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BEAM DESIGN EXCEL
Citation preview
Date:
Prepared by:
Checked by:
Revisions:
l.
227b= 150d= 200
r== 0.008
ll. Check if Steel Yields:
20.7
276b= 0.85
b= 0.85-.008(fc'-30) : But should not be less than 0.65= 0.924
150
= 8,973.45 241,776.00
= 0.037 Steel yields !!!
lll. If steel yields:
w=
= 0.1009
= 10606081.4112 N.m= 10.61 Kn.m
lV. If steel does not yield:
0.003
Ultimate Moment of Beam w/ Given Tension Area, As
Solve for r:
As=
As/bd
f'c=
fy=
if f'c is greater than 30 Mpa, use b as:
0.85 f 'c b (600)
fy (600+fy)
r fy
f'c
Mu= f f'c bd2 w (1-0.59w)
c
d
d-a/2
d-c
b Es = 200000
Solve for fs from the strain diagram:
= 0.0003 d-c c
= c
a =
S 0 (T=C)
=
=
c
600
c =2a
a = 2243.3625b = 136200x = -27240000
c = 83.94 mm= -144.65
= c
= 829.57a =
= 71.35 mm
=== 27849814.0763 N.m= 27.850 Kn.m
C=0.85f'c ab
a
T= As fy e=fs/es
fs/es
fs 600 (d-c)
bc
FH=
As fs 0.85 f'c a b
As fs 0.85 f'c b c b
As 600 (d-c) = 0.85 f'c b c b
As (d-c) = 0.85 b f'c b c2
- b +-√b 2 - 4ax
fs 600 (d-c)
bc
Mu fT(d-a/2)fAsfs(d-a/2)
l. Assume a < t:
b L
a
d d-a/2
T =
design parameters:
324.70 b = 400.00 d = 450.00
f'c = 27.60 fy = 275.00 L = 5,000.00 t = 100.00
50.00
500.00
300.00
b is the smallest of; interior beam end beam
b b'
For interior beams:
1. b = L/4== 1250
2. b =
= 1650
3. b =
= 450
For end beams:
1. b' ==
T-BEAMS (determining tension steel area As w/ known Mu) :
C = 0.85 f'c a b
As fy
Mu =
bw =
S1 =
S2 =
S1 S2 S3 bw'
bw
*Where L is the beam span
16 t + bw
*Where t is the flange thickness
*Where bw is the web thickness
S1 /2 + S2 /2 +bw
*Where S1 and S2 is the clear distance to next web
L/12 + bw' *Where bw' is the web thickness
= 508.33
2. b' =
= 800
3. b' =
= 500
solve for a:
by quadratic equation:
a = 2x
x = 4222.8y = -3800520z = 324700000
a = 804.41 mm95.59 mm Assumption is correct
ll. If a < or = t, assumption is correct:
T = C
3261.81
1.4fy0.0051
0.1450 p > pmin, therefore ok!!!!
lll. If a > or = t, assumption is not correct:b
t a
d
As d-a/2 d-t/2
6 t + bw' *Where t is the flange thickness
*Where bw' is the web thickness
S3 /2 + bw' *Where S3 is the clear distance to next web
Mu = f C (d- a/2)
Mu = f 0.85 f'c a b (d-a/2)
- y +-√y 2 - 4xz
Asfy = 0.85 f'c a b
As = mm2
check also for r min
rmin =
rmin =
r = As
bw dr =
C1 C2
bw T1 = As1 fy T2 = As2 fy
Mu1 Mu2
As = As1 + As2
1 2 1
2985.82
295,596,000.00 N.m
295.60 Kn.m
29.10 Kn.m
by quadratic equation:
a = 2x
x = 4222.8y = -3800520z = 29104000
a = 892.28 mm7.72 mm
263.58
3249.40
0.144 p > pmin, therefore ok!!!!
Mu = Mu1 + Mu2
solve for As2 :
T2 = C2
As2fy = 0.85 f'c a (b-bw) t
As2 = mm2
solve for Mu2 and Mu1 :
Mu2 = f T2 (d- t/2)
Mu2 = f As2 fy (d- t/2)
Mu2 =
Mu2 =
Mu1 = Mu - Mu2
Mu1 =
solve for a and As1:
Mu1 = f C2 (d-a/2)
Mu1 = f 0.85 f'c a b (d-a/2)
- y +-√y 2 - 4xz
T1 = C1
As1fy = 0.85 f'c ab
As1 = mm2
As = As1 + As2
As = mm2
r = As
bw dr =
Page 25file:///tt/file_convert/54552b22b1af9ff92a8b49bd/document.xls
Date:
Prepared by:
Checked by:
Revisions:
l.Mu= 650.00 Kn.m
b= 300.00 mmd= 600.00 mm
414.00 Mpa
30.00 Mpab = 0.85
Ru= Mu
= 6.69 Mpa
ll.
r =
r = 0.0191
lll.
b = 0.85-.008(fc'-30) : But should not be less than 0.65= 0.850
fy (600+fy)
0.031
0.023 Singly Reinforced !!!
lV.
1.4fy
0.0034 Therefore, use p!!!
V. Compute for As:
Required Tension Steel Area (As) of a Beam w/ Given Ultimate Moment (Mu):
Solve for Ru:
fy=
f'c=
f b d2
Solve for r:
0.85f' c 1 - 1 - 2Ru
fy 0.85f'c
Check if the beam needs compression steel by computing rmax:
rb = 0.75 rmax
if f'c is greater than 30 Mpa, use b as:
rb = 0.85 f'c b (600)
rb =
rmax = 0.75 rb
rmax =
Check rmin:
rmin =
rmin =
Page 26file:///tt/file_convert/54552b22b1af9ff92a8b49bd/document.xls
3441.695
d' C2 = As'fs'
a C1
d d-a/2 d-d'
T2 = As2'fy
T1 = As1 fy
l.d' = 100.00 mm
0.023
= 4182.21
ll. Solve for a and c:
a = 226.33 mma =c = 266.27 mm
lll.
== 758,629,743.14 N.m
758.63 Kn.m
-108.63 Kn.m
=
-583.09
lV. Verify if compression steel will yields:
0.003
d'c
c-d'
As = r b dAs = mm2
In case the beam is doubly-reinforced: (r > rmax)
Solve for As1:
rmax =
As1 = rmax b d
mm2
C1 = T1
0.85 f'c a b = As1 fy
b c
Solve for Mu1, Mu2 and As2:
Mu 1 = f T1 (d - a/2)
f As1 fy (d-a/2)
Mu 1 =
Mu 2 = Mu - Mu1
Mu 2 =
Mu 2 = f T2 (d - d')
f As2 fy (d - d')
As2 = mm2
es = fs/Es
Page 27file:///tt/file_convert/54552b22b1af9ff92a8b49bd/document.xls
0.003 c-d' c
c
374.67 Mpa Compression Steel Does Not Yield…….
V. If compression steel yields:
= -583.09
3599.12
Vl. If compression steel does not yield:
-644.30
3599.12
f s/E s =
fs = 600 c-d'
fs =
As' = As2
mm2
As = mm2
C2 = T2
As' fs = As2 fy
As' = As2 fy
fs
As' = mm2
As = As1 + As2 mm2
As = mm2
NEGLECT THIS PORTION IF THE COMPRESSION STEEL WILL NOT YIELD. PROCEED TO PART V1.
Page 28file:///tt/file_convert/54552b22b1af9ff92a8b49bd/document.xls
As) of a Beam w/ Given Ultimate Moment (Mu):
Computing Mu of a Doubly Reinforced Beam w/ Given As and As'
d' C2 = As'fs'
a C1 = 0.85 f'c ab
d d-a/2 d-d'
stress diagram stress diagram
l. Assume compression steel yields:
0.810
400.00 Mpa
35.00 Mpad = 510.00 mmd' = 65.00 mmb = 280.00 mm
5,089.00
2,035.00
b=b= 0.810
= 3054
ll. Solve for a and c:
a = 146.65 mma =c = 181.05 mm
lll. Verify if compression steel will yields:
0.003
As'
T2 = As2'fy
T1 = As1 fy
As
b =
fy =
f'c =
As = mm2
As' = mm2
if f'c is greater than 30 Mpa, use b as:
0.85-.008(f'c-30) : But should not be less than 0.65
fs = fy
As2 = As'
As1 = As -As2
mm2
C1 = T1
0.85f'c ab = As1 fy
b c
Verify first the value of f'c. If it is < or = 30 Mpa, then the value of b is 0.85, but if f'c is > 30 Mpa, make sure that the value to be input is the answer given by the formula: b = 0.85-0.008(f'c-30).
The minimum value for b is 0.65.
d'c
c-d'
0.003 c-d' c
c
1071.23 Mpa fs>fy Therefore compression steel yields !!!
lV. If compression steel yields:
=
== 806,104,599.04 N.m
Mu = 806.10 Kn.m
V.
from the strain diagram:
0.003 c-d' c
c
from the stress diagram:
T
csolve c by quadratic formula:
c= 2a
a = 6,747.30
es = fs/Es
fs/Es =
fs = 600 c-d'
fs =
Mu = Mu1 + Mu2
f T1 (d - a/2) + f T2 (d-d')
f As1 fy (d - a/2) + f As2 fy (d-d')
If fs < fy, the assumption is wrong, compression steel will not yield:
fs/Es =
fs = 600 (c-d')
C1 + C2 =
0.85 f'c a b + As' fs = As fy
0.85 f'c b1c b + As' 600 (c-d') = As fy
- b +-√b 2 - 4ax
Neglect this portion if the compression steel will not yield.
Use the positive (+) value of c.
b= 814600x= (79,365,000.00)
c= 63.76 mm-184.49 mm
solve for fs:
c
-11.69 Mpa fs<fy Therefore compression steel will not yield…….
solve for a:a =
= 51.64 mm
solve for Mu:
=
== 177,931,913.84 N.m
Mu = 177.932 Kn.m
fs = 600 (c-d')
fs =
b c
Mu = Mu 1 + Mu2
fC1 (d-a/2) + f C2 (d-d')
f0.85 f'c a b (d-a/2) + f As' fs(d-d')
Use the positive (+) value of c.
0.003
d'
c-d'
strain diagram
Verify first the value of f'c. If it is < or = 30 Mpa, then the value of b is 0.85, but if f'c is > 30 Mpa, make sure that the value to be input is the answer given by the formula: b = 0.85-0.008(f'c-30).
The minimum value for b is 0.65.
fs>fy Therefore compression steel yields !!!
fs<fy Therefore compression steel will not yield…….