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

We will calculate the design spacing of stirrups using the formula for shear reinforcement in an RCC beam. 
Step 1: Calculate the shear capacity of concrete \(V_c\) The shear capacity provided by the concrete \(V_c\) is calculated using the formula: \[ V_c = \tau_c \times b \times d \] where:
\( \tau_c \) = 0.35 N/mm² (design shear strength of concrete),
\( b \) = 250 mm (width of the beam),
\( d \) = 400 mm (effective depth of the beam).
Substituting the values: \[ V_c = 0.35 \times 250 \times 400 = 35,000 \, {N} = 35 \, {kN} \] Step 2: Calculate the shear force to be resisted by stirrups \(V_s\) The total shear force \(V_u\) is given as 120 kN. The portion of this shear force to be resisted by the stirrups is the remaining portion after the concrete’s contribution. Therefore, \[ V_s = V_u - V_c = 120 \, {kN} - 35 \, {kN} = 85 \, {kN} \] Step 3: Calculate the area of shear reinforcement per leg \(A_v\) The stirrups are two-legged, and each leg has a diameter of 8 mm. The area of one stirrup leg \(A_v\) is given by: \[ A_v = 2 \times \frac{\pi}{4} \times (8)^2 = 2 \times \frac{\pi}{4} \times 64 = 2 \times 50.24 = 100.48 \, {mm}^2 \] Step 4: Use the formula for shear reinforcement capacity The formula for shear reinforcement capacity is: \[ V_s = \frac{A_v \times f_y}{s} \] where:
\( A_v \) = area of one stirrup leg (100.48 mm²),
\( f_y \) = 415 N/mm² (yield stress of steel),
\( s \) = spacing between stirrups (the unknown we need to solve for),
\( V_s \) = shear force resisted by the stirrups (85 kN).
Rearranging the formula to solve for \(s\): \[ s = \frac{A_v \times f_y}{V_s} \] Substituting the 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 (c/c) of the stirrups is 160 mm (rounded to the nearest integer).