Post on 08-Feb-2022
� Cover-plated Beams
� Built-up W Sections
� Plate Girder
� Stiffeners
Timber and Steel DesignTimber and Steel Design
Mongkol JIRAVACHARADET
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Lecture Lecture 1188 BuiltBuilt--up Beamsup Beams
Cover-plated Beams
d
tp
tp
2
22
reqd s
dI I A
= +
22 ( / 2)
/ 2reqd s
s
A dS S
d
S Ad
= +
= +
��������� 17.1 ������������ ���� ������������ 60 � . ��������������������� !"�������#�$% � �&'�"��� �����������(�)(����"��� �$�*+�������%,������������ ��"������������� #�(-�"���� 1,400 ��./� .2 #!(�(�&�� W600�137
�� �6"-��7���( 12 t/m
7 m
����� � �&'�"������ = 180 ��./- &�
M = (12.18)(7)2/8 = 74.6 &�-- &�
Sreqd = (74.6�105)/1,400 = 5,329 � .3
������� W600�137 (I = 103,000 ��.4 S = 3,530 ��.3)
58.2 cm 59.0 cm59.8 cm <
max. 60 cm
����*�"� S �����(��'�:
S = (103,000 + 2(36)(29.5)2)/29.9
= 5,540 � .3 > 5,329 � .2 OK
#$�������� W600 � 137 %��� PL8 � 450 ��. )�*��+,-./
� �&'�"�#!(�6"-��7��� 8 .
Sreqd = Ss + A d
5,329 = 3,530 + A (59.0)
A = 30.5 � .2
+��#$�*0��1�+2/ PL 8 � 450 ��. )�*��+,-./ (A = 36 ��.2)
Built-up wide-flange sections
h
Af = area of flange
tw
tf
Built-up W sections: / 6,360 /w bh t F≤
/ 6,360 /w bh t F>Plate girder:
��������� 17.2 ���������� W %� ������ 1.50 - &���B " ����#!(�6"-��' ���������)����������!"����-����� 20 - &� �������������6"� ���-� � 4 &�/- &� �����������)��&��-�� � �&'�"� �����������(�)(��� �$�*+�������%,���������� #!(-��7� A36 ���-!D�� ���EF�-�& �� ���-!D�� E70
d/t = 150/1.5 = 100 < 107 OK
5,355 5,355107
2,500y
d
t F≤ = =
����� - &+�� ���-HD� �������� ��� �&'�� ������� = 300 ��./- &�
M = (4.3)(20)2/8 = 215 &�-- &�
V = (10)(4.3) = 43 &�
)��)���6"-�����&(������������ Compact section (&������� 6-1)
Min tw = 150/107 = 1.4 � . (#$� 1.5 ��.)
+��#$�*0��1�� 1.5 � 150 ��. (Aw = 225 ��.2)
2
31 22 212
2 2
ffw
reqd
f f
thAt h
Sh h
t t
+ = +
+ +
�4�56�)�����*4�178��:
fv = 43,000/(1.5�150) = 191 ��./� .2
kv = 5.34 or = 0.676 < 0.8
Fv = 2,500/(2.89�0.676)=1,280 ��./� .2 > 191 ��./� .2
Since h/tw < 260 (AISC) and web shear stress < max. value
Stiffeners are not needed
/�41+8�/*0��-./:
Sreqd = 215 x 105 / 1,650 = 13,030 � .3
( )2
3 146 21 2(1.5) 1462 21213,030
146 1462 2
2 2
fA + = +
+ +
� �&' tf = 2 � .
Af = UD����)��%,�����)(�� = 53.7 � .2 (#$� 2××××28 ��.)
1+8�/#$�*0��1�� 1.5××××146 ��. *+,*0��-./ 2××××28 ��.
PLATE GIRDERS
Riveted or bolted
WT
WT
PL
Welded Welded box girder
PL girder (may be for
full depth of story)
Plate Girder Design
Intermediate
stiffeners do not
have to run all the
way to the tension
flange
Stiffeners
Allowable Bending Stresses < 0.66Fy
Allowable Shear Stresses < 0.40Fy
Intermediate Stiffeners
Proportions of Plate Girders
Web Size:Vertical component in
compression flange
Web
Vertical component in
tension flange
1. No transverse stiffeners or stiffeners spacing > 1.5 h
( )160,1
000,984
+≤
yfyfw FFt
h
2. Use transverse stiffeners with spacing 1.5h≤
yfw Ft
h 670,16≤
Depth: about 1/6 - 1/15 of span
Flange:
h
Af
Af
tw
23
212 2
wf
t h hI A
= +
( )23
3
2 / 2/12
/ 2 / 2
6
fw
wf
A ht hS
h h
t hA h
= +
= +
Equating required section modulus to S :
2
6
wf
b
t hMA h
F= +
6
wf
b
t hMA
F h= −
0.60f
y
MA
F h≈
Assume Fb = 0.60Fy:
Allowable Bending Stress
′ ≤b b PG eF F R R
No reduction in Fb due to unbraced length because of large size of PL girder
6,360where 1 0.005 1.0w
PG
f w b
A hR
A t F
= − − ≤
Girder
xb
x
Mf
S=
Increase in compression
flange stress due to lateral
buckling of web
Dotted line shows a linear
stress variation while solid
curved line shows actual variation
( )212 3
1.0
12 2
w
f
e
w
f
A
AR
A
A
α α
+ − = ≤
+
where α = 0.6 Fyw /Fb � 1.0
Tension Field Action
Pratt Truss
Weak tension field
in end panels
Tension field
action
Intermediate Stiffeners
when h/tw > 260 ( ) yv
y
v FCF
F 40.089.2
≤=
where( )
and 0.8 than less is when /
000,165,32 v
wy
v
v CthF
kC =
0.8 than more is when /
585,1v
y
v
w
CF
k
th=
( )and 1.0 than less is when
/
34.500.4
2 h
a
hakv +=
( )1.0 than more is when
/
00.434.5
2 h
a
ha+=
If include tension field action, allowable shear stress may be increased to:
( )y
v
v
y
v Fha
CC
FF 40.0
/115.1
1
89.2 2≤
+
−+=
Panel aspect ratio: 0.3/
260/
2
≤
≤
wthha
Moment of inertia:4
50
≥h
I st
Total area: ( )( )
w
v
st YDhtha
ha
h
aCA
+−
−=
2
2
/1
/
2
1
- D�� Y = ��&���"� Fyw &"� Fy )���6"-��' ������
D = 1.0 ��������6"-��' ������-%V�$"
= 1.8 ��������6"-��' ������-�����-%V#!(-��7�H��
= 2.4 ��������6"-��' ������-�����-%V#!(�6"-��7�
3
34012.0
= yw
vs
Fhf
Shear transfer between stiffener and web (due to tension field action)
Bearing Stiffeners
Width outside of fillets
Chamfer
Filler PLs
Stiffener LStiffener
PLs
Corners are cut to avoid
flange-to-web weld.
Difficult to analyze because support load with web
Portion of web to support load: 12tw at girder ends
25tw at interior concentrated loads
Effective length of bearing stiffener columns: KL = 0.75 h
��������� 17.3 ��������%� ���!"��-����� 15 - &� ����������6" 2.2 &�/- &� �� ������
32 &� ���� � 1/3 !"���� �����������(�)(���������%,����������������������%W'�'�'���� ���
�������� ���-%V��� #!(-��7� A36 �� UX&'��� �� ������
����� � �&'�������� = 300 ��./- &�, -)���6Y$ '���-HD��� - &+���������
32 ton 32 ton
2.5 ton/m
5 m 5 m 5 m50.75t 50.75t
50.75t
50.75t
38.25t
6.25t
38.25t
6.25t
222.5t-m 222.5t-m230.3t-m
6,360 6,360164
1,500w b
h
t F= = =
( )984,000 984,000
3252,500(2,500 1,160)1,160w yf yf
h
t F F≤ = =
++
142w
h
t=
��/*))*0��1��1)8;�����: � �&'��� ���)���� = (1/10)(1500) = 150 � .
� �&'��� ���)��-�� = 150 Z 8 = 142 � .
��&���"� h/tw ��������)��-�����B "&(�����"������������� #�(�D�
tw = 142/164 = 0.866 � .
[(�B " ��6"-��' ���������)�����D�[(� ��&"�"�� ����"� 1.5 -�"�� � � ��"��%,���
tw = 142/325 = 0.437 � .
��� ��(��������-UD��%\������������"� tw = 0.8 � .
+��#$�*0��1�� 1 � 142 ��. (Aw = 142 ��.2)
( )2230.3(100)108.1 cm
(1.5)(142)f
b
MA
F h≈ = = ×USE 3 40 cm
��/*))-./<��1)8;�����:
������������� � = (2�3�40+1�142)(7,850)/1002
= 299.87 � 300 ��./- &� OK
�4�56�)/�4%/��1��,17=�,� ���&������� 6-1 �������)����������&���"���� ��(��&"���� ������������+��������������
( )0.46 0.46
4.05 4.050.414
142/ck
h t= = =-D������ h/t > 70 �����
795 79510.23
/ 2,500 /0.414yf c
b
t F k= = =��������(�&��B "�� ������
406.67 10.23
2 2(3)
f
f
b
t= = < OK��&���"�)���(�&�����#!(
<>?6�)���@���������*+,�4�56�)�����*4�:
142 cm
1 cm
3 cm
148 cm
3 cm
40 cm
142 6,360 6,360Since 142 164 , 1.0
1 1,500PG
w b
hR
t F
= = < = = =
Flange and web are made of A36: Re = 1.0
I = (1/12)(1.0)(142)3+(2)(3x40)(72.5)2
= 1,500,107 � .4
S = 1,500,107/74 = 20,272 � .3
fb �������!"���� = (230.3x105)/20,272
= 1,136 ��./� .2
b b PG e bF F R R F′ = =
����� Fb ������&����������"�-�� �(�)(���D� 0.6Fy = 1,500 ��./� .2
(-D������-��B "�� �������"� Fb ���&(��B "-�' 0.6Fy)
37,173 1050047.39 53.56
10.55 2,500
b
T
CL
r
× ×= = < =
����* Iy �� rT )��%,��� + 1/6 �6"-��
16,000
143.7
Iy = (1/12)(3)(40)3 = 16,000 � .4
Af + 1/6 web = (3)(40) + (1/6)(1)(142) = 143.7 � .2
rT = = 10.55 � .
#!(�"� Cb = 1.0 -D������ - &+ ���������������!"����(230.3 &�-- &�)���-�'�)��
*. ������B " ����������� ��"� ����"� - &+(222.5 &�-- &�)���%����������
*0��164��/��+��*)/����;�����//4,��1-A�5>�:
3 cm
0.6 cm
k=3.6 cm
32,000
( 5 ) 1.0(0 5 3.6)
1,778 ksc 0.66 1,650 ksc
w
y
R
t N k
F
=+ + ×
= > =
Check web yielding for interior loads
Bearing stiffeners are required
�"�����-HD����%�����
fv = 50.75(1,000)/(142x1) = 357 ��./� .2
a/h = 500/142 = 3.52
h/tw = 142 > 3,179 / 2,500 63.58=
/�4��/*))*0��164��/��+��4,�����$���:
���������&���"� a/h > 1.0
( )2 2
4.00 4.005.34 5.34 5.66
3.52/vk
a h= + = + =
( )2 2
3,165,000 3,165,000(5.66)0.355
(2,500)(142)/
vv
y w
kC
F h t= = =
2,500(0.355)
2.89 2.89
307 ksc 357 ksc
y
v v
FF C= =
= < NG
USE intermediate stiffeners
��&����"��6"-��' ������&�����#!(&���� �.3 �� Fv = 357 ksc, h/tw = 142,
a/h = 1.81 a = 1.81(142) = 257 � . (#$� 2.5 1��4)
*0��164��/��+��*0��*4/����5�/-+��<�� 2.5 1��4
a/h = 250/142 = 1.76
&��������� &(������6"-��' ������� ��"��!"��-U'� -&' #!"�����
V = 50.75 – 2.5(2.5) = 44.5 &�
fv = (44.5)(1,000)/(142) = 313.4 ��./� .2
� � ����6"-��' �������6"���[��������������-%V��� = 5 Z 2.5 = 2.5 - &�
( )2 2
4.00 4.005.34 5.34 6.63
1.76/vk
a h= + = + =
( )2 2
3,165,000 3,165,000(6.63)0.416
(2,500)(142)/
vv
y w
kC
F h t= = =
2,500(0.416) 360 ksc 313.4 ksc
2.89 2.89
y
v v
FF C= = = > OK
No need for additional intermediated stiffeners
�����U'���*� Fv ���������&���"� a/h > 1.0
��/*))*0��164��/��+��: #!(�6"-��' ������-�����
22
2
1 0.416 1.761.76 (142)(1.0) 9.53 cm
2 1 1.76stA
−= − =
+
#$�*0��164��/��+��1���� 0.8 x 12 ��. (Ast
= 9.6 ��.2)
����������-HD��� �������"� &���������������������������
fv = (38.25)(1,000)/(142) = 269 ��./� .2
Fb = (0.825 – 0.375(269/360))(2,500)
= 1,362 ��./� .2 > 269 ��./� .2 OK
���4�6���<���/�������<������
12 79515 15.9
0.8yF
= < =
4 4
4142Min 65.05 cm
50 50st
hI
= = =
Actual Ist using depth from web to outside of stiffener
Moment of inertia:
#$�*0��164��/��+��4,�����$��� 0.8××××12××××136 ��.
Ist = (1/3)(0.8)(12.5)3 = 521 � .4 > 65.05 � .4 OK
��� ���(�������� = 142 – 0.6 – 6(1) = 135.4 � .
*0��164��/��+��*)/��: ����������%W'�'�'�����%���
fa = ���%W'�'�'�� / Aeff = 50.75(1,000)/48
= 1,057 ��./� .2 < 1,457 ��./� .2 OK
88.748
979,2=
<����?�����*4����#�*0��164��/��+��
I = (1/12)(1.2)(31)3 = 2,979 � .4
Aeff = (2)(15)(1.2) + (12)(1) = 48 � .2
r = � .
KL = (0.75)(142) = 106.5 � .
KL/r = 106.5/7.88 = 13.52
Fa = 1,457 ��./� .2 ���&������� ).1
Web
12 mm
Aeff shown shaded15 cm
1 cm
15 cm
31 cm
12tw = 12 cm
+��#$� 2 PL 1.2 ×××× 15 ��. ���#�4B-
#$�*0��164��/��+��*)/�� 2 PL 1.2 x 15 x 140 ��.
2.5 m 2.5 m 2.5 m 2.5 m 2.5 m 2.5 m
5.0 m 5.0 m 5.0 m
PL 0.8 x 12 cm2 PL 1.2 x 15 cm 2 PL 1.2 x 15 cm
FINAL DESIGN