QUICK SOLUTION : RCC DESIGN

Some Important Questions

(DCE Semester)

•As per limit state method, which is the maximum strain value of concrete?

✓ The maximum strain value of concrete is 0.0035

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• What does it mean by “M30 grade” of concrete? What is the value of bending stress/σ_cbc ?

✓M30 represents, the concrete mixture that, after 28 days of curing, will give a resistance of 30 N / mm². 

σ_cbc of M30 concrete is 10 N/mm².

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• A RCC column of 3.2m effective height s.t. a concentrated load of 900kN. Using M30 concrete, Fe500 steel, design the column both in shear and compression. Draw the column c/s.

✓ Given Data :

f_ck = 500 N/mm²

f_y = 30 N/mm²

effective length = 3.2 m = 3200 mm

load =  900 kN

So, P_u = 900 × 1.5 = 1350 kN = 1350×10³

(Assuming it a short column)

Total C/S area of column = A

Suppose, 2% steel of C/S area.

So,

Asc = 2% of A = 0.02A

A_c = A – A_sc = 0.98A

Now,

P_u = 0.4f_ckA_c + 0.67f_yA_sc

or, A = 73131.1 mm² (Putting all the values)

Assuming, the column is square, 

breadth or width of column = √A = 270 mm

Provide, (270 mm×270 mm) column section.

Therefore, 

slenderness ratio = effective length/least lateral dimension = 3200÷270 = 11.88⟨12

hence, our assumption is correct.

Again,

P_u = 0.4f_ckA_c + 0.67f_yA_sc

or, Asc = 1471.21 mm₂

Considering  4 nos  of  16 mm  dia bar for longitudinal reinforcement.

so that, 

A_sc – (4×𝜋×162)/4 = A_sc'

or, A_sc' = 666.96 mm²

Remaining reinforcement could be  20 mm  dia,

no. of bar = A_sc'/A_s = 2 nos

Therefore, 4 nos of 16 mm dia bar and 2 nos of 20 mm dia bar as longitudinal reinforcement.

Minimum dia of the lateral bar = (20/4) = 5 mm (OK with I.S. RULE)

Spacing of tie :

1. Least lateral dimension of column = 270 mm,

2. 16×dia of longitudinal bar = 16×20 = 320 mm,

3. 300 mm, whichever is less.

Hence, providing 5 mm dia bar @ 270 mm C/C.

C/S COLUMN DESIGN