CIE 119
Principles of Steel
Design
Topics:
❑ Introduction to Steel Design
❑ Determination of the Cross-sectional
Properties of Structural Members
❑ Analysis and Design of Members for
Bending
Getting to know…
ENGR. ROBERT V. MANAOIS JR.
REGISTERED CIVIL ENGINEER- NOVEMBER 2018
Introduction to
Steel Design
Steel is widely used in construction
especially in high rise building as
structural member because of its
strength, cost, and durability.
31,400 tonnes
of steel
57,000 tonnes
of steel
w/ 18,000
metallic parts
joined by 2.5
million rivets
Structural Steel Shapes
DESIGNATION
Web
W 18 X 50
Flange
S 18 X 70
Weight per
Length , (lb/ft)
WT 18 X 105
L 6 X 6 X 3/4
Length of two legs
and its thickness
(in.)
Moment Inertia of Built-up Section
Sample Problem 1
Compute the moment of inertia about neutral axis both in xaxis and y-axis.
10 mm
16 mm
3
2
16 mm
549 mm
Solution:
➢
At x-axis
533 mm
X
1050 mm
Ix = I1 + 2(I2 + I3)
1
I1 = 12 (10 x 10503) = 964.688x106 mm 4
1
16 mm
16 mm
250 mm
300 mm
Y
By transformed area method,
1
I2 = 12 (300 x 163) + 16(300)(533)2 = 1,363.7296x106 mm4
1
I3 = 12 (250 x 163) + 16(250)(549)2 = 1,205.689x106 mm4
Ix = 964.688x106 mm 4 + 2(1,363.7296x106 mm4
+ 1,205.689x106 mm4)
Ix = 6,103.525x106 mm4
Moment Inertia of Built-up Section
Sample Problem 1
Compute the moment of inertia about neutral axis both in xaxis and y-axis.
16 16
Solution:
16 16
1050 mm
➢
3
2
At y-axis
Iy = I1 + 2(I2 + I3)
10
1
X
250 300
Y
I1 =
1
12
(103 x 1050) = 87.5x103 mm 4
I2 =
1
12
(3003 x 16) = 36x106 mm4
I3 =
1
12
(2503 x 16) = 20.833x106 mm4
Iy = 87.5x103 mm 4 + 2(36x106 mm4
+ 20.833x106 mm4)
Iy = 113.754x106 mm4
Moment Inertia of Built-up Section
Sample Problem 2
Find the ff:
ў 152.5
1.
Location of neutral axis of the built-up section measured from
the top of the flange of the channel
2.
Moment of inertia about neutral axis in x-axis
3.
Section modulus of the built-up section
285.10
NA
2
19.9
1
Built up section
Y
Solution:
Channel
Aў = A1y1 + A2y2
A = 3,929 mm2
X
d
d = 305 mm
Ў = 190.872 mm
Ix = 53.7x106 mm4
Y
INA = I1 + I2 ; use transformed area method
Angular
A = 1,600
mm2
X
y
x = 19.9 mm
x
(3929+1600) ў = 3929(152.5) + 1600(285.10)
Iy = 0.799x106 mm4
INA = 53.7x106 + 3929(190.872-152.5) 2
+ 0.799x106 + 1600(305-190.872-19.9) 2
INA = 74.490x106 mm4
S= 𝐼
𝐶
S = 74.49x106 / 190.872
S = 390.262x103 mm3
Bending Stresses
Y
tf
d
tw
X
bf
Symbols and Notations
Fb -allowable bending stress
bf - base of compression flange
Fy - yield strength of steel
tf - thickness of compression flange
tw - thickness of the web
rt - radius of gyration of
compression flange plus 1/3 of the
compression about y-axis
d – depth of the section
Lb – unbraced or unsupportedlength
Bending Stresses
Process in solving bending stress:
1.
Check if the section is compact using the
ff. criteria:
a.
Its flanges must be continuously
connected to the web.
b.
Flange width-thickness ratio:
bf 170
≤
𝟐t f
√𝑭𝒚
c.
Depth to web thickness ratio:
d 1680
≤
tw √𝑭𝒚
If the section conforms with these criteria,
then the steel member is compact. If not,
Fb = 0.6Fy
2. Check if it is laterally supported
L1 =
𝟐𝟎𝟎bf
√𝑭𝒚
L2 =
𝟏𝟑𝟕,𝟗𝟎𝟎
𝑭𝒚𝒅
bftf
where Lc is the smaller value and Lu is the
larger value
If Lb ≤ Lc, laterally supported, then Fb =
0.66 Fy
Lc < Lb < Lu, partially laterally supported,
then Fb = 0.60Fy
Lb > Lu, laterally unsupported, proceed to
step 3. laterally supported
Bending Stresses
3. For laterally unsupported, the bending
stress will be computed as follows:
value of
𝑳𝒃
rt
Case 1:
𝟕𝟎𝟑𝟐𝟕𝟎𝑪𝒃
𝑭𝒚
Case 2:
<
𝑳𝒃
rt
<
𝟑𝟓𝟏𝟔𝟑𝟑𝟎𝑪𝒃
𝑭𝒚
>
𝟑𝟓𝟏𝟔𝟑𝟑𝟎𝑪𝒃
𝑭𝒚
, Fb is the larger value of
Fb =
, Fb is the larger
Fb = Fy
Fb =
-
𝑭𝒚
𝟏𝟎.𝟓𝟓𝒙𝟏𝟎𝟔 𝑪𝒃
𝟖𝟐𝟕𝟒𝟎 𝑪𝒃
𝑳𝒃𝒅
bft𝒇
But Fb < 0.60 Fy
]
𝟏𝟏𝟕𝟐𝟏𝟎𝟎 𝑪𝒃
𝑳𝒃 𝟐
rt
𝑳 𝟐
rt
𝑳𝒃𝒅
bft𝒇
Fb =
𝟐
[
𝟑
𝟖𝟐𝟕𝟒𝟎 𝑪𝒃
But Fb < 0.60 Fy
where,
𝑴
𝑴
Cb = 1.75 + 1.05 𝟏 + 0.3 ( 𝟏 )2 but not more
𝑴𝟐
𝑴𝟐
than 2.3
𝑀1 - smaller bending moment at the ends of
the unbraced length taken about the strong
axis
Bending Stresses
𝑀2 - larger bending moment at the ends of
the unbraced length taken about the strong
axis
𝑀1
𝑀2
is positive when the beam is on reverse
curvature bending
𝑀1
𝑀2
𝑀1
𝑀2
is negative when the beam is on single
curvature bending
𝑀1
𝑀2
4. For partially compacted section about
170 bf 250
strong axis, where
<
<
√𝑭𝒚
𝟐t f
√𝑭𝒚
Fb = Fy [ 0.79 -0.000762
bf
𝑭𝒚 ]
𝟐t f
5. For compact section bending about weak
axis,
Fb = 0.75 Fy
Bending Stresses
Sample Problem 1
Given:
Solution:
W 360 x 91 Properties
➢
Check for compact section
d = 353 mm
bf = 254 mm
bf 170
≤
𝟐t f
√𝑭𝒚
d 1680
≤
ok!
tw √𝑭𝒚
Therefore, compact section.
➢
Check for laterally supported
tf = 16.4 mm
bf
254
=
= 7.744
2tf 2(16.4)
L1 =
tw = 9.53 mm
170
Fy = 248 MPa
√𝐹𝑦
L2 =
bf 170
<
ok!
𝟐tf √𝑭𝒚
d 1680
≤
tw √𝑭𝒚
L = 7.5 m
Cb =1.0
Required:
Fb if a) Lb is 2.5 m
b) Lb is 3.75 m
c) Lb is 7.5 m
= 10.795
d 353
=
= 37.041
tw 9.53
1680
= 106.68
√𝐹𝑦
mm
200bf
√𝐹𝑦
=
137,900
𝐹𝑦𝑑
bftf
200(254)
√248
=
= 3225.803 mm
137,900
248(353)
= 6561.686
254(16.4)
Therefore, Lc = 3225.803 mm and
Lu = 6561.686 mm
a)
Lb = 2500 mm
Lb < Lc ; laterally supported
Fb = 0.66Fy = 0.66(248) = 163.68
MPa
Y
Bending Stresses
tf
Sample Problem 1
Given:
W 360 x 91 Properties
d = 353 mm
bf = 254 mm
tf = 16.4 mm
tw = 9.53 mm
Fy = 248 MPa
L = 7.5 m
d
Solution:
bf
Y
X
Iy =
𝟏 𝟏
𝟏
(t
𝟏𝟐 f
[ (
bf
𝟏𝟐 𝟑
)(bf)3 +
d−2tf
)][tw]3
𝟐
𝐼𝑦
𝐴
Fb = 0.6Fy = 0.6(248)
rt =
Fb = 148.8 MPa
𝑑−2tf
A = bf tf + tw (
) = 254(16.4) +
6
353−2(16.4)
9.53(
) = 4,474.184 mm2
c) Lb = 7500mm
Required:
➢
c) Lb is 7.5 m
tw
Lc < Lb < Lu ; partially
supported
Lb > Lu ; unsupported
b) Lb is 3.75 m
𝟏 d-2tf
(
)
𝟑
𝟐
b) Lb = 3750 mm
Cb = 1.0
Fb if a) Lb is 2.5 m
tf
Solve for rt,
(1/3 of the compression
about y-axis)
6
1 1 d−2tf
[ (
)][tw]3
12 3
2
Iy =
1
(t
12 f
Iy =
1
(16.4)(254)3
12
)(bf )3 +
+
1 1 353−2(16.4)
[ (
)][9.53]3
12 3
2
Iy = 22.3995x106 mm4
Bending Stresses
Sample Problem 1
Given:
Solution:
W 360 x 91 Properties
𝐼𝑦
𝐴
22.3995x106
=
4,474.184
d = 353 mm
rt =
bf = 254 mm
rt = 69.226
tf = 16.4 mm
tw = 9.53 mm
Fy = 248 MPa
𝐿𝑏
rt
=
Fb = Fy
= 108.341
69.226
703270𝐶𝑏
𝐹𝑦
= 53.252
Required:
Fb if a) Lb is 2.5 m
b) Lb is 3.75 m
c) Lb is 7.5 m
3516330𝐶𝑏
𝐹𝑦
703270𝐶𝑏
𝐹𝑦
-
2
[
3
𝐹𝑦
bft𝑓
rt
10.55𝑥106 𝐶𝑏
-
82740 𝐶𝑏
𝐿𝑏𝑑
𝐿 2
248 108.341
10.55𝑥106
]
2
] = 96.905 MPa
82740
= 7500(353) = 130.184 MPa
254(16.4)
Fb = 130.184 MPa < 0.6Fy ok!
L = 7.5 m
Cb =1.0
Fb = 248
Fb =
7500
2
[
3
= 119.075
<
𝐿𝑏
rt
<
Therefore, Case 1
3516330𝐶𝑏
𝐹𝑦
Seatwork
Located in your Student’s Activity
Sheet Module # 2
•
Will be submitted to our Google
Classroom that I will create later after
the our discussion
•
To be submitted at the end of our
class hour ( 10:30 AM )
Next meeting…
More Examples
in Bending
Stresses
Compression
Members Part 1
Problem Set
SAS Module 4