A Procedure for Determining Maximum and Minimum Dimensions for a Singly Reinforced Rectangular Concrete Beam Loaded to its Flexural Capacity.

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A Procedure for Determining Maximum and Minimum Dimensions for
a Singly Reinforced Rectangular Concrete Beam Loaded to its Flexural
Capacity.

By T. B. Quimby
UAA Civil Engineering Department
Spring 2004


RC beam strength is primarily a function of width (b), height (h), and the amount of
reinforcing (A
s
). There is an infinite combination of these three variables that will result
in a beam that exactly satisfies that flexural strength requirement:

M
u
<
φM
n
(Eq. 1)

This equation can be expanded, for the case of a singly reinforced rectangular section (or
a section that can be treated as such) in which the steel yields to:










−≤
bf
fA
dfAM
c
ys
ysu
)85(.2
φ
(Eq. 2)

Note that d (the effective depth of the steel) is a function of the height, h.

To obtain a particular strength of beam, it can be observed that increases in A
s
allow a
decrease in the size parameters, b and d, and vice versa. Consequently, determining the
size (b x h) of a section can be determined by substituting into equation (2) a
value/expression for A
s
that represents the steel condition that we are interested in
designing for.

There are three cases that are of particular interest:

1. The maximum A
s
which will minimize b and h
2. The minimum A
s
which will maximize b and h
3. The maximum A
s
which will minimize b and h and
for which deflections are not
likely to be a problem.

If we can write expressions for A
s
for each case, the we can substitute the expressions
into equation (2) and solve for relationship for b and d which can be extended to be a
relationship for b and h.

Minimum Size Beams

For this case we want to increase A
s
and still keep the beam “tension controlled” (ACI
318-02, 10.3.3 and 10.3.4). This will be the A
s
that results in a tensile strain of 0.005
RC Beam Sizing, 1/31/2008 page 1
when the beam is fully loaded (i.e. M
u
= φM
n
). To develop this expression, we use the
beam cross section and strain diagram shown in Figure 1.

Note that, in order to solve the equilibrium equations for the pure flexure condition, T
s

must equal C
c
. Writing expressions for both we get:

cccyss
AfCfAT

=
=
= 85. (eq. 3)



Figure 1
Beam Diagrams

Note that by setting the strain in the concrete to 0.003 and the strain in the steel to 0.005
this dictates that c = 0.375d. We also know that a = β
1
c = β
1
(.375d) and A
c
= ba.
Substituting these observations into equation (3) gives us an expression for A
s
in the two
unknowns, b and d, that we can substitute into equation (2):

bd
f
f
A
y
c
s
)375.0(
85.
1
β









= (eq. 4)

Maximum Size Beams

For this case, A
s
is controlled by ACI 318-02 10.5.1. The expression, when modified to
include the limiting state becomes:

RC Beam Sizing, 1/31/2008 page 2
bd
f
f
A
y
c
s
)3,200max(

= (eq. 5)

Like equation (4) this is an expression for A
s
in terms of the two size variables, b and d.
Substituting equation (5) into equation (2) will yield a relationship between b and d that
will result in the largest effective cross sectional size.

Minimum Beam Size for Which Deflections are NOT LIKELY
to be a Problem.

Increasing A
s
while maintaining a given strength, φM
n
, results in a smaller overall beam
size and a smaller moment of inertia. This results in increased deflection, or stiffness, of
the final beam. A rule of thumb for designing a section so that it is large enough so that it
is
not likely
to have deflection problems is to limit the neutral axis depth, c, to about
0.375 of that of the neutral axis location, c
b
, for the balanced condition.

For the balanced condition, the steel strain is exactly equal to its yield strain (f
y
/E
s
=
f
y
/29,000,000, units in psi). From a strain diagram and using similar triangles, this yields:

y
bdefl
f
d
cc
+
=≈
000,87
000,87
375.0375.0
(eq. 6)

Using the same procedure and logic as was used to develop A
s
for the minimum size
beam, we obtain the following expression for A
s
:

bd
ff
f
A
yy
c
s
+









=
000,87
)000,87(375.085.
1
β (eq. 7)

Like equations (4) and (5) this is an expression for A
s
in terms of the two size variables, b
and d. Substituting equation (7) into equation (2) will yield a relationship between b and
d that will result in the smallest effective cross sectional size for which deflection are not
likely to be a problem.

Putting It All Together

Making the substitutions for A
s
into equation (2) as indicated above and simplifying the
resulting expressions yields the following relationships for the size variables, b and d.

For the SMALLEST BEAM SIZE:







−′

2
375.0
1)375)(.85(.
1
1
2
β
βφ
c
u
f
M
bd (eq. 8)

RC Beam Sizing, 1/31/2008 page 3
For the LARGEST EFFECTIVE BEAM SIZE:














)85(.2
)3,200max(
1)3,200max(
2
c
c
c
u
f
f
f
M
bd
φ
(eq. 9)

For the SMALLEST BEAM SIZE NOT LIKELY TO HAVE DEFLECTION
PROBLEMS:









+
⋅−








+


yy
c
u
ff
f
M
bd
000,87
)000,87(375.0
2
1
1
000,87
)000,87(375.
)85(.
1
1
2
β
βφ
(eq. 10)

Now What Do We Do?

The above relationships establish a relationship that you can use to
guide
you in selecting
b and h for the beam you are designing. You first need to decide which criteria (smallest,
largest, smallest w/o deflection problems) that you want to guide your design. Next you
need to solve the appropriate equation for your condition. The next step is to use the
resulting solution to guide you in selecting b and h. One common way of doing this are to
establish an approximate relationship between b and d (say let d = 1.5b) then solve for the
two variables. Another method is to build a table with a column of b values and the
another with the corresponding d values. To determine h, add an appropriate estimated
value to d for the distance from the center of the steel group (which you have not yet
selected) and the tension face of the beam. Finally, select a b and h that is reasonably
close to your results and satisfies other limitations on beam dimensions.

All this work, and all you have is a b and h. You still need to select the steel for the
beam. This can be done by using equation (2). Estimate d, then substitute your values
for b and d into the equation and solve the quadratic for A
s
. Be sure to maintain the
inequality sign so that you will know what way to go when you select real bars to put in
the beam.

Finally select real bars and accurately place them in a scaled drawing of your cross
section, making sure that all spacing and cover requirements are met.

Then, ignoring all prior calculations, compute the actual φM
n
for the beam that you drew
and show that it is greater than M
u
. If it does not satisfy equation (1), then find someway
to adjust A
s
, b, or h so that equation (1) is satisfied.

Your final result should be an accurate scaled drawing/sketch and a calculation that
shows that all the limit states and criteria have been met.

RC Beam Sizing, 1/31/2008 page 4
RC Beam Size Selection
ACI 318-02
by T.B. Quimby
2/4/2004
Mu 253.3 ft-k f'c 3000 psi
phi 0.9 fy 60000 psi
Beta1 0.85
Use approximation equations to select a b and h
Smallest Defl control Largest
Required bd^2 4942.9 7751.9 17575.9 in^3
use b 12 14 18 in
computed d 20.30 23.53 31.25 in
cov + bar 2.88 2.88 2.88 in
est. h 23.17 26.41 34.12 in
Suggested h 24 27 36 in
A
ctual h 24 27 34 in
h/b 2.00 1.93 2.00
Compute a required area of steel
estimated d 21.13 24.13 31.13 in
Quad:a 52941.1765 45378.1513 35294.1176 Note, these are quadratic coefficients
Quad:b -1140750 -1302750 -1680750 to be used in the quadratic
Quad:c 3039600 3039600 3039600 equation
Strength Req'd As 3.11 2.56 1.88 in^2
ACI 10.5.1 Req'd As 0.85 1.13 1.87 in^2
Req'd As 3.11 2.56 1.88 in^2
Select the actual bars and compute the actual strength and other limiting criteria
A
ctual Bars:(4) #8 (2) #10 (2) #9
Actual As 3.16 2.54 2.00 in^2
Actual d 21.63 24.63 31.63 in
Actual pMn 263.5 257.1 272.9 ft-k
Mu/pMn 96% 99% 93% 99% OK
N.A. location 7.29 5.02 3.08 in
Steel Strain 0.00590 0.01171 0.02785
ACI 10.3.3 85% 43% 18% 85% OK
ACI 10.5.1 27% 45% 95% 95% OK
Approximate weight 300.0 393.8 675.0 lbs/ft