Steel Design Plastic Analysis Designing means to fabricate structures which will be able to support the given load conditions. These may be either steel structures or RCC structures. There are three methods of steel design: a) Working Stress method: The worst combination of working (service) loads is ascertained and the members are proportioned on the basis of working stress. The worst combination should not exceed the permissible stresses. The permissible stresses are generally some fraction (factor of safety) of the yield stress. b) Plastic Method: We find that steel can withstand higher loads than the elastic limit. In this method the reserve strength of the structure is used. This method is vased on failure conditions and not working stresses. This method extends the structural usefulness of the material upto the ultimate strength.
. In this method the loads are primarily supported by resistance to bending as deformation is small in case of bending. So continuous beams and rigid frames are analysed using this method. Gen not used for statically determinate frames or pin connected beams. c) Limit State method: this is based on limit strength and serviceability. That is the acceptable limit for the safety and serviceability before failure. The objective is to achieve a structure which will not become unfit for the period design like say 50 years.
Stress Strain curve of steel
Fig 2.1 shows the stress strain curve of annealed(i.e it has been given metallurgical treatment) mild steel specimen in tension. A is the limit of proportionality. B is the elastic limit. C represents the upper yield point at which there is a definite increase in strain without any further increase in stress. C’ is the lower yield point as the stress drops abruptly, then strain increases on constant stress up to D. Beyond D strain increases further and the material is said to be strain hardening at E. F is failure point. 1
The strain scale is enlarged considerably in Fig 2.1 (b), in order to study the yield range. fu and fL are the upper and lower yield point in fig 2.1(b). For the purpose of plastic design the elastic limit and the lower yield point may be assumed to be numerically identical. The yield strength is fixed at fy and the strain hardening region is ignored. Ignoring the strain hardening adds to the safety. In fig 2.1 (c) the stress strain behaviour is idealized and it is assumed that extension is unlimited at constant stress fy So during Bending of beams the following picture illustrates what happens assuming the idealised stress-strain curve:
Collapse takes place when plastic moment is reached at the highly stressed points. Shape factor: Shape factor is the ratio of plastic moment to yield moment of the section. It depends on the cross section of the shape. It is a measure of the reserve strength available in the section after bending. 𝜎𝑦 𝑍𝑝 𝑍𝑝
S=Mp/My =
=
𝜎𝑦 𝑍𝑒 𝑍𝑒
Shape Rectangle
Moment of inertia about Cg 𝑏𝑑3 12 𝑏ℎ3 36 𝜋𝑑4 64
Triangle Circle
Moment of inertia about base 𝑏𝑑3 3 𝑏ℎ3 12 NA
Shape factor of rectangle Shape factor 𝑆 =
𝑀𝑝 𝑀𝑦
=
𝜎𝑦 𝑍𝑃 𝜎𝑦 𝑍𝐸
=
𝑍𝑃 𝑍𝐸
=
𝑏𝑑2 4 𝑏𝑑2
= 1.5
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𝐴
Zp =2 (𝑦 ̅̅̅1 + ̅̅̅); 𝑦2 A is the area of the whole cross-section and ̅̅̅ 𝑦1 and ̅̅̅ 𝑦2 are the distances from the centroidal axis to the centroids of the area in compression and tension respectively. In fig 2.5 C is compression and T is tension.
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Some shape factors are as givrn below
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Load factor: Ratio of collapse load to the working load. F=Pu/Pw=Mp/Mw=fyZp/(f x Ze)=fy/f x S= FOS x S(Shape factor) What is the load factor that is generally assumed in case of steel design? 1.7 to 2 What is the reduction in load factor when the structures are subject to winds? 25 % For any structure to fail a number of hinges need to formed: There is no relation between the number of hinges and no. of redundancies. You have to find by experience. Redundancy is the static indeterminacy of the structure. What is partial collapse of structure? No of plastic hinges in the collapse mechanism are less than r+1 What is complete collapse of structure? No of plastic hinges in the collapse mechanism are equal to r+1 What is over complete collapse of structure? No of plastic hinges in the collapse mechanism are more than r+1 For calculating the failure load there are three theorems: Static or lower bound theorem; Kinematic or upper bound theorem; Uniquness theorem. What is the Static or lower bound theorem? For a given frame and loading if there exists a set of loads which is both safe and statically admissible with a set of load P, then the value of the load P must be less than or equal to the collapse load Pu. Lower limit to the true load What is Kinematic or upper bound theorem?
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For a given frame subjected to a set of loads P, the value of P which is found to correspond to any assumed mechanism must be either greater than or equal to the collapse load Pu. Upper limit to the true load What is Uniqueness theorem? For a given frame if one load is found which is both statically admissible and safe and if in this dist the bending moment is such as to cause fully plastic moment at sufficient cross section to cause failure as a mechanism due to rotation of plastic hinges then the load will be equal to the collapse load. Conditions to be satisfied in the plastic moment of analysis: Equilibrium condition: Summation of all moments and forces are zero Mechanism condition or Continuity condition: Ultimate load is reached when a mechanism forms Yield Condition: Bending moment at any section must not exceed full plastic moment at section.
Note: How the area is calc to find ext. work done.
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For static method you have to know the BMD well. And for kinematic method the principle of virtual work is used. The main aim is to find the maximum Mp and minimum Ultimate load Wu which causes this maximum Mp in the beam. Thus we don’t equate 𝑀𝑃 =
7
𝑊𝐿 12
in the above Fig 2.16 as this
is not the maximum Mp but we equate 2𝑀𝑝 =
𝑊𝐿 . 8
This is because the middle hinge has to form to
complete the mechanism and it requires 2𝑀𝑝 . What is the principle of virtual work? Work done by the external forces equal the work done by the internal forces. We=Wi Find the reaction at A in the following figure. Ans =5wL/12. Hint ∑ 𝑀𝐵 = 0 /m
Hinge length for a point load on a simple beam with rectangular cross-section.
Hinge length for uniformly distributed load with simple beam is L/√3;
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The independent mechanisms are first formed, using only single loads and then these mechanisms are combined. They are combined such that the geometry remains the same, if you can’t keep the geometry same you do not need to do those mechanisms. The geometry remains the same meaning that the mechanisms should be combined such that the geometry of the independent mechs remain (i.e the angles should be same)
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What are the number of independent mechanisms? n=N-r; n = possible no. of mechanisms, N= number of plastic hinges; r= number of redundancies
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A gable is the generally triangular portion of a wall between the edges of a sloping roof. What is the max yield stress of grade of steel for plastic design? 450 Mpa What are the classification of cross Sections and their char? Plastic(class 1): Cross Section which can develop plastic hinges. Only these section are used for plastic analysys Compact(class 2): Plastic moment developed but does not form plastic hinges Semi – Compact(class 3): Extreme fibre in compression can reach yield stress, but due to local buckling doesn’t develop plastic hinges Slender(class 4): Buckle locally even before attainment of yield stress.
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𝑀𝑝 =
5𝑤𝐿 12
Connections The three different types of connections are: Riveted, bolted and pinned connections: These connections are alike while the welded connection is different. There are certain terms associated with riveted connections: These are as explained below: What is the grip of the rivet? The distance between the undersides of the two heads of rivet. What is the nominal diameter? Diameter of shank. What are hot driven field rivets? When the rivets are heated before driving What is the gross diameter of rivets? The diameter of the rivets when it is hot is equal to the diameter of hole is called gross diameter. Where are cold driven rivets used? These are inserted into the rivet hole at room temperature, and very high pressures are required to form the heads. Only limited to applications where there is high pressure Is riveted connection practiced nowadays? No The main advantages of bolted connection over riveted connections are: a) Erection of structure can be speeded up. b) Less skilled persons are required. c) More economical due to lesser equipment and labour costs. Holes can be either punched or drilled. Punched holes are used in materials with yield stress less than 360 MPa and thickness should not exceed 5660/fy mm.
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Disadvantages: 1) 2) 3) 4)
Cost of material is high Tensile strength of bolts is reduced because of area reduction at root of threads Loose fitting so less strength May get loose due to vibrations
Joints may be of different kinds: Lap joint; Butt joint Explain: Load transfer is either by bearing and shearing or by friction. Bearing: Bearing is the stress on the plate due to the bolt being forced onto the plate Shearing: This is the stress on the bolt which tries to shear away or cut away the bolt at the point of contact. This is shown in Fig.1
Fig. 1: Shearing Explain that for bolting certain terms like diameter of bolts, spacing of bolts are reqd to be known. Few large diameter bolts are more economical than a more number of small dia bolts. Pitch: It is the distance between two consecutive bolts measured parallel to the direction of stress.
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Why is a minimum amt of pitch required? 1) Prevent bearing failure of members 2) Permit efficient installation of bolts Why is the maximum pitch required? 1) To reduce length of connection and gusset plate 2) To have uniform stress in bolts. What is the minimum ptch? 2.5 times the nominal diameter. What is the maximum pitch? Tension Members: 16t or 20mm, whichever is less Compression members: 12t or 200mm Pitch of tacking bolts should not exceed 32t or 300mm, but when exposed to weather 16t or 200mm. Tacking bolts are used to make members act in unison also known as stitch bolts. These bolts are not for being subjected to stresses like tension or compression. Gauge: Distance between adjacent bolt lines; The gauge is such as to accommodate the bolts properly as shown in fig 4.8. Edge distance explain Distance at right angle to the direction of stress from the centre of the bolt hole to the adjacent edge of the member. End Distance: From the centre of hole to the end of element along the direction of stress. What is the maximum edge distance? 250
12t𝜖 𝑎𝑛𝑑 ∈= √ 𝑓 and t is the thickness of thinner outer plate 𝑦
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N.B.: While designing generally two or three bolts are provided , even if one bolt hole is required, this is a safety measure,as one bolt may fail to live upto its strength
The minimum of bearing, shearing, tensile strength of bolt and tensile strength of plate is taken as strength of bolted joint, now we calculate them: Tell we use a factor of safety of 𝛾𝑚𝑏 =1.25 for ultimate strength and for yield strength 𝛾𝑚0 = 1.1. The best bolting pattern is “diamond”. How is class designated? 4.6 Ultimate stress 4 x 100 =400N/mm2; Yiels dtrength =0.6 x 400 =240N/mm2 How is the cover plate thickness selected in case of butt joint? Plate thickness not less than 5/8t Cover plates are used in butt joints to connect the two members.
Formulae for bolt connection design Shearing strength of bolts: 𝑉𝑛𝑠𝑏 =
𝑓𝑢𝑏 √3
(𝑛𝑛 𝐴𝑛𝑏 + 𝑛𝑠 𝐴𝑠𝑏 )
𝑉𝑛𝑠𝑏 Nominal shear strength of bolt 𝑓𝑢𝑏 ultimatre tensile stress of bolt 𝑛𝑛 number of shear planes with threads intercepting the shear plane 𝑛𝑠 number of shear planes without threads intercepting the shear plane 𝜋𝑑 2 ; 4
𝐴𝑠𝑏 nominal shank Area of bolt=
d is dia of bolt.
𝐴𝑛𝑏 net tensile stress area. =0.78𝐴𝑠𝑏 Now for long joints and other factors we use some reduction factors: 15
𝑉𝑛𝑠𝑏 =
𝑓𝑢𝑏 √3
(𝑛𝑛 𝐴𝑛𝑏 + 𝑛𝑠 𝐴𝑠𝑏 )𝛽𝑙𝑗 𝛽𝑙𝑔 𝛽𝑝𝑘𝑔 𝑙𝑗
𝛽𝑙𝑗 = reduction factor used in long bolts = 1.075− 200𝑑 ; 0.75 ≤ 𝛽𝑙𝑗 ≤ 1; 𝑙𝑗 is the length of joint between the first and last row of bolts. Used when 𝒍𝒋 > 𝟏𝟓𝒅 8𝑑
𝛽𝑙𝑔 =reduction factor for large grip length = 3𝑑+𝑙 ; Used when total thickness of connected plates 𝑔
exceed five times the nominal dia of bolt𝛽𝑙𝑔 ≯ 𝛽𝑙𝑗 𝑎𝑛𝑑 𝛽𝑙𝑔 ≯ 8𝑑 𝛽𝑝𝑘𝑔 = reduction factor when packing plate thickness exceeds 6mm. =(1 − 0.0125𝑡𝑝𝑘𝑔 ). 𝑡𝑝𝑘𝑔 is the thickness of thicker packing 𝑉𝑠𝑏 ≤
𝑉𝑛𝑠𝑏 𝛾𝑚𝑏
𝛾𝑚𝑏 = 1.25 (𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟) 𝑉𝑠𝑏 is the shear strength of bolt used for designing. When finding bolt value use: 𝑉𝑛𝑠𝑏 𝑉𝑠𝑏 = 𝛾𝑚𝑏
Remember whenever there is ultimate strength involved use partial safety factor of 1.25, when yield strength is involved use 1.1 Bearing strength of bolt: 𝑉𝑛𝑝𝑏 = 2.5𝑘𝑏 𝑑𝑡𝑓𝑢 ; 𝑒
𝑘𝑏 =smaller of 3𝑑 ; 0
𝑝 3𝑑0
− 0.25;
𝑓𝑢𝑏 ; 𝑎𝑛𝑑 𝑓𝑢
1.0;
𝑑 = nominal dia of bolt 𝑑0 = hole dia 𝑡 = aggregate thickness for connected plates (will be explained later) experiencing stress in the same direction. If bolts are countersunk then thickness of plate minus half of the depth of countersunk. 𝑉𝑛𝑝𝑏 =nominal bearing strength of bolt.
𝑉𝑝𝑏 ≤
𝑉𝑛𝑝𝑏 𝛾𝑚𝑏
𝛾𝑚𝑏 = 1.25 (𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)
When finding bolt value use: 𝑉𝑝𝑏 =
𝑉𝑛𝑝𝑏 𝛾𝑚𝑏
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Tensile strength of bolt 𝑇𝑛𝑏 = 0.9𝑓𝑢𝑏 𝐴𝑛𝑏 < 𝑓𝑦𝑏 𝐴𝑠𝑏 𝑇𝑛𝑏 = nominal tensile strength of bolt. 𝑇𝑑𝑏≤
𝑇𝑛𝑏 𝛾𝑚𝑏
Tensile strength of plate: (not in syllabus) 𝐴𝑛 = (𝐵 − 𝑛𝑑ℎ )𝑡; for chain bolting 𝐴𝑛 = [𝐵 − 𝑛𝑑ℎ + ∑
𝑝2 ] 𝑡; 4𝑔
for staggered bolting 𝑇𝑛𝑑 = 0.9𝐴𝑛
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𝑓𝑢 𝛾𝑚1
𝛾𝑚𝑏 𝛾𝑚0
Efficiency of joint (not in syllabus) 𝜂=
𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑙𝑡𝑒𝑑 𝑗𝑜𝑖𝑛𝑡 𝑝𝑒𝑟 𝑝𝑖𝑡𝑐ℎ 𝑙𝑒𝑛𝑔𝑡ℎ × 100 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑝𝑙𝑎𝑡𝑒 𝑝𝑒𝑟 𝑝𝑖𝑡𝑐ℎ 𝑙𝑒𝑛𝑔𝑡ℎ
If pitch is staggered then take two pitch lengths and take them from the middle, such that two rivets are included. Thus two rivets have to fail, but only one hole in a pitch (because it is (p-d) x t and no breadth term is included).
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Which plate thickness to be used? 1) Double cover butt joint. If thickness of cover plates = tp and thickness of thinner member tm then if 2tp>tm use tm Else 2tp 2) If 2 angle sections or I-sections connected back to back . If thickness of gusset plates = tp and thickness of thinner member tm then if 2tm>tp use tp else 2tm
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