Transcription Namin

7/30/2019 Transcription Namin

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DOUBLY REINFORCED RECTANGULAR BEAMS

Ok… ahmmm... We have to classify the problem whether an analysis or a design problem. This is againa section subject to a positive moment wherein you have tension at the bottom and compression at the

top. So you have your tension bar AS and your compression bar A’S, your distances d and d’and of

course the width b although not indicated here. For analysis, the actual situation is we divide this into

two couples (M1 and M2) where the first couple is composed of the force CC which is the concretecompression resultant treated as a rectangle with a corresponding tension force that will have to be equal

to Cc from the equilibrium equation that the summation of forces must be equal to zero. In USD, it is

easy to determine the location of the compression resultant because the stress distribution is rectangularwith a uniform stress of 0.85f C’. This Cc is in the center of the rectangle, so if the height of the rectangle

is “a” CC will be acting at a distance2

a from the top, therefore this will give you a level arm of 2

ad −

. So the force multiplied by this level arm gives you the Nominal Moment M 1. You have to this; youhave a couple made of a compression steel C S with a corresponding tension force TS2. The level arm will

be just the difference of “d” and d’. That level arm multiplied by the force C S will give you the second

couple M2. As usual you have to use the strain diagram. Always, the strain of concrete is 0.003 (ultimatestress design). Meaning this is the limit design and you want to know the capacity of the compression

bar if it will fail. And this will happen when concrete crushes because steel just like “tira-tira” will just

keep on elongating even if it had already yielded. When the strain here becomes 0.003, then you have

the Ultimate stage. The analysis working equations are again taken from the basic equation from thesummation of forces be equal to zero where CS + CC = TS where TS is made up of TS1 and TS2

corresponding to CS and CC. Again, take note from this drawing that TS1= AS1f y and TS2=AS2f y. In USD

what makes it simpler is that the stress in the tension bar is equal to f y so it always have to be underreinforced as prescribed by the code to ensure ductile failure. Therefore, laging mag-yi-yield ang tension

bars. REQUIREMENT NG CODE YAN!!!! SO STRESS ñan is always f y. To continue, TS=ASf y and AS

of course is made up of AS1 and AS2 because the stresses here are equal. CC is 0.85f C’ab, CS=A’S f S. From

similar triangles'

003.0

d cc

SC

−∈

= , so you will have 003.0*'

c

d cSC

−=∈ . If the strain of the steel is less

than the yield strain, it will follow the Hooke’s Law. But if the strain of the steel is greater than the yield

strain, again as I have said, you are in the flat portion; your stress will be equal to f y. You can skip the



Reinforced Concrete Design ( RCD – 1 )

Monday 5:00 – 9:00 / Thursday 6:00 – 9:00 Engr. Alberto S. Cañete - Professor

Ultimate Stress Design (USD)

Lecture Notes

2k6 – 2k7

step of comparing the strain. Instead, what you can do is simply compute this formula for the stress of

the compression steel and then check whether it is greater than the f y or not. If it is greater than f y, thennag-yield siya. Sa excel, min [E*∈ , f y]. Again, according to the procedure that we followed sa WSD, CC

+ CS = 0.85f c’ [ab – A’s] + A’S f sc{ ito ang meaning niya 0.85 fc’ times the concrete area (a*b) minus

the area occupied by the compression bar}. Rearranging, you now have 0.85f c’ (ab) + A’S (f sc-0.85f c’)With this you are now dealing with the solid rectangle and theoretically a solid circle. Therefore the

analysis for this is a de-kahon na procedure, given everything including the section dimension. Take

note, ang first choice ninyo is singly, dahil pareho lang ng design at pareho lng ng bayad. So only when

you are forced na magdodoubly kayo unless, dagdagan ang bayad mo at magba-value engineering kathen you can exert an extra effort. In reality, ang beam ninyo ay may top bars and bottom bars according

to the code, minimum ang 2 sa taas at 2 sa baba otherwise wala kang pagtatalian ng stirrups mo.

)c'f 85.0s(f A'c(ab)0.85f'

f sA's]A'-c[ab0.85f'CC

zone,concretein the barsncompressio by thedisplacedearea/volumconcreteheconsider tto:note

fyf

Ef if

E

fystrain)(yieldAlso,

0.003*cd'-c

:ianglessimilar tr from

fsA'Ccab,0.85f'C fy;AT

:Where

TCC0F

ANALYSIS

sc

scsc

scysc

scsscysc

sy

sc

sscss

scs

−+=

+=+

=→∈∈

∈=→∈∈

=∈

=∈

===

=+→=∑

>

<

Ok... Your first step is to solve for max ρ . You don’t have to compute the min ρ anymore. For this to be a

candidate for doubly, kailangan marami ang tension bars. This is an analysis problem, given ang A S, if

actual ρ < max

ρ then you can analyze it as singly because the tension reinforcements are not that much,

kayang-kaya siyang i-match nung concrete (CC) ∴ hindi na kailangan ng tulong ng A’S para lang

balansehin yung AS. Pag sinuswerte kau, then analyze it as singly. But when we say na iaanalyze sa

singly, it doesn’t mean na hindi na i-che-check ang min ρ , kailangan lagpas xa sa min ρ , otherwise you

still have brittle failure. BUT!!!!!!!!! PAG MINAMALAS-MALAS KAYO and your actual ρ > max

ρ

then you have to analyze it as doubly… ü If it is doubly, you have to compute for the ' ρ which is

equal to bd

AS '

and then check the difference between this and the'

ρ which you have to compare with

max ρ (paki-open ang code please page 4-35 article 410.4.3 last sentence).

“Classmates, yung iba ay self explanatory na lang. Binasa na lang ni sir ungformula at ung iba ay previous transcription natin. Mapapamahal kasi kungmadami ang ipiprint, nakakawindang lang naman basahin.” ÜANALYSIS:






Lecture Notes

2k6 – 2k7

( )

[ ]( ) ( ) 0''600''85.0'600f A'85.0

)('85.0)600)('(''85.0f A

)('85.0)600('

''85.0f A

)'85.0(')('85.0f A

0

600*'

003.0*'

200000

003.0*'

yield)notdo barson(compressif f If 8.)

)d'-c)(d0.85f'-s(fyA')2

a-(dcabf'0.8590.0MnMu

confirmed)isn(assumptioIf 7.)

600*'a

c

c(b)0.85f'

c)0.85f'-s(fyA'-Asfy a

)'85.0(')('85.0

C

c)0.85f'-f f yield, barsncompressio(Assume.)6

AsAsuse!ok )'-(

)bd(As 'use)'-(if 5.)

s/bdA'' 4.):DoublyFor

doubly

singlyif 3.)

As/bd .)2

1.)

:Solution

Mn Mu:Find

cf'fy,s,A'As,,d'd, b,Given

YS

2

1

21YS

1YS

YS

scsc

1sc

ysc

ysc

1

C

yscysc

actmax

maxmaxmax

max

max

max

=−+−−

−−+=

−

−+=

−+=+=

=Σ

−=

−=∈=

=−

=∈

<

+==

∈∈

−==

=

−−=−=

=∈∈=→

+=∴+=→=

→→

=

=

d AccA f Acbc f

cc f d c Abcc f c

cby sidebothmultiplyc f c

d c Abcc f

c f f Aabc f

C C T

F

cd c f

c

d c E f

cac

d c

c

d c f

c f fy s A Asfyabc f

C T

S S S

S

S

scS

S C S

sc

s

SC

S S

β

β

β

β

φ

β

ρ ρ ρ

ρ ρ ρ ρ ρ ρ ρ ρ

ρ

ρ ρ

ρ ρ

ρ

ρ

φ

A

S

ID

E






Lecture Notes

2k6 – 2k7

( )( )

−−+

−==

+±=

=

+−=

=

''85.0'2)'85.0(90.0

2A

4BB c

'A'600C

''85.0'600f AB

'85.0:

2

S

YS

1

d d c f f A

a

d cab f Mn Mu

yield not dobarsncompressio If

AC

d

cA f A

bc f A Let

scS

S S

φ

β

DESIGN :

(Doubly) R

MPaisR of unit:note(Singly) R bd

Mu R 3.)

c'f 7.1

fy*-1fyR 2.)

,1.)

:Solution

sA'&As:Find

d'moment),(designMufy,c,f'd, b,:Given

max

max2

maxmaxmax

max1

>

<φ

=

ρρ=

ρβ

s(fyA'

If c)0.85f'-(fy

fyAs sA'If .)12

Tc

If c)-0.85f'(E

fyAs sA'If .)11

E

fy &0.003

c

d'-c .)10

a c9.)

cb0.85f'

fyAs a.)8

AsAsAs7.)

)d'-fy(d

M As.)6

MMu

M5.)

bdA bdR M4.)

:DoublyFor

sc2

ysc

s

sc

scs

2ysc

s

ysc

1

1

21

22

12

maxs1

2

max1

∈=→∈∈

=

∈

∈

=→∈∈

=∈×=∈

=

=

+=

=

−=

=→=

β

φ

ρ

SAMPLE PROBLEM:

A

SI

DE



2

2

L

M8w

8

wLmaxM

=

=




Lecture Notes

2k6 – 2k7

Determine the maximum WL that the beam can carry if WDL = 10 kN/m.f’c =28Mpa using Grade 60 bars.

Solution:

W W

L

M W

Mu

Mu

Mu

bd

As

u LL

u

y sc

y

sc

8

600

9.0

0

.)7

200

41

2

2266.)

ac

3694a

AA

0'-5.)

300

402'4.)

2sA'3.)

46As2.)

0.851.)

2

1

SS

max

021686.0

(0.75x0.8

1

max

max

WW LM8W 600Mu 9.0Mu 0Mu.)7 20004122266.)

ac

3694a

AA

0.0'-5.)

300x

402.'4.)

42sA'3.)

bd

As

46As2.)

0.851.)

uLL 2u yscysc 1

SS

max

02.0max

(0.7max

1






Lecture Notes

2k6 – 2k7

T-BEAMS

Effective Width, be (NSCP Art. 418.11, p.4-26)

A. Interior Beams1. L/4

2. bw + 16ts

3. c-c spacing

B. Exterior Beams

1. bw + L/12

2. bw + 6ts3. (c-c spacing + bw)

2

IRREGULAR BEAMS






Lecture Notes

2k6 – 2k7

ANALYSIS:

Given: be, bw, d, ts, As, f”c, fy

Find: Mu = ø MnSolution:






Lecture Notes

2k6 – 2k7

1. β1, ρmax

2. Assume a = tsCc=0.85f”c ts be

Ts=Asfy3. If Ts < Cc => a < ts => singly

Ts > Cc => a > ts => irregular

If irregular

4.)

5.) Asw = As - Asf

6.) As = Asf + Asw

Mn = Mf + Mw

7.)

8.) If ρw < ρmax => ok! No reduction in Asw

If ρw > ρmax => Asw = ρmax bwd

9.)w

ysw

cb'f 85.0

f Aa =

10. Mw =

2

a-df A ysw

11. Mu = ø (Mf + Mw)

DESIGN:

Given : be, bw, ts, d, f’c, f y, MuFind: As

1.) β1, ρmax

2.) Assume a = ts

=

2

t-dt bcf'0.85[øMnø s

se

3.) If Mu < ø Mn a < ts singly

( )( )

(

)t

d(f Af

b bc'f 85

b bc'f 85.0f

sysf

e

weysf

−=

−

−=

d b

A

w

sww =



( )

−=

=

−=

2tdCM

fy

CAsf

t b bc'f 85.0C

sf f

f

swef


Monday 5:00 – 9:00 / Thursday 6:00 – 9:00

Engr. Alberto S. Cañete - Professor


Lecture Notes

2k6 – 2k7

If Mu > ø Mn a > ts irregular

If irregular

4.)

5.) f u

w M-M

M

φ

=

6.)c f

fym

'85.0=

7.) yw 2

ww

f d b

MX =

8.)m

2mX-1-1 dreq'

ww =ρ

9.) If ρw > ρmax increase beam size bw,dρw < ρmax ok! Req’d Asw = ρw bwd

10. As = Asf + Asw

SAMPLE PROBLEM:

Design an interior beam of a floor system shown having a simple span of 18m, slab thickness of

100 mm, c-c spacing of 2.5 m, stem width of 300 mm, d= 630 mm, f’c = 20 Mpa, h= 700mm and grade40 bars.

The floor system is to carry super imposed loads listed below.

Dead Loads:

1. Ceiling = 0.5 kPa

2. Floor Finish = 1.8 kPa3. Movable Partition = 1.0 kPa

Live Loads = 4.8 kPa

Solution:

Dead Loads:Ceiling = 0.5 kPa (2.5 m) = 1.25 kN/m

Floor fin = 1.8 kPa (2.5m) = 4.5 kN/m

Movable Par= 1.0 kPa (2.5m) = 2.5 kN/m

Slab = 0.1 (2.5) (24) = 6 kN/mBeam web = 24 (0.3) (0.7-0.1) = 4.32 kN/m

ΣDL = 18.57 kN/m

Live Load = 4.8 ( 2.5) = 12 kN/m :(2.5) is tributary width






Lecture Notes

2k6 – 2k7

Wu = 1.4 (18.57) + 1.7 (12)

Wu = 46.398 kN/m

( )kNm

LWu Mu 12.1879

8

18398.46

8

22

===

Effective Width

a.) m L

be 5.44

18

4===

b.) be = bw + 16ts = 300mm + 16(100mm) = 1.9m

c.) be = 2.5 m.

Therefore.. be = 1.9m (smallest value governs)

1.) β1 = 0.85

( )( )( )

+

=ρ

276600

600

276

2085.085.075.0max

ρmax = 0.026895

2.) ( ) ( )(2

'85.0 tsatsbec f Mn −=φ φ

= 0.9 (0.85) (20) (1900) (100) (630-100/2) x 10-6

= 1686.06 kNm

(compare to Mu)3.) irregular asdesign∴> Mn Mu φ

4.)

( )( )

kNm6.1577102

100630)000,720,2(Mf

mm07.9855276

2720000Asf

000,720,21003001900200.85C

6

2

f

=

−=

==

=−=

−

5.) kNm Mw 31.5106.15779.0

12.1879=−=

6.)

( )

235.15

2085.0

276

'85.0

===

c f

fym

7.)( )

( ) ( )015528.0

276630300

100031.510Xw

2

2

==

8.)( )( )

018224.016.235

015528.016.2352-1-1wdreq' == ρ

9.) ρw < ρmax ok! Req’d Asw = ρw bwd = 0.018224 (300)(630) = 3444.34 mm2

10.)

mm As

Asw As As

241.13299

34.344407.855

=

+=

+=



3-25 mm φ

Wu

0.4 Wu0.4 Wu

m0.1 m0.1m3.0 m3.0




Lecture Notes

2k6 – 2k7

QUIZ SINGLY AND DOUBLY

MPa.414 fymm,6.40d mm,3.20b

rebars.60Gradeand 40MPac f' Wu

===

=

and

Usecarry.can below bamthat theloadmaximumtheDetermineI.

At B:

3-25 mm φ

6-25 mm φ

3-25 mm φ






Lecture Notes

2k6 – 2k7

[ ]

0.0033816min

ρusethenlue,greater vatheuse

0.00338164141.4

fy1.4min

ρ

0.0038492414)(4

40

4fycf'min

ρ

0.028376414)(600

(600)

414

(0.85)(40)0.75(0.85)maxβ

.74785730)(407

0.050.85

1)β1.0

=∴

===

=

×==

=

+

=

=−−=

0.0072ρmin

ρactual

ρmax

ρ

0.0072actual

ρ

(320)(640)

(3)2(25)4

π

bd

Asactual

ρ2.0)

=∴⟩⟩

=

==

m1492.40kN/uW

335.79u0.225W

u0.225W

)u0.65W0.4Wu(22

6

1B

M

u2.8WBR

kN/m335.79φMu

)6-10(1)2

56.04-(6402)(414)0.9(1472.6

)2

aφAsFy(d4.0)φ.0

56.04(320)(0.85)(40)

4)1472.62(41

cb)(0.85)(f'

AsFy 3.0)

=

=

=

+×=

=

=

×

=

−=

=

=

At C:

6-25 mm φ






Lecture Notes

2k6 – 2k7


Monday 5:00 – 9:00 / Thursday 6:00 – 9:00


Ultimate Stress Design (U

Lecture N

2k6

0.014(320)(640)

(6)2(25)4

π

bd

Asactual

ρ

0.0038192min

ρ

0.028376axρ

0.778571

β1.0)

===

=

=

=

178.01kN/3.6

640.84

uW

u3.6W

BR

640.84kN/m

(64)(414)0.9(2945.2

)2

a(dφAsFyφMu

11220)0.85(40)(3

4)2945.24(41a 0.014ρ minρactualρmaxρ

=

=

=

=

−=

=

=∴

⟩

QUIZ 5: DOUBLY REINFORCED RECTANGULAR BEAM & IRREGULAR SECT

1.) Determine the maximum safe live load WL that the beam below can carry. Use f’c = 32 and Grade 60 bars. Assume that the total dead load WD = 25 kN/m.m-kN73.26910*

2

815.47600*414*6.1256*9.0M

mm815.47400*32*85.0

414*6.1256

0052358.0

600*400

6.1256,mm6.12564*)20(

4A

0034160.0

0033816.0414

4.1

0034160.0414*4

32

024367.0414600

600*414

32*85.0*83571.0*75.0

83571.0)3032(7

05.085.0

SLIDER AT CAPACITY MOMENT

6U

min

max22S

min

min

min

max

1

=

−=

==

><

====

===

==

=+=

=−−=

−

a

ρ

ρ ρ

π

ρ

ρ

ρ

ρ

β

( )

30.72210*2

58.140595*414*5.3694*9.0M

mm58.140400*32*85.0

414*5.3694

Singly015523.0595*400

5.3694

mm5.36946*28*4

A

SUPPORT FIXED AT CAPACITY MOMENT

6U

max

22S

=

−=

==

∴<==

==

−

a

ρ ρ

π




Monday 5:00 – 9:00 / Thursday 6:00 – 9:00



Lecture Notes

2k6 – 2k7 kN/m3788.2

7.1

25*1.4-30.956WL

kN/m30.956w

kN/m956.3030.722333.23

333.236667.3)75.1*2(6

6

kN/m562.7373.2696667.3

6667.3)275.1*2(6

2

6667.3)75.12*2(6

22*75.3

75.32*2

275.1R

ANALYSIS

2

2

2

R

−==

=∴

=⇒=

=−+=

=⇒=

=+=

=+−=

=+

=

ww

wwww M

ww

www M

wwww M

www

neg

pos

pos






Lecture Notes

2k6 – 2k7


Monday 5:00 – 9:00 / Thursday 6:00 – 9:00



Lecture Notes

2k6 – 2k7

2.) Determine the ultimate moment capacities of the sections shown below. Use f’c = 23 MPa

and Grade 40 bars.

030929.0276600

600

276

)23(85.0)85.0(75.0

85.0

mm450 b

mm250 b

max

1

w

e

=

+

=

=

=

=

ρ

β

m-kN88.265102

100730)276(67.1416

mm67.1416276

)100)(450250)(23(85.0A

IRREGULAR 1.812887)276)(6()25(

4

488750)100)(250)(23(85.0

ta assume

6

2sf

2

s

−=

−−=

−=−

=

∴>==

==

=

− x M

C T T

C

f

C S S

C

π

m-kN53.477)47.79688.265(9.0

m-kN47.7962

84.136730)276(9.4361

mm84.136)450)(23(85.0

)276(9.4361

Ainreductionnook!

013278.0)730(450

9.4361

9.436167.1416)6()25(4

swmax

2

=+−==

−=

==

∴<

==

=+=

u

w

w

w

sw

M

M

a

A

ρ ρ

ρ

π

)governs!(lowestmm445 b

mm4452

150)120470(

b2

1

spcg)cc(2

1

b b.)

mm600)80(61206t b ba.)

BeamExterior

150)80(6

e

we

swe

=∴=

++=+=

=+=+=

>=eb

2sf

6

2sf

22

s

max

1

mm324.137567.1841991.3216A

m-kN80.477102

80980)276(67.1841

mm67.1841276

)80)(120445)(23(85.0A

IRREGULAR 516.887889)276(991.3216

695980)445)(80)(23(85.0

mm991.3216)4()32(4

ta assume

03093.0)04124.0(75.0

04124.0276600

600

276

)23(85.085.0

85.0

=−=−=

=

−=

=−

=

∴>===

==

==

=

==

=

+

=

=

−

sw

s sw

f

C S sS

C

s

bal

A A A

x M

C T fy AT

C

Aπ

ρ

ρ

β






Monday 5:00 – 9:00 / Thursday 6:00 – 9:00


Lecture Notes

m-kN4.1474

2xm-kN18.737

)29.34180.447(90.0

)(

m-kN29.34110

2

80286.161980)276(324.1375

mm80286.161)120)(23(85.0

)276(324.1375

Ainreductionnook!

011695.0)980(120

324.1375A

6

swmax

sw

=

=

+=

+=

=

−=

==

∴<

===

−

u

u

u

w f u

w

w

ww

M

M

M

M M M

x M

a

d b

φ

ρ ρ

ρ

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Transcription Namin