1. PROJECT : PAGE :
CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
Concrete Beam Design Based on ACI 318-99
INPUT DATA
f 'c = 2.5 ksi Mu = 272.5 ft-k
Main Bar fy = 60 ksi Vu = 46.4 k
Stirrup fy = 60 ksi Tu = 113.0 ft-k
b= 24 in, (ACI 8.10.2)
Top bars 3 # 7 h= 30 in
Bot bars 7 # 7 bw = 24 in
hf = 0 in
Stirrup size ==> # 4 @ 10 in o.c.
No. of legs = 4 d (optional) = in The design is adequate.
DESIGN SUMMARY 2
Main bar top (Compressive Reinf.) : Use 3 # 7 Stirrups : Use # 4 @ 10 in. o.c. (4 legs)
Main bar bottom (Tensile Reinf.) : Use 7 # 7 ( 1 layer )
GOVERNING CASE ANALYSIS
εu = 0.003 β1 = 0.85 ρ max = 0.0134
d = 27.6 in φ = 0.90 ES = 29000 ksi
d' = 2.4 in ρ min = 0.0033 ρw = As = 0.0063
b wd
AS
Case 1 Case 2 Case 3 Case 4
CASE 2 APPLICABLE
FLEXURAL ANALYSIS
Case 1 Analysis :
2M u 0.85ab f C '
a =d − d 2− = in AS = = in2
0.85φb f C ' fy
Case 2 Analysis :
2  Mu  h  0.85 f C ' ( ab w + bh f −b wh f )
a =d − d2−  − ( b − b w ) h f  d − f  = 2.72 in AS = = 2.31 in2
bw  0.85f f C '  2  fy
Case 3 Analysis :
  a 
0.75 β 1ε ud M u −φ 0.85 f C ' ba  d − 
a= =   2 
fy in AS ' = = in2
εu + φ f S ' ( d − d ')
ES
f S ' 0.85 f C ' ba a − β1d '
AS = AS ' + = in2 f S ' = E Sε u = ksi
fy f y a
Case 4 Analysis :
  a  h f 
0.75β 1ε ud M u −φ 0.85 f C ' b wa  d −  + 0.85 f C ' ( b −b w ) h f  d − 2 
a= = in   2   in2
fy AS ' = =
εu + φ f S ' ( d − d ')
ES
f S ' 0.85 f C ' ( b − b w ) h f + a bw 
 = a − β1d '
AS = A S ' + in2 f S ' = E Sεu = ksi
fy fy a

2. SHEAR ANALYSIS (cont'd)
Check section limitation (ACI 11.5.6.8) Determine concrete capacity (ACI 11.3.1.1 or 11.3.2.1)
V u ≤ 10φb wd
'
fc V C = 2b wd f
'
c
= 66.15 k
46.4 < 281.1 k [Satisfactory]
where φ= 0.85
V C = ( 1.9 A + 2500 ρ wB ) b wd = 72.06 k ,<== applicable
Check shear reinforcement (ACI 11.5)

 , for V u < φ c
0
V
where A =MIN ( f
'
c , 100 = ) 54.07
 2
 Av  50
 bw φ c V d 
V B = MIN  u , 1.0  = 0.391
 s  = , for ≤V u ≤φ c
V
2 Mu 
 Re qD  f y
V −φ c
V
 u , for φ c ≤V u
V
 φdf y

Av 
= 0.240 in2 / ft <  s  =0.960 in2 / ft [Satisfactory]
  ovD
Pr
Check spacing limits for shear reinforcement (ACI 11.5.4)
V u −φ C
V
Vs= = -17.49 k
φ
 d
MIN ( 2 , 24) for V s ≤ 4b wd
'
fc

S max, shear = = 13 > S= 10 in
MIN ( d , 12) for V s > 4b wd f
' [Satisfactory]

 4 c
TORSION ANALYSIS
Check section limitation (ACI 11.6.3.1)
2
  T u Ph 
2
 Vu   where φ= 0.85 (ACI 9.3.2.3)
  +  ≤ φ  V C + 8 f 'c  Ph =
 b wd   1.7 A2   b wd  80 in, (perimeter of centerline of outermost closed
   oh    transverse torsional reinforcement.)
Aoh = 391 in2 (area enclosed by centerline of the outermost
0.423 < 0.425 [Satisfactory] closed transverse torsional reinforcement.)
Check if torsional reinforcement required (ACI 11.6.1)
be = MIN(h-hf , 4hf) =
'  A cp 
2 where 0 in, (one side, ACI 6.1.1)
Tu ≤φ f c   Pcp = 108 in, (outside perimeter of the concrete cross
 P cp  section.)
113.0 > 17.0 ft-k Acp = 720 in2 (area enclosed by outside perimeter of
Torsional reinforcement reqD. concrete cross section.)
Check the max factored torque causing cracking (ACI 11.6.2.2)
'  A cp 
2
T u ≤ 4φ f c 
 P cp 
113.0 > 68.0
Torsional reinforcement reqD for full torque.
Determine the area of one leg of a closed stirrup (ACI 11.6.3.6)
At = Tu = Tu = 0.00 in2 / ft
s 2φ A0 f yv 1.7φ A0 h f yv
Determine the corresponding area of longitudinal reinforcement (derived from ACI 11.6.3.7 & 11.6.5.3)
  
25b w  
'
At f yv 5 Acp f c f yv
A L = MAX  P h , − Ph MAX  At ,  = 0.00 in2
s f yL f yL f yL  s f yv  
  

3. (cont'd)
Determine minimum combined area of longitudinal reinforcement
AL, top = As' +0.5AL = 0.00 in2 < actual [Satisfactory]
AL, bot = As +0.5AL = 2.31 in2 < actual [Satisfactory]
Determine minimum diameter for longitudinal reinforcement (ACI 11.6.6.2)
dbL = MAX(S/24, 0.375) = 0.42 in < 0.88 in [Satisfactory]
Determine minimum combined area of stirrups (ACI 11.6.5.2 & 11.6.6.1)
At / S = MAX[(Av+2At)S, 50bw/fyv] = 0.24 in2 / ft < actual [Satisfactory]
Smax, tor = MIN[(Ph/8, 12) = 10 in
SreqD = MIN(Smax,shear , Smax,tor) = 10 in > actual [Satisfactory]