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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