A 14 in x 24 in rectangular reinforced concrete beam support

A 14 in. x 24 in. rectangular reinforced concrete beam supports a factored uniform load, wu = 4.5 kip/ft, including the beam’s self-weight, as shown below. Design the reinforcement required at the supports and the mid span. Use fc’ = 6,000 psi and fy = 60,000 psi.

Solution

Solution:-

Given

b=14 inch

D= 24 inch

d = 22 inch

effective cover = 2 inch

w_u = 4.5 kips/foot

Let length of beam (l) = 13 feet

Moment (M_u) = w_ul²/8

= 4.5 *13²/8

= 55.0625 kips –feet

M_u = 0.87f_yA_st(d – 0.42x_m)

We know that

x_m/d = 0.48

x_m= 0.48*22 = 10.56 inch

55.0625*12*10³ = 0.87*60000*A_st(22 – 0.42*10.56)

A_st = 0.72 inch² = 483.87 mm²

Minimum Area of steel (A₀) = 0.85 bd/f_y

b = 14*25.4 =355.6 mm

d = 558.8 mm

f_y = 415 N/mm² = 60000 psi

A₀ = 0.85*355.6*558.8/415

A₀ = 406.99 mm²

A₀ is less than A_st . So design is safe

Let use 16 mm diameter bars

Total number of bars = A_st/(/4*16²)

                                      = 483.87/(/4*16²)

                                     = 3

So provide 3 number bars of 16 mm diameter.