An RCC beam 200 mm wide and 450 mm deep, is reinforced with Fe415 grade steel of area 1000 mm² on tension side with an effective cover of 50 mm. If the shear reinforcement of 100 mm² stirrups is provided at a spacing of 150 mm centre to centre (c/c). What is the ultimate shear strength (Vus) corresponding to the shear reinforcement?
- A96.3 kNCorrect
- B90.6 kN
- C9.63 kN
- D9.06 kN
Solution
Concept
Shear carried by vertical stirrups in limit state method is Vus = 0.87 fy Asv d / sv.
Formula
Vus = 0.87 fy Asv d / sv
Step-by-step solution
1. Effective depth d = 450 - 50 = 400 mm. 2. Given fy = 415 MPa, Asv = 100 mm², sv = 150 mm. 3. Vus = 0.87 × 415 × 100 × 400 / 150. 4. Vus = 14,442,000 / 150 = 96,280 N. 5. Vus ≈ 96.3 kN.
Explanation
Vertical stirrups carry shear equal to 0.87 fy Asv d / sv. Use effective depth d = overall depth minus effective cover, so d = 400 mm. The result is about 96.3 kN.
Why the other options are wrong
B: 90.6 kN is a wrong calculation or wrong depth. C and D: 9.63 kN and 9.06 kN are unit (decimal) errors.
Shortcut
0.87 × 415 ≈ 361, so 361 × 100 × 400 / 150 ≈ 96.3 kN.
Common mistake
Using overall depth 450 mm instead of effective depth 400 mm, or making a unit conversion error.
Correct answer: A) 96.3 kN