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…….