Design of Compression members - University of Al...

Structural Steel Design 2015-2016 Dr. Haitham A. Bady

2-26

Design Compression members

Design procedure of columns

The design of steel columns is a trial and error process.

1. The factored load is computed.

2. The effective slenderness ratio (kL/r) can be assumed to be generally between (40 to 60) for

lengths between 10ft to 15ft.

3. Based on the assumed (kL/r), an estimated value of the design buckling stress can be

obtained either by using AISC equations or by using AISC tables (4.22, page 4.318 of AISC

Manual).

4. The estimated column area can be obtained by :-

Fcrassumed

PuA esitm φ−

=.)(

5. Using the estimated column area, a trial column section can be selected.

6. The actual effective slenderness ratio can now be computed and the design buckling stress

can be obtained for the selected section and then the column design strength (ØPn) can be

calculated by ØPn = ØFcr ×Ag

7. The design strength ØPn is compared with the factored load and it must be ≥ (Pu)

8. Check the local buckling for the selected section.


2-27

Example 5:

a) Using Fy= 50 ksi, select the lightest W14 section available for the service column loads

PD= 130 k and PL= 210 k. Assume kL= 10 ft.

b) Repeat (a) using the AISC column design tables

Sol.

Factored load, Pu=1.2PD+1.6PL= 1.2(130) +1.6(210) = 492k

a). Assume 50=r

kL , ØFcr from Table 4.22, p 381 of AISC Manual = 37.5ksi

212.135.37

492).( inA atedesitm ==∴

from AISC Manual , selected W14×48 (A=14.1 in2, rx=5.85in, ry=1.91in).

Check the selected section

82.6291.1

1210

min

=×=r

kL From Table 4.22, p 381 of AISC Manual, ØFcr=33.75ksi

ØPn = ØFcr ×Ag=33.75×14.1= 476k < 492k Not Ok

Select W14×53 (A=15.6 in2, rx=5.84in, ry=1.92in).

5.6292.1

1210

min

=×=r

kL , ØFcr=33.85ksi

ØPn = ØFcr ×Ag=33.85×15.6=528k> 492k Ok ∴Use W14×53

b).For kL=10ft, From Table 4.1, p.4.14 of AISC Manual,

W14×48 ØPn=477 < 492k Not Ok

W14×53 ØPn=528k > 492k Ok ∴Use W14×53


2-28

yr

kLx

r

kL)()( >

Check ( width/thickness) ratio for local buckling

For W14×53 (bf=8.06”, tf=.66”, k=1.25”, d=13.9”, tw=0.37”)

1. For unstiffened element (flange)

11.666.02

06.8

2=

×=

f

f

t

b

< yF

E56.0

=13.9 ( case 3 , Table B4.1) OK

2. For stiffened element (web)

81.3037.0

)25.129.13( =×−=wt

h <

yF

E49.1

=35.82 ( case 10 , Table B4.1) OK

∴ No local buckling occurs

• Design of columns if the effective length is different

o The column design tables (4.1 to 4.20) of AISC Manual provides design strength of columns with respect to y-

axis assuming that (r

kL )y is longer than (

r

kL )x.

o For the case of (r

kL )x.> (

r

kL )y, or if (kL)X > (kL)y, two

approaches can be used:

1. Trial and error procedure

� Select a trial section assuming (r

kL )=50

� Compute of (r

kL )x and (

r

kL )y

� Determine ØFcr based on the larger values of (r

kL)

and multiply by Ag to determine ØPn

� Check the ØPn with factored load and select another size if necessary.


2-29

)(

)()( )(

x

yEquivalent

r

rxkL

ykL =

2. AISC Design tables

� A section size is to be selected from table (4.1 to 420 ) based on (kL)y

� Take y

x

r

rvalue from the table for that shape size and calculate kL)y Equivalent and

compare with (kL)y actual.

I. If (kL) y Equivalent < (kL)y actual, then (kL) y actual controls the design and the selected shape is correct.

II. If (kL) y Equivalent > (kL)y actual, then (kL)x controls the design and another section must be selected based on the value of kL) y Equivalent

Example 6: Select the lightest satisfactory W12 for the following conditions: Fy= 50 ksi, PD =

250k, PL = 400k, KxLx= 26 ft, and KyLy = 13 ft.

a. By trial and error procedure.

b. Using LRFD tables.

Sol.

a. Using trial and error:

Factored load, Pu=1.2Pd+1.6PL= 1.2(250)+1.6(400)=940k

a). Assume 50=r

kL ØFcr from Table 4.22, p 381 of AISC Manual =37.5ksi

2067.255.37


from AISC Manual , selected W12×78 (A=25.6 in2, rx=5.38in, ry=3.07in).


2-30

ftyKLft

r

rxkL

ykL

x

yEquivalent 13)(8.14

75.1

26

)(

)()( )( =>===


94.5738.5

1226)( =×=x

r

kL

Control

81.50

07.3

1213)( =×=y

r

kL

From Table 4.22, p 381 of AISC Manual, ØFcr=35.2 ksi

ØPn = ØFcr ×Ag=35.2×25.6=901k< 940k Not Ok

Selected larger section , say W12×96 (A=28.2 in2, rx=5.44in, ry=3.09in).


35.5744.5

1226)( =×=x

r

kL

Control

48.50

09.3

1213)( =×=y

r

kL

from Table 4.22, p 381 of AISC Manual, ØFcr=35.55 ksi

ØPn = ØFcr ×Ag=33.55×28.2=1002.5k> 940k Ok ∴Use W12×96

b). Using LRFD tables (Table 4.1 of AISC Manual)

Based on (KL)y=13ft, Select W 12×87 from Table 4.1, p.4.17 of AISC Manual

From same table, y

x

r

r =1.75. Since (KL)x=26 >( KL)y=13, calculate the equivalent (KL)y from

the equation:

∴Use (kL)y Equivalent to determine the section size

Based on kL)y Equivalent =14.8ft, Select W 12×96 , ØPn=994.5 k > 940k Ok

Use W12×96


2-31

Check (width/thickness) ratio for local buckling

For W12×96 (bf=12.2”, tf=.9”, k=1.5”, d=12.7”, tw=0.55”)


76.69.02

2.12

2=

×=

f

f

t

b

< yF

E56.0

=13.9 ( case 3 , Table B4.1) OK


7.1755.0

)5.127.12( =×−=wt

h <

yF

E49.1

=35.82 ( case 10 , Table B4.1) OK


• Design of built up compression members

Built –up compression members sections are constructed with more than one shape

built-up into a single member. They include:

1. They may consist of parts in contact with each other, such as cover-plated

sections.

2. they may consist of parts in near contact with each other, such as pair of

angles: These pairs of angles may be separated by a small distance from each

other equal the thickness of the end connection or gusset plates between them.

3. They may consist of parts that are spread well apart, such as pairs of channels:


2-32

For long columns, it may be suitable to use built-up sections where the parts of the columns are

widely separated from each other to give higher moment of inertia such as towers.

General Notes

• When a pair angles are used as a compression member, they need to be fastened together so

they will act as a unit. Welds may be used at intervals or they may be connected with bolts.

• The widely spaced parts of these types must be carefully laced or tied together.

• The built -up section must have sufficient connection between its parts to prevent slippage on

each other and produce high moment of inertia.

• Connections are usually placed at column ends and column mid-span.

For Example


2-33

Connections requirements for built up columns

a. Components are in contact with each other (AISC Specification (Sec. E6):

The design strength of compressive built-up members whose components are in contact

with each other is the same as the usual columns with ONE exception as follows:

• If the members tends to buckle about an axis parallel to the direction of connection between

the W-shape and the plates, the connections are subjected to shear force, )(r

kLhave to

modified by AISC equations ( E6-1 and E6-2) as follows:

1. For intermediate connector that are made by high strength bolts

( snug-tight):

22 )()()(i

om r

a

r

kL

r

kL +=

(E6-1)

2. For intermediate connector that are welded or have pre-tensioned

bolts

22

22 )(

)1(82.0)()(

ib

om r

a

r

kL

r

kL

αα+

+=

(E6-2)

mr

kL)( is the modified slenderness ratio of the built-up compressive members.

or

kL)( is the original slenderness ratio of the built-up compressive members in the buckling

direction.

a is the distance between connections ( in the long direction).

ri is the minimum radius of gyration of the individual components

X-axis

Y-


2-34

rib is the radius of gyration of the individual components in the axis parallel to the axis of

buckling ( i.e. if moment about x -axis we take rx and . if moment about y- axis we take ry).

α is the separation ration = ibr

h

2 where h is the distance between centroids of the individual

components perpendicular to the axis of buckling.

(AISC Specification ( Sec . E7):

)(4

3)(

r

kL

r

ka

i

≤ Whole members

Example 6

You are to design a column for PD=750k and PL=1000k, using Fy=50ksi and Kl=14ft. A

W12×120 is on hand. Design cover plates to snug- tight bolted at 6” spacing to the W section.

Sol.

For W12×120, ( A=35.3in2, d=13.1”, bf=12.3”, Ix=1070in4, Iy=345in4).

Pu=1.2×750+1.6×1000= 2500k

Using trial and error:

a). Assume 50=r


267.665.37


Aplates=66.67-35.5=31.36in2 ( Area for each plate =15.68in2)

Try Two cover plates PL1×16


A= 35.3+2(1×16) = 67.3 in2

Ix=1070+2×16×

2)2

0.11.13(

+=2660in4

X-axis

Y-axis

W12×120


2-35

"29.63.67

2660==xr

Iy=345+2× )12

161(

3+

=1027.7in4

"91.33.67

3.1027 ==yr

97.4291.3

1214)( max =×=

r

kL

Calculate the modified slenderness ratio for x-axis ( parallel to the connection plane)

22 )()()(i

om r

a

r

kL

r

kL +=

(Snug- tight bolted connections are used)

a= 6”

ri=rx= A

I x plates ( minimum radius of gyration of individual member)

Ix=4

3

33.112

116in=×

A=16in2 . ri =rx =0.289”

76.20289.0

6 ==∴ir

a, 71.26

29.6

1214)( =×=xr

kL

22 )76.20()71.26()( +=∴ mr

kL= 33.83< 42.97 does not control the design

Check slenderness ratio limit of the plates

23.3297.424

3)(

4

376.20)( ==≤=

r

kL

r

ka

i Ok

For yr

kL)( = 42.97 , from Table 4.22, p 381 of AISC Manual, ØFcr=39.31 ksi


2-36

ØPn = ØFcr ×Ag=39.31 ×67.3=2646 k< 2500k Ok Use W12×120 steel section with two cover plates PL1×16, Fy = 50ksi

b. Components are Not in contact with each other- Design of lacing and tie plates

(AISC Specification (Sec. E6):

1. Design of tie plates ( discussed in tension members)

2. Design of lacing

a. memberwholememebrindivitual r

kL

r −−

≤ )(4

3)

sconnection lacingbetween distance The(

)min(

b. Vu (on the lacing )=0.02Pu (whole compression member)

perpendicular to the compression member

140)( ≤lacingr

kL for single lacing

200)( ≤lacingr

kL for double lacing

If l ≤ 15in , it is ok to use single lacing made with angles.


2-37

Example 7

a) Using AISC LRFD specifications and Select a pair of 12-in standard channels for the

column and load shown using Fy= 50 ksi. For connection purposes, the back-to-back

distance of the channels is to be 12 in. PD=100k and PL=300k.

b) Using Fy =36ksi, design bolted single lacing for the

column assume that ¾” bolts are used.

Sol.

Pu=1.2PD+1.6PL=1.2(100)+1.6(300)=600k

a). Assume 50=r

kL ØFcr from Table 4.22, p 381 of AISC

Manual =37.5ksi

20.165.37


A (required for each section) =8in2

From AISC-Specification( Table 1.-5 , P1-34), Select C12×30 : (A=8.81in2, d=12”,x′ =0.674”,

Ix=162in2, Iy=5.12in2) Check the selected section

A= 17.62in2

Ix=2×162=324in4 Iy=2[5.12+8.81×(2

12-0.674)2]=510in4

lAtotal

Ixr x 63.17

324== =4.29” lAtotal

Iyr y 63.17

510== =5.38”

"38.5min ==∴ yrr , k=1.0 (from table C2.2.1, pinned ends)

94.5529.4

12201)( max =××=

r

kL


2-38

From Table 4.22, p 381 of AISC Manual, ØFcr=35.82 ksi (By interpolation)

ØPn = ØFcr ×Ag=35.82 ×17.62=631k >600k Ok Check ( width/thickness) ratio for local buckling

For C12×30 (d=12”, bf=3.17”, tf=.501”,TW=0.510 k=1.25”)


33.6501.0

17.3 ==f

f

t

b

< yF

E56.0

=13.9 (case 3, Table B4.1) OK


12.19510.0

)25.1212( =×−=wt

h <

yF

E49.1

=35.82 (case 14 , Table B4.1) OK


Use 2C12 × 30 steel sections

b).Design the lacing

� The distance between lines of bolts

=l=8.5”< 15” it is ok to use single lacing

� Check the slenderness limit for individual

member

Assume 60ͦ inclination with longitudinal axis of

the compression member:

∴Length of channel between the lacing connections =2×8.5×tan30=9.81”

� The slenderness limits

memberwholer

kL

r memebrindivitual

−≤−

)(4

3)

sconnection lacingbetween distance The(

)min(

30 ͦ

30 ͦ


2-39

96.41)94.55(4

39.12)

761.0

9.81( =≤= Ok

� Calculation the force on the lacing channel

Vu=0.02×Pu=0.02×631=12.62k ∴2

1Vu=6.31 (shear force on each plane of lacing)

∴Compression force on the lacing section = )30cos(

2

1Vu

=7.28 k

� Calculation the dimensions of the lacing flat bar

ttb

tb

A

Ir 289.0

12/)( 3

=×

×==

Assume r

L= minimum value of the single lacing =140

07.0140

81.9140

=== Lr

∴r= 0.289t= 0.07 t=0.242” Try (¼)” flat bar 1364/1289.0

81.9 =×

=r

L

< 140 ok

From table 4.22 of AISC Manual ØFcr=12.2 ksi for 136=r

L

2597.02.12

28.7)( inA erequired == ∴(¼” ×2.39”) is needed

� Total length of the lacing =length of lacing between connections + 2×minimum edge

distance

=9.81+2×1 ¼ =12.3” say 14” (Min. edge distance for ¾ bolt is 1 ¼ from table J3.4 of AISC manual)

∴Use (¼” ×2 ⅜× 1’ 2”) flat bar

t

b

30 ͦ

P


2-40

Column used in steel frames

• The effective length of a column is a property of whole structure of which the column is a apart. • The most common method of obtaining the effective

length is to use the alignment charts for column braced

or un-braced against sidesway, using Tables C.C2.3

and C.C2.4 of AISC manual part 2 or table 7.2 ( a and

b) Page 201 of textbook.

• To determine k factor by alignment charts, the

following steps are followed:

1. Compute (G) value at each end of the column (GA

and GB) from the following equations:

beams

columns

L

IL

I

forbeamsL

EI

forcolumnsL

EI

G)(

)(

)4

(

)4

(

∑

∑

∑

∑==

2. Select the appropriate alignment chart [sidesway

prevented ( braced ) and sidesway un-prevented ( un-

braced )] and draw a straight line on the chart between

GA and GB values and read k value.

3. For the column bases the following G values can be

used

a) For pinned column, G=10

b) For rigid connection to footing ( fixed ) , G=1.0


2-41

o The most important assumptions based on which the alignment charts are

made are:

1. The members are elastic.

2. All columns buckle simultaneously.

3. For braced frames , θ 1=θ2 for each beam , (single curvature)

4. For un-braced frames , θ 1=- θ2 for each beam , (double curvature)

� For frames at which assumptions 3 and 4 are not satisfy, used the following multiplier can be

used for (G) values such as the following frames

(L

I)BC×1.5 for sidesway prevented ( braced- or pinned)

(L

I)BC×0.5 for sidesway not prevented (un-braced or rolled)

(L

I)BC×2.0 for sidesway prevented ( braced )

(L

I)BC×0.67 for sidesway not prevented (un-

braced or guided support)

M2 M1

θ1 θ2

θ 1=θ2 , single curvature

M2 M1

θ1

θ2

θ 1=-θ2, double curvature

Depending on the direction of ends moments M1 and M2

Depending on the direction of ends moments M1 and M2


2-42

� If the column behaviour is inelastic (assumption 1 is not valid), the column stiffness factor

will be smallerL

IEτ, and the G value will be smaller and k values will be smaller ( i.e high

column resistance)

• For an inelastic column situation

inelasticG = elasticG ×stiffness reduction factor (SRF), aτ

Table 4.21 of AISC Manual, page 4.317, or table 7.2 page 211 of textbook provides values of

stiffness reduction factor aτ for different values of A

Pu

• To design columns to consider inelastic columns behaviour in frames, the following

steps should be followed:

1. Calculate Pu , assume r

kL and select a trial section size.

2. Calculate A

Pu and select aτ from table 4.21 in Manual or table 7.2 in textbook.

3. Compute values of G elastic and multiply by aτ and find k value.

4. Compute the r

kL actual and the find ØFcr then determine Pu= ØFcr×Ag and compare

it with Pu.

Example 8:

a) Select a W section for column AB

of the un-braced frame shown in

Fig. below assuming that we have

elastic behaviour , PD=450k and

PL=700k, Fy=50ksi

b) Repeat part (b) if inelastic column

behaviour is considered.

Assume that columns above and below column AB are the same size as AB


2-43

Sol.

Pu=1.2PD+1.6PL=1.2(450)+1.6(700)=1660k

a). Assume 50=r


2267.445.37


From AISC-Specification( Table 1.1),the following selects are available W14×176 : (A=51.8in2) W12×170 : (A=50.07in2) we have to use a larger section to account for the increasing in k

values

W14×159 : (A=46.7in2) W12×152 : (A=44.7in2)

W12×136: (A=39.9in2)

∴For W12×170 : (A=50in2,,, Ix=1650in4, rx=5.74in) we assume that the columns are bent about x- direction

16.5)

1230

800(2

)1212

1650(2

)(

)(=

×

×==∑

∑

beams

columns

A

L

IL

I

G

GB=GA=5.16 (Same columns and beams dimensions) From alignment chart ( Table C.C2.4) ( un-braced frame) k=2.3

7.5774.5

12123.2)( =××=∴

r

kL

From Table 4.22, p 381 of AISC Manual ØFc=35.29ksi

ØPn = ØFcr ×Ag=35.29 ×50=1764 >1660k Ok b).Inelastic design Because the inelastic resistance of steel column is greater than the elastic resistance, we select a

lighter section than the elastic behavior section.

Try W12×136: (A=39.9in2 , Ix=1240in4 , rx=5.58”)


2-44

6.419.39

1660==A

Puin2 aτ = 0.1974 from table 4.21 in Manual or table 7.2 in

textbook.

765.0)

1230

800(2

)1212

1240(2

=×

×

×= aAG τ

From alignment chart ( Table C.C2.4) ( un-braced frame) k=1.27

77.3258.5

121227.1)( =××=∴

r

kL


ØPn = ØFcr ×Ag=41.7×39.9=1663 >1660k Ok

Example 9:

We desire to select a W14 section for column CD in the figure shown for which PD=300k and

PL=600k, and Fy =50ksi, Only in-plane behaviour is considered. Furthermore, assume that the

columns immediately above and below CD are approximately the same size as CD, and also

that all the other assumptions on which the alignment charts are met.

a) Assume elastic behaviour.

b) Assume inelastic behaviour.

Sol.

Pu=1.2PD+1.6PL=1.2(300)+1.6(600)=1320k

a). Assume 50=r

kL ØFcr from Table 4.22, p 381 of AISC

Manual =37.5ksi

22.355.37


From AISC-Specification( Table 1.1),the following selects are available

36ft 36ft

15ft

W30×99 W30×99

W30×99

Ix=3990 Ix=3990

15ft

15ft

D W30×99

C


2-45

W14×120 : (A=35.3in2, Ix=1380in4, rx=6.24in) Because beam dimensions are large; we expect k value will be small. From AISC Manual and for W30×99, Ix=3999

83.0)

1236

3999(2

)1215

1380(2

)(

)(=

×

×==∑

∑

beams

columns

A

L

IL

I

G

GB=GA =0.83 Same columns and beams dimensions From alignment chart ( Table C.C2.4) ( isedsway uninhibited ( un-braced frame)) k=1.25

055.3624.6

121525.1)( =××=∴ xr

kL


ØPn = ØFcr ×Ag=40.9×35.3=1443.77k >Pu=1330k Ok b).Inelastic design Because the inelastic resistance of steel column is greater than the elastic resistance, we select a

lighter section than the elastic behavior section.

Try W12×109: (A=32in2 , Ix=1240in4 , rx=6.22”)

25.4132

1330==A

Puin2 aτ = 0.217 by interpolation from table 4.21 in Manual

or table 7.2 in textbook.

162.0217.0)

1230

3990(2

)1215

1240(2

=×

×

×=TopG

Gtop=Gbottom=0.162 From alignment chart ( Table C.C2.4) ( un-braced frame) k=1.1

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2-46

31.832 22.6

12151.1)( =××=∴

r

kL


ØPn = ØFcr ×Ag=41.75×32.0=1336k >1330k Ok

Design of Base Plate • Design procedure for column base plate. 1. Determine the required area of the base plate

according to the design bearing strength of

concrete beneath the base plate. The

following equation can be used:

1

2

1

)85.0(A

Acf

PuA

c′

=φ

cφ =0.6,

1

2

A

A≤2.0 ( in this equation only )

Notes: Base plate area must be larger than

the area of the column section (b×d)

2. Determine the base plate dimensions (N×B)

from the following equations:

∆=0.5(0.95d-0.8bf)

N= 1A +∆

B=N

A1

d and bf are the depth and flange width of

the column cross section.

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Note

the critical sections of bending (maximum bending moments)


2-47

Notes: values of B and N are selected to nearest 1“ or 2” so that values of m and n shown

in the figure are roughly equal and the cantilever moment are equals too.

3. Determine the base plate thickness from the following equation:

NBFy

Pult req ×××=

9.0

2

l is the largest values of m , n and n′

m and n are shown in the figure which are the cantilever dimensions from the base edge in x

and y axes respectively.

4

fdbn =′

Example 10: Design a base plate of A36 steel for a W 12×65 column with Fy=50ksi that

supports loads 0f PD=200k and PL=300k. The concrete has a compressive strength of fc’=3ksi

and the footing has dimensions of 9ft×9ft.

Sol.

Pu=1.2×200+1.6×300=720k

For W 12×65 (d=12.1”, bf=12.0”

Assume 1

2

A

A

=2 (the maximum value)

2

1

2

1 3.2352)385.0(6.0

720

)85.0(

in

A

Acf

PuA

c

=××

=′

=∴φ

235.2in2> d×bf=145.2 in2 ok

∆=0.5(0.95d-0.8bf)= 0.5(0.95×12.1-0.8×12)=0.9475”

Haitham

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Haitham

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Haitham

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Haitham

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Haitham

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Haitham

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2-48

N= 1A +∆= 13.235 +0.9475=16.3” say 16”

B=N

A1 = =16

3.23514.7”

Use 16”×16” square base plate.

• Check the bearing strength of the concrete

ØPn=Øc(0.85fc’)A1 1

2

A

A

=0.6×(0.85×3) ×(16)2×2=783.4k > 720 ok

• Determine the thickness of the base plate

m= (16-0.95×d)/2= (16-0.95×12.1)/2=2.25”

n= (16-0.8×bf)/2=(16-0.8×12)/2=3.2”

"01.34

121.12 =×=′n

l = the largest value =3.2”

"38.11616369.0

72022.3 =

×××××=reqt say 1 ½”

Use PL1 ½” ×16”× 1’ 4” A36 base plate

Example 11( Prob 7.15 page 235 of texbook): Design a base plate of A36 steel for a W 14×120

column with Fy=50ksi that supports loads 0f PD=150k and PL=350k. The footing size is 10ft

by 10 ft and fc’=3ksi and

Sol.

Pu=1.2×150+1.6×350=740k

For W 12×65 (d=14.5”, bf=14.7”

Assume 1

2

A

A

=2 (the maximum value)


2-49

2

1

2

1 83.2412)385.0(6.0

740

)85.0(

in

A

Acf

PuA

c

=××

=′

=∴φ

241.83in2> d×bf=213 in2 ok

∆=0.5(0.95d-0.8bf)= 0.5(0.95×14.51-0.8×14.7)=1.01”

N= 1A +∆= 83.241 +1.01=16.55” say 17”

B=N

A1 = =17

83.24114.22” say 15”

Use 17”×15” rectangular base plate.

• Check the bearing strength of the concrete

ØPn=Øc(0.85fc’)A1 1

2

A

A

=0.6×(0.85×3) ×(15×17)×2=780.3k > 740 ok

• Determine the thickness of the base plate

m= (16-0.95×d)/2= (17-0.95×14.5)/2=1.6125”

n= (16-0.8×bf)/2=(15-0.8×14.7)/2=1.2525”

"65.34

7.145.14 =×=′n

l = the largest value =3.65”

"544.11715369.0

740265.3 =

×××××=reqt say 1⅝”

Use PL 1⅝”×15”× 1’ 5” A36 base plate

Design of Compression members - University of Al...

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