Question

Select reinforcement for the T-beams in the simply-supported floor system shown below. The system is 26-foot-long and subjected to its own self-weight and a service live load of 200 psf. Take the compressive strength of the concrete to be fe' = 3 ksi and the yield stress of the rebar to be fy = 60 ksi. Show a sketch of the designed cross-section, including beam dimensions and reinforcement selection. Limit your reinforcement selection to one that incorporates a single layer of rebar. Also, compute the efficiency of your design and confirm that the design meets ACI requirements. • Note: Since the live load is well over 100 psf, live load reduction is not appropriate. • Hint: Use the beam's tributary width, not the effective flange width, to determine the self-weight of the beam. 10"

Answer 1

Given data :

Span of beam L = 26 ft

Slab thickness h_f = 3 in

Breadth of web, b_w = 10 in

Overall depth of beam,h = 24 in

Effective depth, d = 21 in

center to center distance

left side = 6 ft

right side = 6 ft

Strength of concrete, = 3000 psi

Yield strength of steel, =f_y = 60000 psi

Service live load, =200 psf

CALCULATIONS

STEP-1 ;Computation of factored bending moment

Self weight of beam, = 150 pcf x { 6 x 3/12 + 10 x 21/144}

W_DL =444 plf

Uniformly distributed live load,

WLL = 200psf x 6

WlL = 1200plf

Therefore,factored design load,

  

W = 1.2WDL + 1.6W LL

  

W = 2453plf

  

Wu = 2.45kplf

Factored design moment,

71"11 = ""

My = 207.2kip.ft

STEP -2 : Effective flange width,b

Clear distance

10in leftside = 6ft - - 12

  

= 5.17ft

rightside = 6ft - 10in 12

= 5.17ft

a) b shall not exceed one fourth of the span length

b = L/4

= 26 * 12/4

= 78 in

b) The effective overhanging slab width on each side of the web shall not exceed eight times the slab thickness, that is = 8 x h_f = 24 in

b= 24 + 10 + 24

= 58 in

Therefore, b= 58 in ( the smallest of the three values )

STEP - 3: Computation of area of reinforcement, As

Assuming j =0.95,

  

ja = 19.950

Mu As fyja

  

= 207.2 x 12000/(0.9 X 60000 x 19.95)

= 2.31in?

Provide 2 # 10 bars and these will fit into the web in single layer

Area of steel provided,

Ag = 2.45in?

STEP 4 : Check if A_s >= A_s,min

  

Armin = 3v Terud > 206c fy

As min = 0.58 >0.7

OK

As > As,min

CAPACITY CHECK  

STEP-5 : Computation of a

  

a=085fb

2.45 X 60000 0.85 x 3000 x 58

  

urg > 4166'0 = 1

Since a is less than hf, the beam has rectangular beam action

STEP -6 Check whether f_s = f_y

= = 0.04714

  

B1 = 0.85

Note; i) Forfd <= 4000psi, B1 = 0.85

  

ii) For 4000 < d < 8000psi, 31 = 1.05 – 0.05f/100

  

ii) Forfd >= 8000psi, B1 = 0.65

= 87000 [87,000 + 」

  

= 0.50306 > 0.047 = a/d

Therefore,f_s = f_y

Check if the section is tension controlled or not

NOTE :

i) For single layer of reinforcement , d_t = d

ii) For double layer, d_t = d - clear cover stirrup dia-half of bar dia

Since the tension reinforcement is in single layer

d_t = 21 in

a/d_t = 0.04714

ateldt = 0.375 x 81

  

= 0.31875 > 0.047= a/di

Therefore, the section is tension - controlled and   

Φ = 0.90

Note :

i) for tension controlled sections

Φ = 0.9

ii)for compression controlled sections

Φ = 0.7

for transition sections

Q = 0.356 + 0.204/(a/Bd)

STEP-7 :Computation of nominal moment capacity of beam,M_n

Lever arm, j_d = (d-a/2)

= 20.505 in

nominal moment capacity, M_n = A_sf_y(d-a/2)

= 2.45 * 6000*20.51 = 3014235 in.lb

=251.2 ft-kips

The design moment capacity of the beam is obtained as:

OMR = 0.90 x 251.2 = 226.07 ft – kips

> 207.2ft - kips

Hence, it is SAFE

. . 94 'n # 10. 10 in