DESIGN AND ANALYSIS OF TENSION MEMBERS · Steel Design Misan University Fourth Year Engineering...

Steel Design Misan University Fourth Year Engineering College Dr.Abbas Oda Dawood Civil Department

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DESIGN AND ANALYSIS OF TENSION MEMBERS

1. INTRODUCTION

Tension members are structural elements that subjected to axial tensile forces. Tension

members do not buckle. Therefore, steel can be used most efficiently as tension members.

They are used in various types of structures and include truss members, bracing for building

and bridges, cables in suspended roofs, cables in suspension and cable-stayed bridges,

hanger and sag rods, towers and tie rods.

Any cross-sectional configuration may be used, because for any given section, the only

determinant of the strength of a tension member is the cross-sectional area. Different types

of sections used as tension members are shown below. Circular rods and rolled angle shapes

are frequently used. Built up shapes either from plates, rolled shapes or a combination of

plates and rolled shapes are sometimes used when large loads must be resist. The most

common built-up shapes is the double –angle section which available in AISC manual.

Steel cables are constructed of a number of wire ropes or strands have very high

yield strength in the range of 200 to 250 ksi. Thus, cables are particularly suitable for

covering large spans and are used in long-span suspension bridges, cable roofs, and

cable-stayed bridges. Cables, of course, are flexible. To provide stiffness, cable

structures may be stiffened by adding stiffening members.

When the magnitude of tensile force is small in a tension member, solid round or

rectangular bars are used.


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For larger tensile forces or when more stiffness is required, round or rectangular

tubes may be used. Round tubes or pipes might be preferred when the tension

member is exposed to high-wind condition. Connection details for round tube,

however, are cumbersome to construct.

Single angles are commonly used as tension members, for example, as bracing for

carrying lateral forces due to wind or earthquake. Angle end connection is simple but

eccentric to its centroidal axis. The eccentric application of tensile force produces

bending stresses in members which are often ignored in design practice.

Compared with an angle, a channel connected to the joint at its web often produces

less eccentricity, since the centroid of most channels is close to their web.

For carrying a large tensile force, W sections are used.

For a very large tensile force, built-up sections (for example, channels with lacing

bars) or double angles may be used.


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2. NET AND EFFECTIVE NET AREA

The stress in an axially loaded tension member is given by A

Pf , where, P is the magnitude

of load, and A is the cross-sectional area normal to the load. For example, consider an 8 x ½

in. bar connected to a gusset plate and loaded in tension as shown below

2.1 Net Area

When tension members are connected by welding, the total cross-sectional area is available

for transferring the tension. When the connection is done by bolting (or riveting), holes must

be made in the member. These holes evidently reduce the cross-sectional area available for

transferring the tension. Thus, the net area of the section is the gross area minus deductions

for the holes.

Area of bar at section a – a = 8 x ½ = 4 in2

Area of bar at section b – b = (8 – 2 x 7/8 ) x ½ = 3.12 in2

Therefore, by definition (Equation 4.1) the reduced area of section b – b will be subjected to higher stresses The unreduced area of the member is called its gross area = Ag

The reduced area of the member is called its net area =An

Note: Gusset Plate: is a connection element whose purpose is

to transfer the load from the member to a support of another

member

)1(Dww hgn

where

wn = Net width,

wg = Gross width, and

Dh = Hole diameter = nominal hole-diameter + 1/16 in.

Nominal hole diameter = 16/1dd bh , db = Bolt diameter

Dh = Hole diameter for calculating net area = "8/1d16/116/1d16/1d bbh

Multiplying Eq. (1) by the thickness of the member yields

)2(tDtwtw hgn

Since An = wnt and Ag = wgt, equation (2) can be simplified as follows:

)3(tDAA hgn


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2.1.1 Effect of Staggered Holes

Holes are sometimes staggered, as shown above. Staggering of the holes increase the net

area of the section. In figure above the plate may fail along section ABCD or section ABECD.

According to the approximate procedure provided by LRFD B2, the net width (wn) is obtained

by deducing the sum of the diameters of all the holes located on the zigzag line from the

gross width (wg) and the adding for each inclined line such as BE the quantity s2/4g.

)4(g4

sDww

2

hgn

where

s = the center-to-center spacing of the two consecutive holes in the direction of stress (pitch)

g = the transverse center-to-center spacing of the same two holes (gage)

Multiplying Eq. (4) by the thickness and subs. Since An=wnt and Ag = wgt, yields

)5(tg4

stDAA

2

hgn

)5(tg4

st)16/1d(AA

2

hgn

)5(tg4

st)8/1d(AA

2

bgn

When staggered holes are in different elements of the cross section, the shape can be visualized as a plate, like angle or channel or even if it is an I-shape.

OR

OR


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2.2 Effective area

The connection has a significant influence on the performance of a tension member. A

connection almost always weakens the member, and a measure of its influence is called

joint efficiency.

Joint efficiency is a function of: (a) material ductility; (b) fastener spacing; (c) stress concentration at holes; (d) fabrication procedure; and (e) shear lag. All factors contribute to reducing the effectiveness but shear lag is the most important.

Shear lag occurs when the tension force is not transferred simultaneously to all elements of the cross-section. This will occur when some elements of the cross-section are not connected. A consequence of this partial connection is that the connected element becomes overloaded and the unconnected part is not fully stressed as shown in figure below. shear lag can be accounted for by using a reduced or effective net area Ae. Shear lag affects both bolted and welded connections. Therefore, the effective net area concept applied to both types of connections. - For bolted connection, the effective net area is Ae = U An (6) - For welded connection, the effective net area is Ae = U Ag (7) Where, the reduction factor (Shear lag factor) U is given in AISC D3.3 Table 3.1, as following:

1. For any type of tension member except plates and round HSS with 3.1

)8(L

x1U

Where, x is the distance from the centroid of the connected area to the plane of the

connection, and L is the length of the connection.

If the member has two symmetrically

located planes of connection, x is

measured from the centroid of the

nearest one – half of the area.


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- The distance L is defined as the length of the connection in the direction of load.

2. Plates

3. Round HSS with 3.1 0.1U

See AISC Manual Table D3.3 for more cases of HSS

4. Single Angle

For welded connections, it is

measured from one end of the

connection to other.

For welded connections, it is

measured from one end of the

connection to other.

If there are weld segments of

different length in the direction of

load, L is the length of the longest

segment.

3- All bolted 0.1U

1- All welded 0.1U

2- Transverse Weld 0.1U

4- Longitudinal Weld

0.1Uw2For

87.0Uw2w5.1For

75.0Uw5.1wFor

Two or three fasteners per line in the

direction of the load 6.0U

Four or more fasteners per line in the

direction of the load 8.0U


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5. W, M, S, or HP, or Tees Cut from These Shapes

1-Flange connected with three or more fasteners per line in the direction of the load

9.0Ud3

2bf

85.0Ud3

2bf

2. Web connected with four or more fasteners per line in the direction of the load

8.0U

Example 1: Determine the effective net area for the tension member shown

Solution

2n

bgn

in02.52

1*)

8

1

8

5(*277.5A

t)8/1d(AA

From Manual, for angle L6*6*1/2

"67.1x

The length of connection = L = 3+3 =6 "

"7217.06

67.11

L

x1U

623.302.5*7217.0AUA ne

Or From Table D3.3m for angle has three bolts in direction of load U=0.6

012.302.5*6.0AUA ne

Use Larger Value =3.623

Example 2: Determine the effective net area for the tension member shown

Solution

From Manual, for angle L6*6*1/2

,in77.5A 2g "67.1x

The length of weld = L = 5.5 "

"6964.05.5

67.11

L

x1U

2ge in02.477.5*6964.0AUA


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Example 3: Compute the design (smaller) net area for the tension member shown. The holes are for 1"-diameter bolts.

Solution

- For section 1-1 (abde)

2n

bgn

in3125.104

3*)

8

11(*2

4

3*16A

t)8/1d(AA

- For section 2-2 (abcde)

22

n

2

bgn

in14.104

3*

5*4

3*2

4

3*)

8

11(*3

4

3*16A

tg4

st)8/1d(AA

Use the

smaller value 2

n in14.10A

Example 4: Determine the effective net area for the angle-shape shown. The holes are for 7/8"-diameter bolts.

Solution

From Manual, for

angle L8*6*1/2

,in75.6A 2g "5.0t

"75.45.025.23g

- For section 1-1 (abdf)

2n in75.55.0*)

8

1

8

7(*275.6A

- For section 2-2 (abcdeg)

2222

n

in015.55.0*]3*4

5.1

75.4*4

5.1

5.2*4

5.1[

]5.0*)8

1

8

7(*4[75.6A

For section 3-3 (abceg)

22

n in363.55.0*]5.2*4

5.1[]5.0*)

8

1

8

7(*3[75.6A

Each fastener (bolt)transmit 1/10 T, thus due to fastener d resist 0.1T:

At section 3-3 : Tension =0.9

2n in959.5

9.0

363.5A

)Control(in015.5A 2n

Due to both legs of the angle are connected (no shear lag reduction)

2ne in015.5AA

1

1

2

2

3

6"

1

1

8"

g

2

2 1.5

3

3


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Failure Modes of Tension Members

A tension member can fail by reaching one of two limit states:

(1) Excessive deformation (tensile yielding failure)

Excessive deformation can occur due to the yielding of the gross section along the length

of the member

(2) Fracture (tensile rupture failure)

Tensile rupture occurs when the stress on the effective area of the section is large

enough to cause the member to fracture, which usually occurs across a line of bolts

where the tension member is weakest. Namely fracture of the net section can occur if

the stress at the net section reaches the ultimate stress Fu.

3- Shear Block

For some connection configurations, the tension member can fail due to ‘tear-out’ of

material at the connected end. This is called block shear. This failure plane usually occurs

along the path of the centerlines of the bolt holes for bolted connections. This type of

failure could also occur along the perimeter of welded connections.

For certain connection configurations where tensile failure could be accompanied by

shear failure such that a block of the tension member tears away, for example, the single

angle tension member connected as shown in the Figure below is subjected to the

phenomenon of block shear.


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For the case shown above, shear failure will occur along the longitudinal section a-b and

tension failure will occur along the transverse section b-c. Also see below figures

Block shear strength is determined as the sum of the shear strength on a failure path and the tensile strength on a perpendicular segment:

- Block shear strength = net section fracture strength on shear path + gross yielding strength on the tension path

OR - Block shear strength = gross yielding strength of the shear path + net section

fracture strength of the tension path

TENSILE STRENGTH

Tension members according to LRFD designed to resist factored axial load of Pu and

)9(unt PP

The design strength nt P is evaluated as follow:

1- Yielding on gross section

9.0

AFP

t

)10(gyn

Where

Fy = Minimum yield stress, and Ag = Gross area of the tension member.


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2- Fracture in effective net area

75.0

AFP

t

)11(eun

where Fu = Minimum tensile stress, and Ae =Effective area of the tension member.

3- Block Shear

75.0

AFUAF6.0AFUAF6.0P

t

)12(ntubsgvyntubsnvun

Where

Agv = Gross area subjected to shear,

Ant = Net area subjected to tension

Anv = Net area subjected to shear

Ubs = 1.0 for uniform tension stress

The design strength of a tension member is the smaller of limit states (gross yielding, net

section fracture, or block shear failure). The member design strength must be greater than the ultimate factored design load in tension.

Example 4: Calculate the block shear strength of the single angle tension member shown. The single angle L 4 x 4 x 3/8 made from A36 steel is connected to the gusset plate with 5/8 in. diameter bolts as shown below. The bolt spacing is 3 in. center-to-center and the edge distances are 1.5 in and 2.0 in as shown in the Figure below.

Solution:

Assume a block shear path and calculate the required areas

2nt in609.0

8

3*)

8

1

8

5(*5.0

8

3*2A

2gv in813.2

8

3*)5.133(A

2nv in109.2

8

3*)

8

1

8

5(*5.2813.2A

kips7.108609.0*58*0.1109.2*58*6.0

AFUAF6.0P ntubsnvun

kips1.96609.0*58*0.1813.2*36*6.0

AFUAF6.0POR ntubsgvyn

The smaller value controls, so the available strength of the member in block shear is 96.1 kips,

For A36

Fy = 36 ksi

Fu = 58 ksi


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Example 5: Determine the design tension strength for a single channel C15 x 50 connected to a 0.5 in. thick gusset plate as shown in Figure. Assume that the holes are for 3/4 in. diameter bolts and that the plate is made from structural steel with yield stress (Fy) equal to 50 ksi and ultimate stress (Fu)

equal to 65 ksi.

Solution

From AISCM for C15*50 ,in7.14A 2g

"716.0t w , "798.0x

- Yielding Due to Tension

kips6627.14*50*9.0AFP gytnt

- Fracture Due to Tension

- For section 1-1

2n in19.12716.0*)

8

1

4

3(*47.14A

"867.06

798.01

L

x1U

57.1019.12*867.0AUA ne

kips51557.10*65*75.0AFP eutnt

- Block Shear rupture

- 2nt in565.4716.0*)

8

1

4

3(*3716.0*9A

2gv in74.10]716.0*)5.133[(*2A

2nv in61.7]716.0*)

8

1

4

3(*5.237.5[*2A

kips13.445]565.4*65*0.161.7*65*6.0[75.0

]AFUAF6.0[P ntubsnvutnt

kips19.464]565.4*65*0.174.10*50*6.0[75.0

]AFUAF6.0[POR ntubsgvytnt

13.445Pnt

Thus block shear is the critical case and the design tension strength of the member is 445.13 kips,

Example 6: For the member of example 5. The service loads

are 35 kips dead load in additional member weight and 15 kips live load. Investigate this member for compliance with AISC specifications.

Solution

Weight of C15*50 = 50 lb/ft

kips05.35351000

50D

Combination 1

kips07.4905.35*4.1D4.1

Combination 1 kips06.6615*6.105.35*2.1L6.1D2.1

kips06.66Pu

Since 13.44506.66(PP ntu the member is

satisfactory

C 15 * 30

3@3=9

1

1

1.5 3 3


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DESIGN ON TENSION MEMBERS

The design of a tension member involves finding the lightest steel section (angle, wide-lange,

or channel section) with design strength (φPn) greater than or equal to the maximum

factored design tension load (Pu) acting on it.

unt PP

- Pu is determined by structural analysis for factored load combinations

- φt Pn is the design strength based on the gross section yielding, net section fracture, and

block shear rupture limit states.

The design of a tension member can be summarized as follows:

1. Determine the minimum gross area from the tensile yielding failure mode:

uyggytnt PFA9.0AFP

)13(F9.0

PA

y

ug

2. Determine the minimum net area from the tensile fracture failure mode:

UF75.0

PAAUABut,

F75.0

PAPFA75.0AF75.0P

u

unne

u

ueuueeun

where the net area is found from :

)14(AUF75.0

PA holes

u

ug

3. Use the larger Ag value from equations (13) and (14), and select a trial member size

based on the larger value of Ag.

4. For tension members, AISC specification Section D1 suggests that the slenderness

ratio KL/rmin should be greater than or equal 300 to prevent flapping or flutter of the

member,

)15(300r

LK

min

where

K =Effective length factor (usually assumed to be 1.0 for tension members),

L =Unbraced length of the tension member, and

rmin = Smallest radius of gyration of the member.

UF75.0

PAAA

u

uholesgn


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Tables for the Design of Tension Members

Part 5 of the Manual contain tables to assist the design of tension members of various cross-

sectional shapes. The AISC manual tabulates the tension design strength of standard steel

sections - Include: wide flange shapes, angles, tee sections, and double angle sections.

- The net section fracture strength is tabulated for an assumed value of U = 0.75,

obviously because the precise connection details are not known.

- For all W, Tee, angle and double-angle sections, Ae is assumed to be = 0.75 Ag

- The engineer can first select the tension member based on the tabulated gross

yielding and net section fracture strengths, and then check the net section

fracture strength and the block shear strength using the actual connection

details.

Example 7: A tension member with a 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

me2mber with rectangular cross section plate. Use A36 steel and assume connection with one line of 7/8 inch diameter bolts.

Solution

kips8.10452*6.118*2.1L7.1D2.1Pu

Ag For tensile yielding

2

y

ug in235.3

36*9.0

8.104

F9.0

PAquiredRe

Ag For tensile rupture

plateboltedfor0.1U

t41.2

t*)8

1

8

7(

0.1*58*75.0

8.104

AUF75.0

PAquiredRe holes

u

ug

in23.0300

12*75.5

300

Lrmin

235.3A235.391.2A2

1tLet gg

Use PL 1/2 *7

073.012

)2/1(*7I,235.35.32/1*7A

3

min

tincreaseNG23.0in14.05.3

073.0

A

Ir minmin

235.3A235.3035.3A8

5tLet gg

Use PL 5/8 *5.5

NG23.0in18.0r,111.0I,235.344.3A minmin

41.3A235.341.3A1tLet gg

Use PL 1 * 3.5

OK23.0in29.0r,29.0I,41.35.3A minmin

Use a PL 3.5*1 in

Thickness provided Width

Ag Required Ag Available

r Remarks Yielding Fracture

1/2" 7 3.235 2.91 3.5 0.14 Ag Available < Ag Required ---- OK

r < rmin----------NG 5/8" 5.5 3.235 3.035 3.44 0.18 Ag Available < Ag Required ---- OK

r < rmin----------NG 1" 3.5 3.235 3.41 3.5 0.29 Ag Available < Ag Required ---- OK

r > rmin----------OK


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Example 8: Design a member to carry a factored

maximum tension load of 100 kips. Assume that

the member is a wide flange connected through

the flanges using eight ¾ in. diameter bolts in two

rows of four each as shown in the figure below.

The center-to-center distance of the bolts in the

direction of loading is 4 in. The edge distances are

1.5 in. and 2.0 in. as shown in the figure below.

Steel material is A992.

Solution

For steel A992

ksi65Fandksi50F uy

kips100Pu

Ag For tensile yielding

2

y

ug in22.2

50*9.0

100

F9.0

PAquiredRe

Ag For tensile rupture

Shear lag factor, U, is assumed 0.75 for W shape due to flange have bolts less than three thus alternative values of Table D3.3 not applicable.

75.0U

t5.374.2

t*)8

1

4

3(*2*2

75.0*65*75.0

100

AUF75.0

PAquiredRe holes

u

ug

Go to the Table 5.1 of AISC manual, with Pu=100

and Ag = 2.96 and select W 8*10, see the table

For W 8*13 the gross yielding strength = 173 kips, and

net section fracture strength=140 kips

Check selected section for net section fracture

- Ag = 3.84 in2

- An = 3.84 – 4*(7/8+1/8) 0.255 = 2.95 in2

- From dimensions of WT4 x 6.5 "03.1x

"74.04

03.11

L

x1U

- 19.295.2*74.0AUA ne

OK100P7.10619.2*65*75.0fractureforP unt

Check the block shear rupture strength

Identify block shear path

- 2nt in084.1]255.0*)

8

1

4

3(*5.0255.0*5.1[4A

2gv in12.6]255.0*)42[(*4A

2nv in78.4]255.0*)

8

1

4

3(*5.112.6[*4A

kips66.192]084.1*65*0.178.4*65*6.0[75.0

]AFUAF6.0[P ntubsnvutnt

kips55.190]565.0844.1*65*0.112.6*50*6.0[75.0

]AFUAF6.0[POR ntubsgvytnt

OK100P55.190P unt

Therefore, W 8 x 13 is acceptable.

Section tf Ag Required Ag Available

Slenderness effect

Remarks Yielding Fracture

W 8*10 0.205 2.22 3.46 2.96 N/A Ag Available < Ag Required---- NG

W 8*13 0.255 2.22 3.63 3.84 N/A Ag Available > Ag Required---- OK


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TENSION RODS

Rods with a circular cross section are commonly used as tension members when slenderness

is not consideration. Tension rods might be referred to as hanger rods or sag rods. Hangers

are tension members that are hung from one member to support other members. Sag rods

are often provided to prevent a member from deflecting (or sagging) under its own self-

weight. Tension rods are also commonly used as diagonal bracing in combination with a

clevis and turnbuckle to support lateral loads.

The more commonly used threaded rods is a rod

where the nominal diameter is greater than the root

diameter. The tensile capacity is based on the available

cross-sectional area at the root where the threaded portion

of the rod is the thinnest.

The AISC specification does not limit the size of tension rods, but the practical minimum

diameter of the rod should not be less than 5⁄8 in. since smaller diameter rods are more

susceptible to damage during construction.

The design strength of a tension rod is given in the AISC specification as

unt PP

unt

bnn

F75.0F,75.0

AFP

Ab = Nominal unthreaded body area.

)16(butnt A)F75.0(P

For design

)F75.0(75.0

PA

u

ub

Sag rod Hanger Bracing

Threaded road


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The Fu term in the above equations is the minimum tensile stress of the threaded rod.

There are several acceptable grades of threaded rods that are available in AISCM ,

the most common of which are summarized in Table below.

Grades of threaded rods

Example 7: A threaded rod is to be used as a bracing member that must resist a service tensile load 2of 2 kips dead load and 6 kips live load. What size rod is required if A 36 steel is used?

Solution

kips126*6.12*2.1L7.1D2.1Pu

in3678.0)58*75.0(75.0

12

)F75.0(75.0

PrequiredA

u

ub

2b d

4A

684.03678.0*4

d,quiredRe

Use a 3/4 in diameter rod

OK3678.0442.0Ab

HW 1: Design a member to carry a factored maximum tension load of 100 kips. The member is a single angle section connected through one leg using four 1 in. diameter bolts. The center-to-center distance of the bolts is 3 in. The edge distances are 2 in. Steel material is A36.

HW 2: Select an unequal – leg angle tension member 15 feet long to resist a service dead load of 35 kips and a service live load of 70 kips. Use A36 steel. The connection is shown in figure. Bolt diameter 3/4".

H.W 3: Design a 9 ft single angle tension member to support a dead tensile working load of 30 k and a live tensile working load of 40 k. The member is to be connected to one leg only with 7/8" bolts (at least four in a line 3 in on center). Assume that only one bolts is to be located at any cross section. Use A36 steel.


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Example 10: The two C12 * 30 have been

selected to support a dead tensile load of 120 k

and a 240 k working load. The member is 30 ft

long, consists of A36 steel, and has one line of

three 7/8" bolts in each channel flange 3in on

center. Determine whether the member is

satisfactory. Assume centers of bolts holes are

1.75 in from the backs of the channels.

Solution:

From AISC Manual for C15*30

4xf

2g in162I,501.0t,in81.8A

Cofbackfrom674.0axisy

,762.0r,in12.5I y4

y

For A36 steel

ksi65Fandksi50F uy

kips528240*6.1120*2.1L7.1D2.1Pu

- Tensile Yielding Strength

OK5289.570)81.8*2(*36*9.0AFP gytnt

- Tensile Rupture Strength

2n in62.15501.0*)

8

1

8

7(*281.8[*2A

"89.0)33(

674.01

L

x1U

9.1362.15*89.0AUA ne

OK5287.6049.13*58*75.0AFP eutnt

Slenderness Effect

4x in324162*2I

42

y in510]326.5*81.812.5[*2I

in29.481.8*2

324rx

in38.581.8*2

510ry

29.4rrrr xminyx

OK3009.8329.4

12*30

r

L

x

Check Block Shear

HW

DESIGN AND ANALYSIS OF TENSION MEMBERS · Steel Design Misan University Fourth Year Engineering...

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Transcript of DESIGN AND ANALYSIS OF TENSION MEMBERS · Steel Design Misan University Fourth Year Engineering...