Lecture on April 3, 2012

Last Lecture

• Buckling• Radius of gyration• Design steps for columns

Design of Columns - Steps• Given are the loads or forces, Steel strength

• Remember: The Allowable compressive stress Fa is a function of slenderness ratio ( ), which can be determined only once “r” is known. For that we need to select a section , shape and size.

• Assume a suitable allowable stress Fa .

• For A36 steel, Fa = 12 – 16 ksi for columns and 9 – 12 ksi for struts.• Determine area required by using the formula:

Design of Columns - Steps• Select suitable shape and size which gives area,

approximately equal to or greater than required in the previous step.

• For section selected, compute actual ( ) and calculate allowable stress by using formulae / manual.

• If allowable stress is greater than actual gross compressive stress, then section selected is OK, otherwise revise and select another section.

Lacing and Tie Plates

General Discussion• In structural work, particularly bridge work, the system of

slender, diagonal members which connect the two opposite parallel members or flanges of a structural iron or steel beam, column, or strut. In lacing the several members form a single, continuous zigzag line, but do not cross one another

• Single angles are satisfactory for bracings and for light trusses. Top chord members of roof trusses are usually made up of double angles back-to-back. The pair of angles used, has to be connected together, so they will act as one unit. Welds may be used at intervals – with a spacer bar between the connecting legs. Alternately “stitch bolts”, washers and “ring fills” are placed between the angles to keep them at the proper distance apart (e.g. to enable a gusset to be connected). Such connections are called tack connections and the terms tack welding or tacks bolting are used.

General Discussion• Single channels or C-sections are generally not

satisfactory for use in compression, because of the low value of radius of gyration in the weak direction. They can be used if they could be supported in a suitable way in the weak direction.

• Compression members composed of two angles, channels, or tees back-to-back in contact or separated by a small distance shall be connected together by tack riveting, tack bolting or tack welding so that the individual sections do not buckle between the tacks before the whole member buckles

General Discussion

Single Lacing

Double Lacing

Battens

Built up Box Section

Built up I Section

San Francisco Oakland Bridge

Example # 1

• Data– Length of Column = 24 feet– Steel = A36– End condition Top = Pinned– End condition Bot = Pinned

• Assumption– In this example we will assume 0.5 so

Fa = 0.5 x Fy = 0.5 x 36 = 18 ksi

Lacing

2 x C 15 x 33.9

Example # 1 - Contd• We have the sections and lets note down the AISC Specs– Section C 15 x 33.9– Area “A” 9.960 Page 1-40 and 1-41

– Ix 315.000

– Iy 8.130

– x 0.787– Value of K from Table C-C2.1 is 1.00 – Calculating the Allowable load

– Pa = Fa x Ag = 18 x 2 x 9.96 = 358.56 kips

– Now we have to find and for that first find moment of inertia along both axis.

Example # 1 - Contd

2.

7.5”

0.81”

0.787”

15”

0.4”

1/2”

3.4”9”

Example # 1 - Contd• Ixx = ?

– Ixx = 2 x Ix

– Ixx = 2 x 315

– Ixx = 630

• Iyy = ?

– Iyy =

– Iyy = 2 x (8.13 + 9.96 x (4.5 + 0.787)^2)

– Iyy = 2 x (8.13 + 278.41)

– Iyy = 573.08

Example # 1 - Contd• rmin = ?

– rmin = 5.36 in

– So OK

– From Table C-36, Page 3-16, 53 = 18.08, 54 = 17.99

– By interpolating the Allowable Stress = Fa = 18.014 ksi

– Calculating the Allowable load

– Pa = Fa x Ag = 18.014 x 2 x 9.96 = 358.84 kips

Example # 1 - Contd• Lacing: Code Guidelines – Page 5-44– Lacing should be proportioned to resist the shearing

stress normal to axis of the member equal to 2 % of the total compressive stress in the member.

– The ratio l/r for lacing bars arranged in single systems shall not exceed 140.

– The ratio l/r for lacing bars arranged in double systems shall not exceed 200. Double lacing bars shall be joined at their intersections.

– For lacing bars in compression the unsupported length of the lacing bar shall be taken as the distance between fasteners or welds connecting it to the components of the built-up member for single lacing, and 70% of that distance for double lacing.

Example # 1 - Contd• Lacing: Code Guidelines – Page 5-44– The inclination of lacing bars to the axis of the member

shall preferably be not less than 60o. for single lacing and 45o for double lacing.

– When the distance between the lines of fasteners or welds in the flanges is more than 15 in, the lacing preferably shall be double or be made of angles.

Example # 1 - Contd• Lacing– The angle in between the lacing is 60o

– Distance between the fasteners

= 9 + ( 2 x 3.4 ) – ( 2 x 1.75 )

= 12.3 in < 15 in

Hence use single lacing.– Shear stress = 2 % of Load– = 0.02 x 358.84– = 7.177 kips– Force of single lacing = 7.177 ÷ 2 – = 3.588 say 3.6 kips– =

60o

1.75”

9”3.4”

60o

60o 14.2”

14.2”

12.3”

Example # 1 - Contd

• Let us assume that the plate area is A = b x t • The ratio l/r for lacing bars arranged in single systems

shall not exceed 140.• We know that

• Now

Example # 1 - Contd

inch

Example # 1 - Contd

= >

– From Table C-36, Page 3-16, 131 = 8.70, 132 = 8.57

– By interpolating the Allowable Stress = Fa = 8.677 ksi

– Calculating the width b

– Pa = Fa x Ag = > 4.2 = 8.677 x b x 3/8

– b = 1.29 say 1.5 inch– Min edge distance = 1.25 so b = 2 x 1.25 = 2.5 inch– Hence we will use

Example # 2

• Data– Length of Column = 30 feet– Steel = A36– End condition Top = Pinned– End condition Bot = Pinned

• Requirements– Capacity of built up Member– Design of Lacing

• Assumption– In this example we will assume 0.5 so

Fa = 0.5 x Fy = 0.5 x 36 = 18 ksi

Lacing

4 x L 8” x 8” x ½”

21” x 21 “ Column

Example # 2 - Contd• We have the sections and lets note down the AISC Specs– Section L 8” x 8” x ½”– Area “A” 7.750 Page 1-46

– Ix 48.600

– Iy 48.600

– x 2.190– Value of K from Table C-C2.1 is 1.00 – Calculating the Allowable load

– Pa = Fa x Ag = 18 x 2 x 7.75 = 558 kips

– Now we have to find and for that first find moment of inertia along both axis.

Example # 2 - Contd

2.

10.5”

21”

8”5”

1/2”

Example # 2 - Contd• Ixx = ?

–

– Ixx = 4 x ( 48.6 + 7.75 (10.5 – 2.19) ^ 2 )

– Ixx = 2335.139

• Iyy = ?

–

– Iyy = 4 x ( 48.6 + 7.75 (10.5 – 2.19) ^ 2 )

– Iyy = 2335.139

Example # 2 - Contd• rmin = ?

– rmin = 8.68 in

– So OK

– From Table C-36, Page 3-16, 41 = 19.11, 42 = 19.03

– By interpolating the Allowable Stress = Fa = 19.07 ksi

– Calculating the Allowable load

– Pa = Fa x Ag = 19.07 x 4 x 7.75 = 591.17 kips

Example # 2 - Contd• Lacing– The angle in between the lacing is 45o

– Distance between the fasteners

= 21 – ( 2 x 1.75 )

= 17.5 in > 15 in

Hence use double lacing.– Shear stress = 2 % of Load– = 0.02 x 591.17– = 11.82 kips– Force of single lacing = 11.82 ÷ 4 = 2.96 kips – Unsupported length = 0.70 x 24.7 = 17.29”– =

45o

1.75”

21”

45o

24.7”

17.5”

Example # 2 - Contd

• Let us assume that the plate area is A = b x t • The ratio l/r for lacing bars arranged in double systems

shall not exceed 200. Double lacing bars shall be joined at their intersections.

• We know that

• Now

Example # 2 - Contd

Say = 5/16 inch

Example # 2 - Contd

= >

– From Table C-36, Page 3-16, 191 = 4.09, 192 = 4.05

– By interpolating the Allowable Stress = Fa = 4.06 ksi

– Calculating the width b

– Pa = Fa x Ag = > 3.0 = 4.06 x b x 5/16

– b = 2.36 say 2.5 inch– Min edge distance = 1.5 so b = 2 x 1.5 = 3.0 inch

(Refer Table J 3.5 Page 5-76)– Hence we will use 17.5” x 3” x 5/16”

Base Plates

Base Plates• Base Plates: Code Guidelines – Page 3-106– Steel base plates are generally used under columns for

distribution of the column load over a sufficient area of the concrete pier or foundation.

– Unless the m and n dimensions are small, the base plate is designed as a cantilever beam, fixed at edges of a rectangle whose sides are 0.80 bf and 0.95d. The column load P is assumed to be distributed uniformly over the base plate within the rectangle. Letting Fb equal 0.75 Fy. The required thickness of base plate “ tp “ is found as under:

Base Plate

0.95 d

n0.8 bf

n

m

m

B

N

A1 = Base plate area

A2 = Pedestal Area

Base Plates• Base Plates: Code Guidelines – Page 3-106– Dimensions of the base plate are optimized if m = n.

This condition is approached when

– Where – When the values of m and n are small (the base plate is

just large enough in area to accommodate the column profile), a different model may be required. With this type of base plate, the column load is assumed to be distributed to the concrete area as follows

– A = 2 ( d + b – 2 L ) L– Where d = depth of column section, in– b = Flange width of column section,

in

Base Plates• Base Plates: Code Guidelines – Page 3-106– The dimension L can be found from the following

equation, where P = Column load

– Designing for fp = Fp, the required plate thickness may be calculated

Base Plates• Steps in design of Base Plates:– (Page 3-108)

– Find A1 and A2 : Use larger value

– Determine N and B– Determine actual bearing pressure on concrete– Determine m and n– Determine L

– Determine tp using m, n and L : The largest governs

Example # 1

• Data– Bearing Pressure at Sp = 1650 psi– Steel = A36– Column Section = W 12 x 65– Column reaction = 370 kips

• Calculate required area of base plate–

Ag = 225 in2 = 15” x 15”

Actual Bearing pressure is

Example # 1 - Contd• We have the sections and lets note down the AISC Specs– Section W 12 x 65– Area “A” 19.10 Page 1-28 and 1-29

– d 12.12 – b 12.00– Calculate the value of “m” , “n” and “L”. We will use

largest value.–

– m =

Example # 1 - Contd

– Putting the values– L = 46 x 10-3

– So For Plate thickness

– Where fb = 0.75 fy = 0.75 x 36000 = 27000 psi

– Hence Plate size is 15” x 15” x 1- 5/16”