Singly-Reinforced Beam Design Example
CEE 3150 – Reinforced Concrete Design
Design a rectangular reinforced concrete beam for loads given below. The simply-supported beam
has a span ℓ = 18 ft and excessive deflections will cause damage. The superimposed dead load
(SDL) is 1.15 kip/ft with other given quantities below.
Given:
fc′ = 4.5 kip/in2
γc = 150 lb/ft3
wL = 0.8 kip/ft
fy = 60 kip/in2
FLEXURAL DESIGN
(A) Choose the beam depth, h. Because excessive deflections could cause damage, we will not
use Table 9.5(a) from the code. Instead, use h = ℓ/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) Assuming r = b/d = 0.5, b = 0.5d = 0.5(15.5 in) = 7.75 in, use b = 8 in.
(D) Compute Mu and Mn . With our estimated beam height and width, we can calculate the
self-weight, SW , and determine the total dead load, wD .
SW =
=
=
wD =
=
bhγc
(8 in)(18 in)(150 lb/ft3 )(1 foot2 /144 in2 )(1 kip/1000 lb)
0.15 kip/ft
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 live
load, 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 =
2.84 kip/ft(18 ft)2
wu ℓ2
=
= 115 kip·ft
8
8
(2)
Assuming that φ = 0.9, the required nominal strength, Mn , can be found.
Mn =
(E) Compute R =
Mn
bd2
R=
Mu
115 kip·ft
=
= 128 kip·ft
φ
0.90
(3)
and find ρ.
128 kip·ft
Mn
=
(12 in/foot)(1000 lb/1 kip) = 798 psi
2
bd
(8 in)(15.5 in)2
Reading from the chart gives ρ ≈ 0.0152.
1
(4)
(F) Compute As = ρbd.
As = ρbd = 0.0152(8 in)(15.5 in) = 1.88 in2
(5)
(G) Choose bars, make sure they fit. Based on the required As = 1.88 in2 , the following options
are available:
Bar Size Ab (in2 ) # Bars
6
0.44
5
7
0.60
4
8
0.79
3
9
1.00
2
10
1.27
2
11
1.56
2
As (in2 )
2.20
2.40
2.37
2.00
2.54
3.12
Based on these options, choose 2 #9s with a total area of As = 2.00 in2 since it is the smallest
area 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 minimum
beam width, bmin , will be the sum of the cover requirements, the stirrup width (on each
side of the beam), the flexural reinforcement (2 #9s), and the clear spacing between flexural
reinforcing. A #9 bar has db = 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, check φMn ≥ Mu .
(1) Factor loads and compute required Mu . 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 =
=
As,min =
=
√
3 4, 500 lb/in2
8 in(15.56 in)
bw d =
fy
60, 000 lb/in2
0.418 in2 < As = 2.00 in2
200 lb/in2
200 lb/in2
8 in(15.56 in)
bw d =
fy
60, 000 lb/in2
0.415 in2 < As = 2.00 in2
q
3 fc′
(8)
(9)
(10)
(11)
Minimum steel area requirement is met.
(3) Compute T, C, a, and c. Assuming the tension steel has yielded:
T = As fy = 2.00 in2 (60 kip/in2 ) = 120 kip
(12)
C = T = 120 kip
(13)
By equilibrium:
2
We can now solve for a and c.
120 kip
C
=
= 3.92 in
′
0.85fc b
0.85(4.5 kip/in2 )(8 in)
c = a/β1 = 3.92 in/0.825 = 4.75 in
a =
(14)
(15)
(4) Check minimum strain. Because we only have one row, d = dt .
c
c
4.75 in
= =
= 0.305 < 0.375
dt
d
15.56 in
(16)
The ratio dct is less than 0.375, thus the section is tension-controlled. A tensioncontrolled section has εt > 0.005, so it follows that εt > 0.004 as required. Thus,
there is significant yielding of the tension steel, which verifies our assumption from the
previous step that the tension steel had yielded. The minimum strain requirement is
met.
(5) Compute Mn , check φMn ≥ Mu . The section is tension-controlled (from the previous
step) so that φ = 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.
b = 8"
d=
15.6"
h=
18"
2 − #9
3