timber and steel design lecture 9–beam-columneng.sut.ac.th/ce/oldce/courseonline/430432/l09 beam...
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� Combined Bending and Axial Stresses
� AISC Interaction Equations
� Purlin between Joists on Roof Truss
� Equivalent Axial Load Method
Timber and Steel DesignTimber and Steel Design
Lecture Lecture 99 –– Beam-Column
Mongkol JIRAVACHARADET
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Concentric Loads vs. Eccentric Loads
eEccentric
load
(a)
Concentric
load
(b)
e = Eccentricity
P1 AB
P2 CD
��������������� ������������������������� ����������
� �� �� ������������ �� �� �����������
P Mcf
A I= ±
yx
x y
M xM yPf
A I I= ± ±
Bending about single axis:
Bending about both axes:
Combined Axial Combined Axial -- Bending StressesBending Stresses
Superposition of stresses from axial force and bending moment
a
P
Aσ = b
My
Iσ =
P My
A Iσ = +
4L150x100x12
�������������� ���������������������� = 490/1,980 = 24.75 %
170(1,000) 548.8(100)(10)
114.12 1,121
1,490 490 1,980 ksc
P Mcf
A I= + = +
= + =
������� Agross = 114.12 cm2
Ix = 2[229 + 57.06(2.41)2] = 1,121 cm4
��������������� L = 4(22.4) = 89.6 kg/m
M = wl2/8 = (89.6)(7.0)2/8 = 548.8 kg-m
� �������� 9-1 ���������� ������ 7.0 ���� ����� �� ������ !��"� 170 �� �$� %�&�� 4L150×100×12 & �+&��,�-.���� ��� !��"�/� �0�����
0.1≤+b
b
a
a
F
f
F
fBending about single axis:
Bending about both axes: 0.1≤++by
by
bx
bx
a
a
F
f
F
f
F
f
Allowable Combined StressesAllowable Combined Stresses
1.0a b
a b
f f
F F+ =
fa
fb
Fa
Fb
/��,! !� fa/Fa �!�.���.��+ 0.15
#$%�#$%� ffaa//FFaa $����#���$����#��� 0.15 : 0.15 :
��� �+�1$�� ����0&�������.����� ��
P
P
d
M
M
���2,-.���� ��� fbx �$� fby ��������2,��+��:
11.0
1 ( / )a ef F≥
′−
����� eF ′ ���-.���� ����$��0-������������.�����4$��5�� 23/12 :
( )
2
2
12
23 /e
b b
EF
KL r
π′ =
1.0
1 1
my bya mx bx
a a abx by
ex ey
C ff C f
F f fF F
F F
+ + ≤ − − ′ ′
�����46+�����70/��,! fa / Fa > 0.15 :
P
P
d
Mb
Mb
1.0
1 1
my bya mx bx
a a abx by
ex ey
C ff C f
F f fF F
F F
+ + ≤ − − ′ ′
(2) Check Overall Stability: fa / Fa > 0.15
(3) For Maximum End Moment:
1.00.60
bya bx
y bx by
ff f
F F F+ + ≤
(1) Small Axial Load: fa / Fa ≤ 0.15
1.0bya bx
a bx by
ff f
F F F+ + ≤
AISC Interaction EquationsAISC Interaction Equations
Modification Factor Modification Factor CCmm ::
(3) No sidesway, but have transverse loading:
3.1 Restrained ends Cm = 0.85
3.2 Unrestrained ends Cm = 1.0
1
2
0.6 0.4m
MC
M= −
- Cm ≤ 1.0
- Use Cm = 1.0 for Conservative value
(1) Subjected to joint translation or sidesway Cm= 0.85
(2) No sidesway, no transverse loading between joints:
(a) Unbraced frame
(b) Braced frame
(a) Single curvature
P
P
M1
M2
Negative M1/M2
(b) Double curvature
P
P
M1
M2
Positive M1/M2
(c) Pinned ends
P
P
Transverse loading
(d) Fixed ends
P
P
Transverse loading
1
2
0.6 0.4 :m
MC
M= −
1.01 /
m
a e
C
f F≥
′−Check !!!Check !!!
������� -����� W350x159 ( A = 202 cm2, Sx= 2,670 cm3, r
x= 15.3 cm,
ry= 8.9 cm, L
c= 4.48 m, L
u= 12.27 m)
� �������� 9-2 ���&���-����� W350×159 C�� ���D%�-����� "��$�����0�� /�24/E��-$F� A36 �����2./�� !���.�!��������D �������C������ �!�.� K
x= 1.92 �$��.� K
y
�2�4����,/-�� .���� 1.0
100 t
100 t
20 t-m
20 t-m
5 m
fa = 100(1,000)/202 = 495 ��./C�.2
KyLy/ry = (1.00)(500)/8.9 = 56.18
KxLx/rx = (1.92)(500)/15.3 = 62.75 (����-$)
Fa = 1,207 ��./C�.2
fa / Fa = 495/1,207 = 0.41 > 0.15
�� �D��� /E������ !� (9.6) �$� (9.7)
fb = 20(1,000)(100)/2,670 = 749 �.�./C�.2
( )
22
2
122,746 kg/cm
23 /e
x x x
EF
K L r
π′ = =
� �0�����&����2./�$".� !� 1 /E��.� Cm = 0.85
0.851.04 1.00 OK
1 / 1 495/ 2,746
m
a e
C
f F= = >
′− −
!��$� E.� �������: Fb = 0.60Fy = 1,500 �.�./C�.2 ���� &�� Lc < Lb < Lu
( )495 (1.04)(749)
0.929 1.0 OK1 / 1,201 1,500
a m b
a a e b
f C f
F f F F+ = + = ≤
′−
!�4$��� �0�����: Fb = 0.66Fy = 1,650 �.�./C�.2
495 7490.784 1.0 OK
0.60 1,500 1,650
a b
y b
f f
F F+ = + = ≤
/��� �0�$��1� ��2��/� �����-�3�
Purlin Purlin between Joists on Roof Trussbetween Joists on Roof Truss
Purlin
Ridge
Main tie
Internal bracing
Rafter
Purlin
occurs betwe
en joints
Rafter subjected to bending
� �������� 9-4 �%�-����� ����� /�24 -����� W200x30.6 �2�/E��4J&� ��.����� &��&"��.� L
0 �� &"��.� U
3 ���� �0����� %������-$F� A36 & ���&����.�&� ��!�%�$� �����
D%�-��/�24-�����.? C�� /�24 (b) ��� �.��� &� �&�� L0 �� U
1�$�D%�-��
��� "� 5 �����1$&���4 ���"�+/-��!����� ��������� !�4$���$� !���� �$� �������
5 t
32 t
32 t
L0
U1
1.12 m1.12 m
(b)
Compressive stress: fa = 32(1,000)/39.01 = 820 ksc
������� -����� W200x30.6 (A = 39.01 cm2, rx= 8.30 cm, r
y= 3.61 cm,
Sx= 277 cm3, S
y= 67.6 cm3, L
c= 1.91 m)
L0
U1
U2
U3
6 @ 2.0 m = 12.0 m
3.0 m
(a)
For a frame braced against sidesway K = 1.0
KxLx/rx = (1.0)(224)/8.30 = 27
KyLy/ry = (1.0)(112)/3.61 = 31 (Control)
&������ !� �.1 Fa = 1,383 ��./C�.2
fa/Fa = 820/1,383 = 0.59 > 0.15
�� �D��� /E������ !� (9.6) �$� (9.7)
( )
2 6
2
12 (2.1 10 )14,834 ksc
23 27exF
π ×′ = =
���� ��5�����6#7���� 3:
�%�-���� �0����� !��!4$�������D Cm= 0.85
�%�-���� �0����� !��!4$����.�����D Cm= 1.0
�.��M$!���� Cm = (0.85 + 1.0)/2 = 0.925
Mcenterline = PL/4 = (5)(2.24)/4 = 2.80 t-m
Mcenterline = 5PL/32 = 5(5)(2.24)/32
= 1.75 t-m
32 t
32 t5 t
L0
U1
32 t
32 t5 t
L0
U1
32 t
32 t5 t
L0
U1
32 t
32 t5 t
L0
U1
/E��.��M$!���� M = (2.80 + 1.75) / 2 = 2.28 ��-����
0.1979.0
834,14
8201
925.0
1
<=−
=
′−
e
a
m
F
f
CUSE 1.0
fb !�&"���� �$� E.� ������� = 2.28(1,000)(100)/277 = 1,004 ��./C�.2
�.��M$!���� ����0 !�4$���� = (0 + 3PL/16)/2 = 1.05 ��-����
fb !�4$��E.� ������� = 1.05(1,000)(100)/277 = 379 ��./C�.2
820 (1.0)(1,004)1.26 1.0
1,383 1,5001
a m b
a ab
e
f C f
F fF
F
+ = + = > − ′
NG ???
/��� �3$�0�$��1� ��2��/� �����-�3� ???
��������/��� �/��� ��/:������/:����� W200x30.6W200x30.6 ??????
Equivalent axial load method:Equivalent axial load method:
M
ConvertP’ (equivalent axial load)
Peff= P(axial load) + P’
Design as
concentric column
Trial-error procedure:
Select trial section Check interaction eqs.����������-#0�
Approximation:Approximation:
eff x yP P mM mUM= + +
- �.� m �$� U �%�-���-����� W �.� P����2��%��,���/���� !� �.3 /5��1�� �
- ���4����,���D ���&�/E��.� m = 8.5 �$� U = 3
Peff = 100 + (7)(5.7) + (4)(5.7)(2.62) = 200 ��� < 279 ������
������� ����� �� ������ m �!����" 8.5 � #��� U = 3
Peff = 100 + (7)(8.5) + (4)(8.5)(3) = 261.5 ���
� ��������� W400x172 (279 ���) ����� �����""��� �.2
���� ������� W400x172 (U = 2.62, m = 5.7)
"�#���������$%"&''�(� W350x137 (214 ���) (U = 2.61, m = 6.6)
Peff = 100 + (7)(6.6) + (4)(6.6)(2.61) = 215 ��� ,� ���-���" 214 ��� .�,/�0��
� �������� 9-5 & �$���-����� W !���� !��"�������� ��� �� ������� P � .���� 100 �� �$�����0��� M
x= 7 ��-���� �$� M
y= 4 ��-���� ���"�+ KL = 5 ���� C
mx= C
my=
0.85 �$�/E��-$F� A36
fa = 100(1,000)/173.6 = 576 ksc
KxLx/rx = 500/15.2 = 32.89
KyLy/ry = 500/8.84 = 56.56 (*� *+,)
Fa = 1,241 ksc
fa/Fa = 576/1,241 = 0.464 > 0.15
� ����"������ W350x137(A = 173.6 cm2, Sx = 2,300 cm3, Sy = 776 cm
3,
rx = 15.2 cm, ry = 8.84 cm, Lc = 4.46 m, Lu = 10.72 m)
Fby = 0.75Fy = 1,875 ksc, Cmx = Cmy = 0.85
���,/����� (10.6) � # (10.7)
fbx = 7(1,000)(100)/2,300 = 304 ksc
fby = 4(1,000)(100)/776 = 515 ksc
Fbx = 0.60Fy !-�� �/�� = 1,500 ksc �������� Lc < Lunbraced < Lu
Fbx = 0.66Fy = 1,650 ksc !-�4 ��
( )ksc 996,9
89.3223
)101.2(122
62
=×
=′π
exF ( )ksc 380,3
56.5623
)101.2(122
62
=×
=′π
eyF
1.0 Use 0.1902.0
996,9
5761
85.0
1
<=−
=
′−
ex
a
mx
F
f
C0.1025.1
380,3
5761
85.0
1
>=−
=
′−
ey
a
my
F
f
COK
�!���� ,����� (9.6) � # (9.7)
0.1948.0875,1
)515)(025.1(
500,1
)304)(0.1(
241,1
576<=++ OK
576 304 5150.843 1.0
1,500 1,650 1,875+ + = < OK
USE W350x137
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