PROJECT : PAGE :
CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
Concrete Beam Design, for New or Existing, Based on ACI 318-19
INPUT DATA & DESIGN SUMMARY
CONCRETE STRENGTH f 'c = 4 ksi, (28 MPa)
REBAR STRENGTH MAIN fy = 60 ksi, (414 MPa)
STIRRUP fy = 60 ksi, (414 MPa)
FACTORED BENDING MOMENT Mu = 1000 ft-kips, (1356 kN-m)
FACTORED SHEAR FORCE Vu = 230 kips, (1023 kN)
FACTORED TORSIONAL MOMEN Tu = 36.5 ft-kips, (49 kN-m)
SECTION DIMENSIONS bw = 24 in, (610 mm)
h= 48 in, (1219 mm)
hf = 8 in, (203 mm) THE DESIGN IS ADEQUATE.
b= 88 in, (2235 mm), (ACI 318-19 6.3.2.1 & 9.2.4.4)
COMPRESSION REINFORCEMENT 4 # 7
TENSION REINFORCEMENT 6 # 9
SHEAR REINFORCEMENT 4 legs # 4 @ 12 in, (305 mm), o.c.
ANALYSIS
CHECK FLEXURAL CAPACITY max = 0.003 0.85 f c'
A s' f s'
2 0.85 f 'C , E c 57 f C' , E s 29000ksi
o max or
Ec Fc
2
c c , for 0
0.85 f C 2
'
c o Parabolic
fC o o
'
0.85 f C , for c o
A sfy
s E s , for s t
fS
f , for s t
y
Cover = 1.5 in, (ACI 318 20.6.1) rprov'd = 0.0055 < rmax = 0.0119 , (ACI 318 9.3.3.1)
d= 45.44 in > rmin = 0.0033 , (ACI 318 9.6.1)
d' = 2.44 in [Satisfactory]
f= 0.90 , (ACI 318-19 21.2) c= 5.39 in, by pure math method
e c,max = 0.0006 Fc = 334.34 kips
es,max = 0.0050 , (ACI 318-19 21.2.2) dc = 1.85 in
fMn = 1185.14 ft-k > Mu [Satisfactory]
CHECK SHEAR CAPACITY
Check section limitation (ACI 22.5.5 & 22.5.1.2) Determine concrete capacity (ACI 22.5.5.1)
V u 10 b wd f c
'
'
V C 2b wd fc 137.93 kips
230.0 < 517.3 kips [Satisfactory]
where f= 0.75
V C 1.9 A 2500 wB b wd
144.10 kips ,<== applicable
Check shear reinforcement (ACI 22.5)
0 , for V u
V c
where
A MIN f
'
c
, 100 63.25
2
d
Av 50b w 0.75 f 'c b w V c B MIN V u , 1.0
s MAX , , for V u V c Mu 0.871
Re qD f f 2
y y
V u V c , for V c V u
df y
Av
s
= 0.716 in2 / ft < Pr ovD 0.800 in2 / ft [Satisfactory]
Check spacing limits for shear reinforcement (ACI 22.6.9.5)
V u V c
Vs
0.00 kips, (ACI 22.5.1.1)
d '
MIN ( , 24) for V s 4b wd f c
2
S max, shear
MIN ( d , 12) for V s 4b wd f ' = 22 > S= 12 in
4 c [Satisfactory]
(cont'd)
CHECK TORSION CAPACITY
Check section limitation (ACI 22.7.7.1)
2
V u T u Ph
2
V C 8 f 'c where f= 0.75 (ACI 21.2)
b d 1.7 2 b d
w Aoh w Ph = 258 in, (perimeter of centerline of outermost closed
transverse torsional reinforcement.)
Aoh = 1,287 in2 (area enclosed by centerline of the outermost
0.215 < 0.474 [Satisfactory] closed transverse torsional reinforcement.)
Check if torsional reinforcement required (ACI 9.5.4.1)
' A cp
2
be = MIN(h-hf , 4hf) =
Tu f c where 32 in, (one side, ACI 9.2.4.4)
P cp Pcp = 160 in, (outside perimeter of the concrete cross
section.)
36.5 < 36.5 ft-k Acp = 1,216 in2 (area enclosed by outside perimeter of
Torsional reinforcement NOT reqD. concrete cross section.)
Check the max factored torque causing cracking (ACI 22.7.3.2)
' Acp
2
T u 4 f c
P cp
36.5 < 146.1
Reduction of the torsional moment can occur.
Determine the area of one leg of a closed stirrup (ACI 22.7.6.1)
At T u Tu
s 2 A0 f yv 1.7 A0 h f yv
0.00 in2 / ft < actual = 0.2 [Satisfactory]
Determine the corresponding area of longitudinal reinforcement (derived from ACI 22.7.6.1 & 9.6.4.3)
25b w
'
At f yv 5 Acp f c f
A L MAX P h , P h yv MAX At ,
s f yL f yL f yL s f yv
0.00 in2
Determine minimum combined area of longitudinal reinforcement
AL, top = As' +0.5AL = 0.00 in2 < actual [Satisfactory]
AL, bot = As +0.5AL = 5.06 in2 < actual [Satisfactory]
Determine minimum diameter for longitudinal reinforcement (ACI 25.7.1.2)
dbL = MAX(0.042 S, 3/8) = 0.50 in < 0.88 in [Satisfactory]
Determine minimum combined area of stirrups (ACI 9.6.4.2 & 9.7.6.3.3)
(Av+2At) / S = 0.40 in2 / ft > MAX [ 0.75(fc')0.5bw/fyv, 50bw/fyv] = 0.24 in2 / ft
Smax, tor = MIN[(Ph/8, 12) = 12 in [Satisfactory]
SreqD = MIN(Smax,shear , Smax,tor) = 12 in > actual [Satisfactory]
