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RCC Design Test 3
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RCC Design Test 3
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  • Question 1/10
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    If the unsupported length of a column is 3 m and its sides are 240 mm × 360 mm the minimum eccentricity so that the column is designed as an axially loaded column is _______
    Solutions

    Concept:

    Minimum eccentricities:

    exmin=max{x500+D3020mm

    eymin=max{y500+b3020mm

    x, ℓy = Unsupported length

    Calculation:

    x = ℓy = 3m = 3000 mm

    D = 360 mm, b = 240 mm

    exmin=max{3000500+3603020mm=18mm

    = 20 mm

    eymin=max{3000500+2403020mm=14mm

    = 20 mm
  • Question 2/10
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    An R.C square footing of length 2 m and uniform effective depth 200 mm is provided for a 300 mm × 300 mm column. The line of action of the vertical compressive load passes through the centroid of the footing as well as of the column. If the magnitude of the load is 320 kN, the nominal transverse (one way) shear stress in the footing is
    Solutions

    Concept:

    Column load (downward) and load due to soil pressure (upward) will lead to crack 45°.

    Thus, the critical section due to one-way shear is at d from the face of the column.

    τo(onewayshear)=(qo×Bb2d2×B)B×d

    Where

    B = width of square footing

    b = width of column

    d = depth of footing

    qo = load intensity of footing

    Calculation:

    qo=320×10002000×2000=0.08N/mm2

    τo=(0.08×20003002×2002×2000)2000×200=0.26N/mm2
  • Question 3/10
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    A square footing of side 4 m is designed to support a square column of 400 mm size. Effective depth of footing is 500 mm. The safe bearing capacity below the footing is 1.5 MPa. The design bending moment for the footing is________
    Solutions

    The critical section for bending moment is the face of the column:-

    Po = Net upward pressure = 1.5 N/mm2 

    Mu=PoB×[B2b2]×12[B2b2]=PoB2[B2b2]2=1.50×40002[400024002]2=9720KNm

  • Question 4/10
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    The reduction coefficient of a long RC column with effective length of 6 m and size 250 × 400 mm is
    Solutions

    Concept:

    The strength reduction coefficient method is used for the design of a long column. It is valid for Working Stress Method only.

    Strength Reduction coefficient is given as

    Cr=1.25eff48b

    Where, b = least lateral dimension of core and for the circular column with helical reinforcement it is the diameter of the core.

    Calculation:

    eff = 6 m = 6000 mm, b = 250 mm

    Cr=1.25600048×250=0.75
  • Question 5/10
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    A column is subjected to direct compression and bending compression both such that the stress developed in direct compression is 2.5 N/mm2 and the stress developed in concrete in bending compression is 3.6 N/mm2. If M - 20 concrete is used in the column. Then identify which of the following statement is correct
    Solutions

    Concept:

    If σccσcc+σcbcσcbc1 (Design is done on the basis of uncracked section)

    If σccσcc+σcbcσcbc>1 (Design is done on the basis of cracked section)

    σcc⇒ Developed (Calculated) stress in concrete in direct compression

    σcbc⇒ Developed (Calculated) stress in concrete in bending compression

    σcc ⇒ Permissible stress in concrete in direct compression

    σcbc ⇒ Permissible stress in concrete in bending compression

    Calculations:

    For M - 20, concrete

    σcbc = 7 N/mm2

    σcc = 5 N/mm2

    Now, σcc=2.5N/mm2

    σcbc=3.6N/mm2

    2.55+3.67=1.014>1

    Hence design is done on the basis of cracked section
  • Question 6/10
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    An RC square footing of side length 3 m and uniform effective depth of 300 mm supports a square column 600 mm × 600 mm at the centroid of the footing. If the magnitude of factored load acting is 420 kN, the punching shear in the footing is ___(in Mpa up to 3 decimal places)
    Solutions

    Concept:

    Two way/punching shear:

    It shall be checked around the column a perimeter half the effective depth of footing slab away from the face of column or pedestal.

    τvtwoway=w[L×B+(a+d)(b+d)]2[(a+d)+(b+d)]d

    Calculation:

    The critical section for two-way shear is at a distance of d/2 from the face of the column as shown in figure in dotted lines.

    Load = 420 kN, Area of footing = 9 m2

    w = 420×10009×106 = 0.0467 MPa

    τvtwoway=0.0467×(30002(600+300)2)2×(900+900)×300 = 0.35 Mpa

    Important Point:

    Footing is designed as per shear criteria. There are two types of shear:

    1. One way shear
    2. Two way shear

    In one way shear the critical section for shear is:

    1. At a distance d from the face of the wall or column as the case may be.
    2. At a distance d/2 from the face of the wall or column if the footing slab is on piles.
  • Question 7/10
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    Which of the following details are required to plot an interaction diagram for a column?

    I. Grade of concrete

    II. Grade of steel

    III. Shape of section

    IV. Size of section
    Solutions

    Interaction Diagram:

    To draw above interaction diagram, following quantities are required:

    1. Grade of concrete

    2. Grade of steel

    3. Section size (b and D both)

    4. Shape of section (Rectangular & circular)

    5. dDratio

    6. Arrangement of Reinforcement

    7. Area of longitudinal steel and ‘%’ of longitudinal steel.

    Important Points about Interaction Diagram:

    • Point B represents balanced failure of section.
    • Portion AB represents compression failure compression failure because concrete attains its permissible value before yielding of steel.
    • Portion BC represents tension failure because concrete attains its permissible value after yielding of steel.
    • Any load combination failing within interaction curve is safe for section.
  • Question 8/10
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    An isolated footing of 2 m × 3 m along the plane of bending moment is provided under the column. If the net maximum and minimum pressure on the soil under the footing are 95 kN/m2 and 55 kN/m2 respectively and the axial load is 450 kN, then the bending moment at its base will be (The centre of gravity of column and footing coincides)
    Solutions

    Area of footing = 2 × 3 = 6 m2

    Section modulus,

    Z=b×d26=2×326=3m3

    Stress, σ=PA±MZ

    95=PA+MZ -------------- (i)

    55=PAMZ -------------- (ii)

    From (i) and (ii), we get

    40=2×MZM=2×M3

    ⇒ M = 60 kNm
  • Question 9/10
    1 / -0

    The load-carrying capacity of a short RC column, 300 mm × 450 mm reinforced with 6 nos 16 mm diameter Fe 415 bar and casted with M25 concrete considering axial load with minimum eccentricity is ______ (in kN up to 2 decimal places)
    Solutions

    Concept:

    The load-carrying capacity for a short column loaded axially with minimum eccentricity is given as

    Pu = 0.4 fck Ac + 0.67 fy Asc

    Where,

    Ac = Area of concrete after reducing steel area, Asc = Area of steel

    Calculation:

    fck = 25, fy = 415, b = 300 mm, d = 450 mm

    ∴ Load carrying capacity,

    P = 0.4 × 25 (Total area – Area of steel) + 0.67 × 415 × Asc

    =0.4×25(300×4506×π4×162)+0.67×415×6×π4×162

    = 1673.37 kN

    Important Point:

    If, effxD or effyb both are less than 12 and

    exmin=max{x500+D3020mm < 0.05 D

    & eymin=max{y500+b3020mm < 0.05 b

    The column is designed as a short axially loaded column with minimum eccentricity.

    For purely axial load, i.e. eccentricity = 0

    P = 0.45 fck Ac + 0.75 fy Asc
  • Question 10/10
    1 / -0

    A short circular RCC column of diameter 500 mm and effective length 4 m is provided with 10 mm diameter helical reinforcement, the pitch of the helical reinforcement satisfying IS 456:2000 provisions will be ____________.

    (Consider M30 grade concrete and Fe 415 steel, Clear cover = 40 mm).
    Solutions

    Concept:

    Pitch of the helical reinforcement is given by:

    0.36fckfy(AgAc1)VhVc

    Where,

    Ag = Gross area of column=  π4(Dg)2

    Ac = area of core π4(Dg2dc)2

    Vh = Volume of helical reinforcement per pitch

    Vh=1000p×πDh×π4ϕh2

    Vc = volume of core = Ac x 1000

    dc = clear cover

    Dc = core diameter

    As per IS 456:2000, the pitch calculated above should satisfying:

     1) p ≤ 75 mm

    2) p ≤ Dc/6

    3) p ≥ 75 mm

    4) p ≥ 3Φh

    Calculation:

    Dh = Dc - ϕh

    Dc = Dg – 2xdc

    Dc = 500 – 80 = 420 mm

    Dh = 420 – 10 = 410 mm

    Ac=π4(50080)2=138544024mm2

    Vc=138544024×1000mm3

    Vh=1000p×π×410×π4×102=144964749.4p

    0.36×30415(196349.541385440241)144964749.4138544024×1000×p

    p = 96.36 mm, but it should not be greater than one sixth of core diameter of column

    p4206=70mm

    Therefore, the pitch should be 70 mm.

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