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Shailendra
Shailendra

Shailendra

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In what case is a Shear Key provided in Foundation?

  1. Fasi Ur Rahman

    Fasi Ur Rahman

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    Fasi Ur Rahman

    Shear keys are designed and provided in various structures to provide resistance against lateral loads like earthquake loads and sliding forces in various structures such as bridges, retaining walls, the basement of residential buildings, precast buildings and culverts, masonry walls in seismic regiRead more

    Shear keys are designed and provided in various structures to provide resistance against lateral loads like earthquake loads and sliding forces in various structures such as bridges, retaining walls, the basement of residential buildings, precast buildings and culverts, masonry walls in seismic regions, and steel columns.

    The shear key is provided in the following structural components-

    1. Bridge Structure
    2. Retaining Walls
    3. Footing-basement Wall
    4. Steel Column Base and Footing Joint
    5. Connections of Precast Concrete Elements
    6. Masonry Wall

    To know in detail about the shear key, please read the article-
    Shear Key in Construction: Purpose, Functions, and Uses

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Abbas Khan Civil Engineer
Abbas Khan Civil Engineer

Abbas Khan Civil Engineer

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Which are the different types of welding processes in steel structures and which one is the most common and affordable?

  1. RaghavArora

    RaghavArora

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    RaghavArora Learner

    The most commonly used and Affordable welding is Stud Welding because it is cost-efficient. The different types of Welding processes in steel structures are as follows:- 1. Stud Welding:- It involves the electric process of welding a fastener on a metal base by heating both the components with an arRead more

    The most commonly used and Affordable welding is Stud Welding because it is cost-efficient.

    The different types of Welding processes in steel structures are as follows:-

    1. Stud Welding:- It involves the electric process of welding a fastener on a metal base by heating both the components with an arc.

    2. Flux Core Welding:- this involves a consumable tubular electrode containing a flux and a constant voltage welding power supply. It’s A portable and fast welding process, and less-skilled workers can easily undergo this.

    3. Shield metal arc welding (SMAW):- it involves a consumable fixed-length electrode covered with a flux, and an electric power source is used to weld two metals together. After the completion of this process, mineral coating flux Of electrode disintegrates and releases a gas Commonly known as Shielding gas, which saves joint from unfavorable atmospheric conditions.

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Abbas Khan Civil Engineer
Abbas Khan Civil Engineer

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What is the procedure to design the double angle tension member in steel structures with formulas? Kindly elaborate on it?

  1. Amit Bhuriya

    Amit Bhuriya

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    Amit Bhuriya Beginner

    As per Section 6 (Design of Tension Members) of IS 800:2007 – Code of Practice for Construction in Steel, The design strength of the tension member is the minimum of following, Design strength due to yielding of the gross section (Tdg) (Clause 6.2 of IS 800:2007). Design Strength Due to Rupture of CRead more

    As per Section 6 (Design of Tension Members) of IS 800:2007 – Code of Practice for Construction in Steel,

    The design strength of the tension member is the minimum of following,

    1. Design strength due to yielding of the gross section (Tdg) (Clause 6.2 of IS 800:2007).
    2. Design Strength Due to Rupture of Critical Section (Tdn) (Clause 6.3 of IS 800:2007).
    3. Design Strength Due to Block Shear (Tdb) (Clause 6.4 of IS 800:2007).

    Let’s take an example to understand the designing of double angle tension member:

    Data Known:

    Service Load, T = 200 N/mm2

    STEP 1:

    Factored Load, Tu = Tdg = 1.5 x 200 = 300 N/mm2

    Considering the tension member fails due to yielding of gross section, determine the gross area of angles required.

    Tdg = AgFy/ꙋm0 → Ag = Tdgꙋm0/Fy

    Ag = (300 x 103 x 1.1)/250 = 1320 mm2

    The total gross area of tension member (Ag) required is 1320 mm2. Remember this is the area of two angle sections. Therefore,

    Gross area of single angle section (Ag1) = (Ag)/2 = 1320/2 = 660 mm2

    From steel table (SP 6-1), choose an angle having gross area of single angle about 25% to 40% more than computed above.

    Taking Rolled Steel Equal Angle ISA 60x60x8 having following properties,

    Sectional Area, A= 896mm2

    Total Gross Area, Ag0 = 2×896 = 1792 mm2 > 1320 mm2 (O.K)

    STEP 2:

    Designing Connections: – We can provide either bolted or welded connections, so let us provide bolted connections.

    Total thickness of angles having outstanding legs placed back to back,

    ta = 8+8 =16mm

    Let us provide 20mm diameter bolts of grade 4.6 and Steel of grade Fe415,

    Diameter of bolt, d = 20mm

    Diameter of bolt hole, dh = 20 + 2 = 22mm (Table 19 of IS 800:2007)

    Fu = 410 N/mm2

    Fub = 400 N/mm2

    Fy = 250 N/mm2

    1. Edge distance of bolts (e) = 1.5dh = 1.5 x 22 = 33 ≈ 40mm
    2. End distance of bolts = 1.5dh = 1.5 x 22 = 33 ≈ 40mm
    • Minimum pitch (p) = 2.5d = 2.5 x 20 = 50 ≈ 60mm
    1. Kb = least of
    2. e/(3dh) = 40/(3×22) = 0.606
    3. (f/(3dh)) – 0.25 = (60/(3×22)) – 0.25 = 0.659
    4. Fub/Fu = 410/400 = 0.975
    5. 1

    Kb = 0.606

    1. Design strength of Bolt (i.e Bolt Value)
      • Design shearing strength of bolt in double shear

    = 2 x (Fub/√3) x (Anb/ꙋmb) where, Anb = 0.78 x (πd2/4)

    = 2 x (400/√3) x ((0.78 x (π(20)2/4))/1.25)

    = 90.545 KN

      • Design bearing capacity of bolt

    = (2.5 Kb d t Fub)/ ꙋmb

    = (2.5 x 0.606 x 20 x 16 x 400)/1.25

    = 155.136 KN

    Therefore, Bolt Value = Least of (90.545, 155.136) = 90.545 KN

    No. of bolts required, N = (Tu/Bolt Value) = (300/90.545) = 3.31 ≈ 4 nos

    STEP 3:

    Check of Strength due to rupture of critical section,

    The design strength,

    Tdn = 0.9Ancfu/ꙋm1 + βAgofy/ ꙋm0

    Where,

    β = 1.4 – 0.076(w/t)( fy/fu)( bs/Lc) ≤ (fu ꙋm0/fy ꙋm1)

    ≥ 0.7

    w = outstand leg width = 60mm

    t = total thickness of angles = 16mm

    w1 = end distance = 40mm

    bs = shear lag distance = w + w1 – t = 60 + 40 – 16 = 84mm

    Lc = length of end connection = 3 x 60 = 180mm

    β = 1.4 – 0.076(60/16)(250/410)(84/180)

    = 1.4 – 0.081

    = 1.319 ≤ (410×1.1/250×1.25) = 1.4432 (OK)

    ≥ (0.7) (OK)

     

    Anc = (60+60-2 x 22) x 8 = 608mm2

    Ago = (60 x 16) = 960mm2

    Tdn = 0.9Ancfu/ꙋm1 + βAgofy/ ꙋm0

    = ((0.9 x 608 x 410)/1.25) + ((1.319 x 960 x 250)/1.1)

    = 179481.6 + 287781.81

    = 467343.6 N

    = 467.34 KN > 300 KN (O.K)

    STEP 4:

    Check for Strength Due to Block Shear (Tdb),

    Tdb = [(Avgfy)/(√3ꙋm0) + (0.9Atnfu)/( ꙋm1)] or [(0.9Avnfu)/(√3ꙋm1) + (Atgfy)/(ꙋm0)]

    Avg = 220 x 16 = 3520 mm2

    Avn = (220-3×22-(22/2)) x 16 = 2288 mm2

    Atg = 40 x 16 = 640 mm2

    Atn = (40-(22/2)) x 16 = 464 mm2

    Tdb = [((3520×250)/ (√3x1.1)) + ((0.9x464x410/1.25)]

    = 461880.22 + 136972.8

    = 598853.02 N

    = 598.85 KN

    Tdb = [((0.9x2288x410)/ (√3x1.25)) + ((640×250/1.1)]

    = 389952.53 + 145454.54

    = 535407.07 N

    = 535.41 KN

    Tdb = min (535.41, 598.85) = 535.41 KN > 300 KN (O.K)

    Therefore, the selected section is safe.

    So, Provide 2 ISA 60x60x8 angles placed in such a way that outstanding legs are placed back to back and attached with 20mm diameter bolts of grade 4.6. Edge distance and End distance is 40mm and pitch is 60mm.

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Gopal Mishra
Gopal Mishra

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Write short notes on architectural features which govern the design of building against earthquake  force

  1. aviratdhodare

    aviratdhodare

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    aviratdhodare

    Such that the forces generated by the most destructive force that The Mother unleashes upon us human, tenets of Earthquake Resistant design need to be incorporated into all buildings we live and work within. For ensuring the safety of people there, it must be ensured by the designers that the buildiRead more

    Such that the forces generated by the most destructive force that The Mother unleashes upon us human, tenets of Earthquake Resistant design need to be incorporated into all buildings we live and work within. For ensuring the safety of people there, it must be ensured by the designers that the buildings don’t collapse in the event of an earthquake. Towards this end, the technical aspect of endurance is – ductility. We shall see that a little later.

    Before proceeding on any elaboration, let the concerned reduce the contradictions/conflicts that nearly regularly surface between the Architects and the Structural Engineers.

    As far as Architectural features are concerned, these are a few of them that must be incorporated. For example, keep the plan symmetrical along its vertical axis and avoid large void spaces within the buildings.

    If architects can take the structural engineers along with them right from the beginning of the architectural planning process, much of the confrontations and changes at later stages can be obviated. Else, the cost for the same size building can be much higher if the architect insists upon very peculiar shapes. Unfortunately, this seamless coordination between these two prime people doesn’t happen in most cases.

    The challenge is much bigger for the structural designers if the owner and the architects insist upon peculiar designs. The Burj Al Arab Hotel of Dubai is one example. Imagine, everything right from the beginning has been and shall ever remain against the stability of this particular building. Yet it stands. More than the architect, it is the structural designer who needs to be applauded for this magic…

    Structural designers must design buildings such that while the columns and the beams are strong, the roof slabs be as light as is feasible given the architectural shape/design of any building.

    Confining reinforcement at the beam-column junctions ensures the required ductility of the buildings in the event of an earthquake. This MUST be ensured at all costs.

    There are numerous other features of the design. But equally significant part is the execution of the design on the ground. Unfortunately, we in India are not able to control this part very well.

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Aj
Aj

Aj

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What is the best method of solving inclined truss? And how do can I get the magnitude of the forces in the members?

  1. RaghavArora

    RaghavArora

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    RaghavArora Learner

    Hey, To solve the inclined trusses, Methods of joints is the appropriate method. With the help of this, one can easily identify the force in respective member of truss. But before solving the truss, some points should be kept in mind and these are listed below :- 1. If two members are meeting at joiRead more

    Hey, To solve the inclined trusses, Methods of joints is the appropriate method. With the help of this, one can easily identify the force in respective member of truss.

    But before solving the truss, some points should be kept in mind and these are listed below :-

    1. If two members are meeting at joint, with no external force acting on them, then forces in both the members will be zero.

    2. If the magnitude of force will come positive then force will be considered as Tensile And if magnitude of force will come negative then force will be considered Compressive.

    3. If two members are collinear then force in the third member ,connected but not in straight line with two members, will be zero: F3 = zero

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Vivek Patel
Vivek Patel

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What is bundled bar in reinforced concrete?

  1. DevilAVRT

    DevilAVRT

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    DevilAVRT Beginner

    A group of reinforced bars, parallelly set to each other to act as a unit should be limited four in a bundle used as reinforcement in reinforced concrete, enclosed by stirrups or ties. Bars greater than 32mm should not be bundled in beams. The development length of every bar of bundled bars should bRead more

    A group of reinforced bars, parallelly set to each other to act as a unit should be limited four in a bundle used as reinforcement in reinforced concrete, enclosed by stirrups or ties. Bars greater than 32mm should not be bundled in beams.

    The development length of every bar of bundled bars should be that for the individual bar, increased by 10 % for two bars in contact, 20 % for three bars in contact, and 33 % for four bars in contact.

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kyawmin
kyawmin

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Structural Drawing is included

  1. Omprakash Dhawale

    Omprakash Dhawale

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    Omprakash Dhawale Beginner
    This answer was edited.

    Sir, I will share some information related with different drawings: Architectural Drawing can be termed as the mother drawing for all the other drawing. Structural drawing is the backbone drawing of the building.  Electrical drawing represent the details of electric fixtures, location of switches. PRead more

    Sir,

    I will share some information related with different drawings:

    • Architectural Drawing can be termed as the mother drawing for all the other drawing.
    • Structural drawing is the backbone drawing of the building.
    •  Electrical drawing represent the details of electric fixtures, location of switches.
    • Plumbing drawing give the location of sanitary, piping for water supply system.

     

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Vivek Patel
Vivek Patel

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What is use of ILD diagram at site?

  1. Kuldeep Singh

    Kuldeep Singh

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    Kuldeep Singh Learner

    ILD or Influence Line Diagram shows us the effect of a point load at some point on a structural member with respect to the position of the load. So, ILD tells us what and where would be the maximum effect of a load places at what location. So to visualize the worst-case scenario and best-case scenarRead more

    ILD or Influence Line Diagram shows us the effect of a point load at some point on a structural member with respect to the position of the load.

    So, ILD tells us what and where would be the maximum effect of a load places at what location. So to visualize the worst-case scenario and best-case scenario and the variations I between, ILD is a comprehensive aid.

    The most common usage is in Bridges, rail tracks, and roads where loads are always pointed loads (axle loads) and always moving. Knowing when and how much will be the maximum effect like bending moment, shear force, etc. helps us set up the design values.

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