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Concrete Case - Architectural Structures - Lecture Notes, Study notes of Structural Design and Architecture

Alagappa University Structural Design and Architecture

Concrete Case, Building Description, Speculative Rental, Exterior Walls, Lateral Loads, Rigid Perimeter Frame, Live Loads, Office Areas, Corridor and Lobby, Construction Loads. This may not be a lecture notes exactly but it contains important information.

Typology: Study notes

2011/2012

Uploaded on 12/22/2012

bageshri 🇮🇳

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1

Case Study in Reinforced Concrete

Building description

The building is a three-story office building intended for speculative rental. Figure 17.37 presents

a full-building section and a plan of the upper floor. The exterior walls are permanent. The

design is a rigid perimeter frame to resist lateral loads.

Loads (UBC 1994)

Live Loads:

Roof:

20 lb/ft2

Floors:

Office areas: 50 lb/ft2 (2.39 kPa)

Corridor and lobby: 100 lb/ft2 (4.79 kPa)

Partitions: 20 lb/ft2 (0.96 kPa)

Wind: map speed of 80 mph (190 km/h);

exposure B

Assumed Construction Loads:

Floor finish: 5 lb/ft2 (0.24 kPa)

Ceilings, lights, ducts: 15 lb/ft2 (0.72 kPa)

Walls (average surface weight):

Interior, permanent: 10 lb/ft2 (0.48 kPa)

Exterior curtain wall: 15 lb/ft2 (0.72 kPa)

Materials

Use f’c =3000 psi (20.7 MPa) and

grade 60 reinforcement (fy = 60 ksi or 414 MPa).

Structural Elements/Plan

Case 1 is shown in Figure 17.44 and consists of a flat plate supported on interior beams, which in

turn, are supported on girders supported by columns. We will examine the slab, and a four-span

interior beam.

Case 2 will consider the bays with flat slabs, no interior beams with drop panels at

the columns and an exterior rigid frame with spandrel (edge) beams. An example

of an edge bay is shown to the right. We will examine the slab and the drop panels.

For both cases, we will examine the exterior frames in the 3-bay direction.

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Download Concrete Case - Architectural Structures - Lecture Notes and more Study notes Structural Design and Architecture in PDF only on Docsity!

Case Study in Reinforced Concrete

Building description

The building is a three-story office building intended for speculative rental. Figure 17.37 presents

a full-building section and a plan of the upper floor. The exterior walls are permanent. The

design is a rigid perimeter frame to resist lateral loads.

Loads (UBC 1994)

Live Loads:

Roof:

20 lb/ft

2

Floors:

Office areas: 50 lb/ft

2

(2.39 kPa)

Corridor and lobby: 100 lb/ft

2

(4.79 kPa)

Partitions: 20 lb/ft

2

(0.96 kPa)

Wind: map speed of 80 mph (190 km/h);

exposure B

Assumed Construction Loads:

Floor finish: 5 lb/ft

2

(0.24 kPa)

Ceilings, lights, ducts: 15 lb/ft

2

(0.72 kPa)

Walls (average surface weight):

Interior, permanent: 10 lb/ft

2

(0.48 kPa)

Exterior curtain wall: 15 lb/ft

2

(0.72 kPa)

Materials

Use f’c = 3000 psi (20.7 MPa) and

grade 60 reinforcement ( fy = 60 ksi or 414 MPa).

Structural Elements/Plan

Case 1 is shown in Figure 17.44 and consists of a flat plate supported on interior beams, which in

turn, are supported on girders supported by columns. We will examine the slab, and a four-span

interior beam.

Case 2 will consider the bays with flat slabs, no interior beams with drop panels at

the columns and an exterior rigid frame with spandrel (edge) beams. An example

of an edge bay is shown to the right. We will examine the slab and the drop panels.

For both cases, we will examine the exterior frames in the 3-bay direction.

Case 1:

Slab:

The slabs are effectively 10 ft x 30 ft, with an aspect

ratio of 3, making them one-way slabs. Minimum

depths (by ACI) are a function of the span. Using

Table 3-1 for one way slabs the minimum is

l n

with

5 inches minimum for fire rating. We’ll presume the

interior beams are 12” wide, so

ln = 10 ft -1 ft = 9 ft

minimum t (or h) .in

ft in/ ft

45 24

Use 5 in.

dead load from slab =

in/ft

lb/ft in

3 

= 62.5 lb/ft

2

total dead load = (5 + 15 + 62.5) lb/ft

2

+ 2” of

lightweight concrete topping with weight of

18 lb/ft

2

(0.68 KPa) (presuming interior wall

weight is over beams & girders)

dead load = 100.5 lb/ft

2

live load (worst case in corridor) = 100 lb/ft

2

total factored distributed load (ASCE-7) of 1.2D+1.6L:

wu’= 1.2(100.5 lb/ft

2

) + 1.6(100 lb/ft

2

) = 280.6 lb/ft

2

Maximum Positive Moments from Figure 2-3, end span (integral with support) for a 1 ft wide strip:

Mu (positive) =

lb

w w ft ft k

lbft ft u n u n

1000

2 2 /^22

 

= 1.62 k-ft

Pick bars and spacing off Table 3-7. Use #3 bars @ 12 in (As = 0.11 in

2

Check the moment capacity. d is actually 5 in – 0.75in (cover) – ½ (3/8in bar diameter) = 4.06 in

a = Asfy/0.85f’cb = 0.11 in

2

(60 ksi)/[0.85(3 ksi)12 in] = 0.22 in

Mn = Asfy(d-a/2) =  )  ( ) 

. in .(. in )( ksi)(. in

ft 12 in

2

1.96 k-ft > 1.89 k-ft needed

(OK)

Maximum Shear : Figure 2-7 shows end shear that is wu ln/ 2 except at the end span on the interior

column which sees a little more and you must design for 15% increase:

Vu-max = 1.15wu ln /2 =

2 lb/ft

. (. )( )( )

ft ft 

= 1452 lb (for a 1 ft strip)

Vu at d away from the support = Vu-max – w(d) = 1452 lb –

in/ft

ft (. )( )(. in)

2 lb/ft 

= 1357 lb

Check the one way shear capacity:  v Vc =  v 2 f (^) c bd (  v = 0.75):

 v Vc = ) psi( )(. )

in in

0.75( 2 3000 12 406 = 4003 lb

Is Vu (needed) <  v Vc (capacity)? YES: 1357 lb  4003 lb, so we don't need to make the slab

thicker.

Interior Beam (effectively a T-beam):

Tributary width = 10 ft for an interior beam.

dead load = (100.5 lb/ft

2

)(10ft) = 1005 lb/ft (14.7 kN/m)

Reduction of live load is allowed, with an influence area, AI, of 2 panels beside an interior beam,

assuming the girder is 12” wide. The live load is 100 lb/ft

2

L = )

I

o A

L  = )

2

ft ft ft

ft

lb

 = 87.3 lb/ft

2

(Reduction Multiplier = 0.873)

live load = 87.3 lb/ft

2

(10ft) = 873 lb/ft (12.7 kN/m)

Estimating a 12” wide x 2 4 ” deep beam means the additional dead load from self weight ( w  A

in units of load/length) can be included. The top 5 inches of slab has already been included in the

dead load:

dead load from self weight =

2

3  

in

ft inwide indeep

ft

lb

= 237.5 lb/ft (3.46 kN/m)

wu = 1.2(1005 lb/ft+237.5 lb/ft) + 1.6(873 lb/ft) = 2888 lb/ft (4.30 kN/m)

The effective width, bE, of the T part is the smaller of

 n

, bw  16 t , or center-center spacing

bE = minimum{29 ft/4 = 7.25 ft = 87 in, 12 in+16x5 in = 92 in, 10 ft = 120 in} = 87 in

The clear span for the beam is

l n = 30 ft – 1 ft = 29 ft

Maximum Positive Moments from Figure 2-3, end span (integral with support) :

Mu (positive) =

lb

w (^) k

lb ft ft u n

1000

2 / 2

 

= 173.5 k-ft

Maximum Negative Moments from Figure 2-4, end span (integral with support) :

Mu-(negative) =

lb

w (^) k

lb ft ft u n

1000

2 / 2

 

= 242.9 k-ft

The required spacing where stirrups are needed for crack control (  v Vc ≥Vu > ½  v Vc ) is

srequired =

50(12in)

0.22in (60,000psi)

50

2



w

v y

b

A f

=22 in and the maximum spacing is d/2 = 10.75 in. or 24”. Use

10 in.

A recommended minimum spacing for the first stirrup is 2 in. from the face of the support. A

distance of one half the spacing near the support is often used.

Spandrel Girders:

Because there is a concentrated load on the girder, the approximate analysis

can’t technically be used. If we converted the maximum moment (at

midspan) to an equivalent distributed load by setting it equal to wu l

2

/8 we

would then use:

Vu

x

48.2 k

36.7 k

43.0 k

41.9 k

21.2 k

10.6 k

21.2 k

10.6 k

7.2 ft = 86 in

10.84 ft = 130 in

to find distances when w is known: x = (peak – crossing value)/w

ex: 41.9 k – 21.2 k)/2.888 k/ft = 7.2 ft

9.35 ft = 112 in

13.0 ft = 156 in

C

L 14.5 ft^ in

14 @ 10 in 5 @ 10 in^13 @ 9 in

END SPAN STIRRUP LAYOUT

spandrel

girder

Maximum Positive Moments from Figure 2-3, end span (integral with support):

Mu+ =

2 wu  n

Maximum Negative Moments from Figure 2-4, end span (column support):

Mu- =

2 wu  n

(with

2 wu  n

at end)

Column:

An exterior or corner column will see axial load and bending moment. We’d use interaction charts for

Pu and Mu for standard sizes to determine the required area of steel. An interior column sees very little

bending. The axial loads come from gravity. The factored load combination is 1.2D+1.6L + 0.5Lr.

The girder weight, presuming 1’ x 4’ girder at 150 lb/ft

3

= 600 lb/ft

Top story: presuming 20 lb/ft

2

roof live load, the total load for an interior column

(tributary area of 30’x30’) is:

DLroof*: 1.2 x 100.5 lb/ft

2

x 30 ft x 30 ft = 108.5 k

assuming the same live load and materials as the floors

DLbeam 1.2 x 237.5 lb/ft x 30 ft x 3 beams = 25.6 k

DLgirder 1.2 x 600 lb/ft x 30 ft = 21.6 k

LLr: 0.5 x 20 lb/ft

2

x 30 ft x 30 ft = 9.0 k

Total = 164.7 k

Lower stories:

DLfloor: 1.2 x 100.5 lb/ft

2

x 30 ft x 30 ft = 108.5 k

DLbeam 1.2 x 237.5 lb/ft x 30 ft x 3 beams = 25.6 k

DLgirder 1.2 x 600 lb/ft x 30 ft = 21.6 k

LLfloor: 1.6 x (0.873)x100 lb/ft

2

x 30 ft x 30 ft = 125.7 k (includes reduction)

Total = 281.4 k

nd

floor column sees Pu = 164.7+281.4 = 446.1 k

st

floor column sees Pu = 446.1+281.4 = 727.5 k

Look at the example interaction diagram for an 18” x 18 “ column (Figure 5-20 – ACI 318-02) using

f

’

c =^ 4000 psi and^ fy =^ 60,000 psi for the first floor having Pu = 727.5 k, and Mu to the column being

approximately 10% of the beam negative moment = 0.1*242.9 k-ft = 24.3 k-ft: (See maximum

negative moment calculation for an interior beam.) The chart indicates the capacity for the

reinforcement amounts shown by the solid lines.

For Pu = 727.5 k and Mu = 24.3 k-ft, the point plots below the line marked 4-#10 (1.57% area of

steel to an 18 in x 18 in area).

Exterior frame (bent) loads:

H 1 = 195 ( 61 )

lb / ft ft

=11,895 lb = 11.9 k/bent

H 2 =

lbk

lbft ft

/

=14.3 k/bent H 3 =

lbk

lb ft ft

/

=13.8 k/bent

Using Multiframe4D, the axial force, shear and bending moment diagrams can be determined

using the load combinations, and the largest moments, shear and axial forces for each member

determined.

(This is the summary diagram of force, shear and moment magnitudes refer to the maximum

values in the column or beams, with the maximum moment in the beams being negative over the

supports, and the maximum moment in the columns being at an end.)

Axial force diagram: Shear diagram:

y

x

342.078 kip

0.000 kip

Px'

342.078 kip

0.000 kip

Px'

y

x

64.549 kip

0.000 kip

Vy '

64.549 kip

0.000 kip

Vy '

Bending moment diagram: Displacement:

y

x

384.616 kip-f t

0.000 kip-f t

Mz'

384.616 kip-f t

0.000 kip-f t

Mz'

y

x

M = 203.7 k-ft

V = 26.6 k

P = 54.7 k

M =190.8 k-ft

V = 28.0 k

P = 121.6 k

M = 165.1 k-ft

V = 21.6 k

P = 192.8 k

M =102.6 k-ft

V = 8. 5 k

P = 1 85 .7 k

M = 142.7 k-ft

V = 21.2 k

P = 1 20 .3 k

M = 183.9 k-ft

V = 24.8 k

P = 53.5 k

M = 39.3 k-ft

V = 4.7 k

P = 91.0 k

M = 32.4 k-ft

V = 3.7 k

P =91.0 k

M = 70.2 k-ft

V = 9.2 k

P = 215.5 k

M = 51.1 k-ft

V = 7.7 k

P = 215. 7 k

M = 104.4 k-ft

V = 11.6 k

P = 340. 6 k

M = 95.5 k-ft

V = 10.3 k

P = 341.7 k

M = 218.8 k-ft

V = 45.4 k

P = 26.6 k

M = 237.8 k-ft

V = 45.5 k

P = 28.9 k

M = 236.3 k-ft

V = 46.6 k

P = 30.8 k

M = 332.4 - ft

V = 60.5 k

P = 1.9 k

M = 330.9 k-ft

V = 60.0 k

P = 6.4 k

M = 317.9 k-ft

V = 59.7 k

P = 10.4 k

M = 338.5 k-ft

V = 60.9 k

P = 6.5 k

M = 349.5 k-ft

V = 61.0 k

P = 4.1 k

M = 347. 9 k-ft

V = 62.0 k

P = 1. 6 k

Beam-Column loads for design:

The bottom exterior columns see the largest bending moment on the lee-ward side (left):

Pu = 192.8 k and Mu = 165.1 k-ft (with large axial load)

The interior columns see the largest axial forces:

Pu = 341.7 k and Mu = 95.5 k-ft and Pu = 340.6 k and Mu = 104.4 k-ft

Refer to an interaction diagram for column reinforcement and sizing.

Case 2

Slab:

The slabs are effectively 30 ft x 30 ft, making them two-way slabs. Minimum thicknesses (by

ACI) are a function of the span. Using Table 4-1 for two way slabs, the minimum is the larger of

l n/36 or 4 inches. Presuming the columns are 18” wide, ln = 30 ft – (18 in)/(12 ft/in) = 28.5 ft,

h = ln/ 36 = (28.5x12)/36 = 9.5 in

The table also says the drop panel needs to be l/ 3 long = 28.5 ft/3 = 9.5 ft, and that the minimum

depth must be 1.25h = 1.25(9.5 in) = 12 in.

For the strips, l 2 = 30 ft, so the interior column strip will be 30 ft/4+30ft/4 = 15 ft, and the middle

strip will be the remaining 15 ft.

dead load from slab =

in/ft

lb /ft in .

3 

= 118.75 lb/ft

2

drop panel

column

shear

perimeter

tributary area for column

Design as for the slab in Case 1, but provide steel in both directions distributing the reinforcing

needed by strips.

Shear around columns: The shear is critical at a distance d/

away from the column face. If the drop panel depth is 12

inches, the minimum d with two layers of 1” diameter bars

would be 12” – ¾” (cover) – (1”) – ½(1”) =about 9.75 in (to the

top steel).

The shear resistance is  v Vc =  v 4 f (^) c bod,  v  0_._ 75 where bo, is the perimeter length.

The design shear value is the distributed load over the tributary area outside the shear perimeter,

Vu = wu (tributary area - b 1 x b 2 ) where b’s are the column width plus d/2 each side.

b 1 = b 2 = 18” + 9.75”/2 + 9.75”/2 = 27.75 in

b 1 x b 2 =

2 2

12

in

ft

(. in)  ( = 5.35 ft

2

Vu =

lb

lb ft ft ft ft k

1000

/ 2 2

   = 254.7 k

Shear capacity:

bo = 2(b 1 ) + 2(b 2 ) = 4(27.75 in) = 111 in

 v Vc =

in in

Concrete Case - Architectural Structures - Lecture Notes, Study notes of Structural Design and Architecture

Related documents

Partial preview of the text

Download Concrete Case - Architectural Structures - Lecture Notes and more Study notes Structural Design and Architecture in PDF only on Docsity!

Case Study in Reinforced Concrete

Building description

The building is a three-story office building intended for speculative rental. Figure 17.37 presents

a full-building section and a plan of the upper floor. The exterior walls are permanent. The

design is a rigid perimeter frame to resist lateral loads.

Loads (UBC 1994)

Live Loads:

Roof:

20 lb/ft

Floors:

Office areas: 50 lb/ft

(2.39 kPa)

Corridor and lobby: 100 lb/ft

(4.79 kPa)

Partitions: 20 lb/ft

(0.96 kPa)

Wind: map speed of 80 mph (190 km/h);

exposure B

Assumed Construction Loads:

Floor finish: 5 lb/ft

(0.24 kPa)

Ceilings, lights, ducts: 15 lb/ft

(0.72 kPa)

Walls (average surface weight):

Interior, permanent: 10 lb/ft

(0.48 kPa)

Exterior curtain wall: 15 lb/ft

(0.72 kPa)

Materials

Use f’c = 3000 psi (20.7 MPa) and

grade 60 reinforcement ( fy = 60 ksi or 414 MPa).

Structural Elements/Plan

Case 1 is shown in Figure 17.44 and consists of a flat plate supported on interior beams, which in

turn, are supported on girders supported by columns. We will examine the slab, and a four-span

interior beam.

Case 2 will consider the bays with flat slabs, no interior beams with drop panels at

the columns and an exterior rigid frame with spandrel (edge) beams. An example

of an edge bay is shown to the right. We will examine the slab and the drop panels.

For both cases, we will examine the exterior frames in the 3-bay direction.

Case 1:

Slab:

The slabs are effectively 10 ft x 30 ft, with an aspect

ratio of 3, making them one-way slabs. Minimum

depths (by ACI) are a function of the span. Using

Table 3-1 for one way slabs the minimum is

with

5 inches minimum for fire rating. We’ll presume the

interior beams are 12” wide, so

ln = 10 ft -1 ft = 9 ft

minimum t (or h) .in

Use 5 in.

dead load from slab =

= 62.5 lb/ft

total dead load = (5 + 15 + 62.5) lb/ft

+ 2” of

lightweight concrete topping with weight of

18 lb/ft

(0.68 KPa) (presuming interior wall

weight is over beams & girders)

dead load = 100.5 lb/ft

live load (worst case in corridor) = 100 lb/ft

total factored distributed load (ASCE-7) of 1.2D+1.6L:

wu’= 1.2(100.5 lb/ft

) + 1.6(100 lb/ft

) = 280.6 lb/ft

Maximum Positive Moments from Figure 2-3, end span (integral with support) for a 1 ft wide strip:

Mu (positive) =

= 1.62 k-ft

Pick bars and spacing off Table 3-7. Use #3 bars @ 12 in (As = 0.11 in

Check the moment capacity. d is actually 5 in – 0.75in (cover) – ½ (3/8in bar diameter) = 4.06 in

a = Asfy/0.85f’cb = 0.11 in

(60 ksi)/[0.85(3 ksi)12 in] = 0.22 in

Mn = Asfy(d-a/2) =  )  ( ) 

1.96 k-ft > 1.89 k-ft needed

(OK)

Maximum Shear : Figure 2-7 shows end shear that is wu ln/ 2 except at the end span on the interior