KINGDOM OF CAMBODIA
Nation Religion King
iiiYiii
Norton University
College of Science
Year IV, Semester 1
Department of Civil Engineering
Group M3
Assignment
Subject: Steel Design
Submitted to Lecturer: Dr. BIN MOLINNE
Submitted by:
1. HANG MEAS
ID: B20230228
2. KEA LEANGHENG
ID: B20224039
3. KEA SOKHOM
ID: B20223319
4. SOY MENGSRONG
ID: B20222851
Academic Year 2024-2025
Norton University
Tension Members
Assignment I
CH3 Tension Members
Use LRFD with a double-angle tension member, 2L 3 × 2 ×
1
4
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿, of A36 steel is subjected to a dead
load of 12kips and a live Load of 36 kips. It is connected to a gusset plate with 3/4 – inch-Diameter bolts
through the long legs. Does this member have enough strength? Assume that 𝐴𝐴𝑒𝑒 = 0.85𝐴𝐴𝑛𝑛 .
 Solution:
Use A36 Steel Take:
𝐹𝐹𝑦𝑦 =250MPa, 𝐹𝐹𝑢𝑢 = 400𝑀𝑀𝑀𝑀𝑀𝑀
We have: 2L3 × 2 × 1/4 → 2𝐿𝐿76.2 × 50.8 × 6.35
For yielding of the gross section:
𝐴𝐴𝑔𝑔 = 12𝑖𝑖𝑖𝑖2 = 774.2𝑚𝑚𝑚𝑚2
The Beam Self weight: DL= 12kips = 53.376KN
The Beam Live Load: LL = 36kips = 160.128KN
•
Factored Load:
𝑃𝑃𝑢𝑢 = 1.2𝐷𝐷𝐷𝐷 + 1.6𝐿𝐿𝐿𝐿
→ 𝑃𝑃𝑢𝑢 = 1.2 × 53.376 + 1.6 × 160.128 = 320.256𝐾𝐾𝐾𝐾
1. For Tensile Yielding:
𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑡𝑡 𝐹𝐹𝑦𝑦 𝐴𝐴𝑔𝑔
For yielding 𝜙𝜙𝑡𝑡 = 0.9 (LRFD)
For Tensile yielding: 𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 0.9 × 250 × 774.2 = 174.2𝐾𝐾𝐾𝐾
Semester I of 2024~2025
page 1 of 17
Year4 Group Ci4~M3
Norton University
Tension Members
1. For Tensile Fracture:
𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑡𝑡 𝐹𝐹𝑢𝑢 𝐴𝐴𝑒𝑒
The net area of one angle:
𝐴𝐴𝑛𝑛 = 𝐴𝐴𝑔𝑔 − 𝐴𝐴ℎ𝑜𝑜𝑜𝑜𝑜𝑜
= 774.2 − 1(23 × 6.35)
The Effective net area:
= 628.15𝑚𝑚𝑚𝑚2
𝐴𝐴𝑒𝑒 = 𝑈𝑈𝐴𝐴𝑛𝑛
Shear leg factor: 𝑈𝑈 = 0.85
→ 𝐴𝐴𝑒𝑒 = 0.85 × 628.15 = 533.93𝑚𝑚𝑚𝑚2
For Tensile Fracture: 𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 0.75 × 400 × 533.93 = 160.18 𝐾𝐾𝐾𝐾
But fracture of the net section control and the design strength for the two angles:
→ 𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 160.18 × 2 = 320.36 𝐾𝐾𝐾𝐾 > 𝑃𝑃𝑢𝑢 = 320.256 𝐾𝐾𝐾𝐾 (𝑂𝑂𝑂𝑂).
Therefore, Member have enough strength 𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 320.36 𝐾𝐾𝐾𝐾 > 𝑃𝑃𝑢𝑢 = 320.256 𝐾𝐾𝐾𝐾 (𝑂𝑂𝑂𝑂).
Semester I of 2024~2025
page 2 of 17
Year4 Group Ci4~M3
Norton University
Tension Members
Assignment II
CH3 Tension Members
Use LRFD to choose a pipe to be used as a tension member to resist a service dead load of 44.48 𝑘𝑘𝑘𝑘
and a service live load of 111.2 𝑘𝑘𝑘𝑘. The ends will be connected by welding completely around the
circumference of the pipe. The length is 2.44 𝑚𝑚.
Solution :
Given data:
A pipe → ASTM A53-Gr. B → 𝐹𝐹𝑦𝑦 = 35 𝑘𝑘𝑘𝑘𝑘𝑘 = 240 𝑀𝑀𝑀𝑀𝑀𝑀, 𝐹𝐹𝑢𝑢 = 60 𝑘𝑘𝑘𝑘𝑘𝑘 = 415 𝑀𝑀𝑀𝑀𝑀𝑀
𝐷𝐷𝐷𝐷 = 44.48 𝑘𝑘𝑘𝑘,
𝐿𝐿𝐿𝐿 = 111.2 𝑘𝑘𝑘𝑘,
𝐿𝐿 = 2.44 𝑚𝑚
Factored load, 𝑃𝑃𝑢𝑢 = 1.2𝐷𝐷𝐷𝐷 + 1.6𝐿𝐿𝐿𝐿 = 1.2 × 44.48 + 1.6 × 111.2 = 231.30 𝑘𝑘𝑘𝑘
1) Tensile yielding
The design tensile strength
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑡𝑡 𝐹𝐹𝑦𝑦 𝐴𝐴𝑔𝑔
𝑃𝑃𝑢𝑢
231.30 × 103
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 ≥
=
= 1070.83 𝑚𝑚𝑚𝑚2 = 1.66 𝑖𝑖𝑛𝑛2
𝜙𝜙𝑡𝑡 𝐹𝐹𝑦𝑦
0.9 × 240
2) Tensile rupture
The design tensile strength
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑡𝑡 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑡𝑡 𝐹𝐹𝑢𝑢 𝐴𝐴𝑒𝑒
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑒𝑒 ≥
𝑃𝑃𝑢𝑢
231.30 × 103
=
= 743.13 𝑚𝑚𝑚𝑚2 = 1.15 𝑖𝑖𝑛𝑛2
𝜙𝜙𝑡𝑡 𝐹𝐹𝑢𝑢
0.75 × 415
Try 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 3 𝑆𝑆𝑆𝑆𝑆𝑆.
𝐴𝐴𝑔𝑔 = 2.07 𝑖𝑖𝑛𝑛2 = 1335.48 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 = 1070.83 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
Effective area:
𝐴𝐴𝑒𝑒 = 𝐴𝐴𝑔𝑔 = 1335.48 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑒𝑒 = 743.13 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
For welded connection with pipe section,
3) Slenderness limitation
Section properties of 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 3 𝑆𝑆𝑆𝑆𝑆𝑆.
4) Block shear rupture
𝑟𝑟 = 1.17 𝑖𝑖𝑖𝑖 = 29.72 𝑚𝑚𝑚𝑚
𝐿𝐿
2440
=
= 8.13 𝑚𝑚𝑚𝑚 < 𝑟𝑟 = 29.72 𝑚𝑚𝑚𝑚 (𝑂𝑂𝑂𝑂)
300
300
Since, the pipe is welded completely around its circumference.
So, block shear is not a concern in this case.
5) Connection
Section properties of 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 3 𝑆𝑆𝑆𝑆𝑆𝑆.
Semester I of 2024~2025
𝑡𝑡 = 0.201 𝑖𝑖𝑖𝑖 = 5.11 𝑚𝑚𝑚𝑚,
page 3 of 17
𝐷𝐷 = 3.50 𝑖𝑖𝑖𝑖 = 88.9 𝑚𝑚𝑚𝑚
Year4 Group Ci4~M3
Norton University
Tension Members
Since, the pipe is welded completely around its circumference.
→ 𝐿𝐿 = 𝜋𝜋𝜋𝜋 = 𝜋𝜋 × 88.9 = 279.29 𝑚𝑚𝑚𝑚
Assume grade of electrode is 𝐸𝐸70𝑋𝑋𝑋𝑋 because 𝐹𝐹𝑦𝑦 = 240 𝑀𝑀𝑀𝑀𝑀𝑀 < 420 𝑀𝑀𝑀𝑀𝑀𝑀
a) Shear strength of weld
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.707𝑤𝑤𝑤𝑤(0.6𝐸𝐸70𝑋𝑋𝑋𝑋6.9)
231.30 × 103
𝑃𝑃𝑢𝑢
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝑤𝑤 ≥
=
= 5.39 𝑚𝑚𝑚𝑚
𝜙𝜙0.707𝐿𝐿(0.6𝐸𝐸70𝑋𝑋𝑋𝑋6.9) 0.75 × 0.707 × 279.29(0.6 × 70 × 6.9)
b) Base metal shear strength
Yielding:
Rupture:
𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑦𝑦 𝑡𝑡𝑡𝑡 = 1.0 × 0.6 × 240 × 5.11 × 279.29
→ 𝝓𝝓𝑹𝑹𝒏𝒏 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟓𝟓𝟓𝟓 𝒌𝒌𝒌𝒌 < 𝑷𝑷𝒖𝒖 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑵𝑵. 𝑮𝑮. )
𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑢𝑢 𝑡𝑡𝑡𝑡 = 0.75 × 0.6 × 415 × 5.11 × 279.29
→ 𝜙𝜙𝑅𝑅𝑛𝑛 = 266.52 𝑘𝑘𝑘𝑘 > 𝑃𝑃𝑢𝑢 = 231.30 𝑘𝑘𝑘𝑘 (𝑂𝑂𝑂𝑂)
By checking base metal shear strength to be safe, we need to revised our section.
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑦𝑦 𝑡𝑡𝑡𝑡
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝑡𝑡 ≥
Try 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 5 𝑆𝑆𝑆𝑆𝑆𝑆.
𝑃𝑃𝑢𝑢
231.30 × 103
=
= 5.75 𝑚𝑚𝑚𝑚 = 0.226 𝑖𝑖𝑖𝑖
𝜙𝜙0.6𝐹𝐹𝑦𝑦 𝐿𝐿 1.0 × 0.6 × 240 × 279.29
𝐴𝐴𝑔𝑔 = 4.01 𝑖𝑖𝑛𝑛2 = 2587.09 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 = 1070.83 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
𝐴𝐴𝑒𝑒 = 𝐴𝐴𝑔𝑔 = 2587.09 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑒𝑒 = 743.13 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
𝑟𝑟 = 1.88 𝑖𝑖𝑖𝑖 = 47.75 𝑚𝑚𝑚𝑚 >
Section properties of 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 5 𝑆𝑆𝑆𝑆𝑆𝑆.
a) Shear strength of weld
𝐿𝐿
= 8.13 𝑚𝑚𝑚𝑚 (𝑂𝑂𝑂𝑂)
300
𝑡𝑡 = 0.241 𝑖𝑖𝑖𝑖 = 6.12 𝑚𝑚𝑚𝑚,
𝐷𝐷 = 5.56 𝑖𝑖𝑖𝑖 = 141.22 𝑚𝑚𝑚𝑚
→ 𝐿𝐿 = 𝜋𝜋𝜋𝜋 = 𝜋𝜋 × 141.22 = 443.66 𝑚𝑚𝑚𝑚
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.707𝑤𝑤𝑤𝑤(0.6𝐸𝐸70𝑋𝑋𝑋𝑋6.9)
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝑤𝑤 ≥
𝑃𝑃𝑢𝑢
231.30 × 103
=
= 3.39 𝑚𝑚𝑚𝑚
𝜙𝜙0.707𝐿𝐿(0.6𝐸𝐸70𝑋𝑋𝑋𝑋6.9) 0.75 × 0.707 × 443.66(0.6 × 70 × 6.9)
b) Base metal shear strength
Yielding:
Rupture:
𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑦𝑦 𝑡𝑡𝑡𝑡 = 1.0 × 0.6 × 240 × 6.12 × 443.66
→ 𝝓𝝓𝑹𝑹𝒏𝒏 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟗𝟗𝟗𝟗 𝒌𝒌𝒌𝒌 > 𝑷𝑷𝒖𝒖 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑢𝑢 𝑡𝑡𝑡𝑡 = 0.75 × 0.6 × 415 × 6.12 × 443.66
→ 𝜙𝜙𝑅𝑅𝑛𝑛 = 507.06 𝑘𝑘𝑘𝑘 > 𝑃𝑃𝑢𝑢 = 231.30 𝑘𝑘𝑘𝑘 (𝑂𝑂𝑂𝑂)
Semester I of 2024~2025
page 4 of 17
Year4 Group Ci4~M3
Norton University
Tension Members
To economic we will,
Try 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 4 𝑆𝑆𝑆𝑆𝑆𝑆.
𝐴𝐴𝑔𝑔 = 2.96 𝑖𝑖𝑛𝑛2 = 1909.67 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 = 1070.83 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
𝐴𝐴𝑒𝑒 = 𝐴𝐴𝑔𝑔 = 1909.67 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑒𝑒 = 743.13 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
𝑟𝑟 = 1.51 𝑖𝑖𝑖𝑖 = 38.35 𝑚𝑚𝑚𝑚 >
Section properties of 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 4 𝑆𝑆𝑆𝑆𝑆𝑆.
a) Shear strength of weld
𝐿𝐿
= 8.13 𝑚𝑚𝑚𝑚 (𝑂𝑂𝑂𝑂)
300
𝑡𝑡 = 0.221 𝑖𝑖𝑖𝑖 = 5.61 𝑚𝑚𝑚𝑚,
𝐷𝐷 = 4.50 𝑖𝑖𝑖𝑖 = 114.3 𝑚𝑚𝑚𝑚
→ 𝐿𝐿 = 𝜋𝜋𝜋𝜋 = 𝜋𝜋 × 114.3 = 359.08 𝑚𝑚𝑚𝑚
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.707𝑤𝑤𝑤𝑤(0.6𝐸𝐸70𝑋𝑋𝑋𝑋6.9)
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝑤𝑤 ≥
231.30 × 103
𝑃𝑃𝑢𝑢
=
= 4.19 𝑚𝑚𝑚𝑚
𝜙𝜙0.707𝐿𝐿(0.6𝐸𝐸70𝑋𝑋𝑋𝑋6.9) 0.75 × 0.707 × 359.08(0.6 × 70 × 6.9)
b) Base metal shear strength
Yielding:
Rupture:
Therefore,
𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑦𝑦 𝑡𝑡𝑡𝑡 = 1.0 × 0.6 × 240 × 5.61 × 359.08
→ 𝝓𝝓𝑹𝑹𝒏𝒏 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 𝒌𝒌𝒌𝒌 > 𝑷𝑷𝒖𝒖 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
𝜙𝜙𝑅𝑅𝑛𝑛 = 𝜙𝜙0.6𝐹𝐹𝑢𝑢 𝑡𝑡𝑡𝑡 = 0.75 × 0.6 × 415 × 5.61 × 359.08
→ 𝜙𝜙𝑅𝑅𝑛𝑛 = 376.20 𝑘𝑘𝑘𝑘 > 𝑃𝑃𝑢𝑢 = 231.30 𝑘𝑘𝑘𝑘 (𝑂𝑂𝑂𝑂)
𝑼𝑼𝑼𝑼𝑼𝑼 𝒂𝒂 𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷 𝟒𝟒 𝑺𝑺𝑺𝑺𝑺𝑺.
Semester I of 2024~2025
page 5 of 17
Year4 Group Ci4~M3
Norton University
Compression Members
CH4 Compression Members
Use LRFD to select a WT section for the compression member shown in below figure. The load is the
total service load, with a live-to-dead load ratio of 2:1. Use 𝐴𝐴50 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠.
Solution :
Given data,
→ 𝐹𝐹𝑦𝑦 = 345 𝑀𝑀𝑀𝑀𝑀𝑀,
Total service load,
𝐴𝐴50 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Hence,
𝐿𝐿 2
= = 2 → 𝐿𝐿 = 2𝐷𝐷
𝐷𝐷 1
𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 800 𝑘𝑘𝑘𝑘
𝐹𝐹𝑢𝑢 = 450 𝑀𝑀𝑀𝑀𝑀𝑀
𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝐷𝐷 + 𝐿𝐿 = 800 𝑘𝑘𝑘𝑘
𝐷𝐷 + 2𝐷𝐷 = 800 → 𝐷𝐷 =
800
= 266.67 𝑘𝑘𝑘𝑘
3
𝐿𝐿 = 2𝐷𝐷 = 2 × 266.67 = 533.33 𝑘𝑘𝑘𝑘
Factored load,
𝑃𝑃𝑢𝑢 = 1.2𝐷𝐷 + 1.6𝐿𝐿 = 1.2 × 266.67 + 1.6 × 533.33 = 1173.33 𝑘𝑘𝑘𝑘
Assume, an arbitrary choice of two-thirds 𝐹𝐹𝑦𝑦 :
2
2
𝐹𝐹𝑐𝑐𝑐𝑐 = 𝐹𝐹𝑦𝑦 = × 345 = 230 𝑀𝑀𝑀𝑀𝑀𝑀
3
3
Flexural Buckling Strength:
𝑃𝑃𝑢𝑢 ≤ 𝜙𝜙𝑐𝑐 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑐𝑐 𝐹𝐹𝑐𝑐𝑐𝑐 𝐴𝐴𝑔𝑔
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 ≥
Try a 𝑊𝑊𝑊𝑊10.5 × 31:
𝑃𝑃𝑢𝑢
1173.33 × 103
=
= 5668.26 𝑚𝑚𝑚𝑚2 = 8.78 𝑖𝑖𝑛𝑛2
𝜙𝜙𝑐𝑐 𝐹𝐹𝑐𝑐𝑐𝑐
0.9 × 230
𝐴𝐴𝑔𝑔 = 9.13 𝑖𝑖𝑛𝑛2 = 5890.31 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 = 5668.26 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑟𝑟𝑦𝑦 = 1.77 𝑖𝑖𝑖𝑖 = 45 𝑚𝑚𝑚𝑚
The Effective Slenderness Ratio:
Semester I of 2024~2025
page 6 of 17
Year4 Group Ci4~M3
Norton University
Compression Members
𝐿𝐿𝑐𝑐
𝐾𝐾𝐾𝐾 0.7 × 6100
=
=
= 94.89 < 200 (𝑂𝑂𝑂𝑂)
𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚
𝑟𝑟𝑦𝑦
45
1) Flexural Buckling Strength
𝐿𝐿𝑐𝑐 𝐾𝐾𝐾𝐾
𝐸𝐸
2 × 105
=
= 94.89 < 4.71� = 4.71�
= 113.40
𝑟𝑟𝑦𝑦
𝐹𝐹𝑦𝑦
𝑟𝑟
345
𝐹𝐹𝑒𝑒 =
𝜋𝜋 2 𝐸𝐸
2 =
𝐿𝐿
� 𝑐𝑐 �
𝑟𝑟
𝜋𝜋 2 × 2 × 105
= 219.22 𝑀𝑀𝑀𝑀𝑀𝑀
94.892
𝐹𝐹𝑦𝑦
345
=
= 1.5737 < 2.25
𝐹𝐹𝑒𝑒 219.22
𝐹𝐹𝑦𝑦
𝐹𝐹𝑐𝑐𝑐𝑐 = �0.658 𝐹𝐹𝑒𝑒 � 𝐹𝐹𝑦𝑦 = (0.6581.5737 ) × 345 = 178.55 𝑀𝑀𝑀𝑀𝑀𝑀
Flexural Buckling Strength: 𝜙𝜙𝑐𝑐 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑐𝑐 𝐹𝐹𝑐𝑐𝑐𝑐 𝐴𝐴𝑔𝑔 = 0.9 × 178.55 × 5890.31 = 946.54 𝑘𝑘𝑘𝑘
𝝓𝝓𝒄𝒄 𝑷𝑷𝒏𝒏 = 𝟗𝟗𝟗𝟗𝟗𝟗. 𝟓𝟓𝟓𝟓 𝒌𝒌𝒌𝒌 < 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑷𝑷𝒖𝒖 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑵𝑵. 𝑮𝑮. )
Because the initial estimate of 𝐹𝐹𝑐𝑐𝑐𝑐 was so far off, assume a value about halfway between 230 and
178.55 𝑀𝑀𝑀𝑀𝑀𝑀. Try
𝐹𝐹𝑐𝑐𝑐𝑐 =
Try a 𝑊𝑊𝑊𝑊12 × 42:
230 + 178.55
= 204.27 𝑀𝑀𝑀𝑀𝑀𝑀
2
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 ≥
𝑃𝑃𝑢𝑢
1173.33 × 103
=
= 6382.24 𝑚𝑚𝑚𝑚2 = 9.89 𝑖𝑖𝑛𝑛2
𝜙𝜙𝑐𝑐 𝐹𝐹𝑐𝑐𝑐𝑐
0.9 × 204.27
𝐴𝐴𝑔𝑔 = 12.4 𝑖𝑖𝑛𝑛2 = 8000 𝑚𝑚𝑚𝑚2 > 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐴𝐴𝑔𝑔 = 6382.24 𝑚𝑚𝑚𝑚2 (𝑂𝑂𝑂𝑂)
𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑟𝑟𝑦𝑦 = 1.95 𝑖𝑖𝑖𝑖 = 49.53 𝑚𝑚𝑚𝑚
The Effective Slenderness Ratio:
𝐿𝐿𝑐𝑐
𝐾𝐾𝐾𝐾 0.7 × 6100
=
=
= 86.21 < 200 (𝑂𝑂𝑂𝑂)
𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚
𝑟𝑟𝑦𝑦
49.53
1) Flexural Buckling Strength
𝐸𝐸
2 × 105
𝐿𝐿𝑐𝑐 𝐾𝐾𝐾𝐾
=
= 86.21 < 4.71� = 4.71�
= 113.40
𝑟𝑟𝑦𝑦
𝐹𝐹𝑦𝑦
𝑟𝑟
345
𝜋𝜋 2 𝐸𝐸
𝜋𝜋 2 × 2 × 105
𝐹𝐹𝑒𝑒 =
=
= 265.59 𝑀𝑀𝑀𝑀𝑀𝑀
86.212
𝐿𝐿𝑐𝑐 2
� �
𝑟𝑟
𝐹𝐹𝑦𝑦
345
=
= 1.2990 < 2.25
𝐹𝐹𝑒𝑒 265.59
Semester I of 2024~2025
page 7 of 17
Year4 Group Ci4~M3
Norton University
Compression Members
𝐹𝐹𝑦𝑦
𝐹𝐹𝑐𝑐𝑐𝑐 = �0.658 𝐹𝐹𝑒𝑒 � 𝐹𝐹𝑦𝑦 = (0.6581.2990 ) × 345 = 200.31 𝑀𝑀𝑀𝑀𝑀𝑀
Flexural Buckling Strength: 𝜙𝜙𝑐𝑐 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑐𝑐 𝐹𝐹𝑐𝑐𝑐𝑐 𝐴𝐴𝑔𝑔 = 0.9 × 200.31 × 8000 = 1442.23 𝑘𝑘𝑘𝑘
𝝓𝝓𝒄𝒄 𝑷𝑷𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑷𝑷𝒖𝒖 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
2) Flexural-Torsional Buckling Strength
Flexural-Torsional Properties of 𝑊𝑊𝑊𝑊12 × 42:
𝐽𝐽 = 1.84 𝑖𝑖𝑛𝑛4 = 0.766 × 106 𝑚𝑚𝑚𝑚4
𝐶𝐶𝑤𝑤 = 6.90 𝑖𝑖𝑛𝑛6 = 1.853 × 109 𝑚𝑚𝑚𝑚6
𝑟𝑟�0 = 4.89 𝑖𝑖𝑖𝑖 = 124.21 𝑚𝑚𝑚𝑚
𝐻𝐻 = 0.721
𝐺𝐺 = 77200 𝑀𝑀𝑀𝑀𝑀𝑀
𝐸𝐸
2 × 105
𝐿𝐿𝑐𝑐 𝐾𝐾𝐾𝐾
=
= 86.21 < 4.71� = 4.71�
= 113.40
𝑟𝑟
𝑟𝑟𝑦𝑦
𝐹𝐹𝑦𝑦
345
𝐹𝐹𝑒𝑒𝑒𝑒 =
𝜋𝜋 2 𝐸𝐸
𝐿𝐿𝑐𝑐𝑐𝑐
� 𝑟𝑟 �
𝐹𝐹𝑒𝑒𝑒𝑒 = �
=�
𝑦𝑦
2 =
𝜋𝜋 2 × 2 × 105
= 265.59 𝑀𝑀𝑀𝑀𝑀𝑀
86.212
𝜋𝜋 2 𝐸𝐸𝐶𝐶𝑤𝑤
1
+
𝐺𝐺𝐺𝐺�
𝐿𝐿2𝑐𝑐𝑐𝑐
𝐴𝐴𝑔𝑔 ���
𝑟𝑟02
𝜋𝜋 2 × 2 × 105 × 1.853 × 109
1
6
+
77200
×
0.766
×
10
�
(0.7 × 6100)2
8000 × 124.212
𝐹𝐹𝑒𝑒𝑒𝑒 = 480.74 𝑀𝑀𝑀𝑀𝑀𝑀
𝐹𝐹𝑒𝑒 = �
=�
𝐹𝐹𝑒𝑒𝑒𝑒 + 𝐹𝐹𝑒𝑒𝑒𝑒
4𝐹𝐹𝑒𝑒𝑒𝑒 𝐹𝐹𝑒𝑒𝑒𝑒 𝐻𝐻
� �1 − �1 −
2�
2𝐻𝐻
�𝐹𝐹 + 𝐹𝐹 �
𝑒𝑒𝑒𝑒
𝑒𝑒𝑒𝑒
265.59 + 480.74
4 × 265.59 × 480.74 × 0.721
� �1 − �1 −
�
(265.59 + 480.74)2
2 × 0.721
𝐹𝐹𝑒𝑒 = 216.25 𝑀𝑀𝑀𝑀𝑀𝑀
𝐹𝐹𝑦𝑦
345
=
= 1.5954 < 2.25
𝐹𝐹𝑒𝑒 216.25
𝐹𝐹𝑦𝑦
𝐹𝐹𝑐𝑐𝑐𝑐 = �0.658 𝐹𝐹𝑒𝑒 � 𝐹𝐹𝑦𝑦 = (0.6581.5954 ) × 345 = 176.94 𝑀𝑀𝑀𝑀𝑀𝑀
Flexural-Torsional Buckling Strength:
Semester I of 2024~2025
page 8 of 17
Year4 Group Ci4~M3
Norton University
Compression Members
𝜙𝜙𝑐𝑐 𝑃𝑃𝑛𝑛 = 𝜙𝜙𝑐𝑐 𝐹𝐹𝑐𝑐𝑐𝑐 𝐴𝐴𝑔𝑔 = 0.9 × 176.94 × 8000 = 1273.97 𝑘𝑘𝑘𝑘
𝝓𝝓𝒄𝒄 𝑷𝑷𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗 𝒌𝒌𝒌𝒌 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑷𝑷𝒖𝒖 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
3) Local Buckling Strength
Section Properties of 𝑊𝑊𝑊𝑊12 × 42:
𝑑𝑑 = 12.1 𝑖𝑖𝑖𝑖 = 307.34 𝑚𝑚𝑚𝑚,
𝑏𝑏𝑓𝑓 = 9.02 𝑖𝑖𝑖𝑖 = 229.11 𝑚𝑚𝑚𝑚,
𝑏𝑏𝑓𝑓
= 5.86,
2𝑡𝑡𝑓𝑓
𝑡𝑡𝑤𝑤 = 0.47 𝑖𝑖𝑖𝑖 = 11.94 𝑚𝑚𝑚𝑚
ℎ
= 25.7
𝑡𝑡𝑤𝑤
𝑡𝑡𝑓𝑓 = 0.77 𝑖𝑖𝑖𝑖 = 19.56 𝑚𝑚𝑚𝑚
(𝐹𝐹𝐹𝐹):
𝜆𝜆 =
𝑏𝑏𝑓𝑓
𝐸𝐸
2 × 105
= 5.86 < 𝜆𝜆𝑟𝑟 = 0.56� = 0.56�
= 13.48
2𝑡𝑡𝑓𝑓
𝐹𝐹𝑦𝑦
345
(𝑊𝑊𝑊𝑊):
𝜆𝜆 =
ℎ
𝐸𝐸
2 × 105
= 25.7 < 𝜆𝜆𝑟𝑟 = 0.75� = 0.75�
= 18.06
𝑡𝑡𝑤𝑤
345
𝐹𝐹𝑦𝑦
→ Flange is Non-slender. → No Local Buckling!
The Effective Area:
(𝐹𝐹𝐹𝐹):
(𝑊𝑊𝑊𝑊):
→ Web is Slender. → Local Buckling!
𝐴𝐴𝑒𝑒𝑒𝑒 = 𝐴𝐴𝑔𝑔𝑔𝑔 = 𝑏𝑏𝑓𝑓 𝑡𝑡𝑓𝑓 = 229.11 × 19.56 = 4481.39 𝑚𝑚𝑚𝑚2
𝜆𝜆 =
𝐹𝐹𝑦𝑦
345
ℎ
= 25.7 > 𝜆𝜆𝑟𝑟 � = 18.06�
= 25.22
𝑡𝑡𝑤𝑤
𝐹𝐹𝑐𝑐𝑐𝑐
176.94
𝑏𝑏 = ℎ = 𝑑𝑑 − 𝑡𝑡𝑓𝑓 = 307.34 − 19.56 = 287.78 𝑚𝑚𝑚𝑚
𝜆𝜆𝑟𝑟 2
18.06 2
𝐹𝐹𝑒𝑒𝑒𝑒 = �𝑐𝑐2 � 𝐹𝐹𝑦𝑦 = �1.49 ×
� × 345 = 378.23 𝑀𝑀𝑀𝑀𝑀𝑀
𝜆𝜆
25.7
𝐹𝐹𝑒𝑒𝑒𝑒
𝐹𝐹𝑒𝑒𝑒𝑒
378.23
378.23
� � = 287.78 �1 − 0.22�
��
𝐹𝐹𝑐𝑐𝑐𝑐
𝐹𝐹𝑐𝑐𝑐𝑐
176.94
176.94
𝑏𝑏𝑒𝑒 = 𝑏𝑏 �1 − 𝑐𝑐1 �
= 285.41 𝑚𝑚𝑚𝑚
𝐴𝐴𝑒𝑒𝑒𝑒 = 𝑏𝑏𝑒𝑒 𝑡𝑡𝑤𝑤 = 285.41 × 11.94 = 3407.79 𝑚𝑚𝑚𝑚2
→ 𝐴𝐴𝑒𝑒 = 𝐴𝐴𝑒𝑒𝑒𝑒 + 𝐴𝐴𝑒𝑒𝑒𝑒 = 4481.39 + 3407.79 = 7889.18 𝑚𝑚𝑚𝑚2
Local Buckling Strength:
Therefore,
𝜙𝜙𝑃𝑃𝑛𝑛 = 𝜙𝜙𝐹𝐹𝑐𝑐𝑐𝑐 𝐴𝐴𝑒𝑒 = 0.9 × 176.94 × 7889.18 = 1256.32 𝑘𝑘𝑘𝑘
𝝓𝝓𝑷𝑷𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑷𝑷𝒖𝒖 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
𝑼𝑼𝑼𝑼𝑼𝑼 𝒂𝒂 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 × 𝟒𝟒𝟒𝟒.
Semester I of 2024~2025
page 9 of 17
Year4 Group Ci4~M3
Norton University
Beams
CH5 Beams
Use LRFD with 𝐴𝐴992 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝐴𝐴50 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) to select a shape for the beams shown in below figure that has
lateral support at the ends only. The concentrated loads are live loads. Do not check deflections.
Solution :
Use A992 steel (A50 steel),
𝐹𝐹𝑦𝑦 = 345 𝑀𝑀𝑀𝑀𝑀𝑀,
𝐹𝐹𝑢𝑢 = 450 𝑀𝑀𝑀𝑀𝑀𝑀
Neglect the beam weight and check it later. 𝑤𝑤𝐷𝐷 = 0 𝑘𝑘𝑘𝑘/𝑚𝑚
Factored loads,
𝑃𝑃𝑢𝑢 = 1.2𝐷𝐷 + 1.6𝐿𝐿 = 1.2 × 0 + 1.6 × 89 = 142.4 𝑘𝑘𝑘𝑘
The reactions after factored loads,
𝑅𝑅𝐴𝐴 =
𝑅𝑅𝐵𝐵 =
𝑃𝑃
142.4
(𝐿𝐿 − 𝑎𝑎 + 𝑏𝑏) =
(8.5 − 1.5 + 5.5) = 209.41 𝑘𝑘𝑘𝑘
𝐿𝐿
8.5
𝑃𝑃
142.4
(𝐿𝐿 − 𝑏𝑏 + 𝑎𝑎) =
(8.5 − 5.5 + 1.5) = 75.39 𝑘𝑘𝑘𝑘
8.5
𝐿𝐿
I. Check the moment strength
The internal bending moment,
For 0 ≤ 𝑥𝑥 ≤ 1.5 𝑚𝑚
For 1.5 𝑚𝑚 ≤ 𝑥𝑥 ≤ 3 𝑚𝑚
For 3 𝑚𝑚 ≤ 𝑥𝑥 ≤ 8.5 𝑚𝑚
𝑀𝑀(𝑥𝑥) = 𝑅𝑅𝐴𝐴 𝑥𝑥
𝑀𝑀(𝑥𝑥) = 𝑅𝑅𝐴𝐴 𝑥𝑥 − 𝑃𝑃𝑢𝑢 (𝑥𝑥 − 1.5)
𝑀𝑀(𝑥𝑥) = 𝑅𝑅𝐴𝐴 𝑥𝑥 − 𝑃𝑃𝑢𝑢 (𝑥𝑥 − 1.5) − 𝑃𝑃𝑢𝑢 (𝑥𝑥 − 3)
The moments required for the computation of 𝐶𝐶𝑏𝑏 .
For 𝑥𝑥 = 2.125 𝑚𝑚,
For 𝑥𝑥 = 3 𝑚𝑚,
𝑀𝑀𝐴𝐴 = 209.41 × 2.125 − 142.4(2.125 − 1.5) = 355 𝑘𝑘𝑘𝑘. 𝑚𝑚
𝑀𝑀𝐿𝐿.𝑚𝑚𝑚𝑚𝑚𝑚 = 209.41 × 3 − 142.4(3 − 1.5) = 414.63 𝑘𝑘𝑘𝑘. 𝑚𝑚
For 𝑥𝑥 = 4.25 𝑚𝑚,
𝑀𝑀𝐵𝐵 = 209.41 × 4.25 − 142.4(4.25 − 1.5) − 142.4(4.25 − 3)
For 𝑥𝑥 = 6.375 𝑚𝑚,
𝑀𝑀𝐶𝐶 = 209.41 × 6.375 − 142.4(6.375 − 1.5) − 142.4(6.375 − 3)
= 320.39 𝑘𝑘𝑘𝑘. 𝑚𝑚
= 160.19 𝑘𝑘𝑘𝑘. 𝑚𝑚
The lateral-torsional buckling modification factor, 𝐶𝐶𝑏𝑏
𝐶𝐶𝑏𝑏 =
12.5𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚
2.5𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 + 3𝑀𝑀𝐴𝐴 + 4𝑀𝑀𝐵𝐵 + 3𝑀𝑀𝐶𝐶
Semester I of 2024~2025
page 10 of 17
Year4 Group Ci4~M3
Norton University
Beams
𝐶𝐶𝑏𝑏 =
12.5 × 414.63
= 1.34
2.5 × 414.63 + 3 × 355 + 4 × 320.39 + 3 × 160.19
From the charts with unbraced length 𝐿𝐿𝑏𝑏 = 28 𝑓𝑓𝑓𝑓 and bending moment of
𝑀𝑀𝐿𝐿.𝑚𝑚𝑚𝑚𝑚𝑚 414.63
=
= 309.43 𝑘𝑘𝑘𝑘. 𝑚𝑚 = 228.24 𝑘𝑘𝑘𝑘𝑘𝑘. 𝑓𝑓𝑓𝑓
1.34
𝐶𝐶𝑏𝑏
Try 𝑊𝑊12 × 65:
𝑤𝑤𝐷𝐷 = 65 𝐼𝐼𝐼𝐼/𝑓𝑓𝑓𝑓 = 0.95 𝑘𝑘𝑘𝑘/𝑚𝑚
𝑏𝑏𝑓𝑓
= 9.92,
2𝑡𝑡𝑓𝑓
ℎ
= 24.9
𝑡𝑡𝑤𝑤
𝑆𝑆𝑥𝑥 = 87.9 𝑖𝑖𝑛𝑛3 = 1.44 × 106 𝑚𝑚𝑚𝑚3 ,
Account for the beam weight,
𝑍𝑍𝑥𝑥 = 96.8 𝑖𝑖𝑛𝑛3 = 1.586 × 106 𝑚𝑚𝑚𝑚3
𝑤𝑤𝐷𝐷 𝐿𝐿2
0.95 × 8.52
= 414.63 + 1.2
= 424.93 𝑘𝑘𝑘𝑘. 𝑚𝑚
𝑀𝑀𝑢𝑢 = 𝑀𝑀𝐿𝐿.𝑚𝑚𝑚𝑚𝑚𝑚 + 1.2
8
8
1) Check in-plane bending
(𝐹𝐹𝐹𝐹):
𝜆𝜆 =
𝑏𝑏𝑓𝑓
= 9.92,
2𝑡𝑡𝑓𝑓
𝐸𝐸
2 × 105
= 0.38�
= 9.15
𝐹𝐹𝑦𝑦
345
𝜆𝜆𝑝𝑝 = 0.38�
𝐸𝐸
2 × 105
�
𝜆𝜆𝑟𝑟 = 1.0� = 1.0
= 24.08
345
𝐹𝐹𝑦𝑦
→ Flange is Noncompact. → Inelastic LB!
𝜆𝜆𝑝𝑝 = 9.15 < 𝜆𝜆 = 9.92 < 𝜆𝜆𝑟𝑟 = 24.08
(𝑊𝑊𝑊𝑊):
𝜆𝜆 =
ℎ
= 24.9,
𝑡𝑡𝑤𝑤
𝜆𝜆 = 24.9 < 𝜆𝜆𝑝𝑝 = 90.53
Plastic moment,
𝐸𝐸
2 × 105
= 3.76�
= 90.53
𝐹𝐹𝑦𝑦
345
𝜆𝜆𝑝𝑝 = 3.76�
→ Web is Compact.
𝑀𝑀𝑝𝑝 = 𝐹𝐹𝑦𝑦 𝑍𝑍𝑥𝑥 = 345 × 1.586 × 106 = 547.17 𝑘𝑘𝑘𝑘. 𝑚𝑚
Check the capacity based on the limit state of flange local buckling (Inelastic LB):
Nominal flexural strength,
𝑀𝑀𝑛𝑛 = 𝑀𝑀𝑝𝑝 − �𝑀𝑀𝑝𝑝 − 0.7𝐹𝐹𝑦𝑦 𝑆𝑆𝑥𝑥 � �
𝜆𝜆 − 𝜆𝜆𝑝𝑝𝑝𝑝
�
𝜆𝜆𝑟𝑟𝑟𝑟 − 𝜆𝜆𝑝𝑝𝑝𝑝
= 547.17 − (547.17 − 0.7 × 345 × 1.44 × 106 ) �
𝑀𝑀𝑛𝑛 = 536.89 𝑘𝑘𝑘𝑘. 𝑚𝑚
9.92 − 9.15
�
24.08 − 9.15
→ 𝝓𝝓𝒃𝒃 𝑴𝑴𝒏𝒏 = 𝟎𝟎. 𝟗𝟗 × 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟖𝟖𝟖𝟖 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌. 𝒎𝒎 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑴𝑴𝒖𝒖 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟗𝟗𝟗𝟗 𝒌𝒌𝒌𝒌. 𝒎𝒎 (𝑶𝑶𝑶𝑶)
Semester I of 2024~2025
page 11 of 17
Year4 Group Ci4~M3
Norton University
Beams
2) Out of plane bending
Section properties of 𝑊𝑊12 × 65:
𝑟𝑟𝑦𝑦 = 3.02 𝑖𝑖𝑖𝑖 = 76.71 𝑚𝑚𝑚𝑚,
ℎ0 = 11.5 𝑖𝑖𝑖𝑖 = 292.1 𝑚𝑚𝑚𝑚
𝑟𝑟𝑡𝑡𝑡𝑡 = 3.38 𝑖𝑖𝑖𝑖 = 85.85 𝑚𝑚𝑚𝑚
𝐽𝐽 = 2.18 𝑖𝑖𝑛𝑛4 = 0.907 × 106 𝑚𝑚𝑚𝑚4
𝐿𝐿𝑏𝑏 = 8.5 𝑚𝑚
𝐶𝐶𝑤𝑤 = 5780 𝑖𝑖𝑛𝑛6 = 1.55 × 1012 𝑚𝑚𝑚𝑚6
2 × 105
𝐸𝐸
= 1.76 × 76.71�
= 3.25 𝑚𝑚
𝐹𝐹𝑦𝑦
345
𝐿𝐿𝑝𝑝 = 1.76𝑟𝑟𝑦𝑦 �
𝐿𝐿𝑟𝑟 = 1.95𝑟𝑟𝑡𝑡𝑡𝑡
0.7𝐹𝐹𝑦𝑦 2
𝐸𝐸
𝐽𝐽𝐽𝐽
𝐽𝐽𝐽𝐽 2
�
+ ��
� + 6.76 �
�
0.7𝐹𝐹𝑦𝑦 𝑆𝑆𝑥𝑥 ℎ0
𝑆𝑆𝑥𝑥 ℎ0
𝐸𝐸
𝐽𝐽𝐽𝐽
0.907 × 106 × 1.0
=
= 0.00216
𝑆𝑆𝑥𝑥 ℎ0 1.44 × 106 × 292.1
Since,
2 × 105 �
0.7 × 345 2
2
�
0.00216 + 0.00216 + 6.76 �
𝐿𝐿𝑟𝑟 = 1.95 × 85.85 ×
� = 10.71 𝑚𝑚
0.7 × 345
2 × 105
𝐿𝐿𝑝𝑝 = 3.25 𝑚𝑚 < 𝐿𝐿𝑏𝑏 = 8.5 𝑚𝑚 < 𝐿𝐿𝑟𝑟 = 10.71 𝑚𝑚
→ Noncompact. → Inelastic LTB!
The nominal strength for the sections with noncompact flanges (Inelastic LTB):
𝑀𝑀𝑛𝑛 = 𝐶𝐶𝑏𝑏 �𝑀𝑀𝑝𝑝 − �𝑀𝑀𝑝𝑝 − 0.7𝐹𝐹𝑦𝑦 𝑆𝑆𝑥𝑥 � �
𝐿𝐿𝑏𝑏 − 𝐿𝐿𝑝𝑝
�� ≤ 𝑀𝑀𝑝𝑝
𝐿𝐿𝑟𝑟 − 𝐿𝐿𝑝𝑝
= 1.34 �547.17 − (547.17 − 0.7 × 345 × 1.44 × 106 ) �
𝑀𝑀𝑛𝑛 = 545.16 𝑘𝑘𝑘𝑘. 𝑚𝑚 < 𝑀𝑀𝑝𝑝 = 547.14 𝑘𝑘𝑘𝑘. 𝑚𝑚 (𝑂𝑂𝑂𝑂)
8.5 − 3.25
��
10.71 − 3.25
→ 𝝓𝝓𝒃𝒃 𝑴𝑴𝒏𝒏 = 𝟎𝟎. 𝟗𝟗 × 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟏𝟏𝟏𝟏 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟔𝟔𝟔𝟔 𝒌𝒌𝒌𝒌. 𝒎𝒎 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑴𝑴𝒖𝒖 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟗𝟗𝟗𝟗 𝒌𝒌𝒌𝒌. 𝒎𝒎 (𝑶𝑶𝑶𝑶)
II. Check the shear strength
Section properties of 𝑊𝑊12 × 65:
Web area:
𝑑𝑑 = 12.1 𝑖𝑖𝑖𝑖 = 307.34 𝑚𝑚𝑚𝑚,
𝑡𝑡𝑓𝑓 = 0.605 𝑖𝑖𝑖𝑖 = 15.367 𝑚𝑚𝑚𝑚
𝑡𝑡𝑤𝑤 = 0.39 𝑖𝑖𝑖𝑖 = 9.906 𝑚𝑚𝑚𝑚
𝐴𝐴𝑤𝑤 = �𝑑𝑑 − 2𝑡𝑡𝑓𝑓 �𝑡𝑡𝑤𝑤 = (307.34 − 2 × 15.367)9.906 = 2740.06 𝑚𝑚𝑚𝑚2
The required shear strength:
𝑉𝑉𝑢𝑢 = 𝑅𝑅𝐴𝐴 + 1.2
Semester I of 2024~2025
𝑤𝑤𝐷𝐷 𝐿𝐿
0.95 × 8.5
= 209.41 + 1.2
= 214.26 𝑘𝑘𝑘𝑘
2
2
page 12 of 17
Year4 Group Ci4~M3
Norton University
Beams
Check the limit state of web:
𝜆𝜆 =
ℎ
2 × 105
𝐸𝐸
= 24.9 < 2.24� = 2.24�
= 53.93
𝑡𝑡𝑤𝑤
𝐹𝐹𝑦𝑦
345
→ Shear web is yielding.
→ 𝐶𝐶𝑣𝑣1 = 1.0,
The nominal shear strength:
𝜙𝜙𝑣𝑣 = 1.0
𝜙𝜙𝑣𝑣 𝑉𝑉𝑛𝑛 = 𝜙𝜙𝑣𝑣 0.6𝐹𝐹𝑦𝑦 𝐴𝐴𝑤𝑤 𝐶𝐶𝑣𝑣1 = 1.0 × 0.6 × 345 × 2740.06 × 1.0 = 567.19 𝑘𝑘𝑘𝑘
𝝓𝝓𝒗𝒗 𝑽𝑽𝒏𝒏 = 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟏𝟏𝟏𝟏 𝒌𝒌𝒌𝒌 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑽𝑽𝒖𝒖 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
III. Check the deflection
Section properties of 𝑊𝑊12 × 65:
𝐼𝐼𝑥𝑥 = 533 𝑖𝑖𝑛𝑛4 = 221.85 × 106 𝑚𝑚𝑚𝑚4
Since the beam is a Floor Beam type, the deflection limit is given by:
𝐿𝐿
8.5 × 103
=
= 23.61 𝑚𝑚𝑚𝑚
360
360
Analyze to draw diagram MNV of live load,
𝐏𝐏𝐋𝐋
A
0 kN
1.5 m
𝐑𝐑 𝐀𝐀 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟒𝟒𝟒𝟒 𝐤𝐤𝐤𝐤
0
𝐏𝐏𝐋𝐋 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟒𝟒 𝐤𝐤𝐤𝐤
C
1.5 m
5.5 m
67.01
(+)
0
𝐑𝐑 𝐁𝐁 = 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑 𝐤𝐤𝐤𝐤
0 N (kN)
209.41
(+)
B
(-)
0 V (kN)
-75.39
0
(+)
314.115
Semester I of 2024~2025
(+)
(+)
0 M (kN.m)
414.63
page 13 of 17
Year4 Group Ci4~M3
Norton University
Beams
Analyze to draw diagram mnv by put 1 unit force at point C because it’s the maximum deflection.
A
0
𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟
1.5 m
1.5 m
5.5/8.5
0
0
C
5.5 m
(+)
(-)
-3/8.5
33/34
3/8.5
0 N (unit)
5.5/8.5
0
B
(+)
(+)
0 V (unit)
0 M (unit.m)
33/17
By Method Integral de Mohr, the live load deflection at point C is given by:
𝛿𝛿𝐿𝐿 =
=
1 1
1
1
1
� (𝑀𝑀1 𝑀𝑀3 𝐿𝐿1 ) + 𝑀𝑀1 (2𝑀𝑀3 + 𝑀𝑀4 )𝐿𝐿2 + 𝑀𝑀2 (𝑀𝑀3 + 2𝑀𝑀4 )𝐿𝐿 + (𝑀𝑀2 𝑀𝑀4 𝐿𝐿3 )�
6
3
𝐸𝐸𝐼𝐼𝑥𝑥 3
6
1 1
1
33 33
33
33
33
1
� �314.115 ×
× 1.5� + 315.115 �2 + � 1.5 + 414.63 � + 2 � 1.5
𝐸𝐸𝐼𝐼𝑥𝑥 3
6
34 17
34
17
34
6
33
1
× 5.5��
+ �414.63 ×
3
17
1
2436.92 × 1012 𝑁𝑁. 𝑚𝑚𝑚𝑚3
2436.92 × 1012
3]
[2436.92
=
𝑘𝑘𝑘𝑘. 𝑚𝑚 =
=
𝐸𝐸𝐼𝐼𝑥𝑥
𝐸𝐸𝐼𝐼𝑥𝑥
2 × 105 × 221.85 × 106
𝜹𝜹𝑳𝑳 = 𝟓𝟓𝟓𝟓. 𝟗𝟗𝟗𝟗 𝒎𝒎𝒎𝒎 >
𝑳𝑳
= 𝟐𝟐𝟐𝟐. 𝟔𝟔𝟔𝟔 𝒎𝒎𝒎𝒎 (𝑵𝑵. 𝑮𝑮. )
𝟑𝟑𝟑𝟑𝟑𝟑
With checking live load deflection, we need to revised our section.
The required 𝐼𝐼𝑥𝑥 is given by:
𝛿𝛿𝐿𝐿 =
2436.92 × 1012 𝑁𝑁. 𝑚𝑚𝑚𝑚3
𝐿𝐿
≤
= 23.61
360
𝐸𝐸𝐼𝐼𝑥𝑥
→ 𝑅𝑅𝑅𝑅𝑅𝑅. 𝐼𝐼𝑥𝑥 ≥
Try 𝑊𝑊16 × 89:
2436.92 × 1012 2436.92 × 1012
=
= 516.08 × 106 𝑚𝑚𝑚𝑚4 = 1239.89 𝑖𝑖𝑛𝑛4
23.61𝐸𝐸
23.61 × 2 × 105
𝑤𝑤𝐷𝐷 = 89 𝐼𝐼𝐼𝐼/𝑓𝑓𝑓𝑓 = 1.30 𝑘𝑘𝑘𝑘/𝑚𝑚
Semester I of 2024~2025
page 14 of 17
Year4 Group Ci4~M3
Norton University
Beams
𝑏𝑏𝑓𝑓
= 5.92,
2𝑡𝑡𝑓𝑓
ℎ
= 27.0
𝑡𝑡𝑤𝑤
𝑆𝑆𝑥𝑥 = 155 𝑖𝑖𝑛𝑛3 = 2.54 × 106 𝑚𝑚𝑚𝑚3 ,
1) Check the moment strength
𝑍𝑍𝑥𝑥 = 175 𝑖𝑖𝑛𝑛3 = 2.868 × 106 𝑚𝑚𝑚𝑚3
Account for the beam weight,
𝑀𝑀𝑢𝑢 = 𝑀𝑀𝐿𝐿.𝑚𝑚𝑚𝑚𝑚𝑚 + 1.2
a) Check in-plane bending
(𝐹𝐹𝐹𝐹):
(𝑊𝑊𝑊𝑊):
𝜆𝜆 =
𝑤𝑤𝐷𝐷 𝐿𝐿2
1.30 × 8.52
= 414.63 + 1.2
= 428.78 𝑘𝑘𝑘𝑘. 𝑚𝑚
8
8
𝑏𝑏𝑓𝑓
2 × 105
𝐸𝐸
= 5.92 < 𝜆𝜆𝑝𝑝 = 0.38� = 0.38�
= 9.15
𝐹𝐹𝑦𝑦
2𝑡𝑡𝑓𝑓
345
→ Flange is Compact.
ℎ
𝐸𝐸
2 × 105
�
𝜆𝜆 =
= 27.0 < 𝜆𝜆𝑝𝑝 = 3.76� = 3.76
= 90.53
𝑡𝑡𝑤𝑤
𝐹𝐹𝑦𝑦
345
→ Web is Compact.
In this case, we calculate as Yielding,
The nominal flexural strength:
𝑀𝑀𝑛𝑛 = 𝑀𝑀𝑝𝑝 = 𝐹𝐹𝑦𝑦 𝑍𝑍𝑥𝑥 = 345 × 2.868 × 106 = 989.46 𝑘𝑘𝑘𝑘. 𝑚𝑚
→ 𝝓𝝓𝒃𝒃 𝑴𝑴𝒏𝒏 = 𝟎𝟎. 𝟗𝟗 × 𝟗𝟗𝟗𝟗𝟗𝟗. 𝟒𝟒𝟒𝟒 = 𝟖𝟖𝟖𝟖𝟖𝟖. 𝟓𝟓𝟓𝟓 𝒌𝒌𝒌𝒌. 𝒎𝒎 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑴𝑴𝒖𝒖 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟕𝟕𝟕𝟕 𝒌𝒌𝒌𝒌. 𝒎𝒎 (𝑶𝑶𝑶𝑶)
b) Out of plane bending
Section properties of 𝑊𝑊16 × 89:
𝑟𝑟𝑦𝑦 = 2.49 𝑖𝑖𝑖𝑖 = 63.25 𝑚𝑚𝑚𝑚,
ℎ0 = 15.9 𝑖𝑖𝑖𝑖 = 403.86 𝑚𝑚𝑚𝑚
𝑟𝑟𝑡𝑡𝑡𝑡 = 2.88 𝑖𝑖𝑖𝑖 = 73.15 𝑚𝑚𝑚𝑚
𝐽𝐽 = 5.45 𝑖𝑖𝑛𝑛4 = 2.27 × 106 𝑚𝑚𝑚𝑚4
𝐿𝐿𝑏𝑏 = 8.5 𝑚𝑚
𝐶𝐶𝑤𝑤 = 10200 𝑖𝑖𝑛𝑛6 = 2.74 × 1012 𝑚𝑚𝑚𝑚6
2 × 105
𝐸𝐸
= 1.76 × 63.25�
= 2.68 𝑚𝑚
𝐹𝐹𝑦𝑦
345
𝐿𝐿𝑝𝑝 = 1.76𝑟𝑟𝑦𝑦 �
𝐿𝐿𝑟𝑟 = 1.95𝑟𝑟𝑡𝑡𝑡𝑡
0.7𝐹𝐹𝑦𝑦 2
𝐸𝐸
𝐽𝐽𝐽𝐽
𝐽𝐽𝐽𝐽 2
�
+ ��
+
6.76
�
�
�
0.7𝐹𝐹𝑦𝑦 𝑆𝑆𝑥𝑥 ℎ0
𝑆𝑆𝑥𝑥 ℎ0
𝐸𝐸
Semester I of 2024~2025
page 15 of 17
Year4 Group Ci4~M3
Norton University
Beams
𝐽𝐽𝐽𝐽
2.27 × 106 × 1.0
=
= 0.0022
𝑆𝑆𝑥𝑥 ℎ0 2.54 × 106 × 403.86
Since,
2 × 105 �
0.7 × 345 2
2
�
𝐿𝐿𝑟𝑟 = 1.95 × 73.15 ×
0.0022 + 0.0022 + 6.76 �
� = 9.18 𝑚𝑚
2 × 105
0.7 × 345
𝐿𝐿𝑝𝑝 = 2.68 𝑚𝑚 < 𝐿𝐿𝑏𝑏 = 8.5 𝑚𝑚 < 𝐿𝐿𝑟𝑟 = 9.18 𝑚𝑚 → Noncompact. → Inelastic LTB!
The nominal strength for the sections with noncompact flanges (Inelastic LTB):
𝑀𝑀𝑛𝑛 = 𝐶𝐶𝑏𝑏 �𝑀𝑀𝑝𝑝 − �𝑀𝑀𝑝𝑝 − 0.7𝐹𝐹𝑦𝑦 𝑆𝑆𝑥𝑥 � �
𝐿𝐿𝑏𝑏 − 𝐿𝐿𝑝𝑝
�� ≤ 𝑀𝑀𝑝𝑝
𝐿𝐿𝑟𝑟 − 𝐿𝐿𝑝𝑝
= 1.34 �989.46 − (989.46 − 0.7 × 345 × 2.54 × 106 ) �
𝑀𝑀𝑛𝑛 = 874.69 𝑘𝑘𝑘𝑘. 𝑚𝑚 < 𝑀𝑀𝑝𝑝 = 989.46 𝑘𝑘𝑘𝑘. 𝑚𝑚 (𝑂𝑂𝑂𝑂)
8.5 − 2.68
��
9.18 − 2.68
→ 𝝓𝝓𝒃𝒃 𝑴𝑴𝒏𝒏 = 𝟎𝟎. 𝟗𝟗 × 𝟖𝟖𝟖𝟖𝟖𝟖. 𝟔𝟔𝟔𝟔 = 𝟕𝟕𝟕𝟕𝟕𝟕. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌. 𝒎𝒎 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑴𝑴𝒖𝒖 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟕𝟕𝟕𝟕 𝒌𝒌𝒌𝒌. 𝒎𝒎 (𝑶𝑶𝑶𝑶)
2) Check the shear strength
Section properties of 𝑊𝑊16 × 89:
Web area:
𝑑𝑑 = 16.8 𝑖𝑖𝑖𝑖 = 426.72 𝑚𝑚𝑚𝑚,
𝑡𝑡𝑓𝑓 = 0.875 𝑖𝑖𝑖𝑖 = 22.225 𝑚𝑚𝑚𝑚
𝑡𝑡𝑤𝑤 = 0.525 𝑖𝑖𝑖𝑖 = 13.335 𝑚𝑚𝑚𝑚
𝐴𝐴𝑤𝑤 = �𝑑𝑑 − 2𝑡𝑡𝑓𝑓 �𝑡𝑡𝑤𝑤 = (426.72 − 2 × 22.225)13.335 = 5097.57 𝑚𝑚𝑚𝑚2
The required shear strength:
𝑉𝑉𝑢𝑢 = 𝑅𝑅𝐴𝐴 + 1.2
Check the limit state of web:
𝑤𝑤𝐷𝐷 𝐿𝐿
1.30 × 8.5
= 209.41 + 1.2
= 216.04 𝑘𝑘𝑘𝑘
2
2
ℎ
𝐸𝐸
2 × 105
�
𝜆𝜆 =
= 27.0 < 2.24� = 2.24
= 53.93
𝑡𝑡𝑤𝑤
𝐹𝐹𝑦𝑦
345
→ Shear web is yielding.
The nominal shear strength:
→ 𝐶𝐶𝑣𝑣1 = 1.0,
𝜙𝜙𝑣𝑣 = 1.0
𝜙𝜙𝑣𝑣 𝑉𝑉𝑛𝑛 = 𝜙𝜙𝑣𝑣 0.6𝐹𝐹𝑦𝑦 𝐴𝐴𝑤𝑤 𝐶𝐶𝑣𝑣1 = 1.0 × 0.6 × 345 × 5097.57 × 1.0 = 1055.20 𝑘𝑘𝑘𝑘
𝝓𝝓𝒗𝒗 𝑽𝑽𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌 > 𝑹𝑹𝑹𝑹𝑹𝑹. 𝑽𝑽𝒖𝒖 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌 (𝑶𝑶𝑶𝑶)
3) Check the deflection
Section properties of 𝑊𝑊16 × 89:
𝐼𝐼𝑥𝑥 = 1300 𝑖𝑖𝑛𝑛4 = 541.1 × 106 𝑚𝑚𝑚𝑚4
Through the formula solved above, the live load deflection is given by:
Semester I of 2024~2025
page 16 of 17
Year4 Group Ci4~M3
Norton University
Beams
2436.92 × 1012
2436.92 × 1012
𝛿𝛿𝐿𝐿 =
=
= 22.52 𝑚𝑚𝑚𝑚
𝐸𝐸𝐼𝐼𝑥𝑥
2 × 105 × 541.1 × 106
𝜹𝜹𝑳𝑳 = 𝟐𝟐𝟐𝟐. 𝟓𝟓𝟓𝟓 𝒎𝒎𝒎𝒎 ≤
𝑳𝑳
= 𝟐𝟐𝟐𝟐. 𝟔𝟔𝟔𝟔 𝒎𝒎𝒎𝒎 (𝑶𝑶𝑶𝑶)
𝟑𝟑𝟑𝟑𝟑𝟑
The beam is simple beam with uniformly distributed dead load, the dead load deflection is given by
equation:
𝛿𝛿𝐷𝐷 (𝑥𝑥) =
𝑤𝑤𝐷𝐷 𝑥𝑥 3
(𝐿𝐿 − 2𝐿𝐿𝑥𝑥 2 + 𝑥𝑥 3 )
24𝐸𝐸𝐼𝐼𝑥𝑥
We use 𝑥𝑥 = 3 𝑚𝑚, because the maximum deflection of live load is at 𝑥𝑥 = 3 𝑚𝑚, so we calculate the dead
load deflection is at the same place.
𝛿𝛿𝐷𝐷 =
1.30 × 3000
(85003 − 2 × 8500 × 30002 + 30003 ) = 0.73 𝑚𝑚𝑚𝑚
24 × 2 × 105 × 541.1 × 106
The maximum dead load + live load deflection:
𝐿𝐿
8500
=
= 35.42 𝑚𝑚𝑚𝑚
240
240
Check the total deflection with limit total deflection,
Therefore,
∆𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 = 𝜹𝜹𝑳𝑳 + 𝜹𝜹𝑫𝑫 = 𝟐𝟐𝟐𝟐. 𝟓𝟓𝟓𝟓 + 𝟎𝟎. 𝟕𝟕𝟕𝟕 = 𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎 <
𝑳𝑳
= 𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒 (𝑶𝑶𝑶𝑶)
𝟐𝟐𝟐𝟐𝟐𝟐
𝑼𝑼𝑼𝑼𝑼𝑼 𝒂𝒂 𝑾𝑾𝑾𝑾𝑾𝑾 × 𝟔𝟔𝟔𝟔 (for without checking deflection).
𝑼𝑼𝑼𝑼𝑼𝑼 𝒂𝒂 𝑾𝑾𝑾𝑾𝑾𝑾 × 𝟖𝟖𝟖𝟖 (with checking deflection).
Semester I of 2024~2025
page 17 of 17
Year4 Group Ci4~M3