Designing of Flat Plate Corner Panel by DDM

7/27/2019 Designing of Flat Plate Corner Panel by DDM

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Assignment No. 4 [CE 405: REINFORCED CONCRETE DESIGNII]

Muhammad Ahsan Khan|CE119 1

DESIGN OF CORNER FLAT PLATE PANEL

24.5

C1 C2

28.25

C5 C6

l1= 28.25

l2= 25.50

Columns are assumed to be 24 x 24 in2

CHECKING LIMITATIONS FOR USING DIRECT DESIGN

METHOD

1.

There are minimum three continuous spans in each direction OK2.

The panels are rectangular and ratio of longer to the shorter span is 1.154 OK3.

Adjacent spans in each direction do not differ by more than 1/3 of the longer span OK

4.

Columns are not offset from either axis between centerlines of successive columns OK

5.

Loads are uniform, and ratio of unfactored live to unfactored dead load is 0.44 OK

DETERMINING SLAB THICKNESS FOR FLAT-PLATE


2/20



From table 9.5(c) of the ACI Code, the minimum thickness of slab without interior beam,

without drop panel, exterior panel, with edge beam and fy= 60ksi,

h =

ln= (28.25*12 24) = 315 in

h =5

= .55

We take hs= 10 in

CALCULATING FACTORED LOADS

wL= 60 psfAs we take in assignment 1 and 2

wD= 10 psf + 150 pcf *

ft = 135 psf where 10 psf is dead load of vinyl tiles and 150

pcf is unit weight of concrete for slab

wU= 1.2(135) + 1.6(60) = 258 psf = 0.258 ksf

CHECKING TWO-WAY AND ONE-WAY SHEARS

A.Corner Column (C1)a)

Two-Way Shear

Assuming concrete cover and #5 bars, so average effective depth d is 0 0.75 5/8 = 8.625 in

b0= x + y

where,

x = 24 + 8.625/2

y = 24 + 8.625/2

so,

b0= 2(24 + 8.625/2)

b0= 56.25 in

VU= (XY xy) wUVU= (13.25*15.125 2.359375*2.359375)(0.258)VU= 50.27 kips


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VC= (2 +

) b0d

VC= (2 +) (1) 000(56.25) (8.625)

VC= 185.33 kipsOR

VC= b0d

VC= 4(0.75)000(56.25) (8.625)VC= 92.67 kipsOR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(56.25) (8.625)

VC= 117 kips

So VC= 92.67 kips

Therefore VC> VU --- OKb)

One-Way Shear

Consider 1 ft strip at a distance d from the face of the column with the length

x,

x = 15.125 2 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b d

VC= (0.75) (2*1000) 12*8.625VC= 9.819 kips

Therefore VC> VU --- OK

B.Exterior Column (C2)


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a)Two-Way Shear


b0= x + 2y

where,

x = 24 + 8.625

y = 24 + 8.625/2

so,

b0= (24 + 8.625) + 2(24 + 8.625/2)

b0= 89.25 in

VU= (XY xy) wUVU= (24.5*15.125 2.359375*2.71875)(0.258)

VU= 93.95 kips

VC= (2 +

) b0d

VC= (2 +

) (1) 000(93.95) (8.625)

VC= 307.49 kipsOR

VC= b0d

VC= 4(0.75)000(93.95) (8.625)VC= 153.75 kipsOR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(93.95) (8.625)

VC= 182.73 kips

So VC= 153.75 kips


b) One-Way Shear


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x,

x = 15.125 2 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b d

VC= (0.75) (2*1000) 12*8.625VC= 9.819 kips


C.

Exterior Column (C5)a)

Two-Way Shear

Assuming concrete cover and #5 bars, so averageeffective depth d is 10 0.75 5/8 = 8.625 in

b0= 2x + y

where,

x = 24 + 8.625/2

y = 24 + 8.625

so,b0= 2(24 + 8.625/2) + (24 + 8.625)

b0= 89.25 in


VC= (2 +

) b0d

VC= (2 +

) (1) 000(89.25) (8.625)

VC= 292.11 kipsOR

VC= b0d

VC= 4(0.75)000(89.25) (8.625)


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VC= 146 kipsOR

VC= (

+ 2 ) b0d

VC= 0.75 (.

. + 2 ) (1) 000(89.25) (8.625)VC= 178.9 kips

So VC= 146 kips


One-Way Shear


x,

x = 14.125 1 .= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b d

VC= (0.75) (2*1000) 12*8.625

VC= 9.819 kips


D.Interior Column (C6)a)Two-Way Shear


b0= 2x + 2y

where,

x = 24 + 8.625

y = 24 + 8.625

so,

b0= 2(24 + 8.625) + 2(24 + 8.625)

b0= 130.5 in


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VC= (2 +) b0d

VC= (2 +

) (1) 000(130.5) (8.625)

VC= 427.12 kipsOR

VC= b0d

VC= 4(0.75)000(130.5) (8.625)VC= 213.56 kipsOR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(130.5) (8.625)

VC= 248 kips

So VC= 213.56 kips


One-Way Shear


x,

x = 14.125 1 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b dVC= (0.75) (2*1000) 12*8.625VC= 9.819 kips



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CALCULATING THE TOTAL STATIC MOMENTS IN THE LONG

AND SHORT DIRECTION

LONG DIRECTION

M0L=

=

.. .

= 544 k.ft

SHORT DIRECTION

M0S=

=

.. .

= 462 k.ft

WIDTH OF COLUMN STRIP AND MIDDLE STRIP IN THE LONG

AND SHORT DIRECTIONLONG DIRECTION

Width of coumn strip = 0.25*2.5 = 6.25

Width of middle strip = 24.5 2(6.25) = 2.25

SHORT DIRECTION


Width of middle strip = 28.25 2(6.25) = 6

EFFECTIVE DEPTH IN EACH DIRECTION

To calculate effective depth in each direction, assume the steel bars in short direction are

placed on top of the bars in long direction.

Therefore,

LONG DIRECTION

d = 10 0.75 5/16 = 8.9375 in

SHORT DIRECTION

d = 10 0.75 5/8 5/16 = 8.3125 in


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DISTRIBUTION OF TOTAL STATIC MOMENT INTO NEGATIVE

AND POSITIVE SPAN MOMENTS

Using ACI Code, Section 13.6.3

LONG DIRECTION

o Exterior Negative Moment = Mne= 0.26*544 = 141.44 k.ft

o Positive Moment Near Mid Span = Mp= 0.52*544 = 282.88 k.ft

o Interior Negative Moment = Mni= 0.7*544 = 380.8 k.ft

SHORT DIRECTION

o Exterior Negative Moment = Mne= 0.3*462 = 138.6 k.ft

o Positive Moment Near Mid Span = Mp= 0.5*462 = 231 k.ft

o

Interior Negative Moment = Mni= 0.7*462 = 323.4 k.ft

TRANSVERSE DISTRIBUTION OF MOMENTS

LONG DIRECTION

COLUMN STRIP MOMENTS

o Negative Moment at Exterior Support = 0.26*544 = 141.44 k.ft

o Positive Moment = 0.312*544 = 169.728 k.ft

o Negative Moment at Interior Support = 0.525*544 = 285.6 k.ft

MIDDLE STRIP MOMENTS

o Negative Moment at Exterior Support = 0


o Negative Moment at Interior Support = 0.175*544 = 95.2 k.ft

SHORT DIRECTION


o Negative Moment at Exterior Support = 1*138.6 = 138.6 k.ft


o Negative Moment at Interior Support = 0.75*323.4 = 242.55 k.ft


o Negative Moment at Exterior Support = 138.6 138.6 = 0o Positive Moment = 231 138.6 = 92.4 k.fto Negative Moment at Interior Support = 323.4 242.55 = 80.85 k.ft


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REINFORCEMENT DETAILS

Long Direction

Column Strip Middle Strip

Exterior Positive Interior Exterior Positive Interior

Mu (k.ft) 141.44 169.728 285.6 0 113.152 95.2

b(in) 73.5 73.5 73.5 147 147 147

d (in) 8.9375 8.9375 8.9375 8.9375 8.9375 8.9375

Ru (psi) 289.0906 346.9088 583.7407 0 115.6363 97.29012

Steel ratio ( ) 0.005634 0.006837 0.012102 0 0.002183 0.001831

As (sq.in) 3.700723 4.491036 7.950125 0 2.868682 2.405931

Min. As (sq.in) 1.323 1.323 1.323 2.646 2.646 2.646

Bars selected 12 no. 5 15 no. 5 26 no. 5 9 no. 5 10 no. 5 9 no. 5

Spacing 6.1 4.9 2.8 16.3 14.7 16.3

Short Direction


Exterior Positive Interior Exterior Positive Interior

Mu (k.ft) 138.6 138.6 242.55 0 92.4 80.85

b(in) 73.5 73.5 73.5 192 192 192

d (in) 8.3125 8.3125 8.3125 8.3125 8.3125 8.3125

Ru (psi) 327.4868 327.4868 573.1019 0 83.57736 73.13019

Steel ratio ( ) 0.006429 0.006429 0.011853 0 0.001569 0.001371

As (sq.in) 3.928101 3.928101 7.241541 0 2.504863 2.187867

Min. As (sq.in) 1.323 1.323 1.323 3.456 3.456 3.456

Bars selected 13 no. 5 13 no. 5 24 no. 5 12 no. 5 12 no. 5 12 no. 5

Spacing 5.6 5.6 3 16 16 16


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DESIGN OF INTERIOR FLAT PLATE PANEL

24.5

C6 C7

28.25

C13 C14

l1= 28.25

l2= 25.50

Columns are assumed to be 24 x 24 in2

CHECKING LIMITATIONS FOR USING DIRECT DESIGN

METHOD

6.

There are minimum three continuous spans in each direction OK7.

The panels are rectangular and ratio of longer to the shorter span is 1.154 OK8.

Adjacent spans in each direction do not differ by more than 1/3 of the longer span OK

9.

Columns are not offset from either axis between centerlines of successive columns OK

10.

Loads are uniform, and ratio of unfactored live to unfactored dead load is 0.44 OK

DETERMINING SLAB THICKNESS FOR FLAT-PLATE


12/20



From table 9.5(c) of the ACI Code, the minimum thickness of slab without interior beam,

without drop panel, interior panel, and fy= 60ksi,

h =

ln= (28.25*12 24) = 315 in

h =5

= .55

We take hs= 10 in

CALCULATING FACTORED LOADS

wL= 60 psfAs we take in assignment 1 and 2

wD= 10 psf + 150 pcf *

ft = 135 psf where 10 psf is dead load of vinyl tiles and 150

pcf is unit weight of concrete for slab

wU= 1.2(135) + 1.6(60) = 258 psf = 0.258 ksf

CHECKING TWO-WAY AND ONE-WAY SHEARS

A. Interior Column (C6)a)

Two-Way Shear


b0= 2x + 2y

where,

x = 24 + 8.625

y = 24 + 8.625

so,

b0= 2(24 + 8.625) + 2(24 + 8.625)

b0= 130.5 in



13/20



VC= (2 +

) b0d

VC= (2 +

) (1) 000(130.5) (8.625)

VC= 427.12 kipsOR

VC= b0d

VC= 4(0.75)000(130.5) (8.625)VC= 213.56 kipsOR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(130.5) (8.625)

VC= 248 kips

So VC= 213.56 kips


One-Way Shear


x,

x = 14.125 1 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b d

VC= (0.75) (2*1000) 12*8.625VC= 9.819 kips


B. Interior Column (C7)a)

Two-Way Shear



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b0= 2x + 2y

where,

x = 24 + 8.625

y = 24 + 8.625

so,

b0= 2(24 + 8.625) + 2(24 + 8.625)

b0= 130.5 in


VC= (2 +

) b0d

VC= (2 +

) (1) 000(130.5) (8.625)

VC= 427.12 kipsOR

VC= b0d

VC= 4(0.75)000(130.5) (8.625)VC= 213.56 kipsOR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(130.5) (8.625)

VC= 248 kips

So VC= 213.56 kips


One-Way Shear


x,

x = 14.125 1 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips


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VC= (2) b d

VC= (0.75) (2*1000) 12*8.625VC= 9.819 kips


C. Interior Column (C13)a)Two-Way Shear


b0= 2x + 2y

where,

x = 24 + 8.625

y = 24 + 8.625

so,

b0= 2(24 + 8.625) + 2(24 + 8.625)

b0= 130.5 in


VC= (2 + ) b0d

VC= (2 +

) (1) 000(130.5) (8.625)

VC= 427.12 kipsOR

VC= b0d

VC= 4(0.75)000(130.5) (8.625)VC= 213.56 kips

OR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(130.5) (8.625)

VC= 248 kips


16/20



So VC= 213.56 kips


One-Way Shear


x,

x = 14.125 1 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b d

VC= (0.75) (2*1000) 12*8.625VC= 9.819 kips


D. Interior Column (C14)c)Two-Way Shear


b0= 2x + 2ywhere,

x = 24 + 8.625

y = 24 + 8.625

so,

b0= 2(24 + 8.625) + 2(24 + 8.625)

b0= 130.5 in


VC= (2 +

) b0d

VC= (2 +

) (1) 000(130.5) (8.625)

VC= 427.12 kips


17/20



OR

VC= b0d

VC= 4(0.75)000(130.5) (8.625)VC= 213.56 kips

OR

VC= (

+ 2 ) b0d

VC= 0.75 (.

.+ 2 ) (1) 000(130.5) (8.625)

VC= 248 kips

So VC= 213.56 kips

Therefore VC> VU --- OKd)

One-Way Shear


x,

x = 14.125 1 .

= 12.40625 ft

VU= wU(1*x)

VU= 0.258 (1*12.40625)

VU= 3.2 kips

VC= (2) b d

VC= (0.75) (2*1000) 12*8.625VC= 9.819 kips


CALCULATING THE TOTAL STATIC MOMENTS IN THE LONG

AND SHORT DIRECTION

LONG DIRECTION

M0L=

=

.. .

= 544 k.ft

SHORT DIRECTION


18/20



M0S=

=

.. .

= 462 k.ft

WIDTH OF COLUMN STRIP AND MIDDLE STRIP IN THE LONG

AND SHORT DIRECTIONLONG DIRECTION


Width of middle strip = 24.5 2(6.25) = 2.25

SHORT DIRECTION

Width of column strip = 0.25*2.5 = 6.25

Width of middle strip = 28.25 2(6.25) = 6

EFFECTIVE DEPTH IN EACH DIRECTION

To calculate effective depth in each direction, assume the steel bars in short direction are

placed on top of the bars in long direction.

Therefore,

LONG DIRECTION

d = 10 0.75 5/16 = 8.9375 in

SHORT DIRECTION

d = 10 0.75 5/8 5/16 = 8.3125 in

DISTRIBUTION OF TOTAL STATIC MOMENT INTO NEGATIVE

AND POSITIVE SPAN MOMENTS

Using ACI Code, Section 13.6.3

LONG DIRECTION

o Negative Moment = 0.65*544 = 353.6 k.ft


SHORT DIRECTION


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PERCENTAGES OF STATIC MOMENTS IN COLUMN AND

MIDDLE STRIPSLONG DIRECTION






o

Positive Moment = 0.14*544 = 76.16 k.ft

SHORT DIRECTION








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REINFORCEMENT DETAILS

Long Direction


Negative Positive Negative Positive

Mu (k.ft) 266.56 114.24 87.04 76.16

b(in) 73.5 73.5 147 147

d (in) 8.9375 8.9375 8.9375 8.9375

Ru (psi) 544.8247 233.4963 88.95097 77.8321

Steel ratio ( ) 0.011195 0.004503 0.001672 0.00146

As (sq.in) 7.354214 2.957992 2.196569 1.91836

Min. As (sq.in) 1.323 1.323 2.646 2.646

Bars selected 24 no. 5 10 no. 5 9 no. 5 9 no. 5

Spacing 3 7.3 16.3 16.3

Short Direction Column Strip Middle Strip

Negative Positive Negative Positive

Mu (k.ft) 226.38 97.02 73.92 64.68

b(in) 73.5 73.5 192 192

d (in) 8.3125 8.3125 8.3125 8.3125

Ru (psi) 534.8951 229.2408 66.86189 58.50416

Steel ratio ( ) 0.010967 0.004417 0.001252 0.001094

As (sq.in) 6.700279 2.698878 1.998215 1.745976

Min. As (sq.in) 1.323 1.323 3.456 3.456

Bars selected 22 no. 5 9 no. 5 12 no. 5 12 no. 5Spacing 3.3 8.2 16 16

Designing of Flat Plate Corner Panel by DDM

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