Singly-Reinforced Beam Design ExampleCEE 3150 – Reinforced Concrete Design

Design a rectangular reinforced concrete beam for loads given below. The simply-supported beamhas a span = 18 ft and excessive deflections will cause damage. The superimposed dead load(SDL) is1.15 kip/ft with other given quantities below.

Given: f ′c = 4.5 kip/in2 wL = 0.8 kip/ft

γc = 150 lb/ft3 fy = 60 kip/in2

FLEXURAL DESIGN

(A) Choose the beam depth,h. Because excessive deflections could cause damage, we will notuse Table 9.5(a) from the code. Instead, useh = `/12 = 18 ft/12 (12 in/foot) = 18 in.

(B) Assume one row of steel:d = h− 2.5 in = 18 in− 2.5 in = 15.5 in.

(C) Assumingr = b/d = 0.5, b = 0.5d = 0.5(15.5 in) = 7.75 in, useb = 8 in.

(D) ComputeMu andMn. With our estimated beam height and width, we can calculate theself-weight,SW , and determine the total dead load,wD.

SW = bhγc

= (8 in)(18 in)(150 lb/ft3)(1 foot2/144 in2)(1 kip/1000 lb)

= 0.15 kip/ft

wD = SDL + SW = (1.15 kip/ft + 0.15 kip/ft)

= 1.3 kip/ft

Because only dead and live loads are given and the dead load is less than eight times the liveload, only one factored load case need be checked:

wu = 1.2wD + 1.6wL = 1.2(1.3 kip/ft) + 1.6(0.8 kip/ft) = 2.84 kip/ft (1)

Which leads to a factored moment,Mu, for a simply-supported beam of:

Mu =wu`

2

8=

2.84 kip/ft(18 ft)2

8= 115 kip·ft (2)

Assuming thatφ = 0.9, the required nominal strength,Mn, can be found.

Mn =Mu

φ=

115 kip·ft0.90

= 128 kip·ft (3)

(E) ComputeR = Mn

bd2 and findρ.

R =Mn

bd2=

128 kip·ft(8 in)(15.5 in)2

(12 in/foot)(1000 lb/1 kip) = 798 psi (4)

Reading from the chart givesρ ≈ 0.0152.

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(F) ComputeAs = ρbd.

As = ρbd = 0.0152(8 in)(15.5 in) = 1.88 in2 (5)

(G) Choose bars, make sure they fit. Based on the requiredAs = 1.88 in2, the following optionsare available:

Bar Size Ab (in2) # Bars As (in2)6 0.44 5 2.207 0.60 4 2.408 0.79 3 2.379 1.00 2 2.0010 1.27 2 2.5411 1.56 2 3.12

Based on these options, choose 2 #9s with a total area ofAs = 2.00 in2 since it is the smallestarea greater than the initial estimate and thus most efficient.

To get the minimum width, assume 1.5 inches clear cover and a #3 stirrup. The minimumbeam width,bmin, will be the sum of the cover requirements, the stirrup width (on eachside of the beam), the flexural reinforcement (2 #9s), and the clear spacing between flexuralreinforcing. A #9 bar hasdb = 1.128 in so that the minimum clear spacing must be 1.128 in.Adding it all up:

bmin = 2(1.5 in) + 2(0.375 in) + 3(1.128 in) = 7.13 in < b = 8 in (6)

The minimum width,bmin, is less than our actual width so the bars will fit.

(H) Analyze the section, checkφMn ≥ Mu.

(1) Factor loads and compute requiredMu. This was computed above:Mu = 115 kip·ft.(2) Check minimum steel. The depth,d, to the centroid of the steel can now be computed

exactly.d = 18 in− 1.5 in− 0.375 in− 1.128 in/2 = 15.56 in (7)

The minimum steel area is then:

As,min =3√

f ′c

fy

bwd =3√

4, 500 lb/in2

60, 000 lb/in28 in(15.56 in) (8)

= 0.418 in2 < As = 2.00 in2 (9)

As,min =200 lb/in2

fy

bwd =200 lb/in2

60, 000 lb/in28 in(15.56 in) (10)

= 0.415 in2 < As = 2.00 in2 (11)

Minimum steel area requirement is met.

(3) ComputeT, C, a, andc. Assuming the tension steel has yielded:

T = Asfy = 2.00 in2(60 kip/in2) = 120 kip (12)

By equilibrium:C = T = 120 kip (13)

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We can now solve fora andc.

a =C

0.85f ′cb

=120 kip

0.85(4.5 kip/in2)(8 in)= 3.92 in (14)

c = a/β1 = 3.92 in/0.825 = 4.75 in (15)

(4) Check minimum strain. Because we only have one row,d = dt.

c

dt

=c

d=

4.75 in

15.56 in= 0.305 < 0.375 (16)

The ratio cdt

is less than 0.375, thus the section is tension-controlled. A tension-controlled section hasεt > 0.005, so it follows thatεt > 0.004 as required. Thus,there is significant yielding of the tension steel, which verifies our assumption from theprevious step that the tension steel had yielded. The minimum strain requirement ismet.

(5) ComputeMn, checkφMn ≥ Mu. The section is tension-controlled (from the previousstep) so thatφ = 0.9.

φMn = φT (d− a/2) = 0.9(120 kip)(15.56 in− 3.92 in/2)(1 foot/12 in) (17)

= 122.4 kip·ft > Mu = 115 kip·ft (18)

The beam is strong enough (φMn > Mu) and all other code checks were satisfactory.Thus, the beam is adequate for the loads.

(I) Sketch.

d =

b = 8"

15.6"

18"h =

2 − #9

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