fabrication procedure.

Shear lag.


All factors contribute to reducing the effectiveness but shear lag

is the most important.

Shear lag occurs when the tension force is not transferred

simultaneously to all elements of the cross-section. This will

Dr. Maha Moddather

simultaneously to all elements of the cross-section. This will

occur when some elements of the cross-section are not

connected.


A consequence of this partial connection is that the connected

element becomes overloaded and the unconnected part is not

fully stressed.

Lengthening the connection region will reduce this effect

Dr. Maha Moddather

Munse and Chesson (1963) suggests that shear lag can beaccounted for by using a reduced or effective net area Ae.

Shear lag affects both bolted and welded connections.

Thus, the effective net area concept applied to both types of

connections.


For bolted connection, the effective net area is Ae = U AnFor welded connection, the effective net area is Ae = U AgWhere, the reduction factor U is given by:

Dr. Maha Moddather

Where, x is the distance from the centroid of the connected areato the plane of the connection, and L is the length of theconnection.If the member has two symmetrically located planes ofconnection, x is measured from the centroid of the nearest one -half of the area.


The distance L is defined as the length of the connection in

the direction of load.

For bolted connections, L is measured from the center of

the bolt at one end to the center of the bolt at the other end.

Dr. Maha Moddather

the bolt at one end to the center of the bolt at the other end.

For welded connections, it is measured from one end of the

connection to other.

If there are weld segments of different length in the

direction of load, L is the length of the longest segment.


For welded connection is by longitudinal welds at the ends as

shown in the figure below, then Ae = UAg

Dr. Maha Moddather

Where, U = 1.0 for L wU = 0.87 for 1.5 w L < 2 wU = 0.75 forw L < 1.5 wL = length of the pair of welds ww = distance between the welds or width of plate/bar

Design of Tension Members Effective Net Area: Staggered Fasteners

For a bolted tension member, A maximum net area can be

achieved if the bolts are placed in a single line.

The connecting bolts can be staggered for several reasons:

(1) To get more capacity by increasing the effective net area

Dr. Maha Moddather

(2) To achieve a smaller connection length(3) To fit the geometry of the tension connection itself.

For a tension member with staggered bolt holes, therelationship f = P/A does not apply and the stresses are acombination of tensile and shearing stresses on the inclinedportion b-c.


For a tension member with staggered bolt holes, therelationship f = P/A does not apply and the stresses are acombination of tensile and shearing stresses on the inclinedportion b-c.

Dr. Maha Moddather


Net section fracture can occur along any zig-zag or straightline. For example, fracture can occur along the inclined patha-b-c-d in the figure above. However, all possibilities must beexamined. Empirical methods have been developed to calculate thenet section fracture strength.

Dr. Maha Moddather

net section fracture strength.


where, d is the diameter of hole to be deducteds2/4g is added for each gage space in the chain beingconsidered.s is the longitudinal spacing (pitch) of the bolt holes in the

Dr. Maha Moddather

s is the longitudinal spacing (pitch) of the bolt holes in thedirection of loadingg is the transverse spacing (gage) of the bolt holesperpendicular to loading dir.

Net area (An) = net width x plate thickness

Effective net area (Ae) = U An


Compute the smallest net area for the plate shown below: The

holes are for M16 bolts.

Dr. Maha Moddather

Hole Diameter = 1.6 + 0.2 = 1.8 cm


For line a-b-d-e

Dr. Maha Moddather

Net Width = 16 2 x 1.8 = 12.4 cm


For line a-b-c-d-e

Dr. Maha Moddather

Net Width = 16 3 x 1.8 + (32+32)/4x5 = 11.5 cm Governs

Net Area = 11.5 x Thickness of Plate

Design of Tension Members Block Shear

For some connection configurations, the tension membercan fail due to tear-out of material at the connected end.This is called block shear. For example, the single angle tension member connectedas shown in the Figure below is susceptible to thephenomenon of block shear.

Dr. Maha Moddather

phenomenon of block shear.

Design of Tension Members Block Shear

Block shear strength is determined as the sum of the shear

strength on a failure path and the tensile strength on a

perpendicular segment.

Block shear strength = net section fracture strength on shear path

Dr. Maha Moddather

Block shear strength = net section fracture strength on shear path + gross yielding strength on the tension path

OR

Block shear strength = gross yielding strength of the shear path + net section fracture strength of the tension path

Design of Tension Members Block ShearWhich of the two calculations above governs? When Fu Ant 0.6FuAnvv Rn = v (0.6 Fy Agv + Fu Ant) v (0.6 FuAnv + Fu Ant)

When Fu Ant < 0.6Fu Anv;

R = (0.6 F A + F A ) (0.6 F A + F A )

Dr. Maha Moddather

v Rn = v (0.6 Fu Anv + Fy Agt) v (0.6 FuAnv + Fu Ant)

Where, v = 0.70

Agv : Gross area subject to shearAgt : Gross area subject to tensionAnv : Net area subject to shearAnt : Net area subject to tension

Design of Tension Members Example

2

Documents

LRFD