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 Design Timber and Steel Design Lecture Lecture 9 9 Beam-Column Mongkol JIRAVACHARADET S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

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Page 1: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

� 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

Page 2: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

Concentric Loads vs. Eccentric Loads

eEccentric

load

(a)

Concentric

load

(b)

e = Eccentricity

Page 3: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

P1 AB

P2 CD

��������������� ������������������������� ����������

� �� �� ������������ �� �� �����������

Page 4: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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 = +

Page 5: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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�����

Page 6: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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

Page 7: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

#$%�#$%� 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 :

Page 8: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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

Page 9: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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

Page 10: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

(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 !!!

Page 11: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

������� -����� 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)

Page 12: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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�

Page 13: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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

Page 14: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

� �������� 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)

Page 15: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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 ×′ = =

Page 16: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

���� ��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 ��-����

Page 17: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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 ??????

Page 18: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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

Page 19: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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

Page 20: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

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 ��

Page 21: Timber and Steel Design Lecture 9–Beam-Columneng.sut.ac.th/ce/oldce/CourseOnline/430432/L09 Beam Column.pdf · Combined Bending and Axial Stresses ... Timber and Steel Design Lecture

( )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