5.1 definition of beam column - eng.uowasit.edu.iq · beam-columns are structural members which...

Chapter Five---------------------------------------------------------------------------------------------------------------Beam-Column ==========================================================================================

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------------------------------------------------------------------------------------------------------------------------------------------ Assist. Prof. Dr.Thaar S. Al-Gasham , Wasit University, Eng. College 136

5.1 definition of beam column

For many structural members bending and axial effect are acts at the same times.

For example if the beam is statistically indeterminate axial forces developed in

the beam. Also the column in frame is subjected to both axial and bending effects.

Beam-columns are structural members which combine the beam function of

transmitting transverse forces or moments (uniaxial bending or biaxial bending)

with the compression (or tension) member function of transmitting axial forces.

5-2 interaction equation according Of ASIC


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Example 1

The beam–column shown in Figure below is pinned at both ends and is subjected

to the loads shown. Bending is about the strong axis. Determine whether this

member satisfies the appropriate AISC Specification interaction equation.


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LRFD Solution

1- The axial load calculation

The effect length

𝑘𝑦𝐿 = 17(1) = 17 𝑓𝑡

From Table (4-1)

∅𝑐𝑃𝑛 = 405 𝑘𝑖𝑝


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𝑃𝑢 = 1.6𝑃𝑙 + 1.2𝑃𝑑 = 1.6(99) + 1.2(35) = 200.4 𝑘

2- Moment calculations

Since the beam-column bends about the strong axis, the terms related to My=0.

𝑙𝑏 = 17′, 𝑡ℎ𝑒 𝑇𝑎𝑏𝑙𝑒 (3 − 10)𝑖𝑠 𝑢𝑠𝑒𝑓𝑢𝑙.

Or , we can calculated it by using the equations of moment strength as follows,

From Table(3-2)

𝐿𝑃 = 8.97 𝑎𝑛𝑑𝐿𝑟 = 31.6 𝑓𝑜𝑟𝑚 𝑇𝑎𝑏𝑙𝑒 (3 − 2)

Since 𝐿𝑏 < 𝐿𝑏 < 𝐿𝑟 , the beam will fail in inelastic LTB(Zone2)


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From Table (3-2) ∅𝑏𝑀𝑝𝑥 = 227, 𝐵𝑓 = 3.67

Cb=1.32 from Table (3-1)

∅𝑏𝑀𝑛 = 𝐶𝑏[∅𝑏𝑀𝑝𝑥 − 𝐵𝐹(𝐿𝑏 − 𝐿𝑝)] ≤ ∅𝑏𝑀𝑝𝑥

∅𝑏𝑀𝑛 = 1.32[227 − 3.67(17 − 8.97)] = 260.7 > 227 ∴ 𝑢𝑠𝑒 227 𝑓𝑡 − 𝑘

The maximum moment due to a concentrated load is,

𝑀𝑢

𝑃𝑢 𝐿

4=

[1.6(12) + 1.2(5)](17)

4= 107.1 𝑓𝑡 − 𝑘

𝑃𝑢

∅𝑐𝑃𝑛=

200.4

405= 0.4948 > 0.2

Use Eq.(3-6)

𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥

∅𝑏𝑀𝑢𝑥+

𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) ≤ 1.0


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0.4948 +8

9(

107.1

227+ 0) = 0.914 < 1.0

This member satisfies the ASIC specifications according to LRFD method.

ASD Solution

1-The axial load calculation

The effect length

𝑘𝑦𝐿 = 17(1) = 17 𝑓𝑡

𝑃𝑛

Ω𝑐= 270 𝑘𝑖𝑝

𝑃𝑢 = 𝑃𝑙 + 𝑃𝑑 = 99 + 35 = 134 𝑘

2- moment calculations

𝑀𝑛

Ω𝑏= 𝐶𝑏 [

𝑀𝑝𝑥

Ω𝑏− 𝐵𝐹(𝐿𝑏 − 𝐿𝑝)] ≤

𝑀𝑝𝑥

Ω𝑏

According to Table(3-2)

𝑀𝑝𝑥

Ω𝑏= 151, 𝐵𝐹 = 2.44

𝑀𝑛

Ω𝑏= 1.32[151 − 2.44(17 − 8.97)] = 173.5 > 151 ∴ 𝑢𝑠𝑒

𝑀𝑛

Ω𝑏= 151

The maximum moment due to a concentrated load is,

𝑀𝑎

𝑃𝑎 𝐿

4=

[12 + 5](17)

4= 72.25 𝑓𝑡 − 𝑘

𝑃𝑎

𝑃𝑛

Ω𝑐

=134

270= 0.4963 > 0.2


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𝑃𝑎

𝑃𝑛/Ω𝑐+

8

9(

𝑀𝑎𝑥

𝑀𝑛𝑥/Ω𝑏+

𝑀𝑎𝑦

𝑀𝑛𝑦/Ω𝑏) ≤ 1.0

0.4963 +8

9(

72.25

151+ 0) = 0.922 < 1.0 𝑜. 𝑘

This member satisfies the ASIC specifications according to ASD method.

5.3 METHODS OF ANALYSIS FOR REQUIRED STRENGTH The foregoing approach to the analysis of members, subjected to both bending

and axial load, is satisfactory so long as the axial load is not too large. The

presence of the axial load produces secondary moments, and unless the axial

load is relatively small, these additional moments must be accounted for. As

shown in Figure below.

In addition to the secondary moments caused by member deformation (P-d

moments, shown in Figure above), additional secondary moments are present

when one end of the member translates with respect to the other. These moments

are called P-Δ moments and are illustrated in Figure below. In a braced frame,

the member ends do not undergo translation, so only the P-d moments are present.

In an unbraced frame, the additional moment, PΔ, increases the end moment. The

distribution of moments in the member is therefore a combination of the primary

moment, the P-d moment, and the P-Δ moment. An unbraced rigid frame depends


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on the transfer of moments at its joints for stability. For this reason, unbraced

frames are frequently referred to as moment frames. Multistory buildings can

consist of a combination of braced frames and moment frames.

Ordinary structural analysis methods that do not take the displaced geometry into

account are called first-order methods. Iterative analyses that account for these

effects are referred to as second-order methods.

5.4 THE MOMENT AMPLIFICATION METHOD

The moment amplification method entails computing the maximum bending

moment resulting from flexural loading (transverse loads or member end

moments) by a first-order analysis, then multiplying by a moment amplification

factor to account for the secondary moment.


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The amplified moment to be used in design is computed from the loads and

moments as follows (x and y subscripts are not used here; amplified moments

must be computed in the following manner for each axis about which there are

moments)

In addition to the required moment strength, the required axial strength must

account for second-order effects. The required axial strength is affected by the

displaced geometry of the structure during loading. This is not an issue with

member displacement (δ), but it is with joint displacement (Δ). The required axial

compressive strength is given by


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5.5 amplification moment at braced frame

Figure below shows a member subjected to equal end moments producing single-

curvature bending (bending that produces tension or compression on one side

throughout the length of the member). Maximum moment amplification occurs

at the center, where the deflection is largest. For equal end moments, the moment

is constant throughout the length of the member, so the maximum primary

moment also occurs at the center. Thus the maximum secondary moment and

maximum primary moment are additive. Even if the end moments are not equal,

as long as one is clockwise and the other is counterclockwise there will be single

curvature bending, and the maximum primary and secondary moments will occur

near each other.


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That is not the case if applied end moments produce reverse-curvature bending

as shown in Figure below. Here the maximum primary moment is at one of the

ends, and maximum moment amplification occurs between the ends. Depending

on the value of the axial load P, the amplified moment can be either larger or

smaller than the end moment.

The maximum moment in a beam–column therefore depends on the distribution

of bending moment within the member.


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Here, always this reduction factor is taken as 1.0.

In the effective length and first-order methods, the flexural rigidity is unreduced,

and EI* = EI. The moment of inertia I and the effective length factor K1 are for

the axis of bending, and K1 = 1.0 unless a more accurate value is computed (AISC

C3). Note that the subscript 1 corresponds to the braced condition and the

subscript 2 corresponds to the unbraced condition.

5.6 Evaluation of Cm

The factor Cm applies only to the braced condition. There are two categories of

members: those with transverse loads applied between the ends and those with no

transverse loads.

1. If there are no transverse loads acting on the member,


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𝐶𝑚 = 0.6 − 0.4 (𝑀1

𝑀2)

M1/M2 is a ratio of the bending moments at the ends of the member. M1 is the

end moment that is smaller in absolute value, M2 is the larger, and the ratio is

positive for members bent in reverse curvature and negative for single-curvature

bending. Reverse curvature (a positive ratio) occurs when M1 and M2 are both

clockwise or both counterclockwise.

2. For transversely loaded members, Cm can be taken as 1.0. A more refined

procedure for transversely loaded members is provided in the Commentary

to Appendix 8 of the Specification. The factor Cm is given as


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Example 2

The member shown in Figure is part of a braced frame. An analysis consistent

with the effective length method was performed; therefore, the flexural rigidity,

EI, was unreduced. If A572 Grade 50 steel is used, is this member adequate? Kx

= Ky = 1.0.


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LRFD solution

1. Determine the applied axial load

𝑃𝑟 = 𝑃𝑢 = 𝑃𝑛𝑡 + 𝐵2𝑃𝑙𝑡

Since the frame is braced, 𝑃𝑙𝑡 = 0

𝑃𝑟 = 𝑃𝑢 = 1.6𝑃𝐿 + 1.2𝑃𝐷 = 1.6(210) + 1.2(70) = 420 𝑘

2. Determining the column strength ∅𝑐𝑃𝑛 from Table(4-2)

𝐾𝐿 = (1)(14) = 14


3. Selecting either beam-column formula to be used

𝑃𝑢

∅𝑐𝑃𝑛=

420

685= 0.613 > 0.2 , 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑖𝑠 𝑢𝑠𝑒𝑑.

𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥


𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) ≤ 1.0

4. Evaluating the flexural strength of the section ∅𝑏𝑀𝑢𝑥

Referring to Table(3-2)


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𝐿𝑏 = 14 > 𝐿𝑝 = 11.9 𝑎𝑛𝑑 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑡ℎ𝑎𝑛 𝐿𝑟

The section may fail in inelastic zone as well as local buckling in flange since the

section is noncompact as shown in figure

5. Calculating the inelastic LTB


Note , the term ∅𝑏𝑀𝑝𝑥 in Table (3-2) refers to the local buckling in flange

moment for a noncompact section rather than the plastic moment

Calculating the Cb

The smaller moment 𝑀1 = 1.6(36) + 1.2(11) = 70.8 𝑓𝑡 − 𝑘

The larger moment 𝑀2 = 1.6(41) + 1.2(14) = 82.4 𝑓𝑡 − 𝑘


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𝐶𝑏 =12.5𝑀𝑚𝑎𝑥

2.5𝑀𝑚𝑎𝑥 + 3𝑀𝐴 + 4𝑀𝐵 + 3𝑀𝐶

=12.5(82.4)

2.5(82.4) + 3(73.7) + 4(76.6) + 3(79.5)= 1.06

∅𝑏𝑀𝑛 = 𝐶𝑏[∅𝑏𝑀𝑝𝑥 − 𝐵𝐹(𝐿𝑏 − 𝐿𝑝)]

∅𝑏𝑀𝑛 = 1.06[356 − 5.41(14 − 11.9)] = 365.31 < 356 ∴ ∅𝑏𝑀𝑛

= 356 𝑘 − 𝑓𝑡

6. Calculating the applied moment

𝑀𝑟 = 𝑀𝑢 = 𝐵1𝑀𝑛𝑡 + 𝐵2𝑀𝑙𝑡

Since the frame is braced, 𝑀𝑙𝑡 = 0

𝑀𝑛𝑡 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀2 = 82.4 𝑘 − 𝑓𝑡

𝐵1 =𝐶𝑚

1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

≥ 1.0


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𝐶𝑚 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (−

70.8

82.4) = 0.9437

𝑃𝑒𝑙

=𝜋𝐸𝐼

(𝐾1𝐿)2, 𝑡ℎ𝑖𝑠 𝑓𝑜𝑚𝑢𝑙𝑎 𝑖𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 (𝑎𝑏𝑜𝑢𝑡 𝑥

− 𝑎𝑥𝑖𝑠)𝑓𝑜𝑟 𝑏𝑒𝑎𝑚 𝑛𝑜𝑡 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑎𝑏𝑜𝑢𝑡 𝑦

𝑃𝑒𝑙 =𝜋𝐸𝐼

(𝐾1𝐿)2=

𝜋2𝐸𝐼𝑥

(𝐾𝑥𝐿)2=

𝜋2(29000)(533)

(1.0𝑥14𝑥12)2= 5405.13 𝑘𝑖𝑝

𝛼 = 1.0 𝑓𝑜𝑟 𝐿𝑅𝐹𝐷

𝐵1 =𝐶𝑚

1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.9437

1 − 1.0𝑥420

5405.13

= 1.0232 > 1.0 𝑜. 𝑘

𝑀𝑟 = 𝑀𝑢 = 𝐵1𝑀𝑛𝑡 = 1.0232𝑥82.4 = 84.3 𝑓𝑡 − 𝑘

𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥


𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) = 0.613 +

8

9(

84.3

356+ 0) = 0.823 < 1.0

The Section W12x65 satisfies the AISC specifications according to LRFD

method

H.W solve the preceding example using the ASD method.

Example 3

The member shown in Figure is a W12 × 65 of A242 steel. First-order analyses

were performed with reduced member stiffnesses. The approximate second-order

analysis method of AISC Appendix 8 can be used, making this a direct analysis

method. For LRFD, the analysis results for the controlling factored load

combination are Pnt = 300 kips, Mntx = 135 kips, and Mnty = 30 ft-kips. For ASD,

the analysis results for the controlling load combination are Pnt = 200 kips, Mntx


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= 90 ft-kips, and Mnty = 20 ft-kips. Use Ky = 1.0, and investigate this member for

compliance with the AISC Specification.

First, determine the yield stress Fy. From Table 2-4 in Part 2 of the Manual, we

see that A242 steel is available in three different versions. From the dimensions

and properties table in Part 1 of the Manual, a W12 × 65 has a flange thickness

of tf = 0.605 in. This matches the thickness range corresponding to footnote l in

Table 2-4; therefore, Fy = 50 ksi.


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LRFD solution

1. Determining the column strength ∅𝑐𝑃𝑛 from Table(4-1)

𝐾𝐿 = (1)(15) = 15


2. Selecting either beam-column formula to be used

𝑃𝑢

∅𝑐𝑃𝑛=

300


𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥


𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) ≤ 1.0

3. Evaluating the strong axis moment 𝑀𝑢𝑥

∅𝑏𝑀𝑢𝑥

Referring to table(3-2)


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4. Calculating the inelastic LTB




Calculating the Cb


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The smaller moment 𝑀1 = 0 𝑓𝑡 − 𝑘

The larger moment 𝑀2 = 135 𝑓𝑡 − 𝑘

In this case the bending moment diagram has triangular shape, in which

𝑀𝐴 =𝑀𝑚𝑎𝑥

3, 𝑀𝐵 =

𝑀𝑚𝑎𝑥

2, 𝑀𝐶 =

2𝑀𝑚𝑎𝑥

3


2.5𝑀𝑚𝑎𝑥 + 3𝑀𝐴 + 4𝑀𝐵 + 3𝑀𝐶=

12.5

2.5 +33

+42

+2𝑥3

3

= 1.67

∅𝑏𝑀𝑛𝑥 = 𝐶𝑏[∅𝑏𝑀𝑝𝑥 − 𝐵𝐹(𝐿𝑏 − 𝐿𝑝)]

∅𝑏𝑀𝑛𝑥 = 1.67[356 − 5.41(15 − 11.9)] = 566.5 < 356 ∴ ∅𝑏𝑀𝑛

= 356 𝑘 − 𝑓𝑡


𝑀𝑟𝑥 = 𝑀𝑢𝑥 = 𝐵1𝑥𝑀𝑛𝑡𝑥 + 𝐵2𝑥𝑀𝑙𝑡𝑥

Since the frame is braced, 𝑀𝑙𝑡𝑥 = 0

𝑀𝑛𝑡𝑥 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀2 = 135 𝑘 − 𝑓𝑡

𝐵1𝑥 =𝐶𝑚

1 − 𝛼𝑃𝑟𝑥

𝑃𝑒𝑙𝑥

≥ 1.0

𝐶𝑚 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (−

0

135) = 0.6

Since a modified flexural rigidity, EI*, was used in the frame analysis, it must

also

be used in the computation of Pe1.


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𝑃𝑒𝑙𝑥 =𝜋2𝐸𝐼∗

(𝐾1𝐿)2=

𝜋2𝐸𝐼𝑥

(𝐾𝑥𝐿)2=

𝜋2(29000)(533)(0.8)

(1.0𝑥15𝑥12)2= 3766.8 𝑘𝑖𝑝


𝐵1𝑥 =𝐶𝑚


𝑃𝑒𝑙𝑥

=0.6

1 − 1.0𝑥300

3766.8

= 0.652 < 1.0 , ∴ 𝑢𝑠𝑒 𝐵1𝑥 = 1.0

𝑀𝑟𝑥 = 𝑀𝑢𝑥 = 𝐵1𝑥𝑀𝑛𝑡𝑥 = 1.0𝑥135 = 135 𝑓𝑡 − 𝑘

6. Evaluating the strong axis moment 𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦

Table (3-4) gives the flexural strength for W-shape with Fy=50, referring to it

∅𝑏𝑀𝑢𝑦 = 161 𝑘 − 𝑓𝑡


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Important note:- the member bent about minor axis will fail always in plastic

zone, unless the flange of it is non-compact, in this case local flange buckling will

happen.


𝑀𝑟𝑦 = 𝑀𝑢𝑦 = 𝐵1𝑦𝑀𝑛𝑡𝑦 + 𝐵2𝑦𝑀𝑙𝑡𝑦

Since the frame is braced, 𝑀𝑙𝑡𝑦 = 0

𝑀𝑛𝑡𝑥 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀2 = 135 𝑘 − 𝑓𝑡

𝐵1𝑥 =𝐶𝑚


𝑃𝑒𝑙𝑥

≥ 1.0

𝐶𝑚 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (−

0

135) = 0.6


also


𝑃𝑒𝑙𝑦 =𝜋2𝐸𝐼∗

(𝐾1𝐿)2=

𝜋2𝐸𝐼𝑦

(𝐾𝑦𝐿)2 =

𝜋2(29000)(174)(0.8)

(1.0𝑥15𝑥12)2= 1229.7 𝑘𝑖𝑝


𝐵1𝑦 =𝐶𝑚

1 − 𝛼𝑃𝑟𝑦

𝑃𝑒𝑙𝑦

=0.6

1 − 1.0𝑥300

1229.7

= 0.794 < 1.0 , ∴ 𝑢𝑠𝑒 𝐵1𝑦 = 1.0

𝑀𝑟𝑦 = 𝑀𝑢𝑦 = 𝐵1𝑦𝑀𝑛𝑡𝑦 = 1.0𝑥30 = 30 𝑓𝑡 − 𝑘

𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥


𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) ≤ 1.0


===


0.453 +8

9(

135

356+

30

161) = 0.956 < 1.0

The section W12x65 is satisfactory according to LRFD method

H.W solve above example using ASD method.

5.7 MEMBERS IN UNBRACED FRAMES

In a beam–column whose ends are free to translate, the maximum primary

moment resulting from the side-sway is almost always at one end. The maximum

secondary moment from the side-sway is always at the end. As a consequence of

this condition, the maximum primary and secondary moments are usually

additive and there is no need for the factor Cm; in effect, Cm = 1.0. Even when

there is a reduction,


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=========================================================

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Example 4:-

A W12 × 65 of A992 steel, 15 feet long, is to be investigated for use as a column

in an unbraced frame. The axial load and end moments obtained from a first-order

analysis of the gravity loads (dead load and live load) are shown in Figure a. The

frame is symmetrical, and the gravity loads are symmetrically placed. Figure b

shows the wind load moments obtained from a first-order analysis. Both analyses

were performed with reduced stiffnesses of all members. All bending moments

are about the strong axis. If the moment amplification method is used, then we

can consider this to be the direct analysis method, and the effective length factor

Kx can be taken as 1.0. Use Ky = 1.0. Determine whether this member is in

compliance with the AISC Specification.


===


LRFD SOLUTION

All the load combinations given in ASCE 7 involve dead load, and except for the

first one, all combinations also involve either live load or wind load or both. The

load combinations for this question can be summarized as;

𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 1: 1.4𝐷

𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 2: 1.2𝐷 + 1.6𝐿

𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 3: 1.2𝐷 + (0.5𝐿 𝑜𝑟 0.5𝑊)

𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 4: 1.2𝐷 + 1.0𝑊 + 0.5𝐿


𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 6: 0.9𝐷 ± 1.0𝑊

The dead load is less than eight times the live load, so combination (1) can be

ruled out. Load combination (4) will be more critical than (3), so combination

(3) can be eliminated. Combination (5) can be eliminated because it will be less

critical than (2). Finally, combination (6) should be investigated for an

overturning effect. The combinations to be investigated are therefore


===



𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 4: 1.2𝐷 + 1.0𝑊 + 0.5𝐿

𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 6: 0.9𝐷 ± 1.0𝑊

Combination (6) will be investigated first. We will consider compression to be

positive. For a wind direction that produces uplift,

0.9𝐷 – 1.0𝑊 = 0.9(85) – 1.0(56) = 20.5 𝑘𝑖𝑝𝑠

The positive result means that the net load is compressive, and we need not

consider this load combination further. Figure 6.18 shows the axial loads and

bending moments calculated for combinations (2) and (4).

𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑 𝑡𝑎𝑏𝑙𝑒𝑠 (4 − 2), 𝑤𝑖𝑡ℎ 𝐾𝐿 = 1.0(15) = 15 𝑓𝑡, ∅𝑐𝑃𝑛

= 662 𝑘𝑖𝑝𝑠.


===


𝐿𝑜𝑎𝑑 𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 2: 𝑃𝑛𝑡 = 454 𝑘𝑖𝑝𝑠, 𝑀𝑛𝑡 = 104.8 𝑓𝑡 − 𝑘𝑖𝑝𝑠, 𝑎𝑛𝑑 𝑀𝑙𝑡

= 0 (𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦, 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑜 𝑠𝑖𝑑𝑒𝑠𝑤𝑎𝑦 𝑚𝑜𝑚𝑒𝑛𝑡𝑠).

𝑇ℎ𝑒 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟 𝑖𝑠

𝑀𝑟𝑥 = 𝑀𝑢𝑥 = 𝐵1𝑥𝑀𝑛𝑡𝑥 + 𝐵2𝑥𝑀𝑙𝑡𝑥

Since the frame is braced, 𝑀𝑙𝑡𝑥 = 0

𝑀𝑛𝑡𝑥 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀2 = 104.8 𝑘 − 𝑓𝑡

𝐵1𝑥 =𝐶𝑚


𝑃𝑒𝑙𝑥

≥ 1.0

𝐶𝑚 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

90

104.8) = 0.2565


also



===


𝑃𝑒𝑙𝑥 =𝜋2𝐸𝐼∗

(𝐾1𝐿)2=

𝜋2𝐸𝐼𝑥

(𝐾𝑥𝐿)2=

𝜋2(29000)(533)(0.8)

(1.0𝑥15𝑥12)2= 3766.8 𝑘𝑖𝑝


𝐵1𝑥 =𝐶𝑚


𝑃𝑒𝑙𝑥

=0.2565

1 − 1.0𝑥454

3766.8

= 0.292 < 1.0 , ∴ 𝑢𝑠𝑒 𝐵1𝑥 = 1.0

𝑀𝑟𝑥 = 𝑀𝑢𝑥 = 𝐵1𝑥𝑀𝑛𝑡𝑥 = 1.0𝑥104.8 = 104.8 𝑓𝑡 − 𝑘

Evaluating the strong axis moment 𝑀𝑢𝑥

∅𝑏𝑀𝑢𝑥

Referring to table(3-2)





===


Calculating the inelastic LTB




Calculating the Cb

The smaller moment 𝑀1 = 90.0 𝑓𝑡 − 𝑘

The larger moment 𝑀2 = 104.8 𝑓𝑡 − 𝑘



3, 𝑀𝐵 =

𝑀𝑚𝑎𝑥

2, 𝑀𝐶 =

2𝑀𝑚𝑎𝑥

3


2.5𝑀𝑚𝑎𝑥 + 3𝑀𝐴 + 4𝑀𝐵 + 3𝑀𝐶

=12.5(104.8)

2.5(104.8) + 3𝑥41.3 + 4𝑥7.4 + 3𝑥56.1= 2.24

∅𝑏𝑀𝑛𝑥 = 𝐶𝑏[∅𝑏𝑀𝑝𝑥 − 𝐵𝐹(𝐿𝑏 − 𝐿𝑝)]


===


∅𝑏𝑀𝑛𝑥 = 2.24[356 − 5.41(15 − 11.9)] = 759.6 < 356 ∴ ∅𝑏𝑀𝑛

= 356 𝑘 − 𝑓𝑡

𝑃𝑢

∅𝑐𝑃𝑛=

454


𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥


𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) ≤ 1.0

0.6858 +8

9(

104.8

356) = 0.947 < 1.0 (𝑜. 𝑘)

𝐿𝑜𝑎𝑑 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 4: 𝑃𝑛𝑡 = 212 𝑘𝑖𝑝𝑠, 𝑀𝑛𝑡 = 47.6 𝑘𝑖𝑝𝑠, 𝑃𝑙𝑡

= 56 𝑘𝑖𝑝𝑠, 𝑎𝑛𝑑 𝑀𝑙𝑡 = 132 𝑓𝑡𝑘𝑖𝑝𝑠.

For braced part

𝑀𝑛𝑡 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀2 = 47.6 𝑘 − 𝑓𝑡

𝐵1 =𝐶𝑚

1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

≥ 1.0

𝐶𝑚 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

40.5

47.6) = 0.2597

𝑃𝑒𝑙

=𝜋𝐸𝐼

(𝐾1𝐿)2, 𝑡ℎ𝑖𝑠 𝑓𝑜𝑚𝑢𝑙𝑎 𝑖𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 (𝑎𝑏𝑜𝑢𝑡 𝑥

− 𝑎𝑥𝑖𝑠)𝑓𝑜𝑟 𝑏𝑒𝑎𝑚 𝑛𝑜𝑡 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑎𝑏𝑜𝑢𝑡 𝑦

𝑃𝑒𝑙 = 3767 𝑘𝑖𝑝


𝐵1 =𝐶𝑚

1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.2597

1 − 1.0𝑥212

3767

= 0.2751 < 1.0 , 𝑢𝑠𝑒 1.0


===


𝐵2 =1.0

1 − 𝛼𝑃𝑠𝑡𝑜𝑟𝑦

𝑃𝑒𝑙 𝑠𝑡𝑜𝑟𝑦

For the unbraced condition, the amplification factor for side-sway, B2, must be

computed. This requires a knowledge of the properties of all the columns in the

story, as well as H and ΔH, so that Pstory and Pe story can be computed. Since these

quantities are not available in this example, we will assume that the ratio of Pstory

to Pe story is the same as the ratio for the column under consideration. That is,

𝑃𝑠𝑡𝑜𝑟𝑦

𝑃𝑒𝑙 𝑠𝑡𝑜𝑟𝑦=

212 + 56

3767= 0.0711

𝐵2 =1.0

1 − 1.0𝑥0.0711= 1.077

𝑇ℎ𝑒 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑒𝑑 𝑎𝑥𝑖𝑎𝑙 𝑙𝑜𝑎𝑑 𝑖𝑠

𝑃𝑟 = 𝑃𝑢 = 𝑃𝑛𝑡 + 𝐵2𝑃𝑙𝑡 = 212 + 1.077(56) = 272.3 𝑘𝑖𝑝𝑠

𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑒𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑖𝑠

𝑀𝑟 = 𝑀𝑢 = 𝐵1𝑀𝑛𝑡 + 𝐵2𝑀𝑙𝑡 = 1.0(47.6) + 1.077(132)

= 189.8 𝑓𝑡 − 𝑘𝑖𝑝𝑠

Although the moments Mnt and Mlt are different, they are distributed similarly,

and Cb will be roughly the same; at any rate, they are large enough that ᶲbMp =

356 ft-kips will be the design strength regardless of which moment is considered.

𝑃𝑢

∅𝑐𝑃𝑛=

272.3


𝑃𝑢

∅𝑐𝑃𝑛+

8

9(

𝑀𝑢𝑥


𝑀𝑢𝑦

∅𝑏𝑀𝑢𝑦) ≤ 1.0

0.4113 +8

9(

189.8

356) = 0.885 ≤ 1.0 𝑜. 𝑘

This member satisfies the AISC Specification requirements.

H.W:- repeat the example using the ASD method.


===


5.8 DESIGN OF BEAM–COLUMNS

Because of the many variables in the interaction formulas, the design of beam–

columns is essentially a trial-and-error process. A procedure developed by

Aminmansour (2000) simplifies this process, especially the evaluation of trial

shapes.

Part 6 of the Manual, “Design of Members Subject to Combined Loading,”

contains tables based on design aids developed by Aminmansour. The procedure

can be explained as follows. If we initially assume that AISC Equation H1-1a

governs, then


===



===


Example 5:-


===


Select a W shape of A992 steel for the beam–column of Figure below. This

member is part of a braced frame and is subjected to the service-load axial force

and bending moments shown (the end shears are not shown). Bending is about

the strong axis, and Kx = Ky = 1.0. Lateral support is provided only at the ends.

Assume that B1 = 1.0.

Solution

LRFD method.

𝑃𝑛𝑡 = 𝑃𝑢 = 1.6 𝑃𝐿 + 1.2𝑃𝐷 = 1.6(147) + 1.2(54) = 300 𝑘

(There is no amplification of axial loads in members braced against sidesway.)

The factored moment at each end is

𝑀𝑛𝑡𝑥 = 1.6 𝑀𝐿 + 1.2𝑀𝐷 = 1.6(49) + 1.2(18) = 100 𝑘 − 𝑓𝑡

Since B1 = 1.0, the factored load bending moment is


===


𝑀𝑛𝑢𝑥 = 𝐵1𝑀𝑛𝑡𝑥 = 100 𝑘 − 𝑓𝑡

The effective length for compression and the unbraced length for bending are

the same:

𝐾𝑙 = 𝐿𝑏 = 16′

The bending moment is uniform over the unbraced length, so Cb = 1.0. Try a

W10 shape.

From Table (6-1) 𝑝 = 1.89𝑥10−3 𝑎𝑛𝑑 𝑏𝑥 = 3.51𝑥10−3

Determine which interaction equation to use:

𝑝𝑃𝑟 = 𝑝𝑃𝑢 = 300(1.89𝑥10−3) = 0.567 > 0.2


===


𝑝𝑃𝑟 + 𝑏𝑥𝑀𝑟𝑥 + 𝑏𝑦𝑀𝑟𝑦 = 𝑝𝑃𝑢 + 𝑏𝑥𝑀𝑢𝑥 + 𝑏𝑦𝑀𝑢𝑦=0.567 + 100(3.51𝑥10−3) +

0 = 0.918 < 1.0 𝑂. 𝐾

To be sure that we have found the lightest W10, try the next lighter one, a W10

× 54, with



𝑝𝑃𝑟 = 𝑝𝑃𝑢 = 300(2.12𝑥10−3) = 0.636 > 0.2


0 = 1.033 > 1.0 𝑛𝑜𝑡 𝑂. 𝐾


===


Note; - the calculations above shows that W10x60 is the best section to resist

the applied loading within the category of W10. In order to select the lightest

one section, searching in others categories with weight less than 60. For

example try W12x58



𝑝𝑃𝑟 = 𝑝𝑃𝑢 = 300(2.00𝑥10−3) = 0.60 > 0.2


0 = 0.913 < 1.0 𝑂. 𝐾

To be sure that we have found the lightest W12, try the next lighter one, a W12

× 53,


===




𝑝𝑃𝑟 = 𝑝𝑃𝑢 = 300(2.21𝑥10−3) = 0.663 > 0.2


0 = 1.015 > 1.0 𝑛𝑜𝑡 𝑂. 𝐾

Note; - the calculations above shows that W12x58 is the best section to resist the

applied loading within the category of W12. In order to select the lightest one

section, searching in others categories with weight less than 58. For example try

W14x53.


===




𝑝𝑃𝑟 = 𝑝𝑃𝑢 = 300(2.96𝑥10−3) = 0.888 > 0.2


0 = 1.239 > 1.0 𝑛𝑜𝑡 𝑂. 𝐾

There are no other possibilities in the deeper W-shape groups.

From preceding calculations, the following sections are adequate according to

LRFD methods;

W10x60 and W12x58, select the lightest one W12x58.

H.W/ repeat the preceding example using ASD method.

Adjustment for Cb Not Equal to 1.0


===


Example 6:-

A structural member in a braced frame must support the following service loads

and moments: an axial compressive dead load of 25 kips and a live load of 75

kips; a dead load moment of 12.5 ft-kips about the strong axis and a live load

moment of 37.5 ft-kips about the strong axis; a dead load moment of 5 ft-kips

about the weak axis and a live load moment of 15 ft-kips about the weak axis.

The moments occur at one end; the other end is pinned. The member length is 15

feet. The frame analysis was a first-order analysis with reduced member

stiffnesses, so the moment amplification method with Kx = 1.0 can be used. In

the direction perpendicular to the frame, Ky = 1.0. There are no transverse loads

on the member. Use A992 steel and select a W10 shape.

Solution

This example will be solved according to the ASD method, the LRFD method is

H.W

𝑃𝐷 = 25𝑘, 𝑃𝐿 = 75 𝑘, 𝑀𝑛𝑥𝐷 = 12.5 𝑓𝑡 − 𝑘, 𝑀𝑛𝑥𝐿 = 37.5 𝑓𝑡 − 𝑘, 𝑀𝑛𝑦𝐷

= 5 𝑓𝑡 − 𝑘, 𝑀𝑛𝑦𝐿 = 15 𝑓𝑡 − 𝑘

𝑃𝑎 = 𝑃𝐷 + 𝑃𝐿 = 25 + 75 = 100 𝑘𝑖𝑝

𝑀𝑛𝑡𝑥 = 𝑀𝑛𝑥𝐷 + 𝑀𝑛𝑥𝐿 = 12.5 + 37.5 = 50 𝑘𝑖𝑝 − 𝑓𝑡

𝑀𝑛𝑡𝑦 = 𝑀𝑛𝑦𝐷 + 𝑀𝑛𝑦𝐿 = 5 + 15 = 20 𝑘𝑖𝑝 − 𝑓𝑡

The effective length for compression and the unbraced length for bending are

the same:

KL = Lb = 15 ft, Try with W10x49


===


𝐶𝑚𝑥 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

0.0

50) = 0.6

𝑃𝑒𝑙𝑥 =𝜋𝐸𝐼

(𝐾1𝐿)2=

𝜋2(0.8)(29000)(341)

(15𝑥12)2= 2409.88 𝑘𝑖𝑝,

𝛼 = 1.6 𝑓𝑜𝑟 𝐴𝑆𝐷

𝐵1𝑥 =𝐶𝑚𝑥

1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.6

1 − 1.6𝑥100

2409.88

= 0.642 < 1.0 , 𝑢𝑠𝑒 1.0

𝐶𝑚𝑦 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

0.0

20) = 0.6

𝑃𝑒𝑙𝑦 =𝜋𝐸𝐼

(𝐾1𝐿)2=

𝜋2(29000)(116)𝑥0.8

(15𝑥12)2= 819.78 𝑘𝑖𝑝,


𝐵1𝑦 =𝐶𝑚𝑦

1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.6

1 − 1.6𝑥100

819.8

= 0.745 < 1.0 , ∴ 𝑢𝑠𝑒𝐵1𝑦 = 1.0

𝑀𝑎𝑥 = 𝐵1𝑀𝑛𝑡𝑥 = 50 𝑘𝑖𝑝 − 𝑓𝑡

𝑀𝑎𝑦 = 𝐵1𝑀𝑛𝑡𝑦 = 20 𝑘𝑖𝑝 − 𝑓𝑡

from Table (6-1)

𝑝 = 2.71 𝑥10−3, 𝑏𝑥 = 5.19 𝑥10−3, 𝑏𝑦 = 10.2𝑥10−3



3, 𝑀𝐵 =

𝑀𝑚𝑎𝑥

2, 𝑀𝐶 =

2𝑀𝑚𝑎𝑥

3


===



2.5𝑀𝑚𝑎𝑥 + 3𝑀𝐴 + 4𝑀𝐵 + 3𝑀𝐶=

12.5

2.5 +33

+42

+2𝑥3

3

= 1.67

Since 𝐶𝑏 > 1.0 ∴ 𝑠𝑒𝑙𝑒𝑐𝑡 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 (𝑏𝑥

𝑐𝑏, 𝑏𝑥 𝑜𝑓 𝐿𝑏 = 0) →

(5.19 𝑥10−3

1.67, 4.78𝑥10−3) → (3.11 𝑥10−3, 4.78𝑥10−3) ∴ 𝑏𝑥 = 4.78𝑥10−3

𝑝𝑃𝑎 = 100(2.71 𝑥10−3) = 0.271

𝑝𝑃𝑎 + 𝑏𝑥𝑀𝑎𝑥 + 𝑏𝑦𝑀𝑎𝑦 = 0.271 + 50(4.78𝑥10−3) + 20(10.2𝑥10−3)

= 0.714 <≪ 1.0

This section is adequate to resist the applied loading but it is not economical

section, try to select smaller one, try with W10x49.

from Table (6-1)

𝑝 = 3.34 𝑥10−3, 𝑏𝑥 = 6.54 𝑥10−3, 𝑏𝑦 = 12.6 𝑥10−3


===


𝐶𝑚𝑥 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

0.0

50) = 0.6


===



(𝐾1𝐿)2=

𝜋2(0.8)(29000)(272)

(15𝑥12)2= 1922.3 𝑘𝑖𝑝,



1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.6

1 − 1.6𝑥100

1922.3

= 0.654 < 1.0 , 𝑢𝑠𝑒 1.0

𝐶𝑚𝑦 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

0.0

20) = 0.6


(𝐾1𝐿)2=

𝜋2(29000)(93.4)𝑥0.8

(15𝑥12)2= 660.1 𝑘𝑖𝑝,



1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.6

1 − 1.6𝑥100

660.1

= 0.792 < 1.0 , ∴ 𝑢𝑠𝑒𝐵1𝑦 = 1.0


𝑀𝑎𝑦 = 𝐵1𝑀𝑛𝑡𝑦 = 20 𝑘𝑖𝑝 − 𝑓𝑡



(6.54 𝑥10−3

1.67, 5.90𝑥10−3) → (3.92 𝑥10−3, 5.90𝑥10−3) ∴ 𝑏𝑥 = 5.90𝑥10−3

𝑝𝑃𝑎 = 100(3.34 𝑥10−3) = 0.334

𝑝𝑃𝑎 + 𝑏𝑥𝑀𝑎𝑥 + 𝑏𝑦𝑀𝑎𝑦 = 0.334 + 50(5.90𝑥10−3) + 20(12.6 𝑥10−3)

= 0.881 < 1.0

Try the next smaller shape. Try W10 × 45,

from Table (6-1)

𝑝 = 4.53 𝑥10−3, 𝑏𝑥 = 7.63 𝑥10−3, 𝑏𝑦 = 17.6 𝑥10−3


===


𝐶𝑚𝑥 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

0.0

50) = 0.6


(𝐾1𝐿)2=

𝜋2(0.8)(29000)(248)

(15𝑥12)2= 1752.65 𝑘𝑖𝑝,



1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.6

1 − 1.6𝑥100

1752.65

= 0.660 < 1.0 , 𝑢𝑠𝑒 1.0

𝐶𝑚𝑦 = 0.6 − 0.4 (𝑀1

𝑀2) = 0.6 − 0.4 (

0.0

20) = 0.6


(𝐾1𝐿)2=

𝜋2(29000)(53.4)𝑥0.8

(15𝑥12)2= 377.4 𝑘𝑖𝑝,



1 − 𝛼𝑃𝑟

𝑃𝑒𝑙

=0.6

1 − 1.6𝑥100

377.4

= 1.042 > 1.0 , ∴ 𝑢𝑠𝑒𝐵1𝑦 = 1.042


𝑀𝑎𝑦 = 𝐵1𝑀𝑛𝑡𝑦 = (1.042)20 = 20.84 𝑘𝑖𝑝 − 𝑓𝑡



(7.63 𝑥10−3

1.67, 6.49𝑥10−3) → (4.569 𝑥10−3, 6.49𝑥10−3) ∴ 𝑏𝑥 = 6.49𝑥10−3

𝑝𝑃𝑎 = 100(4.53 𝑥10−3) = 0.453

𝑝𝑃𝑎 + 𝑏𝑥𝑀𝑎𝑥 + 𝑏𝑦𝑀𝑎𝑦 = 0.453 + 50(6.49𝑥10−3) + 20(17.6 𝑥10−3) = 1.13

> 1.0

The section W10x45 is not safe

According to ASD method, Select W10x49

5.1 definition of beam column - eng.uowasit.edu.iq · beam-columns are structural members which...

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