DESIGN OF BEAM

Note : Fill-up yellow cells only.

END 1 MIDSPAN END 2

LOAD
V M M V M
from MDM 1.2 DL + 1.6 LL 0 -0.526 26 0 -25.51
from Seismic Analysis EL --- 0 --- 0.000 NOT INCLUDED
1.2DL + 1.6LL 0 -0.526 25.502 0 -25.510

DESIGN OF BEAM REINFORCEMENT

Given:
Wu = 0 kN d' =0 mm
fc' = 28 MPa
fy = 420 MPa
f = 0.9
b1 = 0.85 d= 200 mm
Es = 200000 MPa

d= 200 mm
b= 200 mm
d' = 0 mm
L= 4 m b = 200 mm

cover : 40 mm
using 10 mm f stirrups

Wu = 0 kN

L=4 m

0 Vu at d distance

Shear Diagram in KN
d= 0.2 m

0
25.502 kN-m

Moment Diagram

0.53 kN-m 25.51 kN-m

AT END 1 :

Mu 0.526
Mn = = = 0.58444444 kN-m
f 0.9

Assuming single reinforcement

( 0.85 ) f1 fc' 3
rmax = x
fy 7
(0.85)( 0.85) ( 28) 3
= x
420 7
= 0.02064

As = rmax b d
= ( 0.02064 ) ( 200 ) ( 200 )
= 825.7142857 mm2

1 r fy
Mu = r fy f b d2 1- x
1.7 fc'
1 ( 0.02064 ) ( 420 )
= ( 0.0206 ) ( 420 ) ( 0.9 ) ( 200 ) ( 200 )2 1 - x
1.7 28
= 51053914.29 MN-mm
Mu = 51.05391429 kN-m

Mu 51.053914286
Mn = = = 56.7265714 kN-m
f 0.9
56.72657143 > 0.5844444444 Therefore, it's a single reinforcement only!
AT END 1
Mn1 = 56.72657143 kN-m IF DOUBLY-REINFORCED BEAM
As1 = 825.7142857 mm2 IF SINGLY REINFORCED BEAM
Mn2 = 0.5844444444 - 56.7265714
= -56.1421269841 kN-m Mu = f Rn b d²
0.526 = 0.9 Rn ( 200 ) (
From Mn2 = As2 fy ( d - d' ) 0.0730556 Mpa
Mn2 -56.1421 x10e6
As2 = = ρ= 0.85 1 - 1- 2 Rn
fy ( d -d' ) 420 ( 200 - 0 ) fy 0.85 fc
= -668.358654573 mm2
ρ= 0.85 28 1-
Check if compression steel yields, when Ey < Es' : 420
where:
fy 420
Ey = = = 0.00210 ρ= 0.0001742
Es 200000

and 0.003 Es' 0.003 ( c - d' )

= ; Es' =
c c - d' c
1.4 fc' 1.4
 solving for c: a rmin = or =
c= fy 4 fy 420
f1 2Mn1 = 0.0033333333 or 0.003149704
from Mn1 = As1 fy ( d - a/2) ; a = 2d -
As1 fy (0.85) f1 fc' 3
a= 72.857 mm rmax = x
72.8571429 fy 7
therefore c = = 85.714 mm (0.85)( 0.85) (28)
0.85 = x
Therefore, 420
0.003 ( 85.714 - 0 )
Es' = = 0.00300
85.714 when rmin > r <
Therefore, use 0.00333 for As
Ey < Es' Therefore, compression steel yields!
As = f b d1
= ( 0.00333 ) ( 200) ( 200
IF COMPRESSION STEEL YIELDS: IF COMPRESSION STEEL DOESN'T YIELD: = 133.3333333333 mm2
Es' 0.00300
fs' = fy = 420 MPa fs' = ( fy ) = ( 420 )
Ey 0.00210 minimum edge distance = cover - stirrup
As' = As2 = 600 MPa = ( 40 ) + ( 10 ) + ( 2 x 10 )
= -668.358655 mm2 = 70
Mn2 -56.14x10e6
As' = =
fs' ( d - d' ) 600.00 ( 200 - 0 ) For tension reinforcement:
= -467.85106 mm2 using 16 mm f bars
As
N= =
Therefore, COMPRESSION STEEL YIELDS Ad
As' = -668.35865457 mm2
checking the spacing:
As = As1 + As2 200 - 2( 40 ) - 2( 10 ) - 5( 16 )
= 825.71428571 + -668.35865 S=
= 157.35563114 mm2 4

minimum edge distance = cover - stirrupf - 2 x stirrupf

= ( 40 ) + ( 10 ) + ( 2 x 10 )
= 70 mm

For compression reinforcement:

using 20 mm f bars Ad = 314.15926
As' -668.35865
N= = = -2.127 say -2 pcs.
Ad 314.15926 arrangement: 2 bars horizontally

checking the spacing:

200 - 2( 40 ) - 1( 20 )
S= = 80 mm > 25 mm OK!
1

For tension reinforcement:

using 20 mm f bars Ad = 314.15926
As 157.355631
N= = = 0.501 say 1 pcs.
Ad 314.15926 arrangement: 3 bars horizontally

checking the spacing:

200 - 2( 40 ) - 2( 10 ) - 3( 20 )
S= = 20 mm = 25 mm OK!
2

AT MIDSPAN :

Mu 25.502
Mn = = = 28.3355556 kN-m
f 0.9

Assuming single reinforcement

( 0.85 ) f1 fc' 3
rmax = x
fy 7
(0.85)( 0.85) ( 28) 3
= x
0 7
= 0.02064

As = rmax b d
= ( 0.02064 ) ( 200 ) ( 200 )
= 825.7142857 mm2

1 r fy
Mu = f fy f b d2 1- x
1.7 fc'
1 ( 0.02064 ) ( 420 )
= ( 0.0206 ) ( 420 ) ( 0.9 ) ( 200 ) ( 200 )2 1 - x
1.7 28
= 51053914.29 MN-mm
Mu = 51.05391429 kN-m

Mu 51.053914286
Mn = = = 56.7265714 kN-m
f 0.9
56.72657143 > 28.335555556 Therefore, it's a single reinforcement only!

Mn1 = 56.72657143 kN-m IF DOUBLY-REINFORCED BEAM IF SINGLY REINFORCED BEAM

As1 = 825.7142857 mm2
Mn2 = 28.3355555556 - 56.7265714
= -28.391015873 kN-m
Mu = f Rn b d²
From Mn2 = As2 fy ( d - d' ) 25.502 = 0.9 Rn ( 200
Mn2 -28.3910 x10e6 Rn = 3.54194 Mpa
As2 = =
fy ( d -d' ) 420 ( 200 - 0 ) ρ= 0.85 fc' 1 - 1- 2 Rn
= -337.988284203 mm2 fy 0.85 fc

Check if compression steel yields, when Ey < Es' : ρ= 0.85 28 1-

where: 420
fy 420
Ey = = = 0.00210
Es 200000 ρ= 0.00918 Mpa

and 0.003 Es' 0.003 ( c - d' )

= ; Es' =
c c - d' c
 solving for c: a 1.4 fc'
c= rmin = or =
f1 2Mn1 fy 4 fy
from Mn1 = As1 fy ( d - a/2) ; a = 2d - = 0.0033333333 or 0.0031497039
As1 fy
a= 72.857 mm (0.85) f1 fc'
72.8571429 rmax = x
therefore c = = 85.714 mm fy
0.85 (0.85)( 0.85) (28)
Therefore, =
0.003 ( 85.714 - 0 ) 420
Es' = = 0.00300
85.714
when rmin < r
Ey < Es' Therefore, compression steel yields! Therefore, use 0.00918

As = f b d1
IF COMPRESSION STEEL YIELDS: IF COMPRESSION STEEL DOESN'T YIELDS: = ( 0.00918 ) ( 200)
Es' 0.00300 = 367.0463574 mm2
fs' = fy = 420 MPa fs' = ( fy ) = ( 420 )
Ey 0.00210
As' = As2 = 600 MPa minimum edge distance =
= -337.988284 mm2 =
Mn2 -28.39x10e6 =
As' = =
fs' ( d - d' ) 600.00 ( 200 - 0 )
= -236.5918 mm2 For tension reinforcement:
using 16 mm f bars
As
Therefore, COMPRESSION STEEL YIELDS N=
As' = -337.9882842 mm2 Ad

As = As1 + As2 checking the spacing:

= 825.71428571 + -337.98828 200 - 2( 40 ) - 2( 10 ) - 3( 16 )
= 487.72600151 mm2 S=
2
minimum edge distance = cover - stirrupf - 2 x stirrupf
= ( 40 ) + ( 10 ) + ( 2 x 10 )
= 70 mm

For compression reinforcement:

using 20 mm f bars Ad = 314.15926
As' -337.98828
N= = = -1.076 say -1 pcs.
Ad 314.15926 arrangement: 2 bars horizontally

checking the spacing:

200 - 2( 40 ) - 1( 20 )
S= = 100 mm > 25 mm OK!
1

For tension reinforcement:

using 20 mm f bars Ad = 314.15926
As 487.726002
N= = = 1.552 say 2 pcs.
Ad 314.15926 arrangement: 4 bars horizontally

checking the spacing:

200 - 2( 40 ) - 2( 10 ) - 3( 20 )
S= = 6.67 mm < 25 mm NOT OK!
3

AT END 2 :

Mu 25.510
Mn = = = 28.3444444 kN-m
f 0.9

Assuming single reinforcement

( 0.85 ) f1 fc' 3
rmax = x
fy 7
(0.85)( 0.85) ( 28) 3
= x
420 7
= 0.02064

As = rmax b d
= ( 0.02064 ) ( 200 ) ( 200 )
= 825.7142857 mm2

1 r fy
Mu = f fy f b d2 1- x
1.7 fc'
1 ( 0.02064 ) ( 420 )
= ( 0.0206 ) ( 420 ) ( 0.9 ) ( 200 ) ( 200 )2 1 - x
1.7 28
= 51053914.29 MN-mm
Mu = 51.05391429 kN-m

Mu 51.053914286
Mn = = = 56.7265714 kN-m
f 0.9
56.72657143 > 28.344444444 Therefore, it's a single reinforcement only!

Mn1 = 56.72657143 kN-m IF DOUBLY-REINFORCED BEAM IF SINGLY REINFORCED BEAM

As1 = 825.7142857 mm2
Mn2 = 28.3444444444 - 56.7265714
= -28.3821269841 kN-m
Mu = f Rn b d²
From Mn2 = As2 fy ( d - d' ) 25.510 = 0.9 Rn ( 200
Mn2 -28.3821 x10e6 Rn = 3.54306 Mpa
As2 = =
fy ( d -d' ) 420 ( 200 - 0 ) ρ= 0.85 fc' 1 - 1- 2 Rn
= -337.882464097 mm2 fy 0.85 fc

Check if compression steel yields, when Ey < Es' : ρ= 0.85 28 1-

where: 420
fy 420
Ey = = = 0.00210
Es 200000 ρ= 0.00918 Mpa

and 0.003 Es' 0.003 ( c - d' )

= ; Es' =
c c - d' c

solving for c: a 1.4 fc'

 c= rmin = or =
f1 2Mn1 fy 4 fy
from Mn1 = As1 fy ( d - a/2) ; a = 2d - = 0.0033333333 or 0.0031497039
As1 fy
a= 72.857 mm (0.85) f1 fc'
72.86 rmax = x
therefore c = = 85.714 mm fy
0.85 (0.85)(0.85 ) (28)
Therefore, =
0.003 ( 85.714 - 0 ) 420
Es' = = 0.00300
85.714
when rmin < r
Ey < Es' Therefore, compression steel yields! Therefore, use 0.00918

As = f b d1
IF COMPRESSION STEEL YIELDS: IF COMPRESSION STEEL DOESN'T YIELDS: = ( 0.00918 ) ( 200)
Es' 0.00300 = 367.1726284 mm2
fs' = fy = 420 MPa fs' = ( fy ) = ( 420 )
Ey 0.00210
As' = As2 = 600 MPa minimum edge distance =
= -337.882464 mm2 =
Mn2 -28.38x10e6 =
As' = =
fs' ( d - d' ) 600.00 ( 200 - 0 )
= -236.51772 mm2 For tension reinforcement:
using 16 mm f bars
As
Therefore, COMPRESSION STEEL YIELDS N=
As' = -337.8824641 mm2 Ad

As = As1 + As2 checking the spacing:

= 825.71428571 + -337.88246 200 - 2( 40 ) - 2( 10 ) - 6( 16 )
= 487.83182162 mm2 S=
5
minimum edge distance = cover - stirrupf - 2 x stirrupf
= ( 40 ) + ( 10 ) + ( 2 x 10 )
= 70 mm

For compression reinforcement:

using 20 mm f bars Ad = 314.15926
As' -337.88246
N= = = -1.076 say -1 pcs.
Ad 314.15926 arrangement: 2 bars horizontally

checking the spacing:

200 - 2( 40 ) - 1( 20 )
S= = 100 mm > 25 mm OK!
1

For tension reinforcement:

using 20 mm f bars Ad = 314.15926
As 487.831822
N= = = 1.553 say 2 pcs.
Ad 314.15926 arrangement: 5 bars horizontally

checking the spacing:

200 - 2( 40 ) - 2( 10 ) 4( 20 )
S= = 0 mm < 25 mm NOT OK!
4

DESIGN OF VERTICAL STIRRUPS

Vu = 0 kN

Vud = Vu - Wu d
= 0 - ( 0 ) ( 0.2 )
= 0 kN
= 0 N

1
Vc = fc' bw d
6
1
= 28 ( 200 ) ( 200 )
6
= 35,276.684 N

f Vc 0.85 ( 35,276.684 )
= = 13228.7565 N
2 2

f Vc
Vud < Therefore, the beam doesn't need web reinforcement!
2

determine spacing s of the stirrups, using 10 mm stirrups

Av fy d
s=
Vs
where : Vud
Vs = - Vc
f
0
= - 35,276.684
0.85
= -35276.684 N

f
Av = 2 4 ( 10 )2 = 157.079633 mm2

( 157.08 ) ( 420 ) ( 200 )

s= = -374.03428 mm say -380 mm
-35276.684 ( spacing based from Vud )

determine the length or portion of the beam that needs stirrups,

d = 200mm

Vud = 0 N

f Vs
Vu = 0 N
f Vs = 29,985 N
1/2 fVc = 13,229 N
 x= ### m
stirrups needed here
0N - 0x= 13,229
x= ### m

to provide the needed spacing based from other values of V u to minimize stirrups,

At x = 1 m
Vu = 0 - 0(1)
= 0 N
Vu
Vs = - Vc
f
0
= - 35,276.684
0.85
= -35276.684 N

A v fy d
s=
Vs
( 157.08 ) ( 420 ) ( 200 )
= = -374.03428 mm say -380 mm
-35276.684

check for maximum spacing :

1
compare Vs -35,277
= N to fc' bw d
3
1
? 28 ( 200 ) ( 200 )
3
-35,277 N < 70,553 N
d 200
therefore, max. s = = = 100 mm or 600 mm
2 2
Smax = 100 mm

S < Smax use S = -380 mm

At x = 1.5 m
Vu = 0 - 0 ( 1.5 )
= 0 N
Vu
Vs = - Vc
f
0
= - 35,276.684
0.85
= -35276.684 N

Av fy d
s=
Vs
( 157.08 ) ( 420 ) ( 200 )
= = -374.03428 mm say -380 mm
-35276.684

check for maximum spacing :

1
compare Vs = -35,277 N to fc' bw d
3
1
? 28 ( 200 ) ( 200 )
3
-35,277 N < 70,553 N
d 200
therefore, max. s = = = 100 mm or 600 mm
2 2
Smax = 100 mm

S < Smax use S = -380 mm

OPTIONAL SECTION

At x = 2 m
Vu = 0 - 0(2)
= 0 N
Vu
Vs = - Vc
f
0
= - 35,276.684
0.85
= -35276.684 N

Av fy d
s=
Vs
( 157.08 ) ( 420 ) ( 200 )
= = -374.03428 mm say -380 mm
-35276.684

check for maximum spacing :

1
compare Vs = -35,277 N to fc' bw d
3
1
? 28 ( 200 ) ( 200 )
3
-35,277 N < 70,553 N
d 200
therefore, max. s = = = 100 mm or 600 mm
2 2
Smax = 100 mm

S < Smax use S = -380 mm

At x = 2.5 m
Vu = 0 - 0 ( 2.5 )
= 0 N
Vu
Vs = - Vc
f
 0
= - 35,276.684
0.85
= -35276.684 N

Av fy d
s=
Vs
( 157.08 ) ( 420 ) ( 200 )
= = -374.03428 mm say -380 mm
-35276.684

check for maximum spacing :

1
compare Vs -35,277
= N to fc' bw d
3
1
? 28 ( 200 ) ( 200 )
3
-35,277 N < 70,553 N
d 200
therefore, max. s = = = 100 mm or 600 mm
2 2
Smax = 100 mm

S < Smax use S = -380 mm

At x = 3 m
Vu = 0 - 0(3)
= 0 N
Vu
Vs = - Vc
f
0
= - 35,276.684
0.85
= -35276.684 N

Av fy d
s=
Vs
( 157.08 ) ( 420 ) ( 200 )
= = -374.03428 mm say -380 mm
-35276.684

check for maximum spacing :

1
compare Vs = -35,277 N to fc' bw d
3
1
? 28 ( 200 ) ( 200 )
3
-35,277 N < 70,553 N
d 200
therefore, max. s = = = 100 mm or 600 mm
2 2
Smax = 100 mm

S < Smax use S = -380 mm

At x = 3.5 m
Vu = 0 - 0 ( 3.5 )
= 0 N
Vu
Vs = - Vc
f
0
= - 35,276.684
0.85
= -35276.684 N

Av fy d
s=
Vs
( 157.08 ) ( 420 ) ( 200 )
= = -374.03428 mm say -380 mm
-35276.684

check for maximum spacing :

1
compare Vs = -35,277 N to fc' bw d
3
1
? 28 ( 200 ) ( 200 )
3
-35,277 N < 70,553 N
d 200
therefore, max. s = = = 100 mm or 600 mm
2 2
Smax = 100 mm

S < Smax use S = -380 mm

Number of Stirrups and Spacing :

a. ) along the depth

@ 50 mm spacing = 1 pc

@ -380 mm spacing
200 - 50 = 150 mm
150 mm
= -0.39 say 0 pcs
-380 mm

length occupied by 0 pcs stirrups :

( 0 ) ( -380 ) = 0 mm

length already has stirrups :

0 + 50 = 50 mm

b. ) along the 1 m length

@ -380 mm spacing
1000 - 50 = 950 mm
950 mm
= -2.5 say -2 pcs
 -380 mm

length occupied by -2 pcs stirrups :

( -2 ) ( -380 ) = 760 mm

length already has stirrups :

760 + 50 = 810 mm

c. ) along the 1.5 m length

@ -380 mm spacing
1,500 - 810 = 690 mm
690 mm
= -1.82 say -1 pcs
-380 mm

length occupied by -1 pcs stirrups :

( -1 ) ( -380 ) = 380 mm

length already has stirrups :

380 +810 = 1190 mm

d. ) along the 2 m length

@ -380 mm spacing
2000 - 1190 = 810 mm
810 mm
= -2.13 say -2 pcs
-380 mm

length occupied by -2 pcs stirrups :

( -2 ) ( -380 ) = 760 mm

length already has stirrups :

760 + 1190 = 1950 mm

e. ) along the 2.5 m length

@ -380 mm spacing
2500 - 1950 = 550 mm
550 mm
= -1.45 say -1 pcs
-380 mm

length occupied by -1 pcs stirrups :

( -1 ) ( -380 ) = 380 mm

length already has stirrups :

380 + 1950 = 2330 mm
 200 )²

1- 2( 0.07306 )
0.85 (28)

28
 or
4 x 420
Therefore, use 0.00333

7
3
= 0.02064
7

rmax

cover - stirrupf - 2 x stirrupf

( 40 ) + ( 10 ) + ( 2 x 10 )
mm

Ad = 201.0619264
133.33333333
= 0.663 say 1 pcs.
201.0619264 arrangement: 5 bars horizontally

200 - 2( 40 ) - 2( 10 ) - 5( 16 )
= 5 mm < 25 mm NOT OK!

)( 200 )²

0.85 fc

1- 2( 3.5419444444 )
0.85 (28)
 1.4 28
or
420 4 x 420
0.0031497039 Therefore, use 0.00333

7
7
x = 0.02064
7

< rmax
for As

( 200 )

cover - stirrupf - 2 x stirrupf

( 40 ) + ( 10 ) + ( 2 x 10 )
70 mm

mm f bars Ad = 201.0619264
367.04635744
= = 1.826 say 2 pcs.
201.0619264 arrangement: 3 bars horizontally

200 - 2( 40 ) - 2( 10 ) - 3( 16 )
= 26 mm > 25 mm OK!

)( 200 )²

0.85 fc

1- 2( 3.5430555556 )
0.85 (28)

1.4 28
 or
420 4 x 420
0.0031497039 Therefore, use 0.00333

7
3
x = 0.02064
7

< rmax
for As

( 200 )

cover - stirrupf - 2 x stirrupf

( 40 ) + ( 10 ) + ( 2 x 10 )
70 mm

mm f bars Ad = 201.0619264
367.1726284
= = 1.826 say 2 pcs.
201.0619264 arrangement: 6 bars horizontally

200 - 2( 40 ) - 2( 10 ) - 6( 16 )
= 0.8 mm < 25 mm NOT OK!