Contents ACI 318 14 Example 001

User Manual: ACI 318-14 Example 001

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ACI 318-14 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear design of a rectangular concrete beam is calculated in this
example.
A simply supported beam is subjected to an ultimate uniform load of 9.736 k/ft.
This example is tested using the ACI 318-14 concrete design code. The flexural
and shear reinforcing computed is compared with independent hand calculated
results.
GEOMETRY, PROPERTIES AND LOADING
CL

10"

A

13.5"

2.5"
A

Section A-A

10' = 120"
Material Properties
E=
3600 k/in2
ν=
0.2
G=
1500 k/in2

Section Properties
d = 13.5 in
b = 10.0 in
I = 3,413 in4

Design Properties
f’c = 4 k/in2
fy = 60 k/in2

TECHNICAL FEATURES TESTED
 Calculation of Flexural reinforcement, As
 Enforcement of Minimum tension reinforcement, As,min
 Calculation of Shear reinforcement, Av
 Enforcement of Minimum shear reinforcing, Av,min

ACI 318-14 Example 001 - 1

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RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 6.1 in Notes on ACI 318-14 Building Code.

Output Parameter

ETABS

Independent

Percent
Difference

Design Moment, Mu (k-in)

1460.4

1460.4

0.00%

Tension Reinf, As (in2)

2.37

2.37

0.00%

Design Shear Force, Vu

37.73

37.73

0.00%

Shear Reinf, Av/s (in2/in)

0.041

0.041

0.00%

COMPUTER FILE: ACI 318-14 Ex001
CONCLUSION
The computed results show an exact match for the flexural and the shear
reinforcing.

ACI 318-14 Example 001 - 2

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HAND CALCULATION

Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9, Ag = 160 sq-in
As,min =

=

200
bw d = 0.450 sq-in (Govern)
fy
3 f c'
fy

bw d = 0.427 sq-in
 f c′ − 4000 
0.85
=
 1000 

0.85 − 0.05 
β1 =

0.003
=
d 5.0625 in
0.003 + 0.005

=
cmax

amax = β1cmax = 4.303 in

Combo1
wu = (1.2wd + 1.6wl) = 9.736 k/ft
Mu =

∙

wu l 2
= 9.736 102/8 = 121.7 k-ft = 1460.4 k-in
8

The depth of the compression block is given by:
a = d − d2 −

2Mu
0.85 f c'ϕb

= 4.183 in (a < amax)

The area of tensile steel reinforcement is given by:
Mu
1460.4
=
a
0.9 • 60 • (13.5 − 4.183 / 2 )

ϕ fy d − 
2


As

=

As

= 2.37 sq-in

ACI 318-14 Example 001 - 3

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Shear Design
The following quantities are computed for all of the load combinations:
ϕ

=

0.75
f c′ :

Check the limit of

f c′ = 63.246 psi < 100 psi

The concrete shear capacity is given by:
ϕ Vc =

ϕ2

f c′ bd

= 12.807 k

The maximum shear that can be carried by reinforcement is given by:
ϕ Vs =

ϕ8

f c′ bd = 51.229 k

The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)

= 6.4035 k

(ϕ Vc + ϕ 50 bd)

= 11.466 k

ϕ Vmax = ϕ Vc + ϕ Vs = 64.036 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ ϕ (Vc/2),
Av
= 0,
s

else if ϕ (Vc/2) < Vu ≤ ϕ Vmax
Av
(V − φVc )  Av 
≥ 
= u
φ f ys d
s
 s  min

where:
  b
 Av 
w
  = max 50 
 s min
  f yt

  bw
 , 
  f yt

 3

 4


f c′ 


else if Vu > ϕ Vmax,
a failure condition is declared.

ACI 318-14 Example 001 - 4

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Combo1

∙

Vu = 9.736 (5-13.5/12) = 37.727 k

φ (Vc =
/ 2 ) 6.4035 k ≤=
Vu 37.727 k ≤ φ V
=
64.036 k
max
  10   10  3

 Av 
,
 4, 000 
  = max 50 


 s  min
  60, 000   60, 000  4

in 2
 Av 
=
max
=
0.0083,
0.0079
0.0083
{
}
 
in
 s  min
Av
=
s

(Vu − φVc )
in 2
in 2
= 0.041
= 0.492
φ f ys d
in
ft

ACI 318-14 Example 001 - 5



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