Ultimate Limit State Design of Steel Structures Lecture 3 STR 552
Dr.. Maha Moddather Dr Structural Engineering Department Faculty of Engineering – Cairo University
[email protected] Spring 2013
LAST LECTURE Design Philosophy.
Limit State Design.
Advantages of Limit State Design Method.
General Design Requirements. Dr. Maha Moddather
LAST LECTURE Design Philosophy.
Limit State Design.
Advantages of Limit State Design Method.
General Design Requirements. Dr. Maha Moddather
Outline
Stability Requirements and Calculations.
Second-Order Effects.
Local Buckling & Classification of Sections.
Design of Tension Members. Dr. Maha Moddather
Buckling Length of members with WellDefined End Conditions
Dr. Maha Moddather
Effective Buckling Length of Compression Members
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 taken equal 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 oung s mo u us t cm . Iy : The moment of inertia of the chord member about the Y-Y axis (cm4). a : The distance between the U-frames (cm). δ : The flexibility of the U-frame: the lateral deflection near the midspan at the level of the considered chord’s 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 δ is being calculated. The direction of each unit load shall produce a maximum value for δ (cm). Dr. Maha Moddather
Truss with a Compression Member Laterally Unbraced
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. Dr. Maha Moddather
Truss with a Compression Member Laterally Unbraced In case of symmetrical U-frame with constant moment of inertia for each of the cross girder and the verticals through their own length, δ may be taken from: Where: d1 : The distance from the centroid of the compression chord to the . d2 : The distance from the centroid of the compression chord to the centroidal axis of the cross girder of the U-frame. I1 : The second moment of area of the vertical member forming the arm of the U-frame about the axis of bending. I2 : The second moment of area of the cross girder about the axis of bending. B :The distance between centres of consecutive main girders connected by the U-frame. Dr. Maha Moddather
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 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.
Dr. Maha Moddather
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
columns, triggering collapse before attaining its squash load (full compressive 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 Dr. Maha Moddather
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.
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. Dr. Maha Moddather
Main Failure Modes of Hot Rolled Beams 1.
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 do not buckle locall . Such “stocky” beams will collapse by plastic hinge formation. σy
σy
Dr. Maha Moddather
Main Failure Modes of Hot Rolled Beams 2.
Lateral torsional buckling of long beams which are not suitably
braced
in
the
lateral
direction.(i.e.
“un
restrained” beams) Failure occurs by a combination of lateral deflection and twist. The , applied are all factors, which affect failure by lateral torsional buckling.
Dr. Maha Moddather
Main Failure Modes of Hot Rolled Beams 3. Failure by local buckling of a flange in compression or web due to shear or web under compression due to concentrated loads Unlikely for hot rolled sections, which are generally stocky. Fabricated box sections may require flange stiffening to prevent . girders to prevent shear buckling. Load bearing stiffeners are sometimes needed under point loads to resist web buckling.
Dr. Maha Moddather
Main Failure Modes of Hot Rolled Beams 4.
Local failure by (1) shear yield of web (2) local crushing of web (3) buckling of thin flanges Shear yield can only occur in very short spans and suitable web
stiffeners will have to be designed. Local crushin is ossible when concentrated loads act on
unstiffened thin webs. Suitable stiffeners can be designed.
Dr. Maha Moddather
Main Failure Modes of Hot Rolled Beams 4.
Local failure by (1) shear yield of web (2) local crushing of web (3) buckling of thin flanges Buckling of Thin Flanges: This is a problem only when very wide
flanges are employed. Welding of additional flange plates will
avoided.
Dr. Maha Moddather
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 lanes. Both axiall loaded columns and transversel loaded beams exhibit closely similar failure characteristics due to buckling.
Dr. Maha Moddather
Similarity of Column Buckling and Lateral Buckling of Beams Consider a simply supported and laterally unsupported (except
at ends) beam of “short-span” subjected to incremental transverse load 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 com ression flan e and rotate about the lon itudinal axis.
Dr. Maha Moddather
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 bucklin
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.
Dr. Maha Moddather
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
. 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.
Dr. Maha Moddather
Similarity of Column Buckling and Lateral Buckling of Beams This form of instability is more complex (compared to column
instability) since the lateral buckling problem is 3-dimensional in nature. It involves coupled lateral deflection and twist.
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 called its elastic critical moment (M cr). Dr. Maha Moddather
Similarity of Column Buckling and Lateral Buckling of Beams
Column Buckling
Beam Buckling
Dr. Maha Moddather
Factors Affecting Lateral Stability The elastic critical moment (M cr) is applicable only to a beam of
I 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. 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. Dr. Maha Moddather
Factors Affecting Lateral Stability Support 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 su
orts should either com letel
revent 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.
Dr. Maha Moddather
Factors Affecting Lateral Stability Effective Length: The concept of effective length incorporates the various types of
support conditions. For the beam with simply supported end conditions and no
, 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.
Dr. Maha Moddather
Factors Affecting Lateral Stability Effective 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.
Warping restraint prevents rotation of the flange in its plane.
Dr. Maha Moddather
Buckling Length of Compression Flange of Beam Simply Supported Beams: The effective buckling length of compression flange of simply supported beams shall be considered as follows : 1. Compression Flange With No Intermediate Lateral Support:
Free to rotate in plan at bearings Not free to rotate in plan at bearings Not free to rotate in plan at bearings Dr. Maha Moddather
Restraint against Torsion Web or Flange cleats. Bearing stiffeners acting in conjunction with the bearing of the
beam. Lateral end frames or other external supports to the ends of the
compression flanges. 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.
Dr. Maha Moddather
Buckling Length of Compression Flange of Beam Simply Supported Beams: 2. Compression Flange With Intermediate Lateral Support:
Dr. Maha Moddather
Buckling Length of Compression Flange of Beam Cantilever 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 Beam Cantilever 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 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) relative to the centroid of the beam. Dr. Maha Moddather
Factors Affecting Lateral Stability Level 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
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%. Dr. Maha Moddather
Factors Affecting Lateral Stability Level 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 Beam Cantilever Beams without Intermediate Lateral Supports:
Dr. Maha Moddather
Buckling Length of Compression Flange of Beam At Support:
Dr. Maha Moddather
Buckling Length of Compression Flange of Beam At Tip of Cantilever:
Dr. Maha Moddather
Local Buckling Cross-sections subject to compression due to axial load or bending
moment should be classified into Class 1 compact, Class 2 noncompact, Class 3 slender, depending on their width to thickness ratios of section elements and hence, their susceptibility against local buckling.
Cross-sections should be classified to determine whether local
buckling influences their section capacity, without calculating their local buckling resistance.
Dr. Maha Moddather
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 longitudinal edges and including: Webs comprising internal elements perpendicular to
the axis of bending. Flanges comprising internal elements parallel to the
axis of bending. Dr. Maha Moddather
σy
Local Buckling Class 1: Compact Sections
σy
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: Its flanges must be continuously connected to the web or
webs. The limiting width-to-thickness ratios ( λp) of compression
members must be smaller than a limiting value. The unbraced length should not exceed a certain value. Dr. Maha Moddather
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 ( λr) of compression
members must be smaller than a limiting value. σy
σy Dr. Maha Moddather
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 class 3 cross section. Slender sections shall be designed same as non-compact sections except that the section properties used in design shall (b e) based on the effective width be of compression elements. be = ρ b Dr. Maha Moddather
Local Buckling Class 3: Slender Sections
be = ρ b
Dr. Maha Moddather
Local Buckling Class 3: Slender Sections
Dr. Maha Moddather
Local Buckling
Dr. Maha Moddather
Local Buckling Effective Width and Buckling Factor for Stiffened Compression Elements
Dr. Maha Moddather
Local Buckling Effective Width and Buckling Factor for Unstiffened Compression Elements
Dr. Maha Moddather
Design of Tension Members Tension member are structural elements subjected to axial
tensile forces. Generally they are used in: Truss members
Cables such as: Suspended roof systems. Suspension. Bridges. Dr. Maha Moddather
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.
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. Dr. Maha Moddather
Design of Tension Members The stress in a tension member is uniform throughout the
cross-section except: Near the point of application of load, and At the cross-section with holes for bolts. The cross sectional area will be reduced by amount equal to
t e area remove
y o es.
Tension members are frequently connected at their ends with
bolts. The typical design problem is to select a member with
sufficient cross sectional area: Factored load < factored strength Dr. Maha Moddather
Design of Tension Members Consider an 8 x 0.5 cm bar connected to a gusset plate and loaded in tension.
Anet M16 Bolts
Agross
Area of Bar at section (a-a) = 8 x 0.5 = 4 cm 2 Area of Bar at section (b-b) = [8 – 2x(1.6+0.2)]x0.5 = 2.2 cm 2 Dr. Maha Moddather
Design of Tension Members A 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 < F ) Fracture:
Fracture of the net section can occur if the stress at the net section reaches the ultimate stress F u. To prevent Fracture, the stress at the net sectional area must be smaller than ultimate stress (f < F u). Dr. Maha Moddather
Design of Tension Members The nominal strength in yielding is:
Pn = Fy * Ag And the nominal strength in fracture is:
Pn = Fu* Ae ere
e
s
e e ec ve ne area:
e
n
The resistance factor Φ = Φt is smaller for fracture than for yielding reflecting the more serious nature of reaching the limit state of fracture: For yielding Φt =0.85 For fracture Φt =0.70
Dr. Maha Moddather
Design of Tension Members Why is fracture (& not yielding) the relevant limit state at the
net section? Yielding will occur first in the net section. However, the deformations induced by yielding will be localized around the . excessive deformations in the complete tension member. Hence, yielding at the net section will not be a failure limit state.
Dr. Maha Moddather
Design of Tension Members Example:
FD.L. = 5 ton FL.L. = 10 ton
2.0 m
FW.L. = 3 ton Using St.37, choose a suitable section.
2.0 m
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
Dr. Maha Moddather
Design of Tension Members Example:
FD.L. = 5 ton FL.L. = 10 ton
2.0 m
FW.L. = 3 ton 2.0 m
Tr 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.
Dr. Maha Moddather
Design of Tension Members Example:
FD.L. = 5 ton FL.L. = 10 ton
2.0 m
FW.L. = 3 ton 2.0 m
Try 2 < 60x6 Bolted M16 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
Pu = 28.2 ton > 24.4 ton O.K.
Or = 0.7 x 3.7 x 11.66 = 30.2 ton Dr. Maha Moddather
Design of Tension Members Example:
FD.L. = 5 ton FL.L. = 10 ton
2.0 m
FW.L. = 3 ton 2.0 m
Tr 1 < 100x10 Bolted M20 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. Dr. Maha Moddather
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.
Material ductility. Fastener spacing. Stress concentration at holes. Fabrication procedure. Shear lag. Dr. Maha Moddather
Design of Tension Members Effective Net Area
All factors contribute to reducing the effectiveness but shear lag is the most important. Shear lag occurs when the tension force is not transferred -
.
occur when some elements of the cross-section are not connected.
Dr. Maha Moddather
Design of Tension Members Effectiv tivee Net Area Area Effec
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 Munse and Chesson (1963) suggests that shear lag can be
accounted for by using a reduced or effective effective net area A e. Shearr lag lag af affe fect ctss both both bolt bolted ed and and weld welded ed conn connec ecti tion ons. s. Shea Thus, the effective net area concept applied to both types of connections. Dr. Maha Moddather
Design of Tension Members Effectiv tivee Net Area Area Effec
For bolted bolted connection, connection, the the effecti effective ve net area is A e = U An For welded welded connectio connection, n, the effect effective ive net area is A e = U Ag Where, the reduction factor U is given by:
Where, x is the distance from the centroid of the connected area to the plane of the connection, and L is the length of the connection. If the member has two symmetrically located planes of connection, x is measured from the centroid of the nearest one half of the area. Dr. Maha Moddather
Design of Tension Members Effectiv tivee Net Area Area Effec 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
. 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. Dr. Maha Moddather
Design of Tension Members Effective Net Area
For welded connection is by longitudinal welds at the ends as shown in the figure below, then Ae = UAg
Where, U = 1.0 for L ≥ w U = 0.87 for 1.5 w ≤ L < 2 w U = 0.75 forw ≤ L < 1.5 w L = length of the pair of welds ≥ w w = distance between the welds or width of plate/bar
Dr. Maha Moddather
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 (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, the
relationship f = P/A does not apply and the stresses are a combination of tensile and shearing stresses on the inclined portion b-c. Dr. Maha Moddather
Design of Tension Members Effective Net Area: Staggered Fasteners For a tension member with staggered bolt holes, the
relationship f = P/A does not apply and the stresses are a combination of tensile and shearing stresses on the inclined portion b-c.
Dr. Maha Moddather
Design of Tension Members Effective Net Area: Staggered Fasteners Net section fracture can occur along any zig-zag or straight
line. For example, fracture can occur along the inclined path a-b-c-d in the figure above. However, all possibilities must be examined. Empirical methods have been developed to calculate the net section fracture stren th.
Dr. Maha Moddather
Design of Tension Members Effective Net Area: Staggered Fasteners
where, d is the diameter of hole to be deducted s2 /4g is added for each gage space in the chain being considered. direction of loading g is the transverse spacing (gage) of the bolt holes perpendicular to loading dir. Net area (An) = net width x plate thickness Effective net area (A e) = U An Dr. Maha Moddather
Design of Tension Members Effective Net Area: Staggered Fasteners
Compute the smallest net area for the plate shown below: The holes are for M16 bolts.
Hole Diameter = 1.6 + 0.2 = 1.8 cm Dr. Maha Moddather
Design of Tension Members Effective Net Area: Staggered Fasteners
For line a-b-d-e
Net Width = 16 – 2 x 1.8 = 12.4 cm
Dr. Maha Moddather
Design of Tension Members Effective Net Area: Staggered Fasteners
For line a-b-c-d-e
Net Width = 16 – 3 x 1.8 + (32+32)/4x5 = 11.5 cm Governs
Net Area = 11.5 x Thickness of Plate Dr. Maha Moddather
Design of Tension Members Block Shear For some connection configurations, the tension member
can fail due to ‘tear-out’ of material at the connected end. This is called block shear. For example, the single angle tension member connected as shown in the Figure below is susceptible to the henomenon of block shear.
Dr. Maha Moddather
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 stren th = net section fracture stren th on shear ath + 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
Dr. Maha Moddather
Design of Tension Members Block Shear
Which of the two calculations above governs? When Fu Ant ≥ 0.6FuAnv
φv Rn = φv (0.6 Fy Agv + Fu Ant) ≤ φv (0.6 FuAnv + Fu Ant) When Fu Ant < 0.6Fu Anv;
φv Rn = φv (0.6 Fu Anv + Fy Agt) ≤ φv (0.6 FuAnv + Fu Ant)
Where, φv = 0.70 Agv : Gross area subject to shear Agt : Gross area subject to tension Anv : Net area subject to shear Ant : Net area subject to tension Dr. Maha Moddather
