In addition to its own dead weight which must be estimated t

In addition to its own dead weight, which must be estimated, the beam shown below carries uniform service dead load of 800 lb/ft and uniform service live load of 1500 lb/ft. Design the beam for a 24-ft span, d- 2b Final dimensions should be rounded to the even inch The unit weight of concrete is 150 Ib/ft, fc 4000 psi, fy 60,000 psi.

Solution

Solution:-

Given

Dead load (w_d) = 800 lb/feet

Live load (w_l) = 1500 lb/feet

Span length(l) =24 feet

f_y =60000 psi

f_ck =4000 psi

Total bending moment = w_dl²/8 + w_l *l²/8

                                           = 800*24²/8 + 1500* 24²/8

                                           = 165.6 *10³ lb – feet

Factored bending moment = 1.5*165.6 *10³ lb –feet

= 2.9808*10⁶ lb –inch

We know that

Moment of resistance = 0.36f_ckbx_m*(d – 0.42 x_m)

         For f_y =60000 psi

x_m =0.48 d

M_u = 0.36 *4000 *b*0.48*d(d - 0.42*0.48*d)

2.9808*10⁶ = 551.854 bd²

d =2b (given)

2.9808*10⁶ = 551.85*b *(2b)

b = 11 inch

d =22 inch

Let effective cover =2 inch

Total depth (D) = 24 inch

Area of tensile steel(A_st) = M_u/(0.87f_y(d – 0.42*x_m))

                                           = 2.9808*10⁶/(0.87*60000*(22 – 0.42*10.56))

                                          = 3.25 inch²

Where x_m =0.48 *22 =10.56 inch

Minimum area of steel = 0.85 bd/f_y

= 0.85 *11*22/60000

= 3.428 *10^-3   Inch²  

Here we see that provided area of steel is greater than minimum area of steel.

So this is safe design

Let 1.128 inch diameter bars are used

Total number of bars = 3.25/(/4*(1.128)²)

                                       = 4

Provide 4 bars of 1.128 inch diameter.

Section size b =11 inch , D =24 inch ,d =22 inch