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At the limit state of collapse, an RC beam is subjected to flexural moment 200 kNm, shear force 20 kN, and torque 9 kNm. The beam is 300 mm wide and has a gross depth of 425 mm, with an effective cover of 25mm. The equivalent nominal shear stress ( τve ) as calculated using the design code turns out to be lesser than the design shear strength ( τc ) of the concrete. 
Q.The equivalent shear force (Vc ) is
  • a)
     20 kN
  • b)
     54 kN
  • c)
     56 kN
  • d)
     68 kN
Correct answer is option 'D'. Can you explain this answer?
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Given information:
- Flexural moment (M) = 200 kNm
- Shear force (V) = 20 kN
- Torque (T) = 9 kNm
- Beam width (b) = 300 mm
- Gross depth (h) = 425 mm
- Effective cover (d') = 25 mm

Calculating the equivalent nominal shear stress (ve):
The equivalent nominal shear stress (ve) can be calculated using the formula:

ve = (V + T/bd') * (h - d') / h

Substituting the given values:
ve = (20 + 9/(300*25)) * (425 - 25) / 425
ve = (20 + 9/7500) * 400 / 425
ve = (20 + 0.0012) * 0.9412
ve = 20.024 * 0.9412
ve = 18.85 MPa

Checking if ve is lesser than the design shear strength (c) of the concrete:
To check if ve is lesser than the design shear strength (c) of the concrete, we need to compare ve with the design shear strength (c) of the concrete.

If ve < c,="" then="" the="" design="" is="" />

If ve > c, then the design is unsafe.

Since the question states that ve is lesser than c, we need to calculate the design shear strength (c) of the concrete to further analyze the given options.

Calculating the design shear strength (c) of the concrete:
The design shear strength (c) of the concrete can be calculated using the formula:

c = 0.17 * √(f'c)

where f'c is the characteristic compressive strength of the concrete.

Assuming f'c = 25 MPa (as it is not provided in the question), we can calculate c:

c = 0.17 * √(25)
c = 0.17 * 5
c = 0.85 MPa

Comparing ve and c:
Since ve (18.85 MPa) is greater than c (0.85 MPa), the design is unsafe.

Calculating the equivalent shear force (Vc):
To calculate the equivalent shear force (Vc), we need to rearrange the formula for ve:

ve = (Vc/bd') * (h - d') / h

Substituting the given values of ve (18.85 MPa), b (300 mm), h (425 mm), and d' (25 mm), we can solve for Vc:

18.85 = (Vc/(300*25)) * (425 - 25) / 425
18.85 = (Vc/7500) * 400 / 425
Vc = (18.85 * 7500 * 425) / (400 * 425)
Vc = 68 kN

Therefore, the equivalent shear force (Vc) is 68 kN, which corresponds to option D.
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At the limit state of collapse, an RC beam is subjected to flexural moment 200 kNm, shear force 20 kN, and torque 9 kNm. The beam is 300 mm wide and has a gross depth of 425 mm, with an effective cover of 25mm. The equivalent nominal shear stress ( τve ) as calculated using the design code turns out to be lesser than the design shear strength ( τc ) of the concrete.Q.The equivalent shear force (Vc ) isa)20 kNb)54 kNc)56 kNd)68 kNCorrect answer is option 'D'. Can you explain this answer?
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