Question

A rectangular RCC beam section of 250 mm width and 400 mm effective depth is under a factored Shear Force of 120 kN. The design shear strength (\(\tau_c\)) of concrete is 0.35 N/mm². Two-legged, 8 mm diameter stirrups are used for the shear reinforcement. Assuming the Yield Stress of Steel, \(f_y = 415\) N/mm², the design spacing (c/c) of the stirrups is ________ mm. (rounded off to the nearest integer)

Hint:

When calculating stirrup spacing, ensure that the shear force is properly divided between the concrete and the shear reinforcement. The spacing ensures the stirrups provide adequate shear resistance.

Answer 1

Stirrup design spacing calculation for an RCC beam is performed using the shear reinforcement formula. Step 1: Concrete Shear Capacity Calculation (\(V_c\)) The shear capacity provided by concrete, \(V_c\), is determined by: \[ V_c = \tau_c \times b \times d \] Where: \( \tau_c \) = 0.35 N/mm² (concrete's design shear strength) \( b \) = 250 mm (beam width) \( d \) = 400 mm (beam effective depth) Substituting values: \[ V_c = 0.35 \times 250 \times 400 = 35,000 \, {N} = 35 \, {kN} \] Step 2: Shear Force for Stirrups Calculation (\(V_s\)) The total shear force, \(V_u\), is 120 kN. The stirrups will resist the portion of this force not covered by concrete. \[ V_s = V_u - V_c = 120 \, {kN} - 35 \, {kN} = 85 \, {kN} \] Step 3: Stirrup Leg Area Calculation (\(A_v\)) Stirrups are two-legged, with each leg having a diameter of 8 mm. The area of a single stirrup leg, \(A_v\), is: \[ A_v = 2 \times \frac{\pi}{4} \times (8)^2 = 2 \times \frac{\pi}{4} \times 64 = 2 \times 50.24 = 100.48 \, {mm}^2 \] Step 4: Shear Reinforcement Capacity Formula Application The shear reinforcement capacity formula is: \[ V_s = \frac{A_v \times f_y}{s} \] Where: \( A_v \) = 100.48 mm² (area of one stirrup leg) \( f_y \) = 415 N/mm² (steel yield stress) \( s \) = stirrup spacing (the unknown) \( V_s \) = 85 kN (shear force resisted by stirrups) Solving for \(s\): \[ s = \frac{A_v \times f_y}{V_s} \] Inputting known values: \[ s = \frac{100.48 \times 415}{85,000} = \frac{41,795.2}{85,000} \approx 0.49 \, {m} = 160 \, {mm} \] Conclusion: The design spacing (center-to-center) for the stirrups is 160 mm (rounded to the nearest whole number).