CE 312 Structural Analysis and Design Sessional-I (Lab Manual)

Transcription

1 CE Structural Analysis and Design Sessional-I (ab Manual) Department of Civil Engineering Ahsanullah University of Science and Technology December, 07

2 Preface Structural Analysis and Design Sessional-I (CE ) manual contains the analysis and design of an industrial roof truss and a plate girder. or providing a complete guideline to the students, basic design concepts of roof truss and plate girder are elaborated with examples and detailed drawings in this manual. Design of support and anchorage system is also discussed for a complete understanding of the students. This manual is prepared using AISC and BNBC standards. The main objective of this manual is to provide the students with sufficient fundamental knowledge about analysis and design of steel members and connections. It is designed to familiarize the students with practical problems and also to develop their ability to design steel structural systems. The manual will also introduce the student to design guidelines that are commonly used by practicing structural engineers. Updated by atema-tuz-zahura Assistant Professor Md. Hossain Nadim ecturer Department of Civil Engineering Ahsanullah University of Science and Technology Prepared by atema-tuz-zahura Assistant Professor Urmi Devi ecturer Department of Civil Engineering Ahsanullah University of Science and Technology

3 INDEX Page no. Part Steel undamentals -7 Part Design of an Industrial Steel Roof Truss 8-5. Introduction 8. Assumptions 8. Advantages 8.4 Types of truss 9-.5 Roof Truss Example Data -.6 Design of Purlin for Dead oad -.6. Analysis and Design of Purlin for Dead oad Analysis and Design of Purlin for Dead oad 8- plus Wind oad.7 Analysis and Design of Sagrods.8 Analysis of the Truss -8.9 Design of Truss Members Design of Top Chord Design of Bottom Chord Design of Verticals Design of Diagonals / Web Members -6.0 Design of Bracing Systems Vertical Bracing 6.0. Top Chord Bracing Bottom Chord Bracing 7-8. Design of Truss Joints (Welded Connections) 8-5. Design of Anchorage and Support: 5-5 Part Design of Steel Plate Girder Introduction 54. Advantages and Disadvantages of Plate girder 55

4 Page no.. Types of Plate Girder 56.4 Essential Elements of I-section Plate Girder Post-Buckling Behavior of the Web Plate Requirements for different components of the plate girder Proportions of Plate Girders Requirement for lexural Strength Requirement for Shear Strength Requirements for Intermediate Stiffener Requirements for Bearing Stiffener Design Procedure Design Example Appendix.

5 Part : Steel undamentals Steel Structure: a structure which is made from organized combination of structural steel members designed to carry loads and provide adequate rigidity. Steel structures involve substructure or members in a building made from structural steel Types of Steel Structure: Truss structures (bar or truss members) rame structures (beams and columns) Grids structures Arch Prestressed structures Beam bridge Truss bridge Arch bridge Cable-stayed bridge Suspension bridge igure.(a): Truss Structure igure.(b): rame Structure (Structural steel frame, n.d.) igure.(a): Truss bridge (Trenton through truss bridge, 008) igure.(b): Arc bridge (Old trails arch bridge, n.d.) P a g e

6 Advantages of Steel: a) High strength per unit weight especially when compared to concrete. This can reduce the size of the elements in the structure and increase the living space. b) Uniformity: that reduces the effect of time on steel as compared to concrete that changes throughout its life. c) Elasticity: steel is elastic, that is it follows Hook's aw as long as its stresses do not exceed its yielding stress. So, steel behaves closer to design assumptions as compared to other materials d) Moment of inertia of steel is accurately calculated where as that of concrete changes as the cracks move up towards the neutral axis and past it. e) Durability and performance: If properly maintained the properties of steel do not change appreciably with time. f) Ductility: since steel is a ductile material, it can undergo extensive deformations after which increased stresses are required for failure to occur. This is a property that can save lives. g) It is easier to add to a steel structure than it is to a concrete structure mainly due to connections. h) It is faster to build a steel structure than it is a concrete structure due to its lightness compared to concrete, it requires no curing time, and the members are easily connected (bolted, welded, and riveted). i) Reliability: Steel Structures are very reliable. The reason for this reliability is uniformity and consistency in properties, and better-quality control because of factory manufacture. j) Possible Reuse: Steel sections can be reused after a structure has been disassembled. Steel also has very good scrap value. Disadvantages of Steel: a) Maintenance cost: steel requires maintenance against corrosion. However, this cost may be eliminated by using atmospheric corrosion-resistant steel such as A4 and A588. b) ireproofing costs: steel will not ignite. However, at 00 steel has very little strength. Its temperature should not exceed 800 beyond which its strength is reduced quickly. c) Buckling: can occur when long slender steel members are exposed to compressive loads. To avoid buckling, a larger cross-section is needed which will increase cost. d) atigue: is caused by a large number of repetitive tensile stress variations. This can reduce the strength and ductility of the steel causing a sudden failure. e) Aesthetics: A considerable amount of money has to be spent on steel structure to improve their appearance. P a g e

7 Steel Design Specifications: The specifications of most interest to the structural steel designer are those published by the following organizations. American Institute of Steel Construction (AISC) American Association of State Highway and Transportation Officials (AASHTO) American Railway Engineering and Maintenance-of-Way Association (AREMA) American Iron and Steel Institute (AISI) Design Methodology The design of a structural member entails the selection of a cross section that will safely and economically resist the applied loads. The fundamental requirement of structural design is that the required strength not exceed the available strength; that is, Required strength available strength Design for strength is performed according to the provisions for load and resistance factor design (RD) or to the provisions for allowable strength design (ASD). Allowable Strength Design (ASD): In this method a member is selected that has cross sectional properties such as area and moment of inertia that are large enough to prevent the maximum applied axial force, shear, or bending moment from exceeding an allowable, or permissible, value. This allowable value is obtained by dividing the nominal or theoretical, strength by a factor of safety. This can be expressed as, Allowable strength = oad and resistance factor design (RD) Nominal strength Safety factor In this method load factors are applied to the service loads, and a member is selected that will have enough strength to resist the factored 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 actored load factored strength In this expression, the factored load is actually the sum of all service loads to be resisted by the member, each multiplied by its own load factor. or example, dead loads will have load factors that are different from those for live loads. The factored strength is the theoretical strength multiplied by a resistance factor. Σ(loads load factors ) resistance resistance factor P a g e

8 Structural Steel: Steel Grade: Different grades of structural steel are identified by the designation assigned them by the American Society for Testing and Materials (ASTM). Property A6 A57 Gr. 50 A99 Yield point, min. 6 ksi 50 ksi 50 ksi Tensile strength, min. 58 to 80 ksi 65 ksi 65 ksi Yield to tensile ratio, max Classification of structural steel: Compact: Section reaches its full strength (yield) before local buckling occurs. Strength of section is governed by material strength Non-compact: Only a portion of the cross-section reaches its full strength (yield) before local buckling occurs Slender: Cross-section does not yield before local buckling occurs. Strength is governed by buckling AISC classifies cross-sectional shapes as compact, noncompact, or slender, depending on the values of the width-to-thickness ratios. Classification are given in AISC Table B4. Notation: λ = width / thickness ratio λ p = upper limit for compact category λ r = upper limit for non-compact category If λ λ p and the flange is continuously attached to the web, the shape is compact If λ p λ λ r, the shape is non-compact If λ > λ r, the shape is slender (These values are discussed later in the manual) The category is based on the worst width-to-thickness ratio of the cross section. or example, if the web is compact and the flange is noncompact, the shape is classified as noncompact. Types of Structural Steel: Hot Rolled Steel: Hot rolling is a mill process which involves rolling the steel at a high temperature (typically at a temperature over 700 ), which is above the steel s recrystallization temperature. When steel is above the recrystallization temperature, it can be shaped and formed easily, and the steel can be made in much larger sizes. Hot rolled steel is typically cheaper than cold rolled steel due to the fact that it is often manufactured without any delays in the process, and therefore the reheating of the steel is not required (as it is with cold rolled). When the steel cools off it will shrink 4 P a g e

9 slightly thus giving less control on the size and shape of the finished product when compared to cold rolled. Cold Rolled Steel: Cold rolled steel is essentially hot rolled steel that has had further processing. The steel is processed further in cold reduction mills, where the material is cooled (at room temperature) followed by annealing and/or tempers rolling. This process will produce steel with closer dimensional tolerances and a wider range of surface finishes. The term Cold Rolled is mistakenly used on all products, when actually the product name refers to the rolling of flat rolled sheet and coil products. Built-up Section: Built -up members are obtained by connecting two or more plates or shapes which then act as a single member. Such members may be made necessary by requirement of the area, which can t be provided by a single rolled shape, or by the requirement of rigidity because for the same area, much greater moment of inertia can be obtained with built-up sections compared to single rolled shapes, or by the requirement of suitable connection, where the width or depth of member necessary for proper connection can t be obtained in a standard rolled section. Standard rolled Shapes (Structural steel shapes, n.d.) Symbol Type of shape Description igure W Wide lange lange surfaces are parallel; flange thickness is not necessarily equal to the web thickness. HP Bearing Pile lange surfaces are parallel; flange and web have equal thicknesses. S American Standard Beam The inner flange surface is sloped. C Channel Standard AISC flanges have sloped inner flange surfaces. 5 P a g e

10 Symbol Type of shape Description igure WT ST MT Tee WT shapes are cut from a wide flange. ST shapes are cut from American Standard Beams. MT shapes are cut from nonstandard I-shapes. HSS TS Hollow Steel Section Steel Tube Either nomenclature is acceptable; however, HSS is more common. Angle Angles come in equal leg or unequal leg sizes. The diagram at left shows an unequal leg. Pipe Pipe -- P Plate Very small plates can also be called bars. oads encountered in structural steel design To be able to design a safe, efficient and economical structure, we have to have an accurate idea of the types of loads the structure will be exposed to during its life time, and what combinations of these loads can occur at the same time. Types of oads: Dead oads: have a constant magnitude and a fixed position. That includes the structures own weight and anything fixed to it. However, to estimate the structures weight we have to know that members are being used. Therefore, we assume the members then check our results. The more experience the designer has, the lower the number of member estimates he has to do. 6 P a g e

11 ive loads: change in magnitude and position. If it is not a dead load then it is a live load. ive loads are of types: Moving loads that move by their own power (cars and trucks). Movable loads (furniture). ew examples of live loads are: i. floor loads ii. Snow and ice iii. Rain especially on flat roofs because ponding develops causing deflections. iv. Traffic loads for bridges. v. Impact loads: such as falling objects or sudden car braking. vi. ateral loads: such as wind, which changes with height, geographic location, surrounding structures vii. Earthquakes are another example of impact loads. viii. longitudinal loads: such as sudden stopping of trains or trucks on bridges. ix. Other live loads: soil pressure on walls or foundations, water on dams, explosions, thermal forces due to temperature changes...etc. 7 P a g e

12 Part : Design of an Industrial Steel Roof Truss. Introduction A truss is a structure composed of slender members joined together at their end points. Planar trusses lie in a single plane. Typically, the joint connections are formed by bolting or welding the end members together to a common plate, called a gusset plate. The basic building block of a truss is a triangle. arge trusses are constructed by attaching several triangles together. A new triangle can be added truss by adding two members and a joint. A truss constructed in this fashion is known as a simple truss.. Assumptions: The main assumptions made in the analysis of truss are: Truss members are connected together at their ends only. Trusses are connected together by frictionless pins. The truss structure is loaded only at the joints.. Advantages of Truss: Quick Installation The primary advantage of a truss is that it can be installed quickly and cost-effectively, even without heavy equipment to lift it into place. Most trusses are factory-built, and delivered to the job site as a complete set for the structure to be built. A truss is traditionally leveraged to the top of the wall, and then slid into position and pivoted upright before being fastened in place. Increased Span The unique properties of a triangular object allow trusses to span across longer distances. Where a square-sided roof would tend to shift or twist, a triangular one maintains its shape, preventing shift and sag. As a further advantage, the entire set of trusses combined becomes stable and able to support many times the weight of a non-reinforced straight roof. oad Distribution The shape of a triangle allows all of the weight applied to the sides (or legs) to be redistributed down and away from the center. In trusses, this transfers the entire weight of the roof to the outer walls, and has the advantage of allowing the interior walls to be built arbitrarily, or even moved or omitted. Accessibility Since the bottom rail of a truss is typically the ceiling of the rooms below, the triangular spaces of the trusses themselves form accessible paths for the installation of electric and other utility applications. The central void of a truss system is generally the attic of a home, with the slope of the roof forming the legs of the triangle. 8 P a g e

13 .4 Types of Truss: Bowstring Truss Towne attice Truss Pratt Truss Scissors Truss Pratt Truss Camel Back Pratt Saw-Tooth Truss Pratt Truss 9 P a g e

14 Modified ink Truss Truss with monitor Knee Brace Traverse Bent ink or rench Traverse Bent Traverse Bent an ink Howe Truss 0 P a g e

15 Howe Truss Warren Truss Howe (Pitch) Truss Warren Truss igure.: Different types of truss igure.: Bracing System of truss igure.: Building Plan P a g e

16 .5 Roof Truss Design Design a Pratt type roof truss from the following data: Design Data: Span = 40 feet Span-to-rise ratio (pitch) = 4: Rise = 0 feet Slope (θ) = tan - (0/0) = (degree) Bay distance (truss-to-truss distance) = 5 feet. ocation: Dhaka, Basic wind speed = 0 Km/h. Exposure category: Exposure A Truss is supported on brick wall of height = feet. Design oads: () Dead load: Self-weight of truss = 60 lb per ft. horizontal span of truss. Sag rod + bracing = psf. (approximately known) C.G.I. sheet roofing = psf. (known) Purlin (self-weight) =.5 psf. (assumed) () Wind load = according to BNBC 99 (Bangladesh National Building Code 99). () Snow load = not applicable for our country. Design Method: Design method followed here is AISC/ASD Steel to be used: A6 (Yield stress (y = 6 ksi) Electrode to be used: E60XX (electrode material tensile strength (EXX) = 60 ksi) P a g e

17 igure.4: Pratt type Roof Truss P a g e igure.5: Truss notations (member numbering): Pratt truss (Roof).6 Analysis and Design of Purlins: Analysis and design of purlin for dead load Analysis and design of purlin for dead load plus wind load.6. Analysis and Design of Purlin for Dead oad: Purlins are nothing but beams. They span between the adjacent trusses, i.e. the spacing of the trusses is the span of purlins. Purlins are placed at top chord joint. Since the principal axes of the purlin are inclined, the dead load causes bi-axial bending in the purlins. A component of dead load acts in the negative Y direction and the other component acts in the X direction. or the loads acting along Y- axis, the purlin acts as a simply supported beam (see figure 7) of span 5 feet (bay distance). Due to the presence of sagrods, the midspan deflection is restrained in the X direction. As a result the

18 purlin act as a continuous beam (see figure 8) for bending in the plane of the roof surface (X direction). The dead load coming on the purlin is from the roofing material and the self-weight of the purlin itself. Weights of the sagrods are so small compared to the other loads that we can safely neglect it. Calculation of total dead load on purlin: C.G.I sheet roofing = psf (known) Self-weight of purlins =.5 psf (assumed but will be checked later) Sagrod weight = negligable Total dead load =.50 psf Uniformly distributed load (UD) on purlin, WD=.50 psf purlin spacing Component of WD in X direction, WDx = WD sinθ =.50 psf 7.45 ft. = lb. per feet = sin =.666 lb. per feet Component of WD in Y direction, WDy = WD cosθ = cos =. lb. per feet Purlin span = 5 feet for loading in Y direction (loading perpendicular to the plane of roof surface). Purlin span =.5 feet +.5 feet for loading X direction (loading in the plane of roof surface) igure.6: Bi-axial loading on the purlin 4 P a g e

19 igure.7: Bending moment diagram for loading in Y direction (loading perpendicular to the roof surface, bending parallel to the roof) igure.8: Bending moment diagram for loading in X direction (loading parallel to the roof surface, bending perpendicular to the roof) Calculation of bending moment: M xx W Dy ft lb.88kip - ft M yy w Dx ft lb kip - ft 5 P a g e

20 M xx M yy = moment about X axis (moment in plane of roof surface) = moment about Y axis (moment perpendicular to the plane of roof surface) In the design of purlin, we assume that the purlin has adequate lateral bracing due to the presence of roofing and sag-rod so that pure bending will govern the design. As our first trial, we select the smallest available American Standard Channel C 4.. rom AISC manual Channel S xx (inch ) S yy (inch ) C S xx & S yy = Section modulus about X axis & Y axis respectively. Allowable bending stress, 0.66 b y or A6 steel, 0.66 = ksi =.76 ksi b y Calculation of actual bending stress: Bending stress developed on purlin section, f M (c ) M (c ) xx y yy x I I xx yy Maximum bending stress developed on purlin section, M (c ) M (c ) xx y yy x f I I xx yy f M xx (I /c ) xx y M yy (I /c ) yy x f M xx S xx M yy S yy f ksi Check of bending stress: Actual bending stress (f =.4 ksi) > allowable bending stress (b =.76 ksi) Thus, section is not OK. Select a higher section. 6 P a g e

21 Table.: Criteria for adequacy of the section If, If, If, Criteria f b f b f b Comments Section is OK Section is not OK; select a higher section Section is OK but not economical; select a lower section Section S xx Table.: Purlin section selection for dead load (inch ) S yy (inch ) Actual bending stress (f) in ksi Allowable bending stress ( ) b in ksi Comments C not OK C not OK C not OK C OK C OK but not economical C OK but not economical C OK but not economical Check self-weight of purlin: or C channel, self-weight is 5.4 lb/ft which is equivalent to 5.4 lb/ft = psf ft (distributed load over the roof surface) which is smaller than previously/initially assumed purlin self-weight.50 psf. So, the purlin C is adequate for resisting bending moment (i.e. bending stress) & its self-weight is well-below the previously/initially assumed value. Select a C section for purlin (mind it, this selection is done only for dead load). 7 P a g e

22 .6. Analysis and Design of Purlin for Dead oad plus Wind oad: Wind oad Calculation (according to BNBC 99): igure.9: Plan and elevation of roof truss Different parameters for wind load calculation: Truss location: Dhaka Vb= Basic wind speed in km/h = 0 km/h B = Horizontal dimension of the building, in meters measured normal to wind direction = bay distance (truss-to-truss spacing) = 5 feet = 7.69 meter. = Horizontal dimension of the building, in meters measured parallel to wind direction = span of truss = 40 feet =.95 meter. H = average/mean height of the roof in meters = 7 feet = 5.89 meter. z = Height above the ground in meters θ = Angle of the plane of roof from horizontal, degrees = (degree) Cc= Velocity-to-pressure conversion co-efficient = CI = Structure importance co-efficient (a factor that accounts for the degree of hazard to hazard to human life and damage to property) =.00 for standard occupancy structures 8 P a g e

23 CZ = Combined height and exposure co-efficient = for exposure A *or exposure category A, you can use the table 4 or alternatively you can use the following equation of combined height and exposure co-efficient CZ. Calculation of wind pressure: C z 0.879(z) ; z = height above ground in meters Here, z = h = 5.89 meter (7 feet) C z (5.89) ( 0.68) CG = Gust response co-efficient =.657 exposure A C pe = External pressure co-efficient = for windward side, wind direction normal to ridge C pe = External pressure co-efficient = for leeward side, wind direction normal to ridge q = Sustained wind pressure in kn/m z p z = Design wind pressure in kn/m q C C C V z c I z b q z = ( ) () (0.897) (0) q z Design wind pressure for windward side: Design wind pressure for leeward side: = 0.87 kn/m p C C q z G pe z p z = (.657) (-0.445) (0.87) p = kn/m = psf (suction) z p = (.657) (-0.70) (0.87) z p = kn/m = psf (suction) z Positive value of design wind pressure indicates thrust (compression) & negative value of design wind pressure indicates suction (tension). Design wind pressure always acts normal/perpendicular to the roof surface [ ksf = kn/m ; kn/m = 0.88 psf; MPa = MN/m = N/mm = 45 psi] 9 P a g e

24 psf psf psf igure.0: Wind direction and pressure distribution on windward side &leeward side Calculation of UD on purlin: UD on purlin on windward side = (design wind pressure on the windward side purlin spacing) = psf feet = lb/ft UD on purlin on leeward side = (design wind pressure on the leeward side purlin spacing) = psf feet = lb/ft Since wind load acts perpendicular to the roof surface, these loads will be combined with the Y component of the dead load (WDy) to get the resultant load. It is clear from the above that the leeward side will govern since its magnitude is higher. Resultant load in Y direction, w y w Dy p z (i.e. resultant load perpendicular to roof surface) = +. lb/ft lb/ft w = 0.56 lb/ft y M xx w () y ft lb ft lb M xx = kip-ft oad (w ) in X direction (in plane of roof) remains the same, so moment about Y axis remains Dx the same M = kip-ft yy 0 P a g e

25 Allowable bending stress, 0.66 b y or A6 steel, 0.66 = ksi =.76 ksi b y Bending stress developed on purlin section, f M (c ) M (c ) xx y yy x I I xx yy Maximum bending stress developed on purlin section, f f M (c ) M (c ) xx y yy x I I xx yy M xx (I /c ) xx y M yy (I /c ) yy x f M xx S xx M yy S yy or previously selected channel section (for dead load) C ( inch ) S xx =.0 inch & S yy = 0.0 f ksi >.76 ksi (not OK) Select channel C 6 ( S xx = 5.80 inch & S yy = 0.64 inch ) f ksi <.76 ksi (OK) Check self-weight of purlin: or C 6 channel, self-weight is lb/ft which is equivalent to lb/ft =.744 psf ft (distributed load over the roof surface) which is greater than previously/initially assumed purlin self-weight.50 psf. Although the purlin C 6 is adequate for resisting bending moment (i.e. bending stress) but its self-weight is high. So, not OK & select another section. Select channel C ( S = 6.08 inch xx & S = 0.65 inch yy ) f.97 ksi <.76 ksi (OK) P a g e

26 Check self-weight of purlin: or C channel, self-weight is 9.8 lb/ft which is equivalent to 9 lb/ft =.48 psf ft (distributed load over the roof surface) which is smaller than previously/initially assumed purlin self-weight.50 psf. So, the purlin C is adequate for resisting bending moment (i.e. bending stress) & its self-weight is well-below the previously/initially assumed value. oading Selected channel Check bending section stress Check self-weight Dead load only C OK OK Dead load + wind load C OK OK inally selected channel for purlins: C Analysis and Design of Sagrods: Sagrods prevent the purlin to deflect in the plane of the roof surface at midspan. Thus according to figure 8 the tensile force in the sagrods is equivalent to the midspan reaction. Sagrod force, 5 w 8 Dx plf 5 feet 8 = 8.85 lb. = kip. (tensile) A round bar of /8 inch diameter will be adequate (this is the minimum size). Assuming that the bolts threads will reduce the effective diameter by /6 inch, the net cross-sectional area will be π/4 /8 / inch. If allowable stress in tension is 0.6. or A 6 t y steel, t 0.6 y ksi.60 ksi, then this rod will be able to carry a load of.60 ksi inch = kip, which is well above the actual load (=0.885 kip)..8 Analysis of the Truss: (Dead load calculation & wind load calculation) igure.: Truss notations (member numbering): Pratt truss (Roof) P a g e

27 Dead oad Calculation: C.G.I sheet roofing = psf (known) Self-weight of purlins =.5 psf (assumed but will be checked later) Sagrod + bracing = psf Total dead load = 4.50 psf (Total 4.50 psf dead load) = (4.50 psf) (purlin spacing) (bay) = 4.50 psf feet 5 feet = lb. Self-weight of the truss (assumed) = 60 lb/ft horizontal span of truss The self-weight of the truss will be equally divided among the top chord and bottom chord. Total 60 lb/ft = 0 lb/ft in top chord & 0 lb/ft in bottom chord Point loads at the top chord joint due to self-weight = (self-weight distributed in top chord) (panel spacing along top chord) = 0 lb/ft ft = 00.0 lb. oad in top chord joint (at ridge & internal top chord joint) = lb lb = lb = kip. oad in top chord joint (at support top chord joint) = ( lb)/ + (00.0 lb)/ = lb = kip. Point loads at the bottom chord joint due to self-weight = (self-weight distributed in bottom chord) (panel spacing along bottom chord) = 0 lb/ft ft = 00.0 lb. = kip. oad in bottom chord joint (at internal bottom chord joint) = kip oad in bottom chord joint (at support bottom chord joint) = (0.000 kip)/ = kip oad at support joint = kip kip = kip. See figure for dead load on the truss. P a g e

28 .6 igure.(a): Dead loads on the truss (self-weight distribution).6 igure.(b): Dead loads on the truss (sagrod, bracing, purlin self-weight, roof weight distribution) 4 P a g e

29 .6 5 P a g e igure.: Total dead loads on the truss Wind oad Calculation (wind blows from left to right): Design wind pressure on windward side = psf Wind load on interior top chord joint windward side = (design wind pressure on windward side) (purlin spacing) (bay) = psf feet 5 feet = lb = kip Wind load on exterior & ridge top chord joint windward side = (design wind pressure on windward side) (purlin spacing/) (bay) = ( psf) (7.455/ feet) (5 feet) = lb = kip Design wind pressure on leeward side = psf Wind load on interior top chord joint leeward side = (design wind pressure on leeward side) (purlin spacing) (bay) = psf feet 5 feet = lb = kip Wind load on exterior & ridge top chord joint leeward side = (design wind pressure on leeward side) (purlin spacing/) (bay) = ( psf) (7.455/ feet) (5 feet) = lb = kip See figure 4 for wind loading on the truss (for wind blowing from left to right)

30 igure.4: Wind loads on the truss for wind blowing from left to right Wind oad Calculation (wind blows from left to right): Mirror image of figure 4 because wind direction is change (previously left-to-right, now right-toleft) (previous windward side is now leeward side & previous leeward side is now windward side). Note: igure.5: Wind loads on the truss for wind blowing from right to left Truss loading for oad direction Dead load Vertically downward See figure. Wind load (left-to-right & Perpendicular to roof right-to-left) surface See figure.4 &.5 6 P a g e

31 Truss Analysis: Truss analysis means determination axial force (which may be either compression or tension) of every of its member/bar. Now, the truss is analyzed for dead load, wind load (left-to-right) & wind load (right-to-left) (this can be done manually or by using computer software) Manual analysis for truss: by using method of joint and/or method of section. Computer software for truss analysis: GRASP. ETABS, SAP, STAAD etc. Both manual & computer analysis will have to done & check the bar force value obtained by manual analysis with computer. 7 P a g e

32 Chart: Sample Design orce for truss members Member ength (ft) Dead oad (Kip) Member orce (Kip) Wind (eft-to- Right) (Kip)+ Wind (Right-to- eft) (Kip) Dead oad only (Kip) Dead oad + Wind (eft-to- Right) (Kip) Dead oad + Wind (Right-to- eft) (Kip) Design member forces (Kip) Tension (Kip) Compression (Kip) Top Chord Bottom Chord Verticals Diagonals/Web members 0U UU UU UU U4U U U U U U U U U U U P a g e

33 .9 Design of Truss Members.9. Design of Top Chord: rom the design chart for truss member or top chord (length, = feet), maximum compressive force = 5.7 kip & maximum tensile force = kip. Select an angle section 4 ; (cross sectional area, A =. inch & minimum radius of gyration, r z = 0.58 inch). Check for Compression: Slenderness ratio, K r = E = modulus of elasticity of steel = psi = 9000 ksi y = yield stress of the steel = 6 ksi (for A 6 steel) E C π = c y π 90 = a = allowable stress in compression (ksi) a K/r y C c 5 K/r K/r 8 C 8 C c c if K r C c a if K r C c = ksi a Allowable force in compression, P = a A = ksi. inch a = kip. (which is greater than design compressive force 5.7 kip). 9 P a g e

34 The 4 section is OK for compressive force. Check for Compression: Slenderness ratio, less than 00). t = allowable stress in tension (ksi) K r = (which is 0.58 t = 0.6 y = ksi =.6 ksi Allowable force in tension, P = t than design tensile force kip). The A =.6 ksi. inch t = 8.96 kip. (which is greater 4 section is OK for tensile force. Top chord 4.9. Design of Bottom Chord: rom the design chart for truss member or bottom chord (length, = feet), maximum compressive force =.667 kip & maximum tensile force = kip. Select an angle section 4 gyration, r z = inch). 5 6 Check for Compression: Slenderness ratio, ; (cross sectional area, A =.09 inch & minimum radius of K r = E = modulus of elasticity of steel = psi = 9000 ksi = yield stress of the steel = 6 ksi (for A 6 steel) y E C π = c y π 90 = = allowable stress in compression (ksi) a 0 P a g e

35 a K/r y C c 5 K/r K/r 8 C 8 C c c if K C r c a a = ksi if K r C c Allowable force in compression, P a = a A = ksi.09 inch = kip. (which is greater than design compressive force.667 kip). The 4 OK for compressive force. Check for Compression: Slenderness ratio, less than 00). t = allowable stress in tension (ksi) K r 5 6 section is = (which is 0.58 t = 0.6 = ksi =.6 ksi y Allowable force in tension, P = t than design tensile force 8.47 kip). The A =.6 ksi.09 inch t = kip. (which is greater 5 4 section is OK for tensile force. 6 Bottom chord Design of Verticals: rom the design chart for truss member or verticals U (length, = 0 feet), maximum compressive force = 5. kip & maximum tensile force = kip. P a g e

36 Select an angle section radius of gyration, r z = inch). 6 Check for Compression: Slenderness ratio, ; (cross sectional area, A = 0.90 inch & minimum K r 0.60 = E = modulus of elasticity of steel = psi = 9000 ksi y = yield stress of the steel = 6 ksi (for A 6 steel) E C π = c y π 90 6 = a = allowable stress in compression (ksi) a π E K r K r if K r C c a π E K r K r C c if a = ksi Allowable force in compression, P a = a A = ksi 0.90 inch = kip. (which is greater than design compressive force 5. kip). The OK for compressive force. 6 section is Check for Compression: Slenderness ratio, less than 00). K r 0.60 = (which is = allowable stress in tension (ksi) t = t 0.6 = ksi =.6 ksi y P a g e

37 Allowable force in tension, P t = t than design tensile force 5.90 kip). The A =.6 ksi inch = kip. (which is greater 6 section is OK for tensile force. Verticals Design of Diagonals / Web Members: rom the design chart for truss member or diagonals U / U kip & maximum tensile force = + 5. kip. Select an angle section of gyration, r z = 0.5 inch). (length, = feet), maximum compressive force = 6 Check for Compression: Slenderness ratio, ; (cross sectional area, A = inch & minimum radius K = r 0.5 E = modulus of elasticity of steel = psi = 9000 ksi y = yield stress of the steel = 6 ksi (for A 6 steel) E C π = c y π 90 6 = a = allowable stress in compression (ksi) a π E K r K r if K r C c a π E K r K r C c if P a g e

38 a = ksi Allowable force in compression, P = a A = ksi inch = a kip. (which is greater than design compressive force 7.45 kip). The OK for compressive force. Check for Compression: Slenderness ratio, less than 00). t = allowable stress in tension (ksi) t = 0.6 y Allowable force in tension, = ksi =.6 ksi P t = t K r greater than design tensile force 5. kip). The 6 section is = (which is 0.5 A =.6 ksi inch =.56 kip. (which is 6 section is OK for tensile force. Check whether the selected 6 OK or not for the other diagonal members section for diagonal member U & U 5 4 U & U is 4 or diagonals U / U (length, = feet), maximum compressive force = kip & maximum tensile force = kip. Select an angle section 6 ; (cross sectional area, A = inch & minimum radius of gyration, r = 0.5 inch). z Check for Compression: Slenderness ratio, K = r 0.5 E = modulus of elasticity of steel = psi = 9000 ksi = yield stress of the steel = 6 ksi (for A 6 steel) y E C π = c y π 90 6 = P a g e

39 a = allowable stress in compression (ksi) a K/r y C c 5 K/r K/r 8 C 8 C c c if K r C c a a = ksi if K r C c Allowable force in compression, P a = a A = ksi inch = kip. (which is greater than design compressive force kip). The for compressive force. Check for Compression: Slenderness ratio, less than 00). t = allowable stress in tension (ksi) t = 0.6 = ksi =.6 ksi y Allowable force in tension, P t = t 6 section is OK K = (which is r 0.5 A =.6 ksi inch =.56 kip. (which is greater than design tensile force.004 kip). The 6 section is OK for tensile force. Diagonals 6 5 P a g e

40 Table.: Design summary for truss members Member type Top chord Bottom chord Verticals Web member / Diagonals Design section Design of Bracing Systems:.0. Vertical Bracing: The members of the vertical bracing will be tied to each other at their crossing point. Therefore, half of their length will be considered in determining slenderness ratio K r. We will assume that, effective length factor, K = The length of member of the vertical bracing () = 0 5 K r minimum 400 feet =.4690 feet =.4690 inch r minimum r minimum minimum r > inch rom AISC chart for angles we select inch. for which r = r = r = z minimum.0. Top Chord Bracing: Similar to the vertical bracing, the members of the top chord bracing will also be tied to each other at their crossing point. Therefore, half of their length will be considered in determining 6 P a g e

41 slenderness ratio K r. We will assume that, effective length factor, K = The length of member of the top chord bracing () = ( 7.45) 5 feet = feet = inch. K 400 r minimum r minimum r minimum minimum r > inch rom AISC chart for angles we select inch. 4 for which r = 4 4 r z = r = 0.4 minimum.0. Bottom Chord Bracing: If we consider the length of the struts equal to the bay distance, the K ratio criterion will r result too large section. To economize our design, we will use a lateral tie at the midspan of the struts very similar to the sagrods used for purlins (see figure below). or these lateral ties, we use steel rods same as the rods. The presence of the ties at the midspan will reduce the unsupported length of the struts by 50%. We will assume that, effective length factor, K = The length of member of the bottom chord bracing () = K r minimum r minimum r minimum 00 r minimum > 0.50 inch 5 feet =.5 feet =.5 inch. 7 P a g e

42 rom AISC chart for angles we select inch. 5 for which r = 6 r z = r = 0.90 minimum Ties for the bottom chord struts are arbitrarily chosen to be round steel bars of inch diameter. These will be connected to the bottom chord struts using standard inch nuts in a manner similar to the sagrods. Table.4: Design summary for bracing systems Bracing type Vertical bracing Top chord bracing Bottom chord strut Design section Design of Truss Joints (Welded Connections): There are two types of joints in the truss joints where all members ends (such as 0, U & 6) and joints where there are one continuous members (such as,,, 4, 5, U, U, U4 & U5). Gusset plate thickness (for a joint) = maximum thickness of the angle sections meeting at that joint + /8 inch. Weld length for a member = maximum tensile or compressive force at that member (kip) allowable weld shear (ksi) effective throat size (inch) =? inch Weld Design of Joint 0: Here, two members 0U ( ) & 0 ( ) meets. Gusset plate thickness (for joint 0) = maximum thickness of the angle sections ( 6 5 inch) meeting at that joint + 8 inch = 6 7 inch. 8 P a g e

43 igure.6: Joint 0 of roof truss Weld for 0: Consider, 0 ( ) & gusset plate ( 7 6 inch) 7 t inch and 5 t inch max 6 min 6 Maximum thickness of the part being connected, 7 t inch. So, Minimum fillet weld max 6 size, s inch (from table, chapter: welded connections) min 6 Minimum thickness of the part being connected, 5 t inch. So, Maximum fillet weld min 6 size, 5 4 s inch inch You can choose either max inch or inch. Choose inch fillet weld igure.7: Weld design for 0 9 P a g e

44 Electrode: E60XX (i.e., electrode material tensile strength (EXX) = 60 ksi). Allowable shear in weld ( v ) = 0. E60XX = ksi = 8 ksi. illet weld size chosen, s = 6 inch Effective throat size, ( t ) = s cos 45 0 = e 6 cos 45 0 inch. Weld length required for member 0U, = U 0 P maximum tensile or compressive v t e =.667 kip 8 ksi cos45 0 inch 6 U 0 =.688 inch + = U 0 =.688 inch Taking moment about ( ( t e 6 v, ) (4 // ) = (.667 kip) (.74 // ) cos 45 0 inch 8 ksi) (4 // ) = (.667 kip) (.74 // ) = inch 9.50 inch Taking moment about ( t e, ) (4 v // ) = (8.099 kip) (.6 // ) ( cos 45 0 inch 8 ksi) (4 // ) = (.667 kip) (.6 // ) 6 = 4.84 inch 4.50 inch 40 P a g e

45 Minimum weld length, = 4 s = 4 minimum 6 = 0.75 inch Both & Alternatively, inch & > minimum + = 4.84 inch. ; OK =.688 inch & =.74 inch ; from which,.6 inch = Weld for 0U: Consider, 0U ( ) & gusset plate ( inch) 7 t inch and t inch max 6 min 4 7 Maximum thickness of the part being connected, t inch. So, Minimum fillet weld max 6 size, s inch (from table, chapter: welded connections) min 6 Minimum thickness of the part being connected, t inch. So, Maximum fillet weld min 4 size, s inch inch (from table, chapter: welded connections) max Use 6 inch fillet weld. igure.8: Weld design for 0U 4 P a g e

46 Electrode: E60XX (i.e., electrode material tensile strength (EXX) = 60 ksi). Allowable shear in weld ( v ) = 0. E60XX = ksi = 8 ksi. illet weld size chosen, s = 6 inch Effective throat size, ( t ) = s cos 45 0 = e 6 cos 45 0 inch. Weld length required for member 0U, = U 0 P maximum tensile or compressive v t e = kip 8 ksi cos45 0 inch U = inch = Taking moment about ( ( t e 6 U 0 v, = inch ) (.5 // ) = (8.099 kip) (.46 // ) cos 45 0 inch 8 ksi) (.5 // ) = (8.099 kip) (.46 // ) = inch 5.50 inch Taking moment about, ( t e v ) (.5 // ) = (8.099 kip) (.04 // ) ( cos 45 0 inch 8 ksi) (.5 // ) = (8.099 kip) (.04 // ) 6 =.558 inch.50 inch Minimum weld length, Both & > minimum minimum ; OK = 4 s = 4 6 = 0.75 inch 4 P a g e

47 Alternatively, inch & + =.558 inch. = inch & =.46 inch ; from which,.04 inch = igure.9: Weld design of joint 0 Weld Design of Joint U: Here, two members 0U (.5 ), U (.5.5 /6) & U (.5 /6) meets. 4 Gusset plate thickness (for joint 0) = maximum thickness of the angle sections ( 4 inch) meeting at that joint + 8 inch = 6 6 inches 4 P a g e

48 igure.0: Weld design of joint U Weld for 0UU: Consider, 0UU ( ) & gusset plate ( inch) 6 t inch and t inch max 6 min 4 6 Maximum thickness of the part being connected, t inch. So, Minimum fillet weld max 6 size, s inch (from table, chapter: welded connections) min 6 Minimum thickness of the part being connected, t inch. So, Maximum fillet weld min 4 size, s inch inch (from table, chapter: welded connections) max Use 6 inch fillet weld. We are designing top chord as a continuous member. The length of weld required to hold the bottom chord with the gusset plate at joint depends on the resultant (absolute value) of the axial forces in members 0U and UU. We have to consider three possible equilibrium conditions to determine the resultant force for design. These three equilibrium conditions are () Dead load only, () D + Wind ( R) and () D + Wind (R ). The process of finding the resultant for design is shown in tabular form below 44 P a g e

49 Equilibrium condition 0U (member force, kip) U 0 UU (member force, kip) U U D D + W ( R) D + W (R ) Magnitude of the resultant, kip U 0 U U 5.7 (.587) = ( ) = (+.5) = Observing the last column, we find that the design force is kip. This force will be used to determine the weld length required to hold the top chord member with the gusset plate at joint U. igure.: Weld design of member 0U and UU Electrode: E60XX (i.e., electrode material tensile strength (EXX) = 60 ksi). Allowable shear in weld ( ) = 0. = ksi = 8 ksi. v E60XX illet weld size chosen, s = inch 6 Effective throat size, ( t ) = s cos 45 0 = cos 45 0 inch. e 6 Weld length required for member 0UU, 45 P a g e

50 = U U U U = Taking moment about ( t e P maximum tensile or compressive v = inch U U 0 v, t e = inch ) (.5 // ) = (5.984 kip) (.46 // ) = kip 8 ksi cos45 0 inch 6 ( 6 cos 45 0 inch 8 ksi) (.5 // ) = (5.984 kip) (.46 // ) =.768 inch inch Taking moment about ( ( t e 6 v, ) (.5 // ) = (5.984 kip) (.04 // ) cos 45 0 inch 8 ksi) (.5 // ) = (5.984 kip) (.04 // ) = inch inch Minimum weld length, Both & Alternatively, inch & > minimum ; OK minimum + = inch. Weld for U: Consider, U ( = inch & t inch and t inch max 6 min = 4 s = 4 6 = 0.75 inch = ) & gusset plate ( 6 6 inch).46 inch ; from which,.04 inch = P a g e

51 6 Maximum thickness of the part being connected, t inch. So, Minimum fillet weld max 6 size, s inch (from table, chapter: welded connections) min 6 Minimum thickness of the part being connected, t inch. So, Maximum fillet weld min 6 size, s inch (from table, chapter: welded connections) max 6 Use 6 inch fillet weld. igure.: Weld design of member U Electrode: E60XX (i.e., electrode material tensile strength (EXX) = 60 ksi). Allowable shear in weld ( ) = 0. = ksi = 8 ksi. v E60XX illet weld size chosen, s = 6 inch Effective throat size, ( t ) = s cos 45 0 = cos 45 0 inch. e 6 Weld length required for member U, = U P maximum tensile or compressive v t e = 0.75 kip 8 ksi cos inch 47 P a g e

52 U + = 0.5 inch = U Taking moment about ( ( t e 6 v = 0.5 inch, ) (.5 // ) = (0.75 kip) (.76 // ) cos 45 0 inch 8 ksi) (.5 // ) = (0.75 kip) (.76 // ) = 0.08 inch 0.50 inch Taking moment about ( ( t e 6 v, ) (.5 // ) = (0.75 kip) (0.74 // ) cos 45 0 inch 8 ksi) (.5 // ) = (0.75 kip) (0.74 // ) = inch 0.50 inch Minimum weld length, Both & Alternatively, inch & > minimum ; OK minimum + = inch. = 0.5 inch & = 4 s = 4 6 = 0.75 inch =.46 inch ; from which,.04 inch = 0.08 Weld for U: Consider, U ( 6 t inch and t inch max 6 min 6 ) & gusset plate ( inch) Maximum thickness of the part being connected, size, s min 6 inch 6 t inch. So, Minimum fillet weld max 6 48 P a g e

53 Minimum thickness of the part being connected, size, s max Use 6 6 inch inch fillet weld. t inch. So, Maximum fillet weld min 6 igure.: Weld design of member U Electrode: E60XX (i.e., electrode material tensile strength (EXX) = 60 ksi). Allowable shear in weld ( ) = 0. = ksi = 8 ksi. E60XX illet weld size chosen, s = Effective throat size, ( t e v 6 inch Weld length required for member U, = U U + ) = s cos 45 0 = 6 cos 45 0 inch. P maximum tensile or compressive v = inch = Taking moment about t e = inch U, =.00 kip 8 ksi cos45 0 inch 6 49 P a g e

54 ( ( t e 6 v ) ( // ) = (.00 kip) (. // ) cos 45 0 inch 8 ksi) ( // ) = (.00 kip) (. // ) = inch 4 inch Taking moment about ( ( t e 6 v, ) ( // ) = (.00 kip) (0.888 // ) cos 45 0 inch 8 ksi) ( // ) = (.00 kip) (0.888 // ) = inch.5 inch Minimum weld length, = 4 s = 4 minimum 6 = 0.75 inch Both & > minimum ; OK Alternatively, inch & + = inch. = inch& =. inch ; from which, inch = igure.4(a): Weld design of joint U 50 P a g e

55 igure.4(b): Weld design of joint U. Design of Anchorage and Support: The truss is supported by reinforced concrete columns and footings, their reactions having been calculated earlier for point dead load and wind loads. The connections between the truss and support are designed in this section for the combined design loads. Combination of Support Reactions from Dead oad and Wind oad: The calculation for the design support reactions is carried out in the following tabular form. Since the truss is supported on base plates on concrete pedestals supported by masonry columns, the design in this study deals mainly with the connections between the truss and the columns. The column forces are nominal, therefore a 0 X0 masonry column is chosen. The maximum tensile stress on the column = 9.9/(0*0) = 0.46 ksi, which is within the allowable limit (Tensile strength 00 psi). Assuming the base plate area = Ap and bearing pressure = 0.5 fc =.05 ksi.05ap = 4.7 Ap = 4.7/.05 5 P a g e

56 Ap = 4.0 in Provide 7 X4 base plate (since the bottom cord members are wide) Since the free portion of the base plate is nominal, a thickness of 0.5 is more than adequate. 0 U (.5"X"X/4") Gusset Plate(0.75" thickness) 0 (4"X"X5/6") ¾ Bolt 8" 7"X4" Base Plate (0.5" thickness) 6" Anchorage Concrete Pedestal Masonry Column Hinge Support 0 igure.5: Design of anchorage at hinge support 0 The base plate is supported on a 0 x0 concrete pedestal and connected to the column by four reinforcements to resist the entire tensile and shear force. Allowable tensile stress = 0.5 fy = 0 ksi and allowable shear stress = 0. fy = ksi Required area (based on tensile force) = 9.9/(4*0) = 0.66 in Required area (based on shear force) =.4/(4*) = 0.07 in Provide 4 #6 (i.e., ¾ diameter) anchor bolts (Area = 0.44 in each). Allowable tensile force per anchor = 0.44*0 = 8.8 kips Allowable bond force per unit length = 5 fc = lb/in =.9 k/in Development length = 8.8/.9 = 4.59 Provide anchorage of 6 for each bolt. 5 P a g e

57 The base plate will be connected to the gusset plate by the section similar to the bottom cord (i.e., a 4XX5/6 double angle section), also with ¾ diameter bolts to transfer the maximum support reaction (= 9.9 kips) by shear. Required area = 9.9/ =.44 in, i.e., provide -/4 diameter bolts in double shear. 6 U 5 (.5"X"X/4") Gusset Plate(0.75" thickness) 5 6 (4"X"X5/6") 6" Roller 8" Roller Support 6 igure.6 Design of anchorage at roller support 6 5 P a g e

58 Part : Design of Steel Plate Girder. Introduction When a member is required that is larger than that is available in rolled beams, it is necessary to build up a section which for the sake of economy has a general shape of an I-beam. This built up I-beam is called plate girder in which Section modulus is greater than any available rolled beam. The moment-resisting capacities of plate girders lie somewhere between those of deep standard rolled wide-flange shapes and those of trusses. Plate girders can be welded (ig. 0(a) to 0(d)), riveted, or bolted (ig. 0(e)). Riveted plate girders are practically obsolete. Very few bolted plate girders are designed nowadays. Therefore, in this manual design of welded plate girders has been covered. In this chapter, we consider large flexural members (girders) that are composed of plate elements in particular, with shapes built up from plates, however, both flanges and webs can be compact, noncompact, or slender. These built-up shapes usually are used when the bending moments are larger than standard hot-rolled shapes can resist, usually because of a large span. These girders are invariably very deep, resulting in noncompact or slender webs. Top lange Intermediate Stiffener Bottom lange Bearing Stiffener igure.: Plate girder in a multistory building 54 P a g e

59 Cover Plate ig..(a): Welded plate girder ig..(b): Plate girder with cover plate ig..(c): Built-up girder with T sections ig..(d): Welded box girder. Advantages and Disadvantages of Plate girder: ig..(e): Riveted or bolted plate girder Plate girders are used in both buildings and bridges. In buildings, when large column-free spaces are designed to be used as an assembly hall, for example, the plate girder is often the economical solution. In such cases, the designer must choose between a plate girder and a truss. Plate girders, in general, have the following advantages over trusses: Connections are less critical for plate girders than for trusses. abrication cost of plate girders is less than that of trusses. Plate girders can be erected more rapidly and more cheaply than trusses. Depth of a plate girder is less than the height of a comparable truss. Consequently, plate girders need less vertical clearance than trusses. This makes them very attractive for multilevel highway bridges. Plate girders generally vibrate less than trusses under moving loads. Painting of plate girders is easier than painting of trusses. This means less maintenance cost for plate girders. In contrast, plate girders in general are heavier than trusses, especially for very long spans. 55 P a g e

60 . Types of Plate Girder: There are different types of plate girder that are used in buildings and bridges. Box Girder: Providing improved torsional stiffness for long span bridges. Hybrid Girder: Providing variable material strength in accordance with stresses. In order to reduce the girder weight and possibly achieve maximum economy, hybrid plate girders are sometimes used. In a hybrid girder, flange plates are made of higher strength steel than that of the web Delta girder: Delta girder, may be used for more stability of the compression flange. High strength flange ow strength web Box girder Hybrid girder igure.: Types of plate girder ig. Types of plate girder Delta girder.4 Essential Elements of I-section Plate Girder: In a built-up I section, there are some elements that need to be designed. Top flange Bottom flange Web Intermediate stiffener Bearing stiffener Welding A A igure.4(a): Plate girder without stiffeners (a) 56 P a g e

61 Top and bottom flange plate: Plate girders basically carry the loads by bending. The bending moment is mostly carried by flange plates. Web: A web plate is needed to unify the two flange plates and to carry the shear. Intermediate Stiffener: In addition to flange plates and a web plate, a plate girder often consists of stiffeners. Thin web plates are susceptible to unstable behavior. Thick web plates make the girder unnecessarily heavy. A relatively thin web plate strengthened by stiffeners often yields the lightest plate girder. Therefore, intermediate stiffeners are provided to stiffen the web plate against buckling and to resist compressive forces transmitted from the web during tension-field action. Bearing Stiffener: Bearing stiffeners should always be provided in pairs at the ends of plate girders and if required at points of application of concentrated loads. These bearing stiffeners should extend roughly to the edges of the flange plates. B C EP ITS ITS ITS ITS ITS EP C B igure.4(b): Plate girder with intermediate and bearing stiffeners (b) D S S S S S S EP ITS ITS ITS ITS ITS EP D igure.4(c): Plate girder with intermediate, bearing and lateral stiffeners (c) 57 P a g e

62 S lange plate ITS ITS Web plate EP Section A-A Section B-B 4t w Section C-C 4t w Section D- D ig. orms of plate girder igure.4(d): orms of plate girder.5 Post-Buckling Behavior of the Web Plate: A relatively thin web plate strengthened by stiffeners often yields the lightest plate girder. Stiffened plate girders are designed on the basis of the ultimate strength concept. As the magnitude of the load on the girder is increased, the web panels between adjacent vertical stiffeners buckle due to diagonal compression resulting from shear. If the plate girder has properly designed stiffeners, the instability of the web plate panels, bounded on all sides by the transverse stiffeners of flanges, will not result in its failure. In fact, after the web panels buckle in shear, the plate girder behaves like the Pratt truss shown in ig..5(a). It will then be able to carry additional loads. A stiffened plate girder has considerable post-buckling strength. The Pratt truss of ig..5(a) is subjected to a concentrated load applied at its midspan. In this truss, the vertical members are in compression and the diagonals are in tension. The postbuckling behavior of the plate girder is analogous to the behavior of this truss. As shown in ig. (b), after the shear instability of the web takes place, a redistribution of stresses occurs; the stiffeners behave like axially compressed members, and shaded portions of the web behave like tension diagonals in the truss of ig. (a). This truss-like behavior is called tension-field action in the literature. The post-buckling strength of the web plate may be three or four times its initial buckling strength. Consequently, designs on the basis of tension-field action can yield better economy. A tension field ordinarily cannot be fully developed in an end panel. This can be understood by considering the horizontal components of the tension fields shown in igure.5(b). (The vertical components are resisted by the stiffeners.) The tension field in panel CD is balanced on the left side in part by the tension field in panel BC. Thus, interior panels are anchored by adjacent panels. Panel AB, however, has no such anchorage on the left side. Hence the anchorage for panel BC must be provided on the left side by a beam-shear panel rather than the tension-field panel shown. 58 P a g e

63 P Diagonal Member Top Chord Vertical Member Bottom Chord (a) Portion of web under tension Top flange A B C D P Intermediate stiffener Web Bottom flange (b) igure.5: Analogy between a truss and a stiffened plate girder AISC G. lists all of the conditions under which a tension field cannot be used: a. In end panels b. When. a h < or. a h < ( 60 h/t w ) (Each of these cases corresponds to k v = 5) c. When A w (A fc +A ft ) >.5 d. When h b fc or h b ft > 6 Where, A w =area of the web A fc =area ofthe compression flange A ft =area of the tension flange b fc =width ofthe compression flange b ft =width of the tension flange 59 P a g e

64 .6 Requirements for different components of the plate girder.6. Proportions of Plate Girders: Whether a girder web is noncompact or slender depends on h/t w, the width to thickness ratio of the web, where h is the depth ofthe web from inside face of flange to inside face of flange and t w is the web thickness. rom AISC B4 the web of a doubly symmetric I shaped section is noncompact if.76 E y < h t w 5.70 E y (.) and the web is slender if h t w > 5.70 E y (.) To prevent vertical buckling of the compression flange into the web, AISC. imposes an upper limit on the web slenderness. The limiting value of h/t w is a function of the aspect ratio, a/h, ofthe girder panels, which is the ratio of intermediate stiffener spacing to web depth or a h.5, ( h t w ) max =.0 E y or a h >.5, ( h t w ) max = 0.40E y Where, a is the clear distance between stiffeners. (.) (.4) t w a h Section- ront View Section- Side View igure.6: ront and side view of the plate girder.6. Requirement for lexural Strength The nominal flexural strength M n of a plate girder is based on one of the limit states of tension flange yielding, compression flange yielding or local buckling (B), or lateral torsional buckling (TB). The aspects related to flexural strength are discussed in AISC chapter. Tension lange Yielding AISC 5 gives the nominal flexural strength based on tension flange yielding as M n = y S xt (.5) where S xt =elastic section modulus referred to the tension side. 60 P a g e

65 Compression lange Yielding The compression flange nominal strength is given by, M n = R pg cr S xc (.6) Where, R pg =bending strength reduction factor cr =critical compressive flange stress, based on either yielding or local buckling S xc =elastic section modulus referred to the compression side The bending strength reduction factor is given by a w R pg = ( h c 5.7 E ).0, a w t w y where a w = h c t w b fc t fc 0 (.7) (.8) b fc =width of the compression flange t fc =thickness ofthe compression flange The critical compression flange stress cr depends on whether the flange is compact, noncompact, or slender. The AISC Specification uses the generic notation λ, λ p, and λ r to define the flange width to thickness ratio and its limits. rom AISC Table B4.b, λ = b f t f (.9) λ p = 0.8 E y (.0) λ r = 0.95 k ce k c = 4 h/t w but (0.5 k c 0.76) (.) (.) = 0.7 y for girders with slender webs. If λ λ p, the flange is compact. The limit state ofyielding will control and cr = y, resulting in M n = R pg y S xc 6 P a g e

66 If λ p < λ λ r, the flange is noncompact. Inelastic B will control and cr = y 0. y ( λ λ p λ r λ p ) (.) If λ > λ r, the flange is slender, elastic B will control and cr = 0.9Ek c ( b f t f ) (.4) ateral Torsional Buckling Whether lateral-torsional buckling will occur depends on the amount of lateral support i.e. unbraced length b. If the unbraced length is small enough, yielding or flange local buckling will occur before lateral-torsional buckling. The length parameters are p and r, where p =.r t E y (.5) r = πr t E 0.7 y (.6) Here, r t =radius of gyration about the weak axis for a portion of the cross section consisting of the compression flange and one third of the compressed part of the web. or a doubly symmetric girder, this dimension will be one sixth of the web depth. If b p, there is no lateral torsional buckling. If p < b r, ailure will be by inelastic TB, and (.7) cr = C b y 0. y ( b p r p ) y If b > r, failure will be by elastic TB, and cr = C bπ E ( b r t ) y (.8) C b =factor to account for nonuniform bending within the unbraced length b.if the moment within the unbraced length b is uniform (constant), there is no moment gradient and C b =.0.6. Requirement for Shear Strength: The AISC Specification covers shear strength in Chapter G. In that coverage the constants k v and C v are used. AISC defines k v, which is a plate buckling coefficient, 6 P a g e

67 k v = (a/h) (.9) = 5 if a h > = 5 if a h > ( 60 h/t w ) = 5 in unstiffened webs with h t w < 60 or C v, which can be defined as the ratio of the critical web shear stress to the web shear yield stress, If h t w.0 k ve y, C v =.0 If.0 k ve y < h t w.7 k ve y, C v =.0 k v E/ y h/t w If h t w >.7 k ve y, C v =.5k ve (h/t w ) y (.0) (.) (.) Whether the shear strength is based on web shear yielding or web shear buckling depends on the web width to thickness ratio h/t w. If h.0 k ve t w y the strength is based on shear yielding, and V n = 0.6 y A w (.) where A w =cross-sectional area of the web. If h >.0 k ve, t w y the strength will be based on shear buckling or shear buckling plus tension field action. If tension field behavior exists, AISC Equation G can also be written as C v V n = 0.6 y A w C v y A w.5 + (a/h) (.4) 6 P a g e

68 The first term in the equation gives the web shear buckling strength and the second term gives the post buckling strength. If there is no tension field action, the second term is omitted, resulting in V n = 0.6 y A w C v (.5) Solution of AISC Equations G- (without tension field) and G- (with tension field) is facilitated by curves given in Part of the Manual. Tables -6a and -6b present curves that relate the variables of these two equations for steel with a yield stress of 6 ksi and Tables - 7a and -7b do the same for steels with a yield stress of 50 ksi..6.4 Requirements for Intermediate Stiffener Without a Tension ield: The requirements for stiffeners when a tension field is not present are given in AISC G.. The required moment of inertia of a pair of stiffeners about an axis through the web is I st bt w j (.6) Where, j = (a/h) (.7) b =smaller of a and h. With a Tension ield The requirements for stiffeners where tension field action is used are given in AISC G.. The first requirement is for the proportions of the stiffener. ( b t ) st 0.56 E yst Where, (.8) ( b t ) st =width to thickness ratio of the stiffener cross section yst =yield stress ofthe stiffener The second requirement is for the moment of inertia of the stiffener or pair of stiffeners. I st I st + (I st I st ) V r V c V c V c Where, (.9) I st =required moment of inertia as calculated for the no tension field case 64 P a g e

69 I st = moment of inertia required to develop the buckling plus post buckling shear strength = h4. p st ( yw 40 E ).5 (.0) p st = max ( yw yst, ) yw =yield stress ofthe girder web V r = the larger of the required shear strengths (V u for RD, V a for ASD) on each side of the stiffener; that is, in the adjacent web panels V c = the smaller of the available shear strengths ( (φ v V n for RD, V n /Ω v for ASD) in the adjacent panels, calculated with no tension field action V c = the smaller of the available shear strengths ( (φ v V n for RD, V/Ω v for ASD) in the adjacent panels, calculated with tension field action.6.5 Requirements for Bearing Stiffener Although the web can be proportioned to directly resist any applied concentrated loads, bearing stiffeners are usually provided. If stiffeners are used to resist the full concentrated load, the limit states of web yielding, web crippling, and side-sway web buckling do not need to be checked. The nominal bearing strength of a stiffener is given in AISC J7 as, R n =.8 y A pb (.) or RD, the resistance factor is φ = or ASD, the safety factor is Ω =.00. ull depth stiffeners should be used in pairs and analyzed as axially loaded columns subject to the following guidelines: The cross section of the axially loaded member consists of the stiffener plates and a length of the web This length can be no greater than times the web thickness for an end stiffener or 5 times the web thickness for an interior stiffener. The effective length should be taken as 0.75 times the actual length that is, K = 0.75h. The nominal axial strength is based on the provisions of AISC J4.4, Strength of Elements in Compression which are as follows: or K r 5 (.) P n = y A g This is the squash load for the stiffener that causes compression yielding with no buckling. or RD, the resistance factor for this limit state is φ = 0.90, for ASD, the safety factor is Ω =.67. or K > 5, the usual requirements for compression members apply r 65 P a g e

70 The weld connecting the stiffener to the web should have the capacity to transfer the unbalanced force. Conservatively, the weld can be designed to carry the entire concentrated load. If the stiffener bears on the compression flange, it need not be welded to the flange. Although no width to thickness ratio limit is given in the Specification for bearing stiffeners, the requirement of AISC Equation G- for intermediate stiffeners can be used as a guide in proportioning bearing stiffeners:.7 Design Procedure ( b t ) st 0.56 E yst The primary task in plate girder design is to determine the size of the web and the flanges. If a variable moment of inertia is desired, decisions must be made regarding the method of varying the flange size that is, whether to use cover plates or different thicknesses of flange plate at different points along the length of the girder. A decision about whether to use intermediate stiffeners must be made early in the process because it will affect the web thickness. If bearing stiffeners are needed, they must be designed. inally, the various components must be connected by properly designed welds. The following step-by-step procedure is recommended.. Select the overall depth. The optimum depth of a girder depends on the situation. Some authors recommend an overall depth of to of the span length. Others suggest a range of 0 give procedures for determining the depth that incorporate the required moment 6 to 0 strength and a specified h ratio. As with any beam design, constraints on the maximum depth t w could establish the depth by default.. Select a trial web size. The web depth can be estimated by subtracting twice the flange thickness from the overall depth selected. Of course, at this stage of the design, the flange thickness must also be estimated. The web thickness t w can then be found by using the following limitations as a guide. Once h and t w have been selected, determine whether the web width to thickness ratio qualifies this member as a slender-web flexural member. If so, the provisions of AISC 5 can be used. (If the web is noncompact, AISC 5 can still be used, but it will be conservative). Estimate the flange size: The required flange area can be estimated from a simple formula derived as follows. A f = M nreq t wh hr pg cr 6 If we assume that R pg =.0 and cr = y, the required area of one flange is A f = M nreq h y A w 6 (.) 66 P a g e

71 Where, M nreq =Required nominal flexural strength = M U /φ b for RD = Ω b M a for ASD A w = web area Once the required flange area has been determined, select the width and thickness. If the thickness originally used in the estimate of the web depth is retained, no adjustment in the web depth will be needed. Otherwise changes have to be made regarding previously selected components. At this point, an estimated girder weight can be computed, then M nreq should be recomputed. 4. Check the bending strength of the trial section. 5. Determine intermediate stiffener spacings and check the shear strength of the trial section. The design curves in Part of the AISC Manual can be used for this purpose or AISC Equation G- and G- 6. Design intermediate stiffeners. If there is not a tension field, the intermediate stiffeners must be proportioned to satisfy the moment of inertia requirement of AISC Equation G-7. If there is a tension field, the width to thickness ratio limit of AISC Equation G- and the moment of inertia requirement of AISC Equation G-4 must be satisfied. 7. Design bearing stiffeners. To determine whether bearing stiffeners are needed, check the web resistance to concentrated loads (web yielding, web crippling, and web sidesway buckling). Alternatively, bearing stiffeners can be provided to fully resist the concentrated loads, and the web need not be checked. If bearing stiffeners are used, the following design procedure can be used. a. Try a width that brings the edge ofthe stiffener near the edge of the flange and a thickness that satisfies AISC Equation G-: ( b t ) st 0.56 E yst b. Compute the cross sectional area needed for bearing strength. Compare this area with the trial area and revise if necessary. c. Check the stiffener-web assembly as a compression member. 8. Design the flange-to-web welds, stiffener-to-web welds, and any other connections (flange segments, web splices, etc.) 67 P a g e

72 .8 Design Example Problem Description 0 ft 0 ft ig.7: Beam ayout Design the internal plate girder where, Thickness of the slab = 5 in maximum permissible depth of beam = 65 in Concentrated load, P D = 78 kip & P = 58 kip are applied at the midpoint Steel used = A6 ( y = 6 ksi) Electrode: E70XX (for welding) Assume that the girder has lateral support at the ends and at the point of application of concentrated load. The girder is restrained against rotation at these points. Solution: The tributary are for the internal beam, A = 0.5 (60+40)(0) = 000 ft 60 ft Assume floor finish = 40 psf and ive load on slab= 75 psf Service dead load, w D = (50 5/+40) ( 000 ) ( ) =.70 k/ft Service live load, w = 75 ( 000 ) ( ) =.5 k/ft P D = 78 kip P = 58 kip w D =.70 kip/ft w =.5 kip/ft 0' 0' igure.8: Plate girder to be designed under service loads 68 P a g e

73 actored loads =. Dead oad +.6 ive oad Total concentrated load = = 86.4 kip Total uniform load = = 4.04 kip/ft 86.4 kip 4.04 kip/ft 0' 0' igure.9: actored oads (excluding girder self-weight) Selection of Web Plate: Determine the overall depth..5 in Span ength 0 Span ength 0 = 60() 0 = 7 in. = 60() = 60 in. Use the maximum permissible depth of 65 inches. Try a flange thickness of t f = 5 inches and a web depth of h 65 in h = 65 (.5) = 6 in. To determine the web thickness, first the limiting values of h/t w have to be examined rom Equations. and.4: or a h.5, ( h ) max =.0 E =.0 9,000 = 40.6 t w y 6 min t w = or a h >.5, = 0.8 in. ( h ) max = 0.40E = 0.40(9,000) =. t w y 6 igure.0: Cross-section of the plate girder 69 P a g e

74 min t w = 6. = 0.9 in. Try a web plate. Determine whether the web is slender h = 6 t w 5/6 = E y = ,000 6 = 6.8 < 98.4 The web is slender. So, the AISC provisions for plate girder can be used. Selection of lange Size Determine the required flange size. rom ig..9, the maximum factored load bending moment is M u = 86.4(60) (60) 8 = 464 ft-kips The required flange area is, A f = M nreq h y = M u/φ b h y A w 6 A w 6 = (464 )/0.90 6(6) 6(5/6) 6 = 4. in If the original estimate of the flange thickness is retained, the required width is b f = A f = 4. = 6. in. t f.5 Try a 8 flange plate. The girder weight can now be computed. Web area: 6(5/6) = 9.8 in. lange area: (5 8) = in. Total: 7.8 in Weight = =(490) = 50 lb/ft (Self-weight of structural steel = 490 lb/ft ) The adjusted bending moment is 70 P a g e

75 M u = (. 0.50)(60) 8 = 4749 ft-kips igure. shows the trial section, and igure. shows the shear and bending moment diagrams for the factored loads, which include the girder weight of 50 lb/ft..5 in 6 in 5/6 in 65 in 8 in igure.: Cross-section of the plate girder 86.4 kip 4.4 kip/ft 0' 0'.4 k 9. k 4749 k-ft Check the flexural strength of the trial section. rom igure., the moment of inertia about the axis of bending is 7 P a g e igure.: SD and BMD for the recomputed factored l x = (5/6)(6) and the elastic section modulus is S X = I x c = 60,640 = 866 in.5 + (.5)(8)(.75) = 60,640 in 4

76 An examination of AISC Equations 5-7 and 5-0 shows that for a girder with a symmetrical cross section, the flexural strength will never be controlled by tension flange yielding. Check for compression flange buckling: Determine whether the compression flange is compact, noncompact, or slender. (use Equations.9 to.) λ = b f t f = 8 (5) = 6.0 λ p = 0.8 E y = 0.s 9,000 6 = 0.79 Since λ < λ p, there is no flange local buckling. The compression flange strength is therefore based on yielding, and cr = y = 6 ksi To compute the plate girder strength reduction factor R pg, the value of a w will be needed: a w = h ct w = 6(5/6) = < 0 b fc t fc 8(.5) rom Equation-7 a w R pg = ( h c 5.7 E ).0, a w t w y = (0.776) ( ).0 = rom AISC Equation 5-7, the nominal flexural strength for the compression flange is M n = R pg cr S xc = 0.984(6)(866) = 65,90 in. kips = 5494ft kips and the design strength is φ b M n = 0.90(5494) = 4945 ft-kips > 4749 ft kips (OK) Although this capacity is somewhat more than needed, the excess will compensate for the weight of stiffeners and other incidentals that have not yet been accounted for. Check for lateral torsional buckling: I y = (.5)(8) + (6/6)( 5 6 ) = 79.0 in 4.5 in 6/6 in 8" 5/6 in A = 8(.5) + 0. ( 5 ) = 0. in 6 7 P a g e igure.: Zone corresponding to lateral torsional buckling

77 r t = I y = A 79.0 = 4.9 in 0. Check the unbraced length. b = 0 ft p =.r t E y =.(4.9) 9,000 6 = 5.9 in. =.77 ft r = πr t E = π(4.9) 9,000 = 5.8 in. = 4.6 ft 0.7 y 0.7(6) Since p < b < r, the girder is subject to inelastic lateral torsional buckling. rom Equation- 7, cr = C b y 0. y ( b p r p ) y or the computation of C b, refer to igure., which shows the loading, shear, and bending moment diagrams based on factored loads. The unsupported segment is divided into four equal spaces, we get points A, B, and C located at 7.5 ft, 5 ft and. ft from the left end of the girder. The corresponding bending moments are M A =.4(7.5) 4.4(7.5) / = 55.4 ft-kips M B =.4(5) 4.4(5) / = ft-kips M C =.4(.5) 4.4(.5) / = ft-kips rom AISC Equation -l,.5m max C b =.5M max + M A + 4M B + M C =.5(4749).5(4749) + (55.4) + 4(86.75) + (97.94) =.5 cr = C b y 0. y ( b p r p ) y =.5 6 (0. 6) = ksi Since ksi > y = 6 ksi, use cr = 6 ksi (same as for the other limit states). The nominal flexural strength is,φ b M n = 4945 ft-kips > 4749 ft-kips (OK) Use a web and 8 flanges, as shown in ig.4 7 P a g e

78 Select intermediate stiffener spacing and check the corresponding shear strength The shear is maximum at the support, but tension field action cannot be used in an end panel. Table -6a in Part of the AISC Manual can used to obtain the required size of the end panel. The curves will be entered with values of h/t w and the required φ v V n /A w, Where, h = 98.4 t w A w = 6( 5 ) = 9.8 in. 6 Required φ vv n A w = V u =.4 =.5 ksi A w 9.8 Using h/t w = 98 and φ v V n /A w = ksi, we get a value of a/h of approximately The corresponding intermediate stiffener spacing is a = 0.60h = 0.60(6) = 7. in. Although the required stiffener spacing is a clear distance, the use of center to center distances is somewhat simpler and will be slightly conservative. In addition, because of the approximations involved in using the curves, we will be conservative in rounding the value of a. Use a distance of 6 inches from the center of the end bearing stiffener to the center of the first intermediate stiffener. Now, a = 6 in a h = 6 6 = 0.58 rom Equation-9, k v = (a/h) = 9.86 h t w = k ve = (9000) = 9. y 6.7 k ve = (9000) =7.8 y 6 h t w >.7 k ve y So, from Equation-, C v =.5k ve (h/t w ) y =0.6 As no tension field action occurs at the end panel, from equation-5 V n = 0.6 y A w C v = 0.6 (6) (9.75) (0.6) =55.85 kip 74 P a g e

79 φv n = 0.9 (55.85) = 9.76 kip >.4 kip (OK) Determine the intermediate stiffener spacings needed for shear strength outside the end panels. At a distance of 6 inches from the left end, the shear is V u = ( 6 ) = 0.4 kips Required φ vv n = 0.4 = 0.86 ksi A w 9.8 Tension field action can be used outside the end panels, so the curves of AISC Table -6b will be used. or h/t w = 98 and φ v V n /A w = ksi, a h =.60 The required stiffener spacing is, a =.60h =.60(6) = 99. in. This maximum spacing will apply for the remaining distance to the centerline of the girder. This distance is 0() 6 = 4 in. The number of spaces required is, =.7 Use four spaces. This results in a spacing of, 4 = 8 in Before proceeding, check the conditions of AISC G. to be sure that tension field action can be used for this girder and this stiffener spacing. a. a = 8 =.06 < (OK) h 6 b. a h < ( 60 h/t w ) = ( ) =.77 (OK) (Conditions a and b are automatically satisfied by staying within the boundaries defined by the upper curve and the lower solid curve of Manual Table -6b.) c. A w = (9.8) = <.5 (OK) (A fc +A ft ) (7+7) d. h = h = 6 =.444 < 6 (OK) b fc b ft 8 The following spacings will be used from each end of the girder: one at 6 inches and four at 8 inches, as shown in igure P a g e

80 6 in 4 Size of Intermediate Stiffeners: igure.4: Spacing of Intermediate stiffeners The first stiffener, placed at 6 inches, defines the left boundary of the first tension field panel. This stiffener must therefore be proportioned to satisfy the requirements of AISC G.. To determine a trial width for all stiffeners, consider the available space. The maximum possible width is b f t w = 8 5/6 = 8.84 in. Try b = 4in. Using.8, minimum required thickness: ( b t ) st 0.56 E yst 4 t ,000 6 t 0.5 in. rom.9 the required moment of inertia l st l st + (I st I st ) V r V c V c V c Now, j =.5 (a/h) 0.5 =.5 = 5.45 > 0.5 (6/6) rom.6, The required moment of inertia calculated for the no tension field case, 76 P a g e

81 I st = bt w j = min (a, h)t w j = min (6,6)(5/6) (5.45) = 6 (5/6) (5.45) = in 4 I st = moment of inertia required to develop the buckling plus post buckling shear strength = h4. ρ st 40 ( yw E ).5 p st = max ( yw yst, ) = max ( 6 6, ) = I st = (6)4 (). 40 ( 6 9,000 ).5 = 6.6 in 4 To the left of this stiffener, the stiffener spacing is 6 inches, and to the right, the spacing is 8 in. The longer panel will have the smaller strength for both the tension field and the no tension field cases defined by V c and V c. rom AISC Manual Table -6a (no tension field action), for h/t w = 98.4 and a/h = 8/6 =.06, φ v V n A w = 5 ksi (by interpolation) or A w = ht w = 6(5/6) = 9.8, φ v V n = V c = 5A w = 5(9.8) = 96.9 kips rom Table -6b (tension field action), for h/t w = 98.4 and a/h =.06, φ v V n A w =. ksi (by interpolation) φ v V n = V c =.A w =.(9.8) = 8 kips rom igure., the larger required strength in the two adjacent panels is V r = V u =.4 kips. rom AISC Equation G-4, I st I st + (l st I st ) V r V c V c V c = ( ) Try two /8 4 plates = 5. in4 rom igure.5 and the parallel-axis theorem, I st = Σ(I Ad ) 77 P a g e

82 = (/8)(4) +(/8)(4)( + 5/) stiffeners = 7.9 in 4 > 5. in 4 (OK) We will use this size for all of the intermediate stiffeners. To determine the length of the stiffeners, first compute the distance between the stiffener to web weld and the web to flange weld (see igure.5): Minimum distance = 4t w = 4( 5 ) =.5 in. 6 Maximum distance = 6t w = 6( 5 ) =.875 in. 6 5/ in Web /8 in 4t w or 6t w If we assume a flange to web weld size o 5/6 in. and.5 in. between welds, the approximate length of the stiffener is h weld size.5 = = in. Use 60 in. Use two plates for the intermediate stiffeners. Size of Bearing Stiffeners Bearing stiffeners will be provided at the supports and at midspan. Since there will be a stiffener at each concentrated load, there is no need to investigate the resistance of the web to these loads. If the stiffeners were not provided, the web would need to be protected by providing sufficient bearing length, l b, as required by AISC Equations J0- through J0-7. Try a stiffener width, b of 8 inches. The total combined width will be 78 P a g e Plan View Elevation View igure.5: Plan and Elevation View of Intermediate Stiffeners

83 (8) + 5/6 (the web thickness) = 6. inches, or slightly less than the flange width of 8 inches. rom Equation-8 ( b t ) s E yst t b 0.56 yst = 8 6 = 0.50 in. E ,000 Try two /4 8 stiffeners. Assume a /6 in. web to flange weld and a / in. cut-out in the stiffener. Check the stiffener at the support. 8 in 0.75 in t w Section in 8 in 0.75 in The bearing strength is R n =.8 y A pb =.8(6)(0.75)(8 0.5) = 79.0 kips φr n = 0.75(79.0) = 547 kips >.4 kips (OK) Check the stiffener as a column. Section - igure.6: Bearing Stiffener at the Support The length of web acting with the stiffener plates to form a compression member is times the web thickness for an end stiffener (AISC J0.8). As shown in igure.6, this length is (5/6) =.75 in. If we locate the stiffener centrally within this length, the point of support 79 P a g e

84 (location of the girder reaction) will be approximately.75 =.875 inches from the end of the girder. Use inches, as shown in igure.6, but base the computations on a total length of web of.75 inches, which gives A = (8) ( ) 4 + ( 5 ) (.75) =.7 in. 6 I =.75(5/6) (8) + 8 ( 4 ) (4 + 5 ) = 7. in. 4 in 6 in igure.7: Side view of the plate girder r = I = A 7. = 4.59 in..7 K r = Kh r = 0.7.5(6) = 0.4 < P n = y A g = 6(.7) = 474. kips φp n = 0.90(474.) = 47 kips >.4 kips (OK) Since the load at midspan is smaller than the reaction, use the same stiffener at midspan. So, for bearing stiffeners use two plates of /4 8 5 Connection Design At this point, all components of the girder have been sized. The connections of these elements will now be designed. E70 electrodes, with a design strength of.9 kips/inch per sixteenth of an inch in size, will be used for all welds. lange to web welds or the flange to web welds, compute the horizontal shear flow at the flange to web junction: Maximum V u =.4 kips 80 P a g e

85 Q =flange area y (see igure.) =.5(8)(.75) = 857. in. I x = 60,640 in. 4 Maximum V uq =.4(857.) =.58 kips/in. I χ 60,640 or the plate thicknesses being welded, the minimum weld size, w is /6 in. If intermittent welds are used, their minimum length is min = 4 w.5 in. = 4 ( ) = 0.75 in. Use.5 in. 6 Try /6-in. -in. fillet welds: Weld shear strength: φr n = 0.75(0.707w nw ) (.4) Where, nw in a fillet weld is 0.6 times the tensile strength of the weld metal Weld shear strength per sixteenth of an inch: φr n = 0.75(0.707w nw ) = 0.75 (0.707) (/6) (0.6 70) =.9 kips/inch Capacity per inch =.9 sixteenths welds = 8.5 kips/in. Check the capacity of the base metal (rom AISC manual equations J4- and J4-4). The web is the thinner of the connected parts and controls. The shear yield design strength per unit length is φr n =.0(0.6 y t) = 0.6 y t for a one inch length (.5) φr n = 0.6 y t = 0.6(6)( 5 6 ) = 6.750kips/in. The base metal shear rupture strength per unit length is φr n = 0.45 u t (.6) = 0.45(58)( 5 ) = 8.56 kips/in. 6 The base metal shear strength is therefore kips/in. < 8.5 kips/in. Use a total weld capacity of kips/in. The capacity of a.5 in. length of a pair of welds is 6.750(.5) = 0. kips 8 P a g e

86 To determine the spacing, let 0. s = V uq I x where s is the center-to-center spacing of the welds in inches and s = 0. = 0. =. in. V u Q/I x.58 Using a center-to-center spacing of inches will give a clear spacing of.5 =.5 inches. The AISC Specification refers to spacing of intermittent fillet welds in Section and Section E6. d 0.75 E y t f, but no greater than in. Adapting these limits to the present case yields 0.75 E t f = ,000 (.5) =.9 in. > in. y 6 The maximum permissible clear spacing is therefore inches, and the required clear spacing of.5 inches is satisfactory. There is no minimum spacing given in the Specification, but the AISC publication, Detailing for Steel Construction, (AISC, 009) states that intermittent welds are more economical than continuous welds only if the center to center spacing is more than twice the length of the weld. In this example, the spacing is equal to twice the length, so either type could be used. Although the -inch center to center spacing can be used for the entire length of the girder, an increased spacing can be used where the shear is less than the maximum of.4 kips. Three different spacings will be investigated:. The closest required spacing of inches.. The maximum permissible center to center spacing of +.5 =.5 in.. An intermediate spacing of 5 inches. The 5-inch spacing can be used when V u Q = 0. or V I x s u = 0.I x = 0.(60,640) Qs 857.(5) = 4. kips Refer to igure. and let x be the distance from the left support, giving V u =.4 4.4x = 4. kips x = 8.46 ft 8 P a g e

87 The.5-inch spacing can be used when V u = 0.I x Qs = 0.(60,640) 857.(.5) = 5.08 kips igure., shows that the shear never gets this small, so the maximum spacing never controls. Use /6-in. in. fillet welds for the flange to web welds, spaced as shown in igure.8. " 74 " = 8'-6" 8 5" ='-8" 0'-" or the intermediate stiffener welds: igure.8: Weld between lange and web Minimum weld size = 6 in. (based on the web thickness of 5 6 in.) Minimum length = 4( ) = 0.75 in. <.5 in. Use.5 in. 6 Use two welds per stiffener for a total of four. The capacity per inch for two /6 in. fillet welds per stiffener plate is.9 = 8.5 kips/in. Check the shear strength of the stiffener (the thinner of the two connected parts). rom Equation-5, the shear yield strength per unit length is φr n = 0.6 y t = 0.6(6)( ) = kips/in. 4 rom Equation-6, the base metal shear rupture strength per unit length is φr n = 0.45 u t = 045(58)( ) = 6.55 kips/in. 4 The base metal shear strength is therefore kips/in. per stiffener. This is less than the shear strength of two welds (using two welds for each plate), so use a weld strength of kips/in. or the two plates (four welds), use 8 P a g e

88 = 0.80 kips/in. Proportioning the intermediate stiffeners by the AISC rules does not require the computation of any forces, but a force must be transmitted from the stiffener to the web, and the connection should be designed for this force. Basler (96) recommends the use of a shear flow of f = 0.045h y E kips/in. (.7) f = 0.045h y E (6) = 0.045(6) =.59 kips/in. 9,000 Use intermittent welds. The capacity of a.5 in. length of the 4 welds is.5(0.80) =6.0 kips Equating the shear strength per inch and the required strength gives 6.0 s =.59 kips/in. or s = 4.58 in. rom AISC G., the maximum clear spacing is 6 times the web thickness but no greater than 0 inches, or 6 t w = 6( 5 6 ) = 5 in. Use a center to center spacing of 4 inches, resulting in a clear spacing of = in. < 5 in. (OK) Use /6-in. -in. fillet welds for intermediate stiffeners, spaced as shown in igure.9. /6-in. -in fillet 4.5 c/c Section igure.9: Weld between web and intermediate stiffener or the bearing stiffener welds: Minimum weld size = 6 in. (based on the thickness t w = 5 6 in.) Minimum length = 4( ) = 0.75 in. <.5 in. Use.5 in P a g e

89 Use two welds per stiffener for a total of four. The capacity per inch for two /6 inch fillet welds per stiffener plate is.9 = 8.5 kips/in. fillet 5.5 c/c Check the shear strength of the web. rom the design of the flange to web welds, the base metal shear strength is kips/in. per stiffener. This is less than the shear strength of two welds (using two welds for each plate), so use a weld strength of kips/in. or the two plates (four welds), use =.50 kips/in. The capacity of a.5 in. length of four welds is.5(.50) = 0.5 kips or the end bearing stiffener, the applied load per inch is Reaction =.4 =.66 kips/in ength avaiable for weld 6 (0.5) rom 0. 5/s =.66 s = 5.5 inches. (Note that a smaller weld spacing is required for the intermediate stiffeners) Use /6-in. -in. fillet welds for all bearing stiffeners, spaced as shown in igure.0. /6-in. -in Section igure.0: Weld between web and bearing stiffener 85 P a g e

90 References: American Institute of Steel Construction. 0a. Steel Construction Manual. 4th ed. Chicago. Old trails arch bridge [Online Image], Retrieved December 7, 07, from Segui, W.T., 0. Steel design. Cengage earning. Structural steel frame [Online Image], Retrieved December 8, 07, from Structural steel shapes [Online Image], Retrieved December 7, 07, from Trenton through truss bridge [Online Image], 008, Retrieved December 7, 07, from 86 P a g e

91 Appendix. AISC, Plate Sizes Note to tables I &II: the first length given obtainable from most, and usually from all, of the mills rolling the given width. The second given is the maximum obtainable from any mill, and such lengths are subject to substantial extras. or plates of large sizes, Designers should consult fabricators regarding possibilities of fabrication, shipment and erection. Table I: Plates (available sizes) ength, in feet, of Universal Mill plates obtainable in the respective widths shown Thickness, inches Width, Inches P a g e

92 AISC, Plate Sizes Table II: Plates (available sizes) ength, in feet, of Universal Mill plates obtainable in the respective widths shown Width, Inches Thickness, inches Source: American Institute of steel construction (AISC), 9 th edition, Page P a g e

93 American Standard Channel Section (C- Section) Properties x Y t f t w d ` X t f b f Designation Nominal weight per foot lb. Area A (in ) Depth d (in) lange width b f (in) Web thickness t w (in) lange thickness t f (in) x (in) I x (in 4 ) Axis X-X S x (in ) r x (in) I y (in 4 ) Axis Y-Y C C C C C C C Source: AISC Shape Database, 4 th edition S y (in ) r y (in) 89 P a g e

94 Designation Nominal weight per foot lb. Area A (in ) Depth d (in) lange width b f (in) Web thickness t w (in) lange thickness t f (in) x (in) Axis X-X Axis Y-Y C C C C C C C C C C C C C C C C C C C C C C C C C Source: AISC Shape Database, 4 th edition I x (in 4 ) S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) 90 P a g e

95 ong eg American Standard Angle Section (- Section) Properties Y X Short eg α Z 9 P a g e Designation Nominal weight per foot lb. Area A (in ) x (in) y (in) I x (in 4 ) Axis X-X Axis Y-Y Axis Z-Z 8 8 -/ / / / / / / / / / Source: AISC Shape Database, 4 th edition S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) r z (in) tan α

96 Designation Nominal weight per foot lb. Area A (in ) x (in) y (in) I x (in 4 ) Axis X-X Axis Y-Y Axis Z-Z 8 6 / / / / / / / / / / / / / / / / / / / / / / / / / Source: AISC Shape Database, 4 th edition S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) r z (in) tan α 9 P a g e

97 Designation Nominal weight per foot lb. Area A (in ) x (in) y (in) I x (in 4 ) Axis X-X Axis Y-Y Axis Z-Z 6 4 / / / / / / / / / 5/ / / / / / / / / / / 5/ / / / / / 5/ / / / / / / / / / Source: AISC Shape Database, 4 th edition S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) r z (in) tan α 9 P a g e

98 Designation Nominal weight per foot lb. Area A (in ) x (in) y (in) I x (in 4 ) Axis X-X Axis Y-Y Axis Z-Z 4 4 / / / / / / / / / / 5/ / / / / / / / / -/ / / -/ 7/ / -/ / / -/ 5/ / -/ / / / / 7/ / / / 5/ / / / -/ / / -/ / / -/ 5/ Source: AISC Shape Database, 4 th edition S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) r z (in) tan α 94 P a g e

99 Designation Nominal weight per foot lb. Area A (in ) x (in) y (in) I x (in 4 ) Axis X-X Axis Y-Y Axis Z-Z -/ -/ / / / / / / / / / / 7/ / / / 5/ / / / / / / / / / / -/ / / -/ / / -/ 5/ / -/ / / -/ / / / / 5/ / / / / Source: AISC Shape Database, 4 th edition S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) r z (in) tan α 95 P a g e

100 Designation Nominal weight per foot lb. Area A (in ) x (in) y (in) I x (in 4 ) Axis X-X Axis Y-Y Axis Z-Z -/ -/ / / -/ / / / / / / Source: AISC Shape Database, 4 th edition S x (in ) r x (in) I y (in 4 ) S y (in ) r y (in) r z (in) tan α 96 P a g e

101 Basic Wind Speeds for Selected ocations in Bangladesh (BNBC-99 Table 6..8) ocation Angarpota Bagerhat Bandarban Barguna Barisal Basic Wind Speed (km/h) ocation almonirhat Madaripur Magura Manikganj Meherpur Basic Wind Speed (km/h) Bhola Bogra Brahmanbaria Chandpur Chapai Nawabganj Maheshkhali Moulvibazar Munshiganj Mymensingh Naogaon Chittagong Chuadanga Comilla Cox s Bazar Dahagram Narail Narayanganj Narsinghdi Natore Netrokona Dhaka Dinajpur aridpur eni Gaibandha Nilphamari Noakhali Pabna Panchagarh Patuakhali Gazipur Gopalganj Habiganj Hatiya Ishurdi Pirojpur Rajbari Rajshahi Rangamati Rangpur Joypurhat Jamalpur Jessore Jhalakati Jhenaidah Satkhira Shariatpur Sherpur Sirajganj Srimangal Khagrachhari Khulna Kutubdia Kishoreganj Kurigram St. Martin s Island Sunamganj Sylhet Sandwip Tangail Kushtia akshmipur 5 6 Teknaf Thakurgaon P a g e

102 Structure Importance Coefficients, C I for Wind oads (BNBC-99 Table 6..9) Structure Importance Category Structure Importance Coefficient, C I Essential facilities.5 Hazardous facilities.5 Special occupancy structures.00 Standard occupancy structures.00 ow-risk structures 0.80 Exposure (Defined by BNBC-99) Exposure: This refers to the conditions of the terrain surrounding the building site. The terrain exposure in which a building or structure is to be sited shall be assessed as being one of the following categories: Exposure A: Urban and sub-urban areas, industrial area, wooded areas, hilly or other terrain covering at least 0% of the area with obstructions of 6 meters or more in height and extending from the site at least 500 meters or 0 times the height of the structure, whichever is greater. Exposure B: Open terrain with scattered obstructions having heights generally less than 0 meters extending 800 meters or more from the site in any full quadrant. This category includes air fields, open park lands, sparsely built-up outskirts of towns, flat open country and grasslands. Exposure C: lat and un-obstructed open terrain, coastal areas and riversides facing large bodies of water, over.5 km or more in width. Exposure C extends inland from the shoreline 400 meter or 0 times the height of the structure, whichever is greater. 98 P a g e

103 Combined Height and Exposure Coefficient, C z Height above ground level, z (meter) Exposure A Exposure B Exposure C P a g e

104 Gust Response actors, G h and G z BNBC-99, Table-6.. Height above ground level (metres) G h and G z Exposure A Exposure B Exposure C or main force wind force resistance system, use building or structure height h for z 00 P a g e

105 Overall Pressure Coefficient, C p for Rectangular Building with lat Roofs BNBC-99, Table-6..5 h/b /B Note:. These coefficients are to be used with Method- given in Sec a(ii). Use C p = ±0.7 for roof in all cases.. inear interpolation may be made for intermediate values of h/b and /B. External Pressure Coefficients, C pe for Roof BNBC-99, Sec Wind Direction Normal to ridge Parallel to ridge Windward Side eeward θ (degrees) h/ Side > 60 < * θ -0.7 for -0.9* all θ values θ of h/ > θ and θ h/b or h/.5 h/b or h/> P a g e

106 Minimum Size of illet Weld AISC, Table-J.4 Material Thickness of Thinner Part Joined, in. (mm) Minimum Size of ilet Weld, in. (mm) To /4 (6) inclusive /8 () Over /4 (6) to / () /6 (5) Over / () to /4 (9) /4 (6) Over /4 (9) 5/6 (8) Maximum Size of illet Weld Thickness along edge of the material, Maximum Size of ilet Weld, in. in. (mm) (mm) ess than /4 (6) Thickness of the material /4 (6) Thickness of the material-/6 Approximate Values of Effective ength actor, K AISC, Table-C-A-7. 0 P a g e

107 0 P a g e Available Shear Stress in Plate Girders AISC table, -6 to -7

108 04 P a g e

109 05 P a g e

110 06 P a g e