2 - Tension Member

1

TENSION MEMBER

- LRFD -

Ari Wibowo, Ph.D

Steel Structure

2

Cross-Sectional Shapes for Tension Members

The more common shapes used for tension members.

The solid circular bar is frequently used, either as

threaded rod or welded to other members. The

threaded end provides simple connection but mus take

into account the reduction of the cross sectional area.

3

Cross-Sectional Shapes for Tension Members

Some typical built-up cross section

Square, rectangular and circular tubes have attractive appearance and

ease of maintenance, but the end connections may become

complicated and expensive. The tubes is useful for longer tension

components, where slenderness and related serviceability

consideration may be important.

4

Behavior and Strength of Tension members

Load-Deformation Relationship

The differences between the curves in (a) and (b) may be attributed to a

number of factors:

1. The load P is not applied concentrically to the member

2. The full-size shape has certain built-in or residual stress

3. The tension member is not perfectly straight.

4. The cross-sectional shape and size may vary to some extent along the

member length.

5


Items 1 and 3 imply that there will be a certain amount of bending in the member,

in addition to the effect of the axial load.

Therefore, as the load

increased, the fiber that is

stressed the most in tension

due to bending plus axial

load will yield before any

other fiber.

It is clear that the

moments that are

produced create a

nonuniform stress

distribution in the cross-

section

That fiber reaches the yield plateau of the stress-strain curve before all of the

others, which are still strained in the elastic range.

6


item 2 : Effect of Residual Stress

7


The presence of residual stresses in

compression members results in a

significant lowering of the actual

strength.

For tension members, on the other

hand, the only effect is a shift in the

load-deformation curve that reflects

the larger deformations, but the

strength stays the same.

8


Item 4 : a member of cross-sectional area is not a constant along

the length.

That hole creates a stress concentration. As a result, the

application of an increasing tensile force will cause yielding at

the edges of the hole first.

On the other hand, while stress concentrations may not

reduce static yield strength of the member, they do have

significant effect on the dynamic or fatigue-related structural

performance.

9


Aspect of behavior of the load-deflection curve for the full-size

member

1. Range 0-1 : All fibers in the cross section are stressed to less than yield

stress. The member is therefore fully elastic, and all deformations are

recoverable upon unloading.

2. Range 1-2 : First yield occurs for P = Pprop, which is the proportional

limit.

3. Range 2-3 : Certain fibers that yielded early are bound to reach strain-

hardening.

4. Range 3-4 : a number fibers will already have reached the descending

portion of the stress-strain curve.

10


Eccentricity Load is less dangerous than lateral load case.

11

BATANG TARIK PADA STRUKTUR

nRQ

For tension members, this takes the form

untntu PP PP or

The central problem of all member design,

including tension member design, is to find a

cross section such that the sum of the factored

loads does not exceed the member strenght;

that is :

12


75.0t

where Pu is the sum of the factored loads. To

avoid reaching the limits state of yielding,

y

ugugy

F.

PAPAF.

900or 900

u

ueueu

F.

P A PAF.

750or 750

To avoid fracture,

If Ae = UAn, then,

UF

PA

u

un

75.0

90.0t

13


The slenderness ratio limitation will be satisfied if,

300

Lr

where r is the minimum radius of gyration of the cross

section and L is the member length.

14

Example 1

A bar 5 x ½ of A36 steel is used as a tension

member. It is connected to a gusset plate with four

5/8-in diameter bolts as shown here. Assume that

the effective net area Ae is equal to the actual net

area and compute the design strength.

15

• For the limit state of yielding,

Solution

2

21 .5.25 inAg

The nominal strength is

kipsAFP gyn 90)5.2(36

And the design strength is

kipsPnt 81)90(9.0

• For fracture of the net section,

example) for this(.75.1

.75.175.05.2

25.2

2

2

43

21

inAA

in

holes

AAA

ne

holesgn

16

The nominal strength is

kipsAFP eun 5.101)75.1(58

And the design strength is

kipsPnt 1.76)5.101(75.0

The smaller value control

17

Limit State

In the design of tension members, the following cross-sectional area definitions are important :

1. Gross cross-sectional area, Ag

2. Net Area, An

3. Effective net area, Ae

For s steel plate with width b, thickness t and single hole of diameter d :

and

btAg

tdbAn

The effective net area is a function of the cross section and the

method of end connection

18

Two modes of failure

1. Yielding on the gross area , = 0.9

If the size of the gross area is such that the condition

is reached before the ultimate stress Fu has been developed anywhere in

the member.

The characteristics are : ductile, large deflection/deformation

phenomenon.

2. Fracture on the effective area , = 0.75

If the size of the effective area is such that the condition

is reached before the yielding stress Fy has been developed on the gross

area.

The characteristics are : brittle, small deflection/deformation

phenomenon, occurs suddenly.

y

g

n FA

P

u

e

n FA

P

19

Stress Definition

Stress are defined as

Then can be rewritten as :

and

Two limit states occur simultaneously if

or

u

e

neffy

g

ngross F

A

PF

A

P and

eueeffn

gyggrossn

AFAP

AFAP

gyeu AFAF

u

y

g

e

F

F

A

A

20

The limit state of yielding on the gross section governs if :

or

Conversely, fracture on the effective net section governs

when

or

gyeu

u

y

g

eg

AFAF

F

F

A

AAonYielding

:

gyeu

u

y

g

ee

AFAF

F

F

A

AAonFracture

:

21

Computation of Areas1. Gross Area

No holes or other area reductions can be present where the

section is made

+ plates and bars, Ag = bt

+ solid circular, Ag = d2/4

or in general :

+ angles, can be treated as an equivalent plate

width : we = l1 + l2 – t , thickness : te = t

so : Gross area : Ag = wete = (l1 + l2 –t)t

iig twA

22

Computation of Areas

2. Net Area

Net area is a gross area that reduced by holes resulting from

fasteners (bolts, rivets) and certain type of welds (plug and

slots).

Such welds (fillet, full and partially groove welds) do not

reduce or increase the cross-sectional area, so :

Welded joints : An = Ag

23

Size of Standard Holes

Recognizing the needs for fabrication and erection tolerances, standard

bolt holes are made 1/16 in. larger in diameter than that of the bolt.

The hole making may damage the material, as in case punch holes, the

edges around the hole are deformed. This region discounted and the

effective holes increased by another 1/16 in.

161

161: dddiameterholeEffective e

24

Gross area: Ag = 6(½) = 3.0 in.2

Net Area : Ag = ( b – de ) t

For a ¾-in.-diameter bolt, the effective hole size is

Consequently,

Example 2

.87

161

161

43 inde

1. A single line of standard holes for ¾-in bolts are placed

in a 6½ plate. Determine the gross and net areas

Solution

2

21

87 in.56.20.6 nA

25

Gross area: Ag = 10( ¾ ) = 7.5 in.2

Net Area : An = ( b – de ) t

For a 7/8-in.-diameter bolt, the effective hole size is

Consequently,

.in0.1161

161

87 ed

2. A double line of standard holes for 7/8-in bolts are placed

in a 10¾ plate. Determine the gross and net areas

Solution

2

43 in.00.60.120.10 nA

26

Oversized and Slotted Holes

Section J3.7 LRFD Specs. gives the

measurement for oversized holes, short

slotted and long slotted holes.

Basically, this types of holes are used in

many instances to facilitate the erection of

the structure, or to allow certain rotations or

deformations for specific loading case.

When the tension member is short

connecting element, such as stiffener or a

gusset plate,

An < 0.85 Ag

27

Influence of Hole Placement

The location of the governing net section is not obvious for

either of the two cases shown above. (a) is a plate and (b) is an

angle with pitch s and gages g between the holes.

28

• For the plate, perpendicular section 1-1 and 2-2 give identical An values,

where only a deduction for one hole has been taken.

• For the angle, it is clear that section 2-2, which incorporates one hole

deduction for each of the legs, for a total of two is more critical than

section 1-1.

But the elastic stress distribution on plate sec. 1-1 and 2-2 is different.

If s is small enough (section 1-1 and 2-2 is close enough), the stress will

influence each other and elastic distribution will changed. On this basis, it is

possible that a section or chain such 1-2 on plate and 2-3-2 on angle may be

the governing one.

29

• The influence of the staggered hole could be accounted for by

using the following expression for the net width of the section :

wn = wg – d – xd

where : wn = the net width, wg is the gross width, d is the

diameter of the hole, and xd represents the strength-reducing

influence of the staggered hole.

• xd could be expressed as

• This ratio reduces to s2/4g, if it is assumed that 2d is significantly

less than 4g.

dg

sdxd

24

2

g

sdxd

4

2

30

• So the net width of a two-hole section, with one-hole offset with

a pitch of s and a gage of g, becomes :

or

Equation above states that the net width is found by deducting fully

for all holes in the chain, but since this overestimates the effect

of the staggered holes, a certain width contribution is added.

The generalized form of the net width equation for m holes with

(m-1) staggers for a chain :

g

sddww gn

4

2

g

sdww gn

42

2

i

iegn

g

smdww

4

2

31

Picture (a).

Given :

The hole pattern for an 18-in.-wide plate is loaded in tension.

Determine the net width that governs the design.

Example 3

32

Solution

Step 1. Chain A B C E F

Deduct for 3 holes @ ( ¾ + ⅛ ) = - 2.63 in.

For BC, add s2/4g = 2.02/(4)(4) = + 0.25 in.

For CE, add s2/4g = 2.52/(4)(10) = + 0.16 in.

Total deduction = - 2.22 in.

Step 2. Chain A B C D E F

Deduct for 4 holes @ ( ¾ + ⅛ ) = - 3.50 in.

For BC, add s2/4g = 2.02/(4)(4) = + 0.25 in.

For CD, add s2/4g = 4.52/(4)(6) = + 0.85 in.

For DE, add s2/4g = 2.02/(4)(4) = + 0.25 in.

Total deduction = - 2.15 in.

33

Step 3. Net width = 18 – 2.22 = 15.78 in.

Step 4. Check maximum allowable net area :

Max allowable net width = 18.0(0.85) = 15.3 in.

Since 15.3 is less than 15.78, the maximum allowable

value of step 4 governs : 15.3 in.

34

Picture (b).

Given :

A 6 4 ½ angle is used as a tension member. It has

holes for ⅞-in. bolts that are placed as shown below.

Example 4

35

Step 1. Determine the width of the folded-out angle “plate” :

we = l1 + l2 – t = 9.5 in

Step 2. Gross area of plate:

Ag = we t = 9.50 (0.5) = 4.75 in.2

Step 3. The gages for the holes are shown above. The gage

between the holes closest to the heel of the angle in the two legs is

found as ( g + g1 – t ) = 2.50 + 2.25 – 0.50 = 4.25 in.

Step 4. Computation of net area: The governing net section will be

section 2-2 or 2-1-2.

Section 2-2: An2 = [9.50 – 2 (⅞ + ⅛)] 0.50 = 3.75 in.2

Section 2-1-2:

Solution

222

3 in.5.350.0)25.4(4

50.2

)50.2(4

50.2)1(350.9

nA

36

Step 5. Governing net area :

Since An3 = 3.50 in.2 is less then An2 as well as An,max =

0.85 Ag = 4.04 in.2, it governs the net area of the given

angle and hole configuration.

37

Effective Net Area• A connection always weakens the member, and the

measure of its influence is called the joint efficiency

• This factor is a function of :

– The ductility of the material

– Fastener spacing

– Stress concentration at holes

– Fabrication procedure

– Shear lag phenomenon

Shear Lag occurs when all elements of the cross section are not

connected. This happens, for example, when only one leg of an

angle is bolted to a gusset plate (figure below)

38

Shear Lag

9.01

L

xU

The consequence of this partial connection is that the

connected element becomes overloaded and the unconnected

part is not fully stressed. Lengthening the connected region

relative to the member length will reduce this effect.

Shear Lag is accounted for by using a reduced, or effective, net

area obtained by multiplying the actual net area by the reduction

factor, U :

x

Where :

= the distance from the centroid of the connected area to

the shear plane of the connection.

L = the length of the connection

39

For bolted connection, the effective net area is :

Ae = U An

For welded connection,

Ae = U Ag

If all element of the cross section are connected, then there

is no reduction, and Ae = An

40

Shear Lag – bolted connection

L

x

L

xL

L

LU

1

'

The reduction can be approximated by multiplying the net area

by the factor U :

A typical example calculation of the

shear lag factor is shown for both a

bolted and a welded case.

For the bolted case, N is the

total number of bolts, and s is

the pitch.

41

In case where there are

transverse and

longitudinal weld :

a. Transverse :

• Longitudinal

L

xUL 1

Ag

tdU

ftransverse

T

.

TL

TTLL

dd

dUdUU

..

It is recommended that weighted average U value should

be used for such cases.

Shear Lag – welded connection

42

Simplified method: 3 condition of rules

for bolted conection

1. For W-, M-, and S-shapes that have a flange width to web depth ratio of

at least ⅔ (and tee shapes cut from these) and are connected through

flanges with at least three fasteners per line in the direction of applied

load,

U = 0.9

2. For all other shapes (including built-up shapes) with at least three

fasteners per line,

U = 0.85

3. For all members with only two fasteners per line,

U = 0.75

Based on average values of x/L for various types of bolted tension

member connections, the Commentary to AISC B3 gives values of

reduction factor U :

43

Example : Application of the rules

44

1. For any W-, M-, and S-shapes or tee shapes connected by transverse

weld alone,

Ae = area of connected element

2. For plates or bars connected by longitudinal welds at their ends,

U = 1.0 for l ≥ 2w

= 0.87 for 1.5w ≤ l < 2w

= 0.75 for w ≤ l < 1.5w

The Commentary to AISC B3 gives values of reduction factor U

for welded connection :

Where :

l = length of the pair welds ≥ w

w = distance between the welds

Simplified method: 2 condition of rules

for welded conection

45

Determine the effective net area for the tension member shown

below

Example 5

Solution

An = Ag – Aholes

= 5.75 – ½ (⅝ + ⅛)(2) = 5.00 in.2

Since only one element (one leg) of the cross section is

connected, the net area must be reduced. From the

properties tables, the distance from the centroid to the

outside face of the leg of an L 6x6x½ is

2in.68.1x

46

The length of the connection is

L = 3 + 3 = 6 in.

Ae = U An = 0.720 (5.00) 3.60 in.2

)(9.0720.06

68.111 OK

L

xU

The average value of U from Commentary can also be used.

Since this shape is not a W, M, S, or tee and has more than

two bolts in the direction of the load, the reduction factor U

can be taken as 0.85, and

Ae = U An = 0.85 (5.00) = 4.25 in.2

Either U value is acceptable, but the value obtained from

AISC equation is more accurate. However, the average values

of U are useful during preliminary design, when actual

section properties are not known.

47

Example 6

Solution

If the tension member 66½ is welded as shown, determine

the effective net area

Only part of the cross section is connected, and reduced

effective net area must be used. The connection is made with

a combination of longitudinal and transverse welds, so it is

not one of the special case for welded members.

9.0695.05.5

68.111

L

xU L

522.075.5

6 21

g

ftransv

TA

tdU

48

2mm381.375.5588.0. ge AUA

Therefore, the effective area of welded members is 3.381

mm2.

588.0

65.5

6522.05.5659.0

..

TL

TTLL

dd

dUdUU

49

Block Shear

For certain configuration, a segment or “block” of material at

the end of the member can tear out. For example as shown

below, is called “block shear”.

The shaded block would tend to fail :

By shear along the longitudinal section ab

By tension on the transverse section bc

• The design strength of tension members are not always

controlled by factor of safety or by the strength of the bolts or

welds with which they are connected. They may instead be

controlled by block shear strength.

• In block shear mode, the failure of the member occurs along a

path involving tension on one plane and shear on a

perpendicular plane along the fasteners. When a tensile load

applied to a particular connection is increased, the fracture

strength of the weaker plane approaches.

• This plane does not fail instantly, because it is restrained by the

stronger plane. The load can be increased until the fracture

strength of the stronger plane is reached and during this time,

the weaker plane yields.

• The total strength of the connection equals the fracture strength

of the stronger plane plus the yield strength of the weaker plane.

50

51

The procedure is based on the assumption that one of the two

failure surface fractures and the other yields :

Fracture on the shear surface is accompanied by yielding on

the tension surface, or

Fracture on the tension surface with yielding on the shear

surface

There are two possible failure modes :

• For shear yield and tension fracture

Rn = [0.6Fy Agv + Fu Ant] (AISC equation J4-3a)

• For shear fracture and tension yield

Rn = [0.6Fu Anv + Fu Agt] (AISC equation J4-3b)

In both cases, = 0.75. Since the limit state is fracture, the controlling

equation will be the one which has the larger fracture term.

52

Example 7

Check the block shear design strength of the tension member

shown below. The holes are for ⅞-in. diameter bolts, and A36

steel is used.

Solution

Shear Areas :

Agv = ⅜(7.5) = 2.812 in.2

and since there are 2.5 hole diameters,

Anv = ⅜[7.5-2.5(1)] = 1.875 in.2

53

Tension areas :

Agt = ⅜(1.5) = 0.5626 in.2

Ant = ⅜[1.5 - 0.5 ] = 0.375 in.2

AISC Equation J4-3a :

Rn = [0.6Fy Agv + Fu Ant]

= 0.75[0.6(36)(2.812) + 58(0.375)]

= 0.75 [ 60.74 + 21.75 ] = 61.9 kips.

AISC Equation J4-3a :

Rn = [0.6Fu Anv + Fy Agt]

= 0.75[0.6(58)(1.875) + 36(0.5625)]

= 0.75 [ 65.25 + 20.25 ] = 64.1 kips.

The second equation has the larger fracture term, therefore, the

second equation governs.

54

Block Shear

Case (a)

The resistance to block shear is

primarily afforded by shear. Since

the shear fracture load is larger than

the shear yield load, failure will be

governed by shear fracture in

combination with tensile yield.

Case (b)

A case where tension is the primary

stress resultant; in this case the limit

state is that of combined tensile

fracture and shear yield.

55

Example 8 : Design

A tension member with the length of 5 feet 9 inches

must resist a service dead load of 18 kips and a

service live load of 52 kips. Select a member with a

rectangular cross section. Use A36 steel and

assume a connection with one line of ⅞ -in diameter

bolts. Solution

Pu = 1.2D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kips

2

2

in 409.20.75(58)

104.8

)(0.75

in 3.2350.90(36)

104.8

)(0.90

y

ue

y

ug

F

PA

F

PA

56

Since Ae = An for this member, the gross area

corresponding to the required net area is

t

t

AAA holeng

409.2

8

1

8

7409.2

Try t = 1 in

2in 4093114092 .)(.Ag

57

Rounding to the nearest ⅛ in., try a 13½ cross

section. Check the slenderness ratio :

2in409.31

409.3

t

Aw

g

g

4

3

min .2917.012

15.3inI

Since 3.409 > 3/235, the required gross area is

3.409 in.2, and

A = 1(3.5) = 3.5 in.2

58

Since I = Ar2,

)(3002392887.0

)12(75.5 maximum

in.2887.05.3

2917.0minmin

OKr

L

A

Ir

Answer : Use a bar 3½ 1 in.

59

Example 9

For the gusset plate in the heavy bracing connection that is

shown below, check whether the plate thickness of ½ in. is

adequate to resist block shear. The steel grade is A36 (Fy = 36

ksi, Fu = 58 ksi), and the holes are drilled, for ⅞-in.-diameter

A325 high-strength bolts.

Solution

It is assumed that the total

factored load of 225 kips is

distributed evenly between

all of the bolts.

The loads listed are those

that are in the gusset plate at

each plane.

60

Assume that planes B and D

have already been evaluated

and found to be acceptable.

Plane A

Anv has two shear plane

Agt has one tension plane

Solution

2

21

161

87 in.53.250.000.32 nvA

2

21 in.00.800.2200.62 gtA

Block Shear Capacity

Pbs = 0.75(0.6FuAnv + FyAgt)

= 0.75(0.6(58)(2.53) + 36(8.00)]

= 261 kips > 80.4 kips (OK)

61

Plane C

Anv = 2[9.00 – 2.50(0.938)] ½ = 6.66 in.2

Agt = 8.00 in.2

Block shear capacity :

Pbs = 0.75[0.6(58)(6.66) + 36(8.00)]

= 390 kips > 192.9 kips (OK)

Since all planes are adequate, it is not necessary to

analyze the other limit state, shear yield plus

tension fracture.

62

Example 10

A W1443 wide-flange shape is connected by flange plates, as

shown below. The bolts are ⅞-in.-diameter A490-X, high

strength bolts. For a single flange the design strength of the

bolts in shear is 211 kips; and the design strength of the bolts in

bearing is 548 kips. Determine the design strength of the

member.

63

Solution

Yield Limit State of Shape :

Gross area of W1443 = Ag = 12.60 in.2

Design yield strength of the member = FyAg

= 0.9(36)(12.6) = 408 kips

Fracture Limit State of Member

• Step 1 Hole area to be deducted in the flange at each connection

L = (N/2 – 1)s = (6/2 -1)(3.00) = 6.00 in.

• Step 2 Determine x and L to be used in the shear lag calculation

The W1443 is

considered as two tee

section WT721.5

64

The x of each WT is found in the LRFD Manual as 1.31 in.

The value of L is :

L = (N/2 – 1)s = (6/2 – 1)(3.00) = 6.00 in.

• Step 3 Calculate the shear lag factor U :

U = 1 – x/L = 1 – 1.31/6.00 – 0.78

The LRFD specification, section B3, notes that for this case, a value of

U = 0.85 may be used.

• Step 4 The design strength of the member is twice the capacity of

each flange connection. The gross area of tee is 6.31 in.2

Design fracture strength = 2AnFuU

= 2(0.75)(6.31 – 1.06)(58)(0.85) = 388 kips

65

Block Shear of Section Flanges :

The design strength is provided

by the larger of the two basic

block shear combination :

(1) Shear yield + Tension fracture

(2) Tension yield + Shear fracture

Step 1 Fracture on shear plane :

Pbs = 2(0.6)FuAnv = kips20953.050.200.8586.021615

Step 2 Yield on tension plane :

Pbs = 2FyAg = 2(36)(2)(0.53) = 76.3 kips

Step 3 Design Strength for condition 1:

(Pbs)total = (Shear + Tension) = 0.75(209+76.3) = 214 kips

66

Step 4 Fracture on tension plane :

Pbs = 2FuAn = 2(58)[2-(½)(15/16)](0.53) = 94 kips

Step 5 Yield on shear plane:

Pbs = 2(0.6)FyAg = 2(0.6)(36)(8.00)(0.53) = 183 kips

Step 6 Design Strength for condition 2:

(Pbs)total = (Tension + Shear) = 0.75(94+183) = 208 kips

Step 7 Governing block shear strength:

The largest capacity from steps 3 and 6 is condition 1; thus

Block shear design strength = 214 kips.

67

Summary For Connection

Bolt design shear strength (given) 211 kips

Connection bolt bearing capacity (given) 548 kips

Gross section yielding of member 408 kips

Effective net section fracture of member 388 kips

Block shear capacity of flanges 214 kips

Conclusion :

The bolt design shear strength of the connection of 211 kips governs

68

Example 11

A WT5 11 in A36 steel is

connected to a gusset plate

with the ends completely

welded and each side

welded along 6 in.

Determine the effective net

area of the member. From

the LRFD Manual, Ag = 3.24

in.2,

in.36.0 and in.07.1 ftx

69

Solution

For the 6-in. welds, the value of U = 1 – 1.07/6.00 = 0.822. For the

end weld, the value of U = 5.75(0.36)/3.24 = 0.639.

The effective reduction coefficient is now found as the weighted

average of the above.

76.075.5)00.6(2

)75.5(639.0)12(822.0

effU

The effective net area is then

Ae = UeffAn = 0.76(3.24) = 2.46 in.2

70

Example 12

The truss diagonal

member consist of a pair

of angles L4 3 ⅜ that

are loaded in tension.

Determine the design

strength T of one angle.

The bolts that will be

used are ¾-in. A325-N,

and the steel is A36.

The bolt design strengths

for the connection in one

angle are 46.5 kips in

shear and 88.1 kips in

bearing

71

Solution

Step 1 Determine the angle design strength for the limit state of

yielding on the gross section :

Ag = 2.48 in.2

T = FyAg = 0.9(36)(2.48) = 80.4 kips

Step 2 Determine the angle design strength in tension fracture

Ag = 2.48 - ⅜(¾+⅛) = 1.23 in.2

Reduction coefficient :

U = 1 – x/L = 1 – 1.28/6.00 = 0.787

T = UFuAn = 0.787(0.75)(58)(1.23) = 42.1 kips

Step 3 Check angle design strength in block shear, for tension

yield and shear fracture combination

72

Tension yield load :

Pbs = FyAgt = 0.75(36)(⅜)(1.5) = 15.2 kips

Shear fracture load :

Pbs = (0.6)FuAnv

= 0.75(0.6)(58)(⅜)[7.25-2.5(⅞+1/16)]

= 48.0 kips

Total block shear capacity :

(Pbs)total = 15.2 + 48.0 = 63.2 kips

Since the lowest tensile design strength is that of angle in tension

fracture, the load 42.1 kips controls. Further, since the larger of the

two block shear mode controls, additional computations are not

needed.

Conclusion : The design tension strength of the angle is 42.1 kips.

Documents

2 - Tension Member