PROBLEM:

A simple supported beam with

a 24 ft span,carries a uniform factored load of 2.25 kips/ft.

The beam will be exposed to weather.The specifications call for Billet, A615-81Grade 60 Rebar.The maximum reinforcement ratio shall be 0.0204.Only 5000 psi concrete is available.Consider flexure only and design a rectangular cross section according to ACI Code latest edition.

1. PROBLEM ANALYSIS:

a. A simple supported beam

b. with a 24 ft span

c. carries a uniform factored

d. load of 2.25 kips/ft.

e. The beam will be exposed

to weather.

f. The specifications call for Billet, A615-81

Grade 60 Rebar.

g. The maximum reinforcement ratio shall be 0.0204

h. Only 5000 psi concrete is available.

i. Consider flexure only and design

j. a rectangular cross section

k. according to ACI Code latest edition.

2. ASKED:

a. b = width of beam

b. h = height of beam

1. rebar sizes

3. GIVEN:

a. length of beam

l = 24 ft

b. factored load per unit length

or per unit area

U = 1.4 D + 1.7 L

U = required strength

D = Design Load effect

L = Live Load effect

wu = 2.25 kips/ft

c. Exterior exposure

impacts required

min. concrete coverage

& max. allow.crack width

3. GIVEN-CONTINUOUS:

d. Yield point of non- prestressed reinforcement

fy = 50 kips/sq.in

e. maximum reinforcement ratio 3/4 pb = 0.0204

f. 28-day compression strength of concrete

fc' = 3 kips/sq.in

4. DIAGRAM

A. ASKED

A rectangular beam b x h

with steel reinforcement

b

h

4. DIAGRAM

B. GIVEN

A simple beam rests on a support at

each end, the ends are being free

to rotate.

5C. b= width (in) & d depth (in):

Eq. 3.56 from Leet book p. 104

= (phi) Stress reduction factor

= 0.9 for flexure, axial tension

and combination of

flexure & tension

p = 0.018 selected by student

0.012 would be more on the safe side

Determine b & d:

For short spans the ratio of d/b will usually run

between 1.5 and 2.

Determine h (overall height of beam):

h = d + coverage + radius of rebar

Minimum required concrete coverage

see Table 3.3 Page 105 Leet

Formed Concrete No 6 - No 18 2 inch

exposed to earth No 5 1.5 inch

or weather

Minimum concrete coverage 2 inches.

For preliminary calculation purpose add

approximately 2.5 inches (2.4 in also ok)

to d to receive h.

h = d (in) + 2.5 (in)

h = 15.6 in + 2.5 in

h = 18.1 in

h = 18 in

fulfills the minimum 2 in coverage

h = 18 in > h min 16.2 in OK

fullfills h min requirement

round off h to nearest inch.

large beams round off to nearest even inch

h depth

radius

coverage

Try a second possible configuration:

h = d + 2.5 in

h = 17.1 + 2.5 in

h = 19.6 in

h = 20 in

h = 20 in > h min 16.2 in O.K.

fullfils h min requirement

2.9 in minus radius of rebar fulfills

2.9 in - r rebar in > = 2 in O.K.

2 in minimum concrete coverage

Although the deeper beam is more economical

20 in x 10 in = 200 sq in

12 in x 18 in = 216 sq in

We will use b=12 in & h=18 in to maximize headroom.

5 D. Steel required for reinforcement:

Select Rebar Table 7.2 page 38 Leet

# 10 1.27 sq in

# 10 1.27 sq in

# 9 1.00 sq in

Total 3.54 sq in

As actual = 3.54 sq in > 3.37 in As required

O. K.

If we add

h = d + radius rebar + concrete coverage

h = 15.6 in + 0.635 in + 2 in

h = 18.235 in

Although a depth of 18 in will result in a slightly

smaller d if 2 in of coverage is maintained, the

moment capacity will still be adequate since

As supplied is greater than the As required.

Normally the designer can neglect small differences of this magnitude.

The built-in safety factor of the design process

will take care of it.

5 E. Check z for control of crack width

As a reinforced concrete beam deflects, the tension side of the beam cracks wherever the low tensile strength of the concrete is exceeded.

The maximum crack width the designer should allow depends on exposuure conditions:

Exposure Max. Crack Width

seawater or

wet dry cyles 0.006 in (0.15 mm)

= exposed 0.008 in (0.20 mm) max

protected 0.016 in (0.41 mm)

are permitted by ACI code but may not be

acceptable for the architect who desires a

smooth monolithic concrete surface.

for steel with a yield point up to 40 kips/in

use Gergely-Lutz Eq. 3.28 p. 81 Leet

w = maximum width of crack in 1000th/in

For steel with a yield point over 40 kips/in

use modified Gergely-Lutz Eq. 3.29 p. 81 Leet

z = w / 0.091

fs = stress in steel due to service load

fs = 0.6 x fy

fy = yield strength of rebar

fs = 0.6 x 50 kips/sq in

fs = 30 kips/sq in

dc = distance from tension surface to center

of the row of reinforcing bars closest

to the outside surface

2.4 in was assume

2 in + radius 0.635in = 2.635 in

A = effective tension area of concrete divided

by the number of reinforcing bars

A = AREA OF CONCRETE IN TENSION

NUMBER OF BARS

A = (COVERAGE +RADIUS)x 2 x b

As / cross sectional area of bar

A = (2.4 in) (2) (12 in)

3.53 in / 1.27 in

A = 20.7 sq in

z = 110 kips/in

ACI code (p. 81 Leet) specifies that z not to exceed

exterior exposure 145 kips/in

0.013 in (0.33mm) crack

interior exposure 175 kips/in

0.016 in (0.41mm) crack

z = 110 kips/in <145 kips/in OK

5 F. Check Rebar spacing:

Spacing between bars must not be less than

1 inch or 1 bar diameter .

Allowing 2 inch of side cover on each side.

The available spacing between bars equals.

b - ( side cover (2) + n rebar diameters) = s

n - 1

12 - (2 in (2) + 2 (10/8) + 1(9/8) = 2.2 in

2 spaces

s = 2.2 in > 10/8 > 1 inch OK

5G. Shear Reinforcement:

In addition to the flexural steel, U-shaped reinforcement, called stirrups, is provided for shear.

To ensure that the reinforcement will remain

in position when the concrete is placed, a rigid cage is formed by wiring the stirrups to the main flexural steel at the bottom and to small diameter bars, used to anchor the stirrups, at the top. The top steel is not designed but arbitrarily selected by the designer.