design and analysis of tension members · steel design misan university fourth year engineering...
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Steel Design Misan University Fourth Year Engineering College Dr.Abbas Oda Dawood Civil Department
Lecture 2 ....... Page 1
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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.
Steel Design Misan University Fourth Year Engineering College Dr.Abbas Oda Dawood Civil Department
Lecture 2 ....... Page 18
Lec
2
Lec
2
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