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Asked: July 17, 2020In: Structural Engineering

What is effect of steel and aluminium, if we use it as reinforcement in structure?

Himanshu joshi
Himanshu joshi

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What is the effect of steel and aluminum, if we use it as reinforcement in structure?

  1. vivek gami

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    Added an answer on July 18, 2020 at 2:47 pm

    Both have different properties & a separate cost which describes below. steel As compared aluminium,steel have less corrosion resistance Steel are chiper than aluminium Steel give lesser malleability and elastic properties than aluminium In weight steel have higher weight Steel is tough & haRead more

    Both have different properties & a separate cost which describes below.

    steel

    • As compared aluminium,steel have less corrosion resistance
    • Steel are chiper than aluminium
    • Steel give lesser malleability and elastic properties than aluminium
    • In weight steel have higher weight
    • Steel is tough & harder than alluminium
    • Generally we use steel as a reinforced member in concrete structures
    • Esay to weld with other structural components.

    Aluminum

    • Higher corrosion resistance
    • Higher cost
    • Higher
    • malleability and elastic properties
    • Lesser weight
    • Generally we not use in concrete
    • Difficult to weld
    • Aluminum is soft metal compare to steel.

     

    We are generally use steel as a reinforcement.

    We doesn’t use aluminium because of soft metal property

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Asked: June 18, 2020In: Structural Engineering

Under what situation shall engineers use jacking at one end only and from both ends in prestressing work?

Rohan Chaugule
Rohan Chaugule

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Under what situation shall engineers use jacking at one end only and from both ends in prestressing work?

  1. nikeetasharma

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    Added an answer on November 28, 2020 at 4:18 pm

    Prestressing is now a must to have phenomenon in every mega project whether it includes flyovers or dam or anything. During prestressing operation at one end, frictional loses will occur and the prestressing force decreases along the length of tendon until reaching the other end. These frictional loRead more

    Prestressing is now a must to have phenomenon in every mega project whether it includes flyovers or dam or anything.

    During prestressing operation at one end, frictional loses will occur and the prestressing force decreases along the length of tendon until reaching the other end. These frictional losses include the friction induced due to a change of curvature of tendon duct and also the wobble effect due to the deviation of duct alignment from the centerline.

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Asked: May 18, 2020In: Structural Engineering

What is the Minimum Grade of Concrete for Precast Construction?

poojan
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What is the minimum grade of concrete to be used for any precast element? Are there any standards for the same?  

  1. Madeh Izat Hamakareem

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    Added an answer on May 18, 2020 at 11:01 pm

    It depends on the structural member. For example, The American Concrete Institute (ACI 318-19) specifies minimum concrete strength for precast pile as 28MPa. The code also provides minimum strength for other structural members. You can also visit: https://theconstructor.org/concrete/design-requiremeRead more

    It depends on the structural member. For example, The American Concrete Institute (ACI 318-19) specifies minimum concrete strength for precast pile as 28MPa. The code also provides minimum strength for other structural members. You can also visit:

    https://theconstructor.org/concrete/design-requirements-concrete/39004/

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Asked: May 31, 2020In: Structural Engineering

What is the Procedure to Design the Double Angle Tension Member in Steel Structures with formulas?

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

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    Added an answer on June 20, 2020 at 11:19 pm

    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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Asked: December 28, 2018In: Structural Engineering

Reinforcement in columns of a 5 storey building

Kumar Ekanan
Kumar Ekanan

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What is the required number of bars and it’s diameter in column for 5 storey  building.

  1. Emmanuel HABIMFURA

    Emmanuel HABIMFURA

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    Added an answer on July 8, 2019 at 9:52 pm

    First of all you have to design that story building to provide reinforcement for all members such as slab ,beam,etc. by using different code such as IS etc.

    First of all you have to design that story building to provide reinforcement for all members such as slab ,beam,etc. by using different code such as IS etc.

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Asked: August 15, 2020In: Structural Engineering

Which is the most economical way of roofing for rural areas?

AdityaBhandakkar
AdityaBhandakkar

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Which is the most economical way of roofing for rural areas?

  1. nikeetasharma

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    Added an answer on November 23, 2020 at 11:41 am

    The following are the most economical way of roofing for rural areas :- 1. Partial roofs 2. Slate roofs 3. Porches 4. Stone roofs 5. Galvanized roofs 6. Tile stone coated roofs

    The following are the most economical way of roofing for rural areas :-
    1. Partial roofs
    2. Slate roofs
    3. Porches
    4. Stone roofs
    5. Galvanized roofs
    6. Tile stone coated roofs

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Asked: July 22, 2020In: Structural Engineering

What is the benefit of smaller dia steel instead of larger dia steel in slab and beam?

vivek gami
vivek gami

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What is the benefit of smaller dia steel instead of larger dia steel in slab and beam?

  1. nikeetasharma

    nikeetasharma

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    Added an answer on December 18, 2020 at 6:40 pm

    Having smaller diameter rebar distributed throughout the structure also distributes the load from the concrete to the steel more effectively as there is a greater area of contact for adhesion between the two.

    Having smaller diameter rebar distributed throughout the structure also distributes the load from the concrete to the steel more effectively as there is a greater area of contact for adhesion between the two.

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Asked: July 17, 2020In: Structural Engineering

Which pressure us higher in retaining wall, water or soil?

vivek gami
vivek gami

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Which pressure us higher in retaining wall, water or soil?

  1. AdityaBhandakkar

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    Added an answer on September 7, 2020 at 6:12 pm

    In technical terms, Above the Water table, Pressure exerted by water on retaining wall = 0.5 K.H^2 K = coefficient of earth pressure H= vertical height of wall. Total stress = effective stress + pore water pressure. Earth pressure due to soil: Total stress = effective stress.  

    In technical terms,

    Above the Water table,

    Pressure exerted by water on retaining wall = 0.5 K.H^2

    K = coefficient of earth pressure

    H= vertical height of wall.

    Total stress = effective stress + pore water pressure.

    Earth pressure due to soil:

    Total stress = effective stress.

     

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