design of 1 way concrete slab

8/10/2019 Design of 1 Way Concrete Slab

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L c Hi Long-BKHCM-81101897

A. Data conversion:

- Shorter span L1=2.3m=7.5ft

-

Longer span L2=5.8m=19ft

- Dimensions of columns: 300x300mm=1x1ft

- Tensile strength of CII, AII bar: Rs=280Mpa=40000psi

- Compressive strength of concrete: Rb=11.5Mpa=1667psi

- CII, AII bars are equivalent to Grade 40 bars designated by ASTM.

- Average unit weight of reinforced slab: c=25kN/m3=156.07 lb/ft3=slab

- Unit weight of mortar bed: v= 20kN/m3=124.86 lb/ft3mor

- Thickness of mortar bed: 2-3cm=0.79-1.18in => choose 1in for the thickness of

mortar bed.

- Load of brick layer: gc=0.4kN/m2=0.058psi=8.354psf

- Live load: 9kN/m2=1.3psi=187.92psf

B. Slab

1.

Slab classification:

- The ratio of longer span to shorter span is 1

2

192.5 2

7.5

L

L so this slab should

be considered as one-way slab.

2. Trial determination of slab thickness:

- According to ACI code table 9.5(a), because the deflection is ignored, the

minimum thickness of slab is: (0.4 )21 100000

yfLh

where: L is span of slab.

fy is tensile strength of reinforcing bar.


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fyis tensile strength of reinforcing bar.


- Because a trial width of beam of 8in will be used, the clear span of slab is ln=l1-

bbeam= 7.5-

8

12 =6.83ft.

Maximum negative moment at face of all supports:

M1= 21

12 u n

w l = 21

390.3 6.83 1517.312

x x ft lb (=2.1kN/m)

Maximum positive moment at end span:

M1= 21

14

u nw l =

21390.3 6.83 1300.5

14


Maximum positive moment at interior span:

M1= 21

16 u n

w l = 21

390.3 6.83 113816


4. Determination of steel area:

Assume that the effective depth of slab is 2.5inch (=63.5mm)

Unit Face of

support

First span Second span

Mu ft-lb 1517.3 1300.5 1138

Assumed 0.9 0.9 0.9

R(2) psi 269.7 231.2 202.3

7.55x10-3 6.35x10-3 5.48x10-3

Asper foot (4) in2 0.2265 in2 0.1905 in2 0.1644 in2

ar diameter mm 8 8 6&8


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(1): is strength reduction factorand varies linearly between 0.65 and 0.9 based

on net tensile strain t. In this case, should be assumed to be equal to 0.9.

(2): R is flexural resistance factor and is calculated as: R= u2

M

.b.d

Where Mu factored load moment

is strength reduction factorand is assumed to be 0.9.b and d are the width and effective depth of one strip the slab which are

12in and 2.5in

(3): is reinforcement ratio and is, in this case, taken from the equation:

R=fy(1-0.59'

fy

cf

Where R is flexural resistance factorcalculated abovefy is tensile strength of steel, which is equal to 40000psi.

fc is compressive strength of concretewhich is equal to 1667 psi.

sis steel area per 1 foot of the slab and is computed as s=.b.d

Where is reinforcement ratio, taken by solving the equation above

b and d are the width and effective depth of one strip of the slab which are

12in and 2.5in

(5): Spacing between bars is calculated as s= s of one bar

s per one foot

A

A

(6): According to ACI 10.5.4 and ACI 10.6.4, maximum spacing of bars is not more

than 3 times the slab thickness 18in or the value [15x(40000

) 2 5c ] (where c is


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(7) As.minis provided by ACI code 10.5.1 as equal to

'

2.min

16673 3 8 18.2 0.44

40000c

s w

y

fA b d x in

f

Where fy is tensile strength of steel, which is equal to 40000psi.

fc is compressive strength of concrete which is equal to 1667 psi. bwis the web width of the beam.

d is the effective depth of the beam.

(8): according to ACI code 7.7.1(c)the minimum concreted cover is 3/4in,therefore

the effective depth of slab is taken as d=h-3/4-

a is the depth of compressive stress block and is calculated as

'a

0.85

s y

c

A f

f b

Where As is steel areafy is tensile strength of steel, which is equal to 40000psi.

fc is compressive strength of concrete which is equal to 1667 psi. b is width of a strip of the slab, which is equal to in

(10) c is the distance to neutral axis, calculated as c= a

0.85

Where a is the depth of compressive stress block computed above.

(11) t is net tensile strain and is calculated ast

t u

d c

c

Where uis compressive strain limit, which is equal to 0.003.


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where Vc and Vnare contbeamutions of concrete and steel

in shear resistance of beam, relatively.

However, shear reinforcement is not possible in slab, it leads to that Vn=Vcor

VuVn VuVc.

- ACI code 11.2.1provides that Vc=2 '

cf bwd

Where, is light-weight concrete modification factor and is equal

to 1 in case (because the

normal-weight concrete is used).fc is compressive strength of concreteis equal to 1667 psi in thiscase.

bw, d are the width and effective depth of slab.

Vc=2x1x 1667 xx2.6=2547lb

- According to ACI code moment coefficients:

1 1

1.15 1.15 390.3 6.83 15322 2u u nV w l x x lb

- The design strength of concrete slab Vc=0.75x2547=1910 lb (=8.5kN)is well

above the total shear force, so the slab is safe from shear cracking.

6. Temperature and shrinkage bars:

-ACI code 7.12.2specifies the minimum temperature and shrinkagereinforcementratio to gross concrete area (concrete area calculated based on total depth rather

than effective depth) is 0.002.

Design ith bars no 6


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7. Bars cut-off:

a. Bar cut-off at spans:

Unit Exterior span Interior span

Cut off bars

Percentage of remaining bars (1)

Theoretical cut-off point (2) ft 1.57 1.78

Actual cutoff point (3) ft 1.25 1.5

Development length (4) ft 0.65 0.36

Actual cutoff point (5) ft 2.77 3

First/second requirement controls ft First First

Chosen cutoff point ft 1.25 1.5

Location of cut off poin t is taken f rom face of support

(1)ACI code 12.11.1provides that at least one-fourth of the positive moment

steel be continued uninterrupted along the same face of a beam a distance at

least 6in. in to the support. This means that three-fourth of positive moment

steel can be cut-off.

(2)Use reference 1, graph A.3, Appendix A to determine the theoretical cut-off

point.

(3)ACI code 12.10.3 expresses that bars must extend beyond the point at which

it is no longer required to resist flexure for a distance equal to d (d is the

effective depth of slab) or 12db (db is diameter of bar), whichever is greater. In


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t is reinforcement location factor. There are not more than 12in

of fresh concrete beneath the horizontal bars (there is no horizontal

bars in slab and the slab thickness is less than 12in.) sot

is equal

to 1.

e is coating factor. The bars are uncoated so

e is equal to 1.

s is reinforcing size factor. The bars are smaller than bars

number 19 sos

is equal to 0.8.

is light-weight aggregate factor. The normal-weight concrete is

used so is equal to 1.

cbis concrete cover and is equal to 3/4 in.

dbis diameter of bar and is equal to 3.14in.

Ktris, to be simplified, permitted to be zero.

b. Bar cut-off at support

Unit Support

Cut off bars

Percentage of remaining bars

Theoretical cut-off point ft 1.57

Actual cutoff point ft 1.25

Development length ft 0.65

Actual cutoff point ft 2.77

First/second requirement controls ft First


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(1)ACI code 12.12.3states that at least one-third the total tensionreinforcement provided for negative moment at a support shall have an

embedment length beyond the point of inflection not less than d, 12db, orln/16, whichever is greater.Using reference 1, graph A.3, Appendix A to determine the theoretical

cut-off point

(2)ACI code 12.10.3 expresses that bars must extend beyond the point at

which it is no longer required to resist flexure for a distance equal to d (d

is the effective depth of slab) or 12db (dbis diameter of bar), whichever is

greater. In this case, 12db=12x0.314=3.768in=0.314ft is chosen.

(3), (4) However, ACI code 12.10.2provides that full development length

musts be provided beyond critical section (critical section is the point of

maximum stress or the point at which the bars are terminated).

+ The development length, according to ACI code 12.2.3, is taken from:

'

3

40 (

y t e sd b

b trc

b

fl d

c Kf

d

In this case:


of fresh concrete beneath the horizontal bars (there is no horizontal

bars in slab and the slab thickness is less than 12in.) sot

is equal

to 1.


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c. Bars anchorage and hook:

-ACI code12.12.1 states that negative moment reinforcement in acontinuous, restrained, or cantilever member, or in any member of a rigidframe, shall be anchored in or through the supporting member byembedment length, hooks, or mechanical anchorage.

- The development length of hooked bars is taken from ACI code 12.5.2

that

'

0.02 0.02 1 400000.314 6.15

1 1667

e y

dh b

c

f x xl d in

xf

A development length of 6.5in is chosen.

- The hook is designed according to ACI code 7.1.1 and 7.2.1that the bars

are bent 180oplus 4dbextension at the free end but not less than 2.5in andthe minimum diameter of bend is not less than 6db.

- An extension of 2.5in and a minimum diameter of 2in are chosen.

C. Secondary beam:

Choose a cross-sectional dimension of beam of 8x22in

1. Load computation:

Dead load from weight of beam:

cxbbeamx(hbeam-hslab)=

22 3.5156.07 0.67 ( )

12 12 =161.21lb/ft

Dead load transmitted from slab:

74.687.5 560 /

1 1

slabDL xl x lb ft


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Combination 1

Combination 2

Combination 3

Combination 4


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Moment diagram:

Shear force diagram


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3. Determination of steel area:

a. Determination of steel area at span:

Assume that the effective depth of beam is 20.2in:

Unit First span Second span Third span

Mu ft-kip 61.02 74.91 77.95

bf(1) in 57 57 57

Trial As (2) in2 1.10 1.35 1.41

Trial a (3) in 0.54 0.66 0.70

Location of neutral axis(4) In flange In flange In flange

Assumed t(5) 0.02 0.02

Redistributed Mu(6) ft-kip 59.93 62.36

Assumed 0.9 0.9 0.9

R(8) psi 35 34 36

8.86x10-4 8.60x10-4 9.12x10-4

As (10) in2 1.02 0.99 1.05

Practicals in2

As>As.min(11) yes yes Yes

Verified d (12) in 20.23 20.23 20.23


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ACI code 18.2provides that:

+ The flange should not exceed span length,

+ The effective overhanging flange should not exceed 8 times the slab thickness+ The effective overhanging flange should not exceed the clear distance to the

next beam.

In this case, the flange width is the smallest of:

1

2

1 119 4.75 57

4 4

min 8 2 8 2 3.5 8 64

2 7.5 902

f slab w

xl x ft in

b x xh b x x in

l x ft in

The width of 57in is chosen.

(2), (3), (4) To determine whether the neutral axis is in flange or in web:

+ Assume that the stress-block depth is equal to the slab thickness

+ Take the steel area Asfrom the equation:

( )2

us

y

MA

af d

Where:

Muis factored moment.

is strength reduction factor, and is assumed to be equal to 0.9

fy is tensile strength of steel, equal to 40000psi, in this case.D is effective depth of beam.

A is the stress-block depth of the beam, and was assumed to beequal to the slab thickness.


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maximum of 20 percent, provided that tat the section where moment is reduced is not

smaller than 0.0075 and moment is calculated by exact method.

(7): is strength reduction factorand varies linearly between 0.65 and 0.9 based

on net tensile strain t. In this case, should be assumed to be equal to 0.9.

(8): R is flexural resistance factor and is calculated as: R= u2

M

.b.d

Where, Mu is factored load moment

is strength reduction factorand is assumed to be 0.9.b and d are the width and effective depth of the beam which are 57in and

18.2in

(9): is reinforcement ratio and is, in this case, taken from equation:

R=fy(1-0.59'

fy

cf

Where R is flexural resistance factoras calculated abovefy is tensile strength of steel, which is equal to 40000psi.

fc is compressive strength of concrete which is equal to 1667 psi.

sis steel area of the beam and is computed as s=.b.d

Where is reinforcement ratio, taken by solving the equation above

b and d are the width and effective depth of beam, which are 57in and

18.2in

(11)As.minis provided by ACI code 10.5.1 as equal to


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'a

0.85

s y

c

A f

f b

Where As is steel area

fy is tensile strength of steel, which is equal to 40000psi.

fc is compressive strength of concrete which is equal to 1667 psi. b is width of beam, which is equal to 57in.

(14) c is the distance to neutral axis, calculated as c= a

0.85

Where, a is the depth of compressive stress block computed above.

(15),(16) t is net tensile strain and is calculated ast

t u

d c

c

Where uis compressive strain limit, which is equal to 0.003.

td is he depth to the centroid of the farthest steel layer of the beam.

c is the distance to neutral axis as calculated above.

ACI code 10.3.5expresses a minimum net tensile strain t of 0.004 to ensure thebeam is underreinforced.

(17) is exactly calculated to verify its assumed value. In this case, tis greater than

0.005, therefore, is verified as 0.9.

(18) Spacing between bars is calculated as s=(bw-2cc-ndb)/(n-1)

Where bwis web width of beam.

ccis thickness of concrete cover.

dbis bar diameter.n is the number of bars.

(19) ACI d 7 6 1 t t th t th i i l i b t ll l b i


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Unit First support Second

support

Third

support

Factored load moment Mu ft-kip 117.66 105.77 113.12

Assumed 0.9 0.9 0.9

R psi 490 441 471

0.0157 0.0137 0.0149

max (1) No No no

As1(2) in2

0.716( 0.477( 0.477(

Moment resistance contributed

by compression steel (3)

ft-kip 40.22 26.80 26.80

Remaining moment (4) ft-kip 77.44 78.97 86.32

Assumed 0.9 0.9 0.9

R kip 323 329 360

9.30x10-3 9.50x10-3 0.0106

As2 in2 1.488 1.52 1.70

Tensile steel area (5) in2

Actual As in

2

As>As min yes Yes Yes


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- Determination of negative moment reinforcement area is done in the same way asthat above.

(1): The maximum reinforcement ratio maxis:'

max 1

1667 0.003=0.85 0.85x0.85 0.0129

0.004 40000 0.003 0.004

c u

y u

f

f

Where:1

is equal to 0.85

fy is tensile strength of steel, which is equal to 40000psi.fc is compressive strength of concrete which is equal to 1667 psi.

u is compressive strain limit, and is equal to 0.003.

If is greater than max, the beam must be considered as doubly reinforced.

(2): Some bars at span should continue to support to become compression

reinforcement and the area of those bars is As1.

(3) Moment resistance contributed by compression steel is calculated as:'

1 1= A ( )s yM f d d

Where:1

As

is compression steel area.

fy is tensile strength of steel, which is equal to 40000psi.

d is effective depth of beam

dis distance from centroid of compression steel to compressive

face of beam.

(4) Total moment is partly resisted by compression reinforcement, therefore the

remaining moment is2 1=

uM M M


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b is width of beam, which is equal to 8in.

(1)Strain of compression steel could be calculated as:'

'

s u

c d

c

Where: c is distance from compressive face of concrete to neutral axis.

d distance from compressive face of concrete to closest layer of

compression steel.

uis strain of compression face of concrete.

Compression steel can be considered to be yielded if its strain is greater than

400000.0138

29000000

y

y

s

f

E

Where: yis yield strain of steelfy yield strength of steel, which is equal to 40000psi

Esis modulus of elasticity of steel, which is equal to2900000psi

c. Bars cut-off

Cut-off of bars at span:


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Unit First span Second span third span

left right

Cut off bars 2 1 2 1 2 1 2

Remaining bars 3 2

2

2 2

2

2 2

2

2

Remaining As in2 0.716 0.828 0.477 0.828 0.477 0.828 0.477

a after bars being cut-off in 0.35 0.41 0.24 0.41 0.24 0.41 0.24

d after bars being cutoff in 20.73 20.09 20.73 20.09 20.73 20.09 20.73

Mnof remaining reinforcement ft-kip 44.15 49.74 29.49 49.74 29.49 49.74 29.49

Theoretical

cut off point

Left side of the span ft 7.42 7.28 5.4 6.62 4.86

Right side of the span ft 5.5 3.74 6.56 4.68 6.62 4.86

Actual

cutoff point



Development length ft 1.48 1.48 1.48 1.48 1.08 1.48 1.08

Actual

cutoff point



First/second requirement controls first First first first first

First first First first first first

Chosen

cutoff point

Left side of the span ft 5.5 5.5 3.5 4.5 3

Right side of the span ft 3.5 2 4.5 2.5 4.5 3

Location of cut off poin t is taken f rom center line of support

ACI code 12.11.1provides that at least one-fourth of the positive moment steel


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ACI code 12.10.3 expresses that bars must extend beyond the point at

which it is no longer required to resist flexure for a distance equal to d (d

is the effective depth of beam) or 12db (dbis diameter of bar), whichever isgreater. The distance of d (the effective depth of the beam) controls in this

case.

+ However, ACI code 12.10.2also provides that full development length

must be provided beyond critical section (critical section is the point of

maximum stress or the point at which the bars are terminated).

+ The development length, according to ACI code 12.2.3, is taken from:

In this case:


of fresh concrete beneath the horizontal bars (the concrete cover is

just 1in) sot

is equal to 1.

e

is coating factor. The bars are uncoated so e

is equal to 1.

s is reinforcing size factor. The bars are smaller than bars

number 19 sos

is equal to 0.8.

'

3

40 (

y t e sd b

b trc

b

fl d

c Kf

d


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Cut-off bars at support

Unit First support Second support Third supportBars being cut off 1 2 1 2 1 2

Remaining bars 2

+2

2 2

+2

2 2

+2

2

Remaining steel area in2 1.762 0.973 1.597 0.973 1.762 0.973

a depth after bars being cut-off In 6.28 3.43 5.64 3.43 6.28 3.43

d after bars being cut-off in 19.82 20.61 19.94 20.61 19.82 20.61

Moment capacity of remainingreinforcement ft-kip 88.17 55.15 82.02 55.15 88.17 55.15

Theoretical cut

off point

Left side of support ft 0.86 1.83 0.84 1.98

Right side of support ft 0.96 2.08 0.8 1.71 0.83 1.94

Actual cutoffpoint


Right side of support ft 2.61 3.80 2.46 3.79 2.48 3.66

Development length ft 3.03 2.46 3.03 1.94 3.03 2.46

Actual cutoff

point


Right side of support ft 3.53 3.42 3.53 2.74 3.53 3.2First/second requirement controls second first second first

second first second first second first

Chosen cutoff

point

Left side of support ft 3.5 4 3.5 4

Right side of support ft 3.5 4 3.5 4 3.5 4

Inflection point Left side of support ft 4.57 4.5

Right side of support ft 4.5 4.06 4.36Cut-off point

beyond inflectionpoint

Left side of support ft 6.29 6.22

Right side of support ft 6.22 5.78 6.08

Actual cutoff

point of remaining

bars


Right side of support ft 5.11 4.74 4.97


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ACI code12.12.1 states that negative moment reinforcement in a continuous,restrained, or cantilever member, or in any member of a rigid frame, shall be

anchored in or through the supporting member by embedment length, hooks, ormechanical anchorage.The development length of hooked bars is taken from ACI code 12.5.2that

'

0.02 e ydh b

c

fl d

f

ACI code 12.5.3permits development length of hooked bar to be multiplied with

the modification factors.

In this case, being inside the spandrel beam, the side cover of those bars is well

greater than 2.5in and the 180 degree hook is used so ACI code 12.5.3(a) is

applicable, which means that the development length of hooked bar can be

multiplied with 0.7.

(*) Theoretically, 2 bars no.20 at top layer can be cut off, however, they should,

practically, continue through the span in order to keep stirrups in place and

provide additional negative moment capacity.

Design of hook:

The hook design is controlled by ACI code 7 1 1 and 7 2 1 that the bars are bent


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Unit Left sideof first

span

Right sideof first

span

Left sideof second

span

Rightside of

second

span

Leftside of

third

span

Maximum shear force kip 33.26 30.58 32.53 32.48 32.98

Location where Vu=0 ft 10.74 9.8 10.4 10.4 10.6

Modified maximum shear

force (1)

kip 28.10 25.38 27.32 27.27 27.79

Strength reduction factor 0.75 0.75 0.75 0.75 0.75

Vc(3) kip 13.07 13.07 13.07 13.07 13.07

smax (5) in 10 10 10 10 10

Vuof smax kip 19.04 19.04 19.04 19.04 19.04

Location where smax can be

applied

ft 4.4 3.7 4.3 4.3 4.5

s1(4) in 5.00 5.93 5.27 5.29 5.14

Practical s1 in 5 5.5 5 5 5

Length of distribution of s1 ft 3 3 3 3 3

Number of stirrups 8 7 8 8 8


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where Vuis total shear force due to factored loads.

Vnis nominal shear strength and is taken as Vn=Vc+Vs.

where Vc and Vsare contributions of concrete andsteel in shear resistance of beam, relatively.

If shear force is greater than contributions of concrete in shear resistance, web

reinforcement is required.

(3)Contribution of concrete to moment resistance is, according to ACI code 11.2.1.1,

calculated as:

'

w2c cV f b d

Where:

is light weight concrete factor, and is equal to 1 (according to ACI code

8.6.1)

fcis compressive strength of concrete, and is equal to 1667.bw, d are the width and effective depth of beam.

(4)Spacing between web reinforcement is, according to ACI code 11.1.7.2, computed as:

v yt v yt

s u c

A f d A f ds

V V V

Where: is strength reduction factor.

Av,minis total cross-section area of 1 stirrup.fyt is yield strength of web steel.

d is the effective depth of beam.

V and V are contributions of concrete and steel in shear resistance of


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fcis compressive strength of concrete, and is equal to 1667.

bwis the width of beam.

fyt is yield strength of web steel.

s is spacing of web reinforcement.

Therefore, spacing of web reinforcement provided by Av,minis:

,min ,min

' 500.75

v yt v yt

wc w

A f A fs

bf b

The maximum spacing between web reinforcement is the smallest of:

,min ,min

' 500.75

/ 2

24

v yt v yt

wc w

A f A f

bf b

d

in

15.4

10

24

in

in

in

smax =10in is chosen.

At the regions of beam where VuVc, according to ACI code 11.4.6.1, the stirrups are

not required.

However, in order to keep the longitudinal bars in place and provide additional shear

capacity, it is reasonable to continue the stirrups through those regions at the maximumspacing of 10in.


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ACI code 8.11.2 limits various combinations of load to 2 cases:+ Factored dead load on all spans with full factored live load on two adjacent

spans; and+ Factored dead load on all spans with full factored live load on alternate spans.

Combination 1:

Combination 2:

Combination 3:


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Location of cross-section unit first span second span

Mu ft-kip 226.90 227.70

Flange width in 67.5 67.5

Trial As in2 2.90 2.91

Trial a in 1.21 1.22

Location of neutral axis in In flange In flange

Assumed t 0.02

Redistributed Mu ft-kip 182.16

Assumed 0.9 0.9

R psi 58 46

1.48x10-3 1.17x10-3

As in2 2.77(=1735mm2) 2.19(=1426mm2)

Practical s in2

As>As.min yes Yes

Verified d in 27.9 28

a in 1.17 0.92

c in 1.39 1.08


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Factored load moment Mu ft-kip 380.81 339.01

Assumed 0.9 0.9

R psi 559.5 498

0.0192 0.0161

max No no

As1 in2

1.58( 1.25(

Moment resistance contributedby compression steel

ft-kip 123.74 98.04

Remaining moment ft-kip 257.07 240.97

R psi 377.7 354

0.0112 0.0104

As2 in2 3.696 3.432

As in2 mm2) mm2)

Actual As in2

As>As.min yes Yes

Validated d 27.67 27.7a in 8.98 8.14

c in 10 57 9 58


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Cut off bars 2 2 2 2 2 2

Remaining bars 2

4

4 2 2

4

4 2

Remaining steel area in2 2.2 1.58 0.79 1.72 1.24 0.62

a after bars being cut-off in 5.17 3.72 1.86 4.05 2.92 1.46

d after bars being cutoff in 28.17 28.65 28.65 28.24 28.69 28.69

Moment capacity of remaining

reinforcement

ft-kip 168.86 127.00 65.70 135.27 101.30 52.01

Theoretical cut

off point

Left side of the span ft 8.83 7.10 6.23 6.02 5.47 4.68

Right side of the span ft 6.4 5.61 4.45

Actual cutoff

point



Development length ft 1.94 1.94 2.46 1.48 1.48 1.94

Actual cutoff

point



First/second

requirement

controls

Left side of the span ft First first first first first first

Left side of the span ft First First first

Chosen cutoff Left side of the span ft 6 4.5 3.5 3.5 3 2


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Unit First support Second support

Bars being cut off 2 2 2 2

Remaining steel area in2 4.22 3.04 3.54 2.36

a after bars being cut-off in 9.93 7.15 8.38 5.55

d after bars being cut-off in 27.97 28.51 27.95 28.57

Moment capacity of remaining

reinforcement

ft-kip 291.24 227.41 252.33 182.63

Theoretical cut

off point


Right side of support ft 1.25 2.14 1.28 2.31

Actual cutoff

point



Development length ft 4.59 4.59 4.59 4.59

Actual cutoff

point



Requirement

controls

Left side of support second second

Right side of support second second second second

Chosen cutoff

point


Right side of support ft 5.5 6 5.5 5.5

Inflection point Left side of support ft 5.3

Right side of support ft 4.5 5Cut-off point beyond inflection point ft 7.68

ft 6.88 7.38


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Bar number: 22 25

Development length with hook in 16.97 19.20

Factored development lengthwith hook(*) in 9.50 10.75

Chosen development length in 9.5 11

Diameter of bend in 5.5 6

Length of extension in 3.5 4

(*) The development length of hooked bar in this case is too long that multiplying

with modification factor 0.7 does not make it small enough to place in girder,

therefore, stirrups should be provided perpendicularly to bars with spacing not

less than 3dbto make modification of 0.8 of ACI code 12.5.3appliciable.

d. Design of stirrups:

No.10 double-legs stirrups should be used.

Unit Left side of

first span

Right side of

first span

Left side of

second span

Maximum shear force kip 71.61 64.06 67.90

Strength reduction factor 0.75 0.75 0.75

Vc kip 27.39 27.39 27.39

s1 in 4.00 4.70 4.32


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No design method is given by ACI code, however, ref. 5 (reinforcement concrete

mechanics and design 6e) proposes a method from Canada concrete Code, that is:

In addition to the stirrups provided for shear in beam, hanger stirrups with atensile capacity of:

should also be provided within a length of bw2+h1/2+2(h1-h2) in main beam and

dbeam/4 in beams adjacent to each face of secondary beam.

Where, Ah is area of hanger reinforcement adjacent to one face of main

beam,

h1 is total depth of main beam

h2 is total depth of secondary beam

d is effective depth of secondary beam.

Vu2is total factored shear.

Choose bars no.10 for stirrups.

Shear force kip 62.198

Area of hanger stirrups for beams in2 1.52

Number of stirrups 7

Length of distribution in 35

spacing in 5

Area of hanger stirrups for beam in2 0.76Number of stirrups 4

Length of distribution in 5.5

i i 1 3

2

ribh yt u

beam

hA f V

h


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References:

(1)

ACI code 318-08 Building code requirements for Structural concrete.(2)ACI detailing manual 2004.

(3)Arthur H.Nilson, David Darwin, Charles W.Dolan,Design of concrete structure, 7th

edition.

(4)Jack C.McCormac, James K.Nelson,Design of reinforced concrete, 7thedition.

(5)James K.Wright, James G.McGregor,Reinforced concrete mechanics and design, 6th

edition.

(6)

V B Tm, Hc Duy, n mn hc kt cu b tng, sn sn ton khi loibn dm.

(7)Nguyn nh Cng, Tnh ton thc hnh Cu kin b tng ct thp.


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design of 1 way concrete slab

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