Republic of Yemen.

University of Aden.
Faculty of Engineering.
Civil Engineering Department.

STEEL DESIGN 1.

B4C / E.

Eng: Mohammed Nageb Haza’a Mohammed.

 Purlins.

Types and uses:

The purlin is a beam and it supports roof decking on flat roofs or cladding on
sloping roofs on industrial buildings.
‫المدادة عبارة عن عارضة وتدعم سطح السقف على األسطح المستوية أو الكسوة على األسطح المائلة في المباني‬
.‫الصناعية‬

Members used for purlins are shown in figure, these are cold-rolled sections
are now used on most industrial buildings.
‫ وهي أقسام ملفوفة يتم استخدامها اآلن في معظم‬،‫يتم عرض األعضاء المستخدمة في المدادة في الشكل‬
.‫المباني الصناعية‬

1 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Type of section productions:

1- Hot Finished.
2- Cold Finished.

Type of Roof:

1- Accessible Roof. )‫(سقف يسهل الوصول اليه‬

2- Non Accessible Roof. )‫(سقف ال يسهل الوصول اليه‬

Purlins Imposed loading for roof is specified in BS 6399:

𝟏 − 𝐀𝐜𝐜𝐞𝐬𝐬𝐢𝐛𝐥𝐞 𝐑𝐨𝐨𝐟.
( 𝒘 = 𝟏. 𝟓 𝑲𝑵/𝒎𝟐 ) 𝒐𝒓 ( 𝑷 = 𝟏. 𝟖 𝑲𝑵 )
1- Flat roof (𝜽 < 𝟏𝟎° ) =
𝟐 − 𝐍𝐨𝐧 𝐀𝐜𝐜𝐞𝐬𝐬𝐢𝐛𝐥𝐞 𝐑𝐨𝐨𝐟.
( 𝒘 = 𝟎. 𝟔 𝑲𝑵/𝒎𝟐 ) 𝒐𝒓 ( 𝑷 = 𝟎. 𝟗 𝑲𝑵 )
{

𝟏) 𝜽 < 𝟑𝟎°
( 𝒘 = 𝟎. 𝟔 𝑲𝑵/𝒎𝟐 ) 𝒐𝒓 ( 𝑷 = 𝟎. 𝟗 𝑲𝑵 )

𝟐) 𝟑𝟎° ≤ 𝜽 < 𝟔𝟎°

2- Slope roof (𝐈𝐧𝐜𝐥𝐢𝐧𝐞𝐝) =
(𝟔𝟎−𝜽)
( 𝒘 = 𝟎. 𝟔 [ 𝟑𝟎 ] 𝑲𝑵/𝒎𝟐 )

𝟑) 𝜽 ≥ 𝟔𝟎° ( 𝑵𝒐 𝑨𝒍𝒍𝒐𝒘𝒆𝒅 )
{

2 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

- Loading on Slope roof (Purlins):

Loading on plane are have to convert it:

Load on slope = Load on plane * cos (𝜽)

Wind Loads:

1) Uplift or Pressure. (Negative)

2) Suction. (Positive)

Loading Combinations:

1) w = 1.4 D.L + 1.6 L.L

2) w = 1.2 D.L + 1.2 L.L + 1.2 W.L
3) w = D.L + 1.4 W.L

Design of Purlins:

1- Hot Finished section:

a) Generally Method.
b) Empirical Method.

2- Cold Finished section.

3 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

 a) Generally Method.

Flat roof: Slope roof:

1) Design Load. 1) Design Load.

𝐰 𝐋𝟐 𝐰 𝐋𝟐
2) Design Moment ( ) 2) Design Moment ( )
𝟖 𝟖

𝑴𝑫 𝑴𝑫
3) 𝒁𝒓𝒆𝒒 = 3) 𝒁𝒓𝒆𝒒 =
𝑷𝒚 𝑷𝒚

4) 𝒁𝒑𝒓𝒐 → (From blue book) 4) 𝒁𝒑𝒓𝒐 → (From blue book)

(𝒁𝒑𝒓𝒐 > 𝒁𝒓𝒆𝒒 ) (𝒁𝒑𝒓𝒐 > 𝒁𝒓𝒆𝒒 )

𝐒𝐩𝐚𝐧
5) Check for deflection. ( )
𝟑𝟔𝟎

4 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Example (1): (Design of purlin for a flat roof)

The roof consists of steel decking with insulation board, felt and rolled-steel,
Joist purlins with a ceiling on the underside. The total dead load is 0.9 KN/m2
and the imposed load is 1.5 KN/m2. The purlins span 4 m and are at 2.5 m
centers. The roof arrangement and loading are shown in figure. Use steel
grade S275 steel.

- Given data: :‫مالحظة‬

D.L = 0.9 KN / 𝐦𝟐 ‫إذا لم يذكر لنا الزاوية في‬
L.L = 1.5 KN / 𝐦𝟐 ‫السؤال نعتبر السطح‬
Span = 4 m , (2.5 c/c) , S275 .)Flat roof(

- Solution:

1) Design load:
D.L = 0.9 * 2.5 = 2.25 KN / m
L.L = 1.5 * 2.5 = 3.75 KN / m
w = 1.4 D.L + 1.6 L.L = 1.4 * 2.25 + 1.6 * 3.75 = 9.15 KN / m

𝐰 𝐋𝟐 𝟗.𝟏𝟓 ∗ 𝟒𝟐
2) 𝑴𝑫 = = = 18.3 KN / m
𝟖 𝟖

𝑴𝑫 𝟏𝟖.𝟑 ∗ 𝟏𝟎𝟑
3) 𝒁𝒓𝒆𝒒 = = = 66.55 c𝐦𝟑
𝑷𝒚 𝟐𝟕𝟓

5 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

4) 𝒁𝒑𝒓𝒐 → (From blue book part (B)) → joists → Provide ( 127 * 76 * 16 )
𝒁𝒑𝒓𝒐 = 90 c𝐦𝟑 > 𝒁𝒓𝒆𝒒 = 66.55 c𝐦𝟑

5) Check for deflection: 𝐈𝐱−𝐱 = 571 c𝐦𝟒

𝟓 ∗ 𝐰 ∗ 𝐋𝟒 𝟓 ∗ 𝟏.𝟓 ∗ 𝟒𝟒
𝜹𝒂𝒄𝒕 = = 𝟑 −𝟖 = 4.271 * 𝟏𝟎−𝟑 m = 4.271 mm
𝟑𝟖𝟒 ∗ 𝐄 ∗ 𝐈 𝟑𝟖𝟒 ∗ 𝟐𝟎𝟓𝟎𝟎𝟎 ∗ 𝟏𝟎 ∗ 𝟓𝟕𝟏 ∗ 𝟏𝟎

𝐒𝐩𝐚𝐧 𝟒𝟎𝟎𝟎
𝜹𝒑𝒆𝒓 = 𝟑𝟔𝟎
=
𝟑𝟔𝟎
= 11.11 mm

𝜹𝒑𝒆𝒓 > 𝜹𝒂𝒄𝒕 → (Safe for deflection)

Example (2): (Design of an angle purlin for a slope roof)

Design an angle for a roof with slope 1 in 2.5. The purlins are simply
supported and span 5.0 m between roof trusses at spacing of 1.6 m. The total
dead load, including purlin weight, is 0.32 KN/m2 and wind load is 0.7 KN/m2
on the slope and the imposed load is 0.6 KN/m2 on plan. Use grade S275 steel.
The arrangement of purlins on the roof slope and loading are shown in the
figure.

6 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

- Given data:
D.L = 0.32 KN / 𝐦𝟐 (on slope) :‫مالحظة‬
L.L = 0.6 KN / 𝐦𝟐 (on plan) ‫عند عدم ذكر حالة الحمل‬
W.L = 0.7 KN / 𝐦𝟐 (on slope) (on plane or on slope)
Span = 5 m , (1.6 c/c) , S275 .)on slope( ‫نعتبره‬
𝟏
𝒕𝒂𝒏−𝟏 ( ) = 𝟐𝟏. 𝟖°
𝟐.𝟓
- Solution:

1) Design load:
D.L = 0.32 * 1.6 = 0.512 KN / m
L.L = 0.6 * 1.6 * cos(𝟐𝟏. 𝟖° ) = 0.891 KN / m
W.L = 0.7 * 1.6 = 1.12 KN / m

Loading Combinations:

1) w = 1.4 D.L + 1.6 L.L = 1.4 * 0.512 + 1.6 * 0.891 = 2.14 KN / m

2) w = 1.2 D.L + 1.2 L.L - 1.2 W.L = 1.2 * 0.512 + 1.2 * 0.891 - 1.2 * 1.12
w = 0.34 KN / m
3) w = D.L - 1.4 W.L = 0.512 – 1.4 * 1.12 = - 1.056 KN / m

Take (w = 2.14 KN / m)

𝐰 𝐋𝟐 𝟐.𝟏𝟒 ∗ 𝟓𝟐
2) 𝑴𝑫 = = = 6.69 KN / m
𝟖 𝟖

𝑴𝑫 𝟔.𝟔𝟗 ∗ 𝟏𝟎𝟑
3) 𝒁𝒓𝒆𝒒 = = = 24.33 c𝐦𝟑
𝑷𝒚 𝟐𝟕𝟓

4) 𝒁𝒑𝒓𝒐 → (From blue book part (B)) → Angle → Provide ( 127 * 75 * 8 )

𝒁𝒑𝒓𝒐 = 29.6 c𝐦𝟑 > 𝒁𝒓𝒆𝒒 = 24.33 c𝐦𝟑

5) No need Check for deflection.

7 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

 b) Empirical Method.
1) Find the loads unfactored.
𝑾𝑷 = D.L + L.L (KN)
𝑾𝒒 = W.L – D.L (KN)

2) Find the value of (𝒁𝑷 ) and (𝒁𝒒 ) from table in code

(Table 27 – Empirical values of purlins).
and select (𝒁𝒓𝒆𝒒 ) the greater value.

3) Find (𝑫𝒓𝒆𝒒) from table in code

(Table 27 – Empirical values of purlins).

4) Find (𝑩𝒓𝒆𝒒) from table in code

(Table 27 – Empirical values of purlins).

5) 𝒁𝒑𝒓𝒐 → (From blue book)

(𝒁𝒑𝒓𝒐 > 𝒁𝒓𝒆𝒒 ) and (𝑫𝒑𝒓𝒐 > 𝑫𝒓𝒆𝒒) and (𝑩𝒑𝒓𝒐 > 𝑩𝒓𝒆𝒒 )

8 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Example (3): (Design using empirical method)

Design an angle for a roof with slope 1 in 2.5. The purlins are simply
supported and span 5.0 m between roof trusses at spacing of 1.6 m. The
total dead load, including purlin weight, is 0.32 KN/m2 and wind load is 0.7
KN/m2 on the slope and the imposed load is 0.6 KN/m2 on plan. Use grade
S275 steel. The arrangement of purlins on the roof slope and loading are
shown in the figure.

- Given data:
D.L = 0.32 KN / 𝐦𝟐 (on slope)
L.L = 0.6 KN / 𝐦𝟐 (on plan)
W.L = 0.7 KN / 𝐦𝟐 (on slope)
Span = 5 m , (1.6 c/c) , S275
𝟏
𝒕𝒂𝒏−𝟏 ( ) = 𝟐𝟏. 𝟖°
𝟐.𝟓
- Solution:

1) Design load:
D.L = 0.32 * 5 * 1.6 = 2.56 KN
L.L = 0.6 * 5 * 1.6 * cos(𝟐𝟏. 𝟖° ) = 4.46 KN
W.L = 0.7 * 5 * 1.6 = 5.6 KN

9 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Loading Combinations:

𝑾𝑷 = D.L + L.L = 2.56 + 4.46 = 7.02 KN

𝑾𝒒 = W.L – D.L = 5.6 – 2.56 = 3.04 KN

2) (𝒁𝑷 ) and (𝒁𝒒 ) from table in code (Table 27 – Empirical values of purlins)

𝐖𝐏 ∗ 𝐋 𝟕.𝟎𝟐 ∗ 𝟓𝟎𝟎𝟎
𝒁𝑷 = = = 19.5 c𝐦𝟑 :‫مالحظة‬
𝟏𝟖𝟎𝟎 𝟏𝟖𝟎𝟎
)mm( ‫يتم ادخال االطوال بال‬
𝐖𝐪 ∗ 𝐋 𝟑.𝟎𝟒 ∗ 𝟓𝟎𝟎𝟎
𝒁𝒒 = = = 6.76 c𝐦𝟑 )KN) ‫واألحمال بال‬
𝟐𝟐𝟓𝟎 𝟐𝟐𝟓𝟎

Take 𝒁𝒓𝒆𝒒 = 19.5 c𝐦𝟑

𝐋 𝟓𝟎𝟎𝟎
3) 𝑫𝒓𝒆𝒒 = = = 111.11 mm
𝟒𝟓 𝟒𝟓

𝐋 𝟓𝟎𝟎𝟎
4) 𝑩𝒓𝒆𝒒 = = = 83.33 mm
𝟔𝟎 𝟔𝟎

5) 𝒁𝒑𝒓𝒐 → (From blue book part (B)) → Angle → Provide ( 120 * 120 * 8 )

𝒁𝒑𝒓𝒐 = 29.5 c𝐦𝟑 > 𝒁𝒓𝒆𝒒 = 19.5 c𝐦𝟑

𝑫𝒑𝒓𝒐 = 120 mm > 𝑫𝒓𝒆𝒒 = 111.11 mm
𝑩𝒑𝒓𝒐 = 120 mm > 𝑩𝒓𝒆𝒒 = 83.33 mm

10 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

2- Cold Finished section.

1) Design loads.

𝑾𝒈𝒓𝒂𝒗𝒊𝒕𝒚 = 1.4 D.L + 1.6 L.L (KN)

𝑾𝒖𝒑𝒍𝒊𝒇𝒕 = D.L – 1.4 W.L (KN)

2) Select section from figure below.

𝑾𝒈𝒓𝒂𝒗𝒊𝒕𝒚 (𝒕𝒂𝒃𝒍𝒆) > 𝑾𝒈𝒓𝒂𝒗𝒊𝒕𝒚

𝑾𝒖𝒑𝒍𝒊𝒇𝒕 (𝒕𝒂𝒃𝒍𝒆) > 𝑾𝒖𝒑𝒍𝒊𝒇𝒕

11 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Example (4):

Select a cold-formed purlin to meet the above requirements example (3).

- Given data:
D.L = 0.32 * 5 * 1.6 = 2.56 KN
L.L = 0.6 * 5 * 1.6 * cos(𝟐𝟏. 𝟖° ) = 4.46 KN
W.L = 0.7 * 5 * 1.6 = 5.6 KN
Span = 5 m

1) Design loads:

𝑾𝒈𝒓𝒂𝒗𝒊𝒕𝒚 = 1.4 D.L + 1.6 L.L = 1.4 * 2.56 + 1.6 * 4.46 = 10.72 KN

𝑾𝒖𝒑𝒍𝒊𝒇𝒕 = D.L – 1.4 W.L = 2.56 – 1.4 * 5.6 = - 5.28 KN

From table figure above → Provide ( P145130 )

𝑾𝒈𝒓𝒂𝒗𝒊𝒕𝒚 (𝒕𝒂𝒃𝒍𝒆) = 11.76 KN > 𝑾𝒈𝒓𝒂𝒗𝒊𝒕𝒚 = 10.72 KN

𝑾𝒖𝒑𝒍𝒊𝒇𝒕 (𝒕𝒂𝒃𝒍𝒆) = 9.41 KN > 𝑾𝒖𝒑𝒍𝒊𝒇𝒕 = 5.28 KN

12 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

 Sheeting Rails.

Sheeting rails support cladding on walls and the sections used are the same
as those for the purlins.
.‫تدعم قضبان الصفائح الكسوة على الجدران واألقسام المستخدمة هي نفسها المستخدمة في المدادات‬

Loading:
Sheeting rails carry a horizontal load from the wind and a vertical one from
self-weight and the weight of the cladding. The cladding materials are the
same as used for sloping roofs (metal sheeting on insulation board). Wind
loads are estimated using BS 6399: Part 2. The wind may act in either direction
due to pressure or suction on the building walls.
ً
‫ مواد التكسية هي نفسها‬.‫وحمال رأسيًا من الوزن الذاتي ووزن الكسوة‬ ‫تحمل قضبان الصفائح حمالً أفقيًا من الرياح‬
BS 6399: ‫ يتم تقدير أحمال الرياح باستخدام‬.)‫المستخدمة في األسطح المنحدرة (الصفائح المعدنية على لوح العزل‬
.‫ قد تعمل الرياح في أي اتجاه بسبب الضغط أو الشفط على جدران المبنى‬.2 ‫الجزء‬

13 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Design of Sheeting Rails.:

1- Hot Finished section:

a) Generally Method:
- Flat roof.
- Slope roof (𝐈𝐧𝐜𝐥𝐢𝐧𝐞𝐝)
b) Empirical Method.

2- Cold Finished section.

a) Generally Method:

1) Vertical factored load = 1.4 D.L

2) Horizontal factored load = 1.4 W.L

3) Design Moments:
𝑾𝑽.𝑳 ∗ 𝐋𝟐
𝑴𝒙 = ( 𝟖
)
𝑾𝑯.𝑳 ∗ 𝐋𝟐
𝑴𝒚 = (
𝟖
)

4) Select section from blue book.

5) Moment Capacity:
𝑴𝒄𝒙 = 𝑷𝒚 * 𝒁𝒙
𝑴𝒄𝒚 = 𝑷𝒚 * 𝒁𝒚

6) Buckling moment = (𝑴𝒃 = 0.8 𝑴𝒄𝒙 )

𝑴𝒙 𝑴𝒚
7) +𝑴 <1
𝑴𝒃 𝒄𝒚

14 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Example (5): (Design of an angle sheering rail)

A simply supported sheeting rail spans 5m. The rails are at 1.5m centers. The
total weight of cladding and self-weight of rail is 0.32 kN/m2. The wind loading
on the wall is ±0.5 kN/m2. The wind load would have to be carefully estimated
for the particular building and the maximum suction and pressure may be
different. The sheeting rail arrangement is shown in Figure. Use Grade S275
steel.

- Given data:
D.L = 0.32 KN / 𝐦𝟐 , W.L = ± 0.5 KN / 𝐦𝟐
Span = 5 m , (1.5 c/c) , S275
Loading:
D.L = 0.32 * 1.5 = 0.48 KN / m
W.L = 0.5 * 1.5 = 0.75 KN / m

1) Vertical factored load = 1.4 D.L = 1.4 * 0.48 = 0.672 KN / m

2) Horizontal factored load = 1.4 W.L = 1.4 * 0.75 = 1.05 KN / m

15 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

3) Design Moments:
𝑾𝑽.𝑳 ∗ 𝐋𝟐 𝟎.𝟔𝟕𝟐 ∗ 𝟓𝟐
𝑴𝒙 = ( 𝟖
) = = 2.1 KN.m
𝟖
𝑾 ∗ 𝐋𝟐 𝟏.𝟎𝟓 ∗ 𝟓𝟐
𝑴𝒚 = ( 𝑯.𝑳 ) = = 3.28 KN.m
𝟖 𝟖
4) Select section:
𝑴 𝟑.𝟐𝟖 ∗ 𝟏𝟎𝟑
𝒁𝒓𝒆𝒒 = = = 11.93 c𝒎𝟑
𝑷𝒚 𝟐𝟕𝟓

Angle ( 100 * 100 * 10 ) → 𝒁𝒑𝒓𝒐 = 24.6 c𝒎𝟑

5) Moment Capacity: (𝒁𝒙 = 𝒁𝒚 )

𝑴𝒄𝒙 = 𝑷𝒚 * 𝒁𝒙 = 275 * 24.6 * 𝟏𝟎−𝟑 = 6.765 KN.m

𝑴𝒄𝒚 = 𝑷𝒚 * 𝒁𝒚 = 275 * 24.6 * 𝟏𝟎−𝟑 = 6.765 KN.m

6) Buckling moment → 𝑴𝒃 = 0.8 𝑴𝒄𝒙 = 0.8 * 6.765 = 5.412 KN.m

𝑴𝒙 𝑴𝒚 𝟐.𝟏 𝟑.𝟐𝟖
7) +𝑴 <1 → + 𝟔.𝟕𝟔𝟓 = 0.87 < 1
𝑴𝒃 𝒄𝒚 𝟓.𝟒𝟏𝟐

16 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

 b) Empirical Method.

1) Vertical unfactored load = D.L (KN)

2) Horizontal unfactored load = W.L (KN)

3) Find the value of (𝒁𝟏 ) and (𝒁𝟐 ) from table in code

(Table 28 – Empirical values of side rails).
and select (𝒁𝒓𝒆𝒒 ) the greater value.

𝒁𝟏 = Load in Horizontal (KN) , 𝒁𝟐 = Load in Vertical. (KN)

4) Find (𝑫𝒓𝒆𝒒) from table in code. (Table 28 – Empirical values of side rails).

5) Find (𝑩𝒓𝒆𝒒) from table in code. (Table 28 – Empirical values of side rails).

6) 𝒁𝒑𝒓𝒐 → (From blue book). (𝒁𝒑𝒓𝒐 > 𝒁𝒓𝒆𝒒 ) and (𝑫𝒑𝒓𝒐 > 𝑫𝒓𝒆𝒒 ) and (𝑩𝒑𝒓𝒐 > 𝑩𝒓𝒆𝒒 ).

17 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Example (6):

Design using empirical method above requirements example (5).

- Given data:
D.L = 0.32 KN / 𝐦𝟐 , W.L = ± 0.5 KN / 𝐦𝟐
Span = 5 m , (1.5 c/c) , S275
Loading:

1) Vertical unfactored load = 0.32 * 5 * 1.5 = 2.4 KN

2) Horizontal unfactored load = 0.5 * 5 * 1.5 = 3.75 KN

𝐖𝐇 ∗ 𝐋 𝟑.𝟕𝟓 ∗ 𝟓𝟎𝟎𝟎
𝒁𝟏 = = = 8.33 c𝐦𝟑
𝟐𝟐𝟓𝟎 𝟐𝟐𝟓𝟎

𝐖𝐯 ∗ 𝐋 𝟐.𝟒 ∗ 𝟓𝟎𝟎𝟎
𝒁𝟐 = = = 10 c𝐦𝟑
𝟏𝟐𝟎𝟎 𝟏𝟐𝟎𝟎

Take 𝒁𝒓𝒆𝒒 = 10 c𝐦𝟑

𝐋 𝟓𝟎𝟎𝟎
3) 𝑫𝒓𝒆𝒒 = = = 111.11 mm
𝟒𝟓 𝟒𝟓

𝐋 𝟓𝟎𝟎𝟎
4) 𝑩𝒓𝒆𝒒 = = = 83.33 mm
𝟔𝟎 𝟔𝟎

5) 𝒁𝒑𝒓𝒐 → (From blue book part (B)) → Angle → Provide ( 120 * 120 * 8 )

𝒁𝒑𝒓𝒐 = 29.5 c𝐦𝟑 > 𝒁𝒓𝒆𝒒 = 19.5 c𝐦𝟑

𝑫𝒑𝒓𝒐 = 120 mm > 𝑫𝒓𝒆𝒒 = 111.11 mm
𝑩𝒑𝒓𝒐 = 120 mm > 𝑩𝒓𝒆𝒒 = 83.33 mm

18 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

 2- Cold Finished section.

1) Horizontal factored load = 1.4 W.L (KN)

2) Select section from figure below.

Ultimate pressure > Horizontal factored load.

Ultimate Suction > Horizontal factored load.

19 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Example (7):

Select a cold-rolled sheeting rail to meet the above requirements example (6).

- Given data:
D.L = 0.32 KN / 𝐦𝟐 , W.L = ± 0.5 KN / 𝐦𝟐
Span = 5 m , (1.5 c/c) , S275

1) Horizontal factored load = 1.4 W.L = 1.4 * 0.5 * 1.5 * 5 = 5.25 KN

2) Select from table figure above → Provide ( P145130 )

Ultimate pressure = 11.763 KN > 5.25 KN

Ultimate Suction = 9.410 KN > 5.25 KN

20 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Steps for solution Compression member:

1) If design select section

𝐅𝐜
𝐀𝐫𝐞𝐪 = 𝐏𝐲
2) Section classification.

3) Determine effective length.

𝑳𝑬𝒚 ≥ 𝑳𝑬𝒙 → 𝑪𝒉𝒆𝒄𝒌 𝒐𝒏𝒍𝒚 (𝒚 − 𝒚)

4) 𝐈𝐟 = {
𝑳𝑬𝒚 < 𝑳𝑬𝒙 → 𝑪𝒉𝒆𝒄𝒌 𝒃𝒐𝒕𝒉 (𝒙 − 𝒙) 𝒂𝒏𝒅 (𝒚 − 𝒚)
Take value smaller.

5) Axial capacity → 𝐏𝐜 = 𝐩𝐜 ∗ 𝐀 𝐠

21 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Examination – Test:

The column shown in figure is pin-ended about both axis and has no
intermediate restraint. Design the column with S275 for the factored load
shown.

- Solution:

1) section selection

𝐅𝐜 𝟐𝟓𝟎𝟎 ∗ 𝟏𝟎𝟑
𝐀𝐫𝐞𝐪 = = = 9090.90 m𝐦𝟐
𝐏𝐲 𝟐𝟕𝟓
𝐀𝐫𝐞𝐪 = 90.9090 c𝐦𝟐

Blue book Try (356 * 368 * 129)UC

𝐀𝐠 = 164 c𝐦𝟐

- Properties:
𝐀𝐠 = 164 c𝐦𝟐 , b/T = 10.5 , d/t = 27.9 , 𝐫𝐲 = 9.43 , 𝐫𝐱 = 15.6 , T = 17.5 mm

2) Section classification:
T = 17.5 mm > 16 mm ( reduction 𝐏𝐲 = 265 N/m𝐦𝟐 )

𝟐𝟕𝟓 𝟎.𝟓 𝟐𝟕𝟓

𝛆=( ) = ( )𝟎.𝟓 = 1.02
𝐏𝐲 𝟐𝟔𝟓

Flange → b/T = 10.5 > 9 𝛆 = 9 * 1.02 = 9.18 … Class 1.

> 10 𝛆 = 10 * 1.02 = 10.2 … Class 2.
< 15 𝛆 = 15 * 1.02 = 15.3 … Class 3.
∴ The flange class 3.

22 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Web → d/t = 27.9 (1) Not applicable. (N.A)

(2) Not applicable. (N.A)

𝐅 𝟐𝟓𝟎𝟎 ∗ 𝟏𝟎𝟑
𝐫𝟐 = =
𝟏𝟔𝟒 ∗ 𝟏𝟎𝟐 ∗ 𝟐𝟔𝟓
= 0.58
𝐀𝐠 ∗ 𝐏𝐲

𝟏𝟐𝟎 𝛆 𝟏𝟐𝟎 ∗ 𝟏.𝟎𝟐

(3) = = 56.7 > 27.9 (web is class 3)
𝟏+𝟐 𝐫𝟐 𝟏+ 𝟐 ∗ 𝟎.𝟓𝟖

∴ The section class 3.

3) Effective length:
𝑳𝑬𝒙 = 1 L = 1 * 6 m = 6 m = 600 cm
𝑳𝑬𝒚 = 1 L = 1 * 6 m = 6 m = 600 cm

𝑬𝒚 𝟔𝟎𝟎 𝒄𝒎 𝑳
4) 𝑳𝑬𝒚 = 𝑳𝑬𝒙 → 𝑪𝒉𝒆𝒄𝒌 𝒐𝒏𝒍𝒚 (𝒚 − 𝒚) ,  = = = 𝟔𝟑. 𝟔
𝒓𝒚 𝟗.𝟒𝟑 𝒄𝒎
Strut Curve → UC → ≤ 40 , (Table 24 , C ,  < 110)

 𝒑𝒄 = 265
𝟔𝟐 −𝟔𝟑.𝟔 𝟏𝟗𝟏− 𝒙
62 191 = = x = 𝒑𝒄 =188.6 N/m𝐦𝟐
𝟔𝟐 − 𝟔𝟒 𝟏𝟗𝟏 − 𝟏𝟖𝟖
63.6 X
64 188

5) 𝐏𝐜 = 𝐩𝐜 ∗ 𝐀 𝐠 = 188.6 * 𝟏𝟎−𝟑 * 164 * 𝟏𝟎𝟐 = 3039.04 KN > 𝐅𝐜 = 2500 KN

(Safe)

23 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Examination – Test:

The column shown in figure has length as shown and axial loading as shown
in fig with addition of tie at the mid high providing restraint (y-y)

- Solution:

1) section selection

𝐅𝐜 𝟐𝟓𝟎𝟎 ∗ 𝟏𝟎𝟑
𝐀𝐫𝐞𝐪 = = = 9090.90 m𝐦𝟐
𝐏𝐲 𝟐𝟕𝟓
𝐀𝐫𝐞𝐪 = 90.9090 c𝐦𝟐

Blue book Try (305 * 305 * 97)UC

𝐀𝐠 = 123 c𝐦𝟐

- Properties:
𝐀𝐠 = 123 c𝐦𝟐 , b/T = 9.91 , d/t = 24.9 , 𝐫𝐲 = 7.69 , 𝐫𝐱 = 13.4 , T = 15.4 mm

2) Section classification:
T = 15.4 mm < 16 mm ( NO reduction 𝐏𝐲 = 275 N/m𝐦𝟐 )

𝟐𝟕𝟓 𝟎.𝟓 𝟐𝟕𝟓

𝛆=( ) = ( )𝟎.𝟓 =1
𝐏𝐲 𝟐𝟕𝟓

Flange → b/T = 9.91 > 9 𝛆 = 9 * 1 = 9 … Class 1.

< 10 𝛆 = 10 * 1 = 10 … Class 2.

∴ The flange class 2.

24 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.

Web → d/t = 24.9 (1) Not applicable. (N.A)

(2) Not applicable. (N.A)

𝐅 𝟐𝟓𝟎𝟎 ∗ 𝟏𝟎𝟑
𝐫𝟐 = =
𝟏𝟐𝟑 ∗ 𝟏𝟎𝟐 ∗ 𝟐𝟕𝟓
= 0.74
𝐀𝐠 ∗ 𝐏𝐲

𝟏𝟐𝟎 𝛆 𝟏𝟐𝟎 ∗ 𝟏
(3) = = 48.39 > 24.9 (web is class 3)
𝟏+𝟐 𝐫𝟐 𝟏+ 𝟐 ∗ 𝟎.𝟕𝟒

∴ The section class 3.

3) Effective length:
𝑳𝑬𝒙 = 1 L = 1 * 6 m = 6 m = 600 cm
𝑳𝑬𝒚 = 1 L = 1 * 3 m = 3 m = 300 cm

4) 𝑳𝑬𝒚 < 𝑳𝑬𝒙 → 𝑪𝒉𝒆𝒄𝒌 𝒇𝒐𝒓 𝒃𝒐𝒕𝒉 𝒂𝒙𝒊𝒔

𝑳𝑬𝒚 𝟑𝟎𝟎 𝒄𝒎 𝑳𝑬𝒙 𝟔𝟎𝟎 𝒄𝒎

𝒚 = = = 𝟑𝟗. 𝟎𝟏 𝒙 = = = 𝟒𝟒. 𝟕𝟕
𝒓𝒚 𝟕.𝟔𝟗 𝒄𝒎 𝒓𝒙 𝟏𝟑.𝟒 𝒄𝒎
UC → ≤ 40 , (Table 24 , C ,  < 110) UC → ≤ 40 , (Table 24 , b ,  < 110)
(y-y) (x-x)

𝒑𝒄 = 275  𝒑𝒄 = 275
 𝟑𝟓 −𝟑𝟗.𝟎𝟏 𝟐𝟒𝟕− 𝒙 𝟒𝟒 −𝟒𝟒.𝟕𝟕 𝟐𝟒𝟓− 𝒙
= 44 245 =
35 247 𝟑𝟓 − 𝟒𝟎 𝟐𝟒𝟕 − 𝟐𝟑𝟖 𝟒𝟒 − 𝟒𝟔 𝟐𝟒𝟓 − 𝟐𝟒𝟐

39.01 X 44.77 X
40 238 x = 𝒑𝒄𝒚 =239.78 N/m𝐦𝟐 46 242 x= 𝒑𝒄𝒙 =243.85 N/m𝐦𝟐

Take 𝒑𝒄𝒚 = 239.78 N/m𝐦𝟐

5) 𝐏𝐜 = 𝐩𝐜 ∗ 𝐀 𝐠 = 239.78 * 𝟏𝟎−𝟑 * 123 * 𝟏𝟎𝟐 = 2949.04 KN > 𝐅𝐜 = 2500 KN

(Safe)

25 Eng: Mohammed Nageb Haza’a Mohammed. B4C / E.