Concrete structures miscellaneous
Direction: At the limit state of collapse, an R.C. beam is subjected to flexural moment 200 kN-m, shear force 20 kN and torque 9 kN-m. The beam is 300 mm wide and has gross depth of 425 mm, with an effective cover of 25 mm. the equivalent nominal shear stress (τve) as calculated by using the desing code turns out to be lesser than the design shear strength (τc) of the concrete.
- The equivalent flexural moment (Mcl) for designing the longitudinal tension steel is
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Equivalent flexural moment (mef) for designing the longitudinal steel is 200 kNm
Correct Option: B
Equivalent flexural moment (mef) for designing the longitudinal steel is 200 kNm
- As per IS 456:2000 for M20 grade concrete and plain barsin tension the design bond stress τbd = 1.2 MPa. Further, IS 456:2000 permits this design bond stress value to be increased by 60% for HSD bars. The stress in the HSD reinforcing steel barsin tension, σs 360 MPa. Find the required development length, Ld, for HSD bars in terms of the bar diameter, Φ
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Ld = Φσst = Φ × 360 4τbd 4 × 1.2 × 1.60
= 46.875 ΦCorrect Option: A
Ld = Φσst = Φ × 360 4τbd 4 × 1.2 × 1.60
= 46.875 Φ
- A rectangular concrete beam 250 mm wide and 600 mm deep is pre-stressed by means of 16 high tensile wires, each of 7 mm diameter, located at 200 mm from the bottom face of the beam at a given section. If the effective pre-stress in the wires is 700 MPa, what is the maximum sagging bending moment (in kNm) (correct to 1-decimal place) due to live load that this section of the beam can with stand with out causing tensile stress at the bottom face of the beam? Neglect the effect of dead load of beam.
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Since the tensile stress at bottom face of the beam is zero.P - M y + Pe y = 0 A I I P = 700 × π (7)2 × 16 4
= 431.0265 × 103N∴ P - M + Pe = 0 A Z Z
Since the prestressing force is located at 200 mm from the bottom face of the beam.
∴ Eccentricity = 300 – 200 = 100 mm∴ = 413.0265 × 103 - M × 6 250 × 600 250 × 6002 + 413.0265 × 103 × 100 × 6 = 0 250 × 6002
⇒ 2.8735 + 2.87351 = 6.66 × 10–8M
⇒ M = 86.205 kNmCorrect Option: D
Since the tensile stress at bottom face of the beam is zero.P - M y + Pe y = 0 A I I P = 700 × π (7)2 × 16 4
= 431.0265 × 103N∴ P - M + Pe = 0 A Z Z
Since the prestressing force is located at 200 mm from the bottom face of the beam.
∴ Eccentricity = 300 – 200 = 100 mm∴ = 413.0265 × 103 - M × 6 250 × 600 250 × 6002 + 413.0265 × 103 × 100 × 6 = 0 250 × 6002
⇒ 2.8735 + 2.87351 = 6.66 × 10–8M
⇒ M = 86.205 kNm
- While designing, for a steel column of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in N/mm2) in limit state method of design as per IS:456-2000 is _____________.
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Permissible bearing stress = 0.45 × fck
= 0.45 × 20
= 9 N/mm2Correct Option: A
Permissible bearing stress = 0.45 × fck
= 0.45 × 20
= 9 N/mm2
- For a beam of cross-section, width = 230 mm and effective depth = 500 mm, the number of rebars of 12 mm diameter required to satisfy minimum tension reinforcement requirement specified by IS: 456-2000 (assuming grade of steel reinforcement as Fe500) is _________.
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Ast min = 0.85 bd fy Ast = 0.85 × 230 × 500 mm2 500 n = Ast 
π d2 
4 = 0.85 × 230 ≈ 2 bars π (12)2 4 Correct Option: A
Ast min = 0.85 bd fy Ast = 0.85 × 230 × 500 mm2 500 n = Ast π d2 4 = 0.85 × 230 ≈ 2 bars π (12)2 4
