Class Lectures 2- Tension Members

5/28/2018 Class Lectures 2- Tension Members

1/9

T. 1

Net width = smallest width across a

possible failure line, either strait or zigzag

Hole dia,

Actual hole dia = bolt dia + "161 (normally)

For net area calculation

hole dia = actual hole dia + "161

= bolt dia + "81 (normally)

Flat Single L Double Ls Channel I Bar Tube Tube

pitch

a. regular holes

gage

pitch

S

gage

g

S

g

a. staggered holes

Fractureat the hole section

Yielding

TENSION MEMBERS

Types of Tension Members

Design Strength

Design strength ntP= ( =nP nominal tensile strength)

This must be smaller of the two values

i. gyAF9.0 yielding of gross section

ii. euAF75.0 fracture of net section

=gA gross area of member

=e

A effective net area

=yF minimum yield strength of steel

=uF tensile strength (ultimate strength ) of steel

Gross Area ( gA )

Full cross-sectional area of a member without any deduction

Net Area ( nA )

Net area, =nA thicknessnet width of element= gross area hole area

tdb )( =

For a welded connection where there is no hole,

ng AA = welded connection with no holes

Calculation of nA for Bolted Connection


2/9

T. 2

2"

3"

3"

2"

2" 21/4" 2" 7/8"bolts

1/2"thickness

A E

C

DB

7/8"bolts

4"

1/2 x 4

g

g

1

2

For strait path failure

Net width =nw gross width, gw hole diatwA nn =

For zig-zag failure path

=nw gw hole dia

+ gageeachfor

g

S

4

2

= gn AA hole area

+ t

g

S

4

2

Example # 1Find the net area of the plate shown

hole dia "0.1""81

87 =+=

Net width, "0.314 ==nw 2

21 5.13 inAn ==

Example # 2Find the net area of the member shown

hole dia "0.1""81

87 =+=

1. Strait path (line A-B)"0.70.129 ==nw

2. Zig-zag path (line A-C-B)

"5.734

25.2

24

25.20.139

22

=

+

+=nw

3. Zig-zag path (line E-C-B)

"92.634

25.2

24

0.20.139

22

=

+

+=nw

4. Zig-zag path (line A-C-D)

"97.634

0.224

25.20.139

22

=++=nw

"92.6=nw ,2

21 46.392.6 inAn ==

Angles:

gross width = leg width- thicknessex: L64

21 , "46.346

21 =+=gw

Distance between holes in two legs (i.e. g)

tgg += 21


3/9

T. 3

3/4"bolts

L6x4x1/2

2" 2"

2"

21/2"

21/2"

5.4222

1

2

1

2

1 =+

L6x4x1/2transverse weld

6"

longitudinal weld

w

L

A=6x1/2

=3"

Example # 3Find the net area of the angle shown

25.9 inwg = , hole dia """ 87

81

34

=+=

strait: 287 75.725.9 inwn ==

zig-zag:

222

87

5.44

2

24

235.9

+

+=nw

254.7 in= 2

21 72.354.7 inAn ==

Effective Net Area ( eA )

Two cases:1. All elements of cross section connected by bolts or welded.2. Not all elements are connected.

Case 1: all elements connected (bolted or welded)

en AA =

Case 2: not all elements connected (bolted or welded)

a) For bolted connection (not all elements connected)

UAA ne = ( 9.0U )

=U a reduction factor Alternative values of U (P.16.1-177 AISC-commentary)

for all members as indicated

a) W, M, or S (width32 depth) & tees cut from them: with

minimum 3bolts/line, 9.0=U b) All other shapes, including build-up shapes, with minimum

3bolts/line, 85.0=U c) All other members with only 2bolts/line, 75.0=U use alternative values

b) For welded connection (not all elements connected)i) When transverse load is transmitted by transverse weld:

UAAe = , 0.1=U

=A area of directly connected partAAe =

ii) Load transmitted to a plate only by longitudinal weld along bothedge at the end,

UAA ge =

0.1=U for wl 2

87.0=U for wlw 5.12 > 75.0=U for wl>5.1


4/9

T. 4

7/8"bolts

W8x28

C9x15

7/8"bolts

3/4"bolts

L3x3x3/8

W8x28

7/8"bolts

L= length of weld on each sidew= distance between longitudinal welds

Example

Calculate the eA values of the following sections

i) width >= 3254.6 depth 06.8=

three bolts/line, 9.0=U 224.8 inAg =

=nA gross area hole area267.7285.0)0.12(24.8 in==

290.667.79.0 inUAA ne ===

ii) Only three bolts/line, 75.0=U 241.4 inA

g = 284.3285.0)0.12(41.4 inAn ==

288.284.375.0 inAe ==

iii) Only three bolts/line, 85.0=U 211.2 inA

g =

328.011.2)(111.283

81

43 =+=nA

2782.1 in= 2515.1782.185.0 inUAA ne ===

iv) All side connected, 0.1=U 271.9 inAg =

290.00.12435.00.1471.9 =nA 2

39.758.074.171.9 in== 239.7 inUAA ne ==


5/9

T. 5

21/2"

51/2"

2"

2L 6x4x1/2

1" bolts13/4"13/4"

1/2"

s

5.4222

1

2

1 =+

Example of Strength Calculation (Capacity)

ExampleFind the maximum tensile capacity of a member consisting 2L26x4x

21

can carry for two cases:a) Welded connection

b) Bolted connection (1" dia bolts), ksiFy 60= and ksiFu 70=

a) Welded connectionNet area = gross area (all sides connected)

250.9 in= Yielding kAFF gyt 51350.9609.09.0 ===

Fracture kAFF eut 53450.97575.075.0 ===

Then the tension capacity, kPnt 513= (yielding controls)

b) Bolted connectionConsider one L

nA calculation:

=gw gross width "5.946 21 =+=

Strait section: "25.7)1(25.981 ==nw

Zig-zag section:44

75.1

5.24

75.1)1(35.9

22

81

+

+=nw

"62.6= 2

21 31.362.6 inAn == for one L

For 2L2, 262.6231.3 inAn ==

All sides connected, 0.1=U , 262.6 inUAA ne ==

tF calculation

Yielding kAFF gyt 51350.9609.09.0 ===

Fracture kAFF eut 37262.67575.075.0 ===

Thus, kPnt 372=


6/9

T. 6

Y

Y Z

Z

X X

X

Y

Y

X

X

Y

Y

X

Design Consideration1. design strength:

required gross area, net area, strength or capacity factor load effects, uP

un PP

2. slenderness ratio:

maximumr

lpreferably 300 (does not apply to tension rod, wires)

=l length of the member

=r minimum radius of gyrationgA

I min=

Example

Calculate maximum slenderness of the following members

i) single angle, L6x4x21 ; ftl 14=

A single angle has no axis of symmetry;therefore, Principal axes are neither x-x nor y-y axis.

z-axis is the minor principal axis. ZI is the minimum.

Thus, minr is about z-axis.

"870.0=Zr (compare "91.1=xr , "15.1=yr )

193870.0

1214=

=

r

l

ii) 2L26x4x

21 with "

83 long legs b/b (back to back)

ftl 16=

One axis of symmetry; x-x and y-y are principal axes

"91.1=xr , "64.1"64.1 min == rry

11764.1

1216=

=

r

l

iii) Tee section ST 10x33; ftl 20=

"10.3=xr , "19.1"19.1 min == rry

20219.1

1220=

=

r

l

note:for single L, Zr is the minimum radius of gyration,

i.e. z-axis controls slenderness


7/9

T. 7

Design ProcedureIn real situations, design is a "trial & error" process, as the connection

detail is not known beforehand. Based on assumed connection, the selectionof a member is made and then, when the connection is finalized, the sectionis checked again:

i) Find required gA , eA and nA from given connection. From required nA ,

calculate required gA .

ii) From the two values of gA , take the larger value.

iii) Find the required minr to satisfy 300min

=r

l.

iv) Select a section that satisfy (ii) and (iii).

v) Once connection is finalized, recheck the member must satisfy strength

criteria andr

l.

Example #1Select a single angle tension member to carry kips40 DL and

kips20 LL . Member is ft15 long and will be connected by single line of bolts

of "87 dia in one leg. Assume single line of bolt holes. A-36 steel.

Step 1: Find the required uP

LLDLPu

6.12.1 += or DLPu

4.1=

206.1402.1 += 404.1 = = k80 controls k56=

Required kPu 80=

Step 2: Find required gA & nA .

Required 247.2369.0

80in

F

PA

y

u

g =

==

Required =

== 284.1589.0

80in

F

PA

u

u

e

controls

One side is connected; assume more than 3 bolts /lineSimplified value 85.0=U

Required 261.285.0

84.1in

U

AA en ===

Step 3: From an assumed connection (which would work), calculateRequired gA .

tAtAA ggn 0.1)(1 81

87 =+=

Thus, ttAA ng 0.116.20.1 +=+=

Step 4: Calculate required minr

"60.0300

1215300

min === lr


8/9

T. 8

7/8"bolts

2MC10"

X

Y

X

Y

Step 5: Select a suitable shape(It is possible to find the best (least weight) section, by changing tand

getting a section. This will be discussed in class)In practice, we use a suitable t. Say, we choose "

43=t

From step 3, required gg AinA >=+= 2

83 53.20.116.2 from step 2

Thus, required 253.2 inAg =

SelectionL483

213 ( 22 53.268.2 ininAg >=

"60.0"719.0min >==rrZ

Example #2Select a pair of MC as sown to carry a factored ultimate load of

kips490 in tension. Assume connection as shown. Steel ksiFy 50= , ksiFu 65=

(A572, grade 50). Length ft30=

1. kPu 490= ; per channel kPu 245=

2. required 244.5509.0

245in

F

PA

y

u

g =

==

required 203.5659.0

245in

F

PA

u

u

e =

==

required 203.50.1

03.5in

U

AA en ===

3. as trial, assume flange thickness in5.0 and web thickness in3.0 60.13.00.125.00.12 == ggn AAA

Thus, =+=+= inAA ng 63.660.103.560.1 controls

4. required "2.1300

1230

300min =

==

lr ( as built-up section "2.1 xr )

5. try MC 10x25; 235.7 inAg = ; itw 38.0= & "575.0=ft

"78.3=xr

6. check capacity244.538.00.12575.00.1235.7 inA

n == 244.5 inA

e =

i) Yielding kAFP gynt 6.661)35.72(509.09.0 ===

ii) Fracture kAFP eunt 4.530)44.52(7575.075.0 ===

kkPnt

4904.530 >= OK

Use 2MC 10x25


9/9

T. 9

2MC 10x25

1/4x5

15'

15'

Tie PlatesFor built-up members, tie plates are required to make members behave

as unit.

Between tie plates, each member behaves as single. Therefore,r

l

between tie plates corresponds to that for a single member.

For single C, "0.1min == yrr

Maximum ftftl 300.2512

0.1300

Class Lectures 2- Tension Members

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